Download Single-Time-Constant Circuits

Lecture 1
Single-Time-Constant (STC) Circuits
Objectives
„ To analyze and understand STC circuits with emphasis on time constant
calculations
„ To identify different STC configurations: low pass and high pass
„ To study the switching performance of the STC circuits
Introduction
Circuits that are composed of, or can be reduced to, one reactive component and one
resistance are known as single-time-constant (STC) circuits.
Time Constant (τ) is defined as the time required to charge a reactive element (capacitor
or inductor) to 63 percent (actually 63.2 percent) of full charge or to discharge it to 37
percent (actually 36.8 percent) of its initial value.
For an RC circuit, the value of one time constant is expressed mathematically as τ = RC.
For RL circuit τ = R/L.
The importance of the STC is appreciated, when we know that the analysis of a complex
amplifier circuit can be usually reduced to the analysis of one or more simple STC
circuits.
Time Constant Evaluation
Example 1:
Find the time constant of the circuit shown in Figure 1.
Figure 1 STC circuit for example 1.
Solution:
Apply Thévenin’s theorem to find the resistance seen by the capacitor. The solution
details are as follow:
Step 1:
Apply Thévenin’s theorem to the circuit that contains R1, R2, and vI to obtain Rth1 and
Vth1 as shown in Figure 2..
Figure 2 First reduction to the STC in Figure 1 using Thévenin’s theorem.
Rth1 is calculated by reducing vI to zero. That means both R1 and R2 will have the same
two nodes.
∴ Rth1 = R1 || R2 .
Vth1 is the open circuit voltage across R2 which may be calculated using voltage division.
∴Vth1=vixR2/(R1+R2) ∴
Vth1 = vi × R2 /( R1 + R2 )
Step 2:
The STC in Figure 1 may be simplified using Thévenin’s circuit in Figure 2 to obtain the
simplified circuit in Figure 3.
Figure 3 First application of Thévenin’s theorem to simplify the STC circuit in Figure 1
Step 3:
Figure 4 Second reduction to the STC circuit in Figure 1 using Thévenin’s theorem.
Rth is calculated by reducing vth1 to zero. That means both Rth1 and R3 will be in series and
the resultant resistance will be in parallel with R4.
∴Rth= (Rth1+R3)|| R4 = ((R1|| R2)+R3)|| R4.
Vth is the open circuit voltage across R4 which may be calculated using voltage division.
∴Vth=Vth1*R4/(Rth1+R3+R4) = vI*R2/(R1+R2) *R4/( (R1|| R2) +R3+R4)
Step 4:
∴ Vth =Vth1 × R4 /(Rth1 + R3 + R4) = vi × R2 /(R1 + R2)× R4 /(R1 || R2 + R3 + R4)
The STC in Figure 2 may be simplified using Thévenin’s circuit in Figure 4 to obtain the
simplified circuit in Figure 5.
Figure 5 Second application of Thévenin’s theorem to simplify the STC circuit in Figure 1
This will result in the resistance seen by the capacitor C as:
[( R1 || R2 ) + R3 ] || R4 .
Thus, τ = {[( R1 || R2 ) + R3 ] || R4 }C
Comment: Since the time constant is independent of the sources, first of all set all
sources to zero. That means, short-circuit all voltage sources and open circuit all current
sources. Then, reduce the circuit as shown in Figure 6.
Figure 6 Reduction of STC circuits to a single reactive element and a single resistance
From Figure 6, the time constant will equal to ReqCeq for Figure 5-a, or Leq/Req for Figure
5-b.
Example 2: Find the time constant of the circuit shown in Figure 7.
Figure 7 STC Circuit for Example 2
Solution:
Set the source to zero (i.e short circuit vI ) to obtain the circuit in Figure 8-a.
The circuit in Figure 8-a may be further simplified using resistance reduction to obtain
the circuit in Figure 8-b.
From Figure 8-b, the resistance seen by the capacitor C will equal to R = R || R
.
eq
1
2
Thus, the time constant will equal to τ = ( R || R )C
1
2
Figure 8 Reduction of the STC circuit in Figure 7 for time constant calculation
Example 3: Find the time constant of the circuit shown in Fig. 9.
Figure 9 STC circuit for Example 3
Solution:
Set the source to zero (i.e short circuit vI ) to obtain the circuit in Figure 10-a.
The circuit in Figure 10-a may be further simplified using capacitance reduction to obtain
the circuit in Figure 10-b.
From Figure 10-b, the capacitance seen by the resistor R will equal to C = C +.C
eq
1
2
Thus, the time constant will equal to τ = {C + C }R
1
2
C1
C2
(a)
R
Ceq
R
(b)
Figure 10 Reduction of the STC circuit in Figure 9 for time constant calculation
Example 4:
Find the time constant of the circuit shown in Figure 11.
Figure 11 STC circuit for Example 4
Solution:
Set the source to zero (i.e short circuit vI ) to obtain the circuit in Figure 12-a.
The circuit in Figure 12-a may be further simplified by noting that R1 and R2 are parallel
and C1 and C2 are parallel to obtain the circuit in Figure 12-b.
From Figure 12, the equivalent capacitance is C eq = C1 + C 2 , and the equivalent
resistance is R = R ||. R
eq
1
2
Thus, the time constant will equal to
τ = {C1 + C2 }{R1 || R2 }
Figure 12 Reduction of the circuit in Figure 11 for time constant calculation
Example 5:
Find the time constant of the circuit shown in Figure 13.
Figure 13 STC circuit for example 5
Solution:
Set the source to zero (i.e short circuit vI ) to obtain the circuit in Figure 14-a.
The circuit in Figure 14-a may be further simplified by noting that L1 and L2 are parallel
to obtain the circuit in Figure 10-b.
From Figure 10, the inductance seen by the resistor R will equal to L = L || L. Thus,
eq
1
2
the time constant will equal to
τ=
{L1 || L2 }
R
Figure 14 Reduction of the STC circuit in Figure 13 for time constant calculation
Classification of STC Circuits
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STC circuits may be classified into two categories, low-pass (LP) and high-pass
(HP) types, with each category displaying distinctly different signal responses.
Low-Pass circuits pass dc (signal with zero frequency) and attenuate high
frequencies, with the transmission being zero at ω=∞.
High-Pass circuits pass high frequencies and attenuate low frequencies, with the
transmission being zero at dc (ω=0).
To classify the STC as a LP or HP we may test the output response at either ω=0
or ω=∞.
At ω=0 capacitors should be replaced by open circuits and inductors should be
replaced by short circuits. If the output is zero, the circuit is of the high-pass type
otherwise, if the output is finite, the circuit is of the low-pass type.
At ω=∞ capacitors should be replaced by short circuits and inductors should be
replaced by open circuits. If the output is finite, the circuit is of the high-pass type
otherwise, if the output is zero, the circuit is of the low-pass type.
Table 1 provides a summary to the classification test procedure.
Table 1Rules for finding the types of STC circuits
Test at
Replace
ω=0
C by open circuit & L by short circuit
Output is finite
Output is zero
ω=∞
C by short circuit & L by open circuit
Output is zero
Output is finite
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Circuit is LP if Circuit is HP if
Figure 15 shows examples of low-pass STC circuits, and Figure 16 shows
examples of high-pass STC circuits.
Please note that each circuit could be either HP or LP depending on the input and
output variable.
As an exercise please verify the circuits type in Figures 15 and 16using the rules
in table 1.
Detailed frequency response of the STC circuits will be addressed in the next
lecture. While the time response to various test signals will be discussed in the
next section.
Figure 15 STC circuits of the low-pass type
Figure 16 STC circuits of the low-pass type
Time response of STC circuits : Step response of STC Circuits
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The first order differential equation describing the system may be solved given
the problem initial conditions to yield the required current or voltage.
Alternatively, we may obtain the required current or voltage without solving any
differential equation by finding the following quantities:
a. the initial value of the capacitor voltage or the inductor current “y0+”
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b. the final value of the capacitor voltage or the inductor current (at t=∞) “y∞”
c. the time constant “τ”
Then use the equation: y(t) = y∞ - (y∞ – y0+).e(-t/τ)
The step response will depend on the STC type (HP, or LP)
Figure 17 shows the step response of both HP and LP types.
For the low-pass circuit
o y0+ = 0 (sudden change is considered very high frequency)
o y∞ = S (DC signal represents zero frequency)
o y(t)=S(1- e(-t/τ) ) which is shown in Figure 17-b
o The slope at t=0 is S/τ
For the high-pass circuit
o y0+ = S (sudden change is considered very high frequency)
o y∞ = 0 (DC signal represents zero frequency)
o y(t)=S.e(-t/τ) which is shown in Figure 17-c
o The slope at t=0 is S/τ
Figure 17-a A step-function signal of height S.
Figure 17-b The output y(t) of a low-pass STC circuit excited by a step of height S.
Figure 17-c The output y(t) of a high-pass STC circuit excited by a step of height S.
Pulse response of STC Circuits
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Figure 18 shows a pulse signal with a height "P" and width "T".
The pulse function may be treated as the sum of two step functions one with a
height “P” at time “t=0” and the other with a height “-P” at time “t=T”
Similar to the step response, the pulse response will depend on the STC type (LP
or HP)
Figure 18 A pulse signal with a height "P" and width "T"
a) Low Pass circuit response
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The response will depend on the ratio between the pulse width “T” and the
time constant “τ”.
Figure 19 shows the response for three different time constant ratio cases
Since low pass circuit passes the DC faithfully, the area under the curve
will be constant equal to T times P as the input signal.
If τ<<T, the output will be similar to the input with smoothed edges
(remember that sudden changes represent very high frequencies)
As shown in Figure 19-a, the rise time is defined as the time required for
the output to rise from 10% to 90% of the pulse height. Similarly, the fall
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time is defined as the time required for the output to fall from 90% to 10%
of the pulse height.
From the exponential equation we can easily prove that tr=tf=2.2*τ
An interesting case is when τ >>T where the circuit will act as an
integrator.
Figure 19 The output y(t) of a low-pass STC circuit excited by a pulse in Figure 18
b) High Pass circuit response
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Similar to the low pass circuit. The response will depend on the ratio
between the pulse width “T” and the time constant “τ”.
Figure 20 shows the response for three different time constant ratio cases
Since high pass circuit has an infinite attenuation to DC, the area under the
curve will equal to zero.
If τ>>T, the output will be similar to the input with small tilt and zero
average (positive area will equal to the negative area)
The tilt ΔP may be calculated by noting the slope of the step response at
t=0 is P/τ. That means ΔP may be approximated by the slope times T =
PT/τ.
An interesting case is when τ <<T where the circuit will act as a
differentiator.
Figure 20 The output y(t) of a high-pass STC circuit excited by a pulse in Figure 18
Lecture 2
S-Domain Analysis, and Bode Plots
Objectives
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„
„
To study the frequency response of the STC circuits
To appreciate the advantages of the logarithmic scale over the linear scale
To construct the Bode Plot for the different STC circuits
To draw the Bode Plot of the amplifier gain given its transfer characteristics
Introduction
In the last lecture we examined the time response of the STC circuits to various test
signals. In that case the analysis is said to be carried in the time-domain.
The analysis and design of any electronic circuit in general or STC circuits in particular
may be simplified by considering other domains rather than the time domain. One of the
most common domains for electronic circuits' analysis is the s-domain. In this domain the
independent variable is taken as the complex frequency "s" instead of the time.
As we said the importance of studying the STC circuits is that the analysis of a complex
amplifier circuit can be usually reduced to the analysis of one or more simple STC
circuits.
S-Domain Analysis
The analysis in the s-domain to determine the voltage transfer function may be
summarized as follows:
• Replace a capacitance C by an admittance sC, or equivalently an impedance 1/sC.
• Replace an inductance L by an impedance sL.
• Use usual circuit analysis techniques to derive the voltage transfer function
T(s)≡Vo(s)/ Vi(s)
Example 1: Find the voltage transfer function T(s) ≡ Vo(s)/Vi(s) for the STC network
shown in Figure 1?
Figure 1 STC circuit to be analyzed using s-Domain
Solution: Step 1:
Replace the capacitor by impedance equal to 1/SC as shown in Figure 2. Note that both
Vi and Vo will be functions of the complex angular frequency (s)
Figure 2 The STC in figure 1 with the capacitor replaced by an impedance 1/SC
Step 2: Use nodal analysis at the output node to find Vo(s),
Vo ( s) Vo ( s) Vo ( s) − Vi ( s)
+
+
=0
1
R2
R1
sC
⎛
1
1 ⎞ V ( s)
⇒ Vo ( s ) ⎜ sC + + ⎟ = i
R1 R2 ⎠
R1
⎝
1
Vo ( s )
R1
⇒ T ( s) =
=
Vi ( s ) ⎛
1
1 ⎞
⎜ sC + + ⎟
R1 R2 ⎠
⎝
∴T ( s) =
1
s+ 1
CR1
C ( R1 || R2 )
As an exercise try to use the impedance reduction, and the voltage divider rule, or any
other method to calculate T(s).
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In most cases T(s) will reveal many useful facts about the circuit performance.
For physical frequencies s may be replaced by jw in T(s). The resulting transfer
function T(jw) is in general a complex quantity with its:
o Magnitude gives the magnitude (or transmission) response of the circuit
o Angle gives the phase response of the circuit
Example 2: For Example 1 assuming sinusoidal driving signals; calculate the magnitude
and phase response of the STC circuit in Figure 1?
Solution:
Step 1:
Replace s by jw in T(s) to obtain T(jω)
1
∵T (s) =
s+ 1
CR1
C ( R1 || R2 )
1
∴T ( jω ) =
jw + 1
CR1
C ( R1 || R2 )
Step 2: The magnitude and angle of T(jω) will give the magnitude response and the phase
response respectively as shown below:
1
T ( jω ) =
CR1
ω + ⎛⎜ 1C ( R || R ) ⎞⎟
1
2 ⎠
⎝
2
2
θ ( jω ) = ∠ (T ( jω ) ) = 0 − arctan[ω C ( R1 || R2 )]
Poles and Zeros
In General for all the circuits dealt with in this course, T(s) can be expressed in the form
T (s) =
N ( s)
D( s)
where both N(s) and D(s) are polynomials with real coefficients and an order of m and n
respectively
• The order of the network is equal to n
• For real systems, the degree of N(s) (or m) is always less than or equal to that of
D(s)(or n). Think about what happens when s → ∞.
An alternate form for expressing T(s) is
s→∞
( s − Z1 )( s − Z 2 )...( s − Z m )
( s − P1 )( s − P2 )...( s − Pm )
where am is a multiplicative constant; Z1, Z2, …, Zm are the roots of the numerator
polynomial (N(s)); P1, P2, …, Pn are the roots of the denominator polynomial (D(s)).
T ( s) = am
Poles — roots of D(s)=0 {P1,P2,…, Pn} are the points on the s-plane where |T| goes to ∞.
Zeros — roots of N(s)=0 {Z1,Z2,…, Zm} are the points on the s-plane where |T| goes to 0.
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The poles and zeros can be either real or complex. However, since the polynomial
coefficients are real numbers, the complex poles (or zeros) must occur in
conjugate pairs.
A zero that is pure imaginary (±jωz) cause the transfer function T(jω) to be
exactly zero (or have transmission null) at ω=ωz.
Real zeros will not result in transmission nulls.
For stable systems all the poles should have negative real parts.
For s much greater than all the zeros and poles, the transfer function may be
approximated as T(s) ≅ am/sn-m . Thus the transfer function have (n-m) zeros at
s=∞.
Example 3: Find the poles and zeros for the following transfer function T(s)? What is the
order of the network represented by T(s)? What is the value of T(s) as s
approaches infinity?
Solution:
Poles : –2 ±j3 and –10 which are the points on s-plane where |T| goes to ∞.
Zeros : 0 and ±j10 which are the points on s-plane where |T| goes to 0.
The network represented by T(s) is a third order which is the order of the denominator
lim s→∞ T ( s ) = 1
Plotting Frequency Response:
Problem with scaling
As seen before, the frequency response equations (magnitude and phase) are usually
nonlinear— some square within a square root, etc. and some arctan function!
The most difficult problem with linear scale is the limited range as illustrated in the
following figure
Figure 3 Linear scale range limitation
If the x-axis is plotted in log scale, then the range can be widened.
Figure 4 Log scale representation
As we can see from Figure 4, the log scale may be used to represent small quantities
together with large quantities. A feature not visible with linear range.
Bode technique: asymptotic approximation
The other problem now is the non-linearity of the magnitude and phase equations. A
simple technique for obtaining an approximate plots of the magnitude and the phase of
the transfer function is known as Bode plots developed by H. Bode.
The Bode technique is particularly useful when all the poles and zeros are real.
To understand this technique let us draw the magnitude and the phase Bode plots of a
STC circuit transfer function given by T(s)=1/(1+s/ωp); where ωp = 1/CR.
Please note that this transfer function represent a low-pass STC circuit. Also, T(s)
represents a simple pole.
Simple Pole Magnitude Bode Plot Construction
Replace s in T(s) by jω to obtain T(jω)
⇒ T ( jω ) =
1
1 + jω / ω p
Find the magnitude of T(jω)
⇒ T ( jω ) =
1
1 + (ω / ω p ) 2
Take the log of both sides and multiply by 20
(
⇒ 20 log ( T ( jω ) ) = −10 log 1 + (ω / ω p ) 2
Define y=20log|T(ω)| and x=log(ω)
The unit of y is the decibel (dB)
In terms of x and y we have
(
y = −10 log 1 + (ω / ω p ) 2
)
)
Now for large and small values of ω, we can make some approximation
• ω>>ωp : y ≅-10 log (ω/ωp)2 = -20 log (ω/ωp) = -20x +20 log(ωp)
which represent a straight line of slope = -20. The unit of the slope will be
dB/decade (unit of y axis per unit of x axis)
• ω<<ωp : y≅-10 log (1) = 0
Which represents a horizontal line.
≅
Finally, if we plot y versus x, then we get straight lines as asymptotes for large and
small ω . Note that the approximation will be poor near ωp with a maximum error of 3
dB (10 log(2) at ω=ωp)
20 log|T|[dB]
Log(
P)
log
[dec]
Slope = -20 dB/dec
Figure 5 Bode Plot for the magnitude of a simple pole
Simple Pole Phase Bode Plot Construction
Replace s in T(s) by jω to obtain T(jω)
∴T ( jw ) =
1+
1
jω
ωp
Find the angle of T(jω)
⇒ θ (ω ) = ∠ (T ( jω ) ) = 0 − arctan (ω / ω p )
Now for large and small values of ω, we can make some approximation
• ω>>ωp : θ(ω) ≅-arctan(∞) = -π/2 = -90°
we can assume that much greater (>>) is equal to 10 times
• ω<<ωp : θ(ω) ≅-arctan(0) = -0°
we can assume that much less (<<) is equal to 0.1 times
∞
•
For the frequencies between 0.1 ωp and 10 ωp we may approximate the phase
response by straight line which will have a slope of -45°/decade with a value
equal to -45° at ω=ωp
Figure 6 Bode Plot for the phase of a simple pole
Bode Plots: General Technique
Since Bode plots are log-scale plots, we may plot any transfer function by adding
together simpler transfer functions which make up the whole transfer function.
The standard forms of Bode plots may be summarized as shown in Table 1.
Table 1 Bode plots standard forms
Form
Equation
Simple Pole
1
1+ s / p
Magnitude Bode Plot
Phase Bode Plot
Form
Equation
Magnitude Bode Plot
Phase Bode Plot
θ
90
Simple Zero
1+ s
z
45
0
Integrating
Pole
Differentiating
Zero
Constant
1
s/ p
s
z
A
z
10
z
10z
log
Example 4: A circuit has the following transfer function Sketch the magnitude and phase
Bode plots.
Solution:
By referring to Table 1 we can divide H(s) into four simpler transfer functions as shown
s ⎞
⎛
1000 ⎜ 1 +
⎟
⎝ 200π ⎠
H ( s) =
below:
s ⎛
s
⎞
⎜1 +
⎟
2π ⎝ 60000π ⎠
The total Bode plot may be obtained by adding the four terms as shown in Figure 7.
Please note that f=ω/2π.
80
|H| dB
20 dB
dec
60
40
20
Log scale
0
-20
-40
-60
1
10
100
1k
10k 100k
1M
10M
f [Hz]
20 dB
dec
-80
Figure 7 Magnitude and Phase Bode Plots of Example 4
Lecture 3B
MOSFET High Frequency Model and Amplifier Frequency Response
Objectives
„ To review the small signal BJT models at low frequencies
„ To study the high frequency BJT models
„ To estimate the BJT unity-gain frequency
Introduction
In the last two lectures we examined the time and frequency response of the STC circuits.
As we said the importance of studying the STC circuits is that the analysis of a complex
amplifier circuit can be usually reduced to the analysis of one or more simple STC
circuits. The frequency response of the amplifier circuits will be explored starting next
lecture. Before starting this study it will be constructive to review in this lecture the BJT
small-signal models, and examining the high-frequency models. Also, the transistor cutoff frequency which is considered a figure of merit at high frequency operation will be
estimated for both transistors.
The BJT small-signal model
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The small signal model of the BJT amplifier is shown in figure 5. Figures 5-a,b
are for the π-model, where Figures 5-c,d are for the T-model.
These models are valid for both NPN and PNP transistors.
For the same operating point, the BJT has higher transconductance and higher
output resistance that the MOSFET.
The small-signal parameters are controlled by the Q-point (operating point).
gm Vπ
Figure 1 small signal-models of the BJT
•
The BJT small-signal parameters may be summarized in Table 3
Table 1 BJT small signal parameters
Symbol
Parameter
Value
I
gm = C
VT
Transconductance
gm
rπ
re
VT is the thermal voltage = kT/q, which
equals 25mV at room temperature.
k is Boltzman's constant
T is the absolute temperature in
Kelvins
q is the electron charge
Base input resistance
⎛V ⎞ β
VT
=β⎜ T ⎟=
IB
⎝ IC ⎠ g m
β is the common-emitter current gain
Emitter input resistance
⎛V ⎞ α
VT
=α ⎜ T ⎟ =
IE
⎝ IC ⎠ g m
α is the common-base current gain
rπ =
re =
Symbol
ro
Parameter
Output resistance
Value
ro =
V A +V CE
IC
VA
IC
VA is the early voltage.
The BJT high-frequency model:
Figure 2 The high-frequency hybrid- π model of the BJT
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The high frequency hybrid-π model for the BJT is shown in Figure 6.
This model is useful for signal frequencies up to a several tens of megahertz, after
which a more detailed model becomes necessary.
•
Typically, the base-emitter junction capacitance Cπ is in the range of few pF to
few tens of pF, while the collector-base junction capacitance Cμ is in the range of
fraction of pF to few pF
The base resistor rx is added partly to account for the comparatively long internal
connection from the base external connection and the actual internal base
connection.
A representative resistance value for this lumped resistor is in the range of 50Ω to
perhaps 200Ω.
This resistor ordinarily can be neglected for hand estimates.
Note that rx becomes the dominant input resistance for frequencies so high that Cπ
effectively short-circuits rπ.
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A second base-width modulation effect, characterized by a resistor connected
between the base and collector is omitted; its influence is dominated by the
collector junction reverse-bias capacitance Cμ.
The emitter junction (diffusion) capacitance Cπ represents the charge store to
support the current flow across the base.
The BJT Cutoff frequency:
•
As defined earlier, it is the frequency at which the current gain of the transistor
becomes one. (i.e. no more active element). It is calculated by finding the short
circuit collector current in terms of the base current.
•
Using the high frequency model of BJT we can draw the circuit to estimate the
cut-off frequency of the BJT as shown in Figure 7.
sC μ V π
I c = ( g m − s C μ )V π
Figure 3 Circuit used to estimate the BJT cutoff frequency
•
Applying nodal analysis at the input and output nodes as we did earlier. We can
estimate the cut-off frequency as follows:
Vπ
+ s C πV π + s C μ V π
rπ
V
I b = π + s (C π + C μ ) V π
rπ
g m − sC μ
I
hfe ≡ c =
1
Ib
+ s (C π + C μ )
rπ
Ib =
Assuming
g m >> sC μ ⇒
Assuming
g m >> sC μ ⇒
hfe ≅
βo
1+
s
where ω p =
ωp
g m rπ
1 + s (C π + C μ )rπ
g m rπ
hfe ≅
1 + s (C π + C μ )rπ
hfe ≅
1
(C π + C μ )rπ
∴ Unity gain bandwidth
(ωT ) = βo ω p =
hfe ≡
hfe ≅
hfe ≅
gm
(C π + C μ )
g m − sC μ
Ic
=
1
Ib
+ s (C π + C μ )
rπ
g m rπ
1 + s (C π + C μ )rπ
βo
1+
s
ωp
assuming g m >> sC μ
where ω p =
1
(C π + C μ )rπ
∴ Unity gain bandwidth (ωT ) = βo ω p =
•
gm
(C π + C μ )
We can observe from the last analysis that the common-emitter current gain (hfe)
frequency response is similar to a simple pole with ωp as the pole frequency. This
may be drawn as shown in Figure 8
|hfe| [dB]
3 dB
βo
-6 dB/octave
[log scale]
0 dB
β
T
Figure 4 Bode plot of |hfe|
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•
As we can see from the last equation. Higher ωT means higher gm and lower
internal BJT capacitances which means better amplifier operation.
Typically, fT is ranging from about 100MHz to Tens of GHz.
Lecture 4a
Common Source Amplifier Frequency Response
Objectives
„ To understand different frequency bands in the amplifier frequency response
„ To analyze the CS amplifiers
Introduction
In the last lecture we examined the transistor high frequency model.
We used this model to calculate the cut-off frequency of the transistor.
As we know, one of the major applications of the transistor is the amplifier.
In this lecture, the frequency response of the amplifier circuits will be discussed.
Also, the frequency response of the common source amplifier will be calculated and
the dominant poles will be determined.
Amplifier Transfer Function
In general, any amplifier exhibits the transfer function A(s) shown below.
|Av(jω)|
[dB]
FH(jω)
FL(jω)
Amid
Midband
region
P1
2
P2
P3
L
H
P4
[log scale]
Figure 1 A typical amplifier frequency response
The transfer function may be written as:
N (s )
A (s ) =
v
D (s )
a + a s + a2s 2 + ... + am s m
A (s ) = 0 1
v
b0 + b1s + b 2s 2 + ... + b n s n
A (s ) = A
F (s )F (s )
v
mid L
H
Amid is the midband gain between the upper and the lower cutoff frequencies or the
3dB frequencies 
and L respectively).
The amplifier bandwidth
The amplifier bandwidth is defined as the difference between
and L.
Normally, the amplifier is designed so that its bandwidth coincides with the spectrum
of the signals that it is required to amplify.
Otherwise, signal distortion will occur.
The lower and upper functions FL(s) and FH(s) may be written as:
F (s ) =
L
( s + ω )( s + ω ) ...( s + ω )
( s + ω )( s + ω ) ...( s + ω )
L
Z1
L
Z2
L
Zk
L
P1
L
P2
L
Pk
⎛
s ⎞⎛
s ⎞ ⎛
s ⎞
⎜ 1 + H ⎟ ⎜1 + H ⎟ ... ⎜1 + H ⎟
ω Z 1 ⎠ ⎝ ω Z 2 ⎠ ⎝ ω Zl ⎠
F (s ) = ⎝
H
⎛
s ⎞⎛
s ⎞ ⎛
s ⎞
⎜1 + H ⎟⎜ 1 + H ⎟ ... ⎜ 1 + H ⎟
⎝ ω P 1 ⎠⎝ ω P 2 ⎠ ⎝ ω Pl ⎠
For the lower frequency band (frequencies much less than
response may be calaculated as follows:
F ( jω)
H
→ 1
for
H).
The frequency
ω ω HZi , ω HPi
,i =1…l
∴
A (s ) ≅ A
F (s )
vL
mid L
For the higher frequency band (frequencies much higher then
response may be calaculated as follows:
F ( jω)
L
∴
→ 1
for
ω ω LZj , ω LPj
L).
The frequency
,j =1…l
A (s ) ≅ A
F (s )
vH
mid H
Amplifier Low-Frequency Response
The midband gain and the upper and the lower cutoff frequencies that define the
bandwidth of the amplifier are of more interest than the complete transfer function of
the amplifier.
The low cutoff frequency may be calculated using FL(s). If FL(s) can be approximated
to the following form
s
F (s ) ≅
L
s + ωP 2
∴
ωL ≅ ωP 2
Where the pole P2 is called the dominant low-frequency pole (> all other poles) and
zeros are at frequencies low enough to not affect L.
If there is no dominant pole at low frequencies, poles and zeros interact to determine
L.
To explain that if there is no dominant pole at low frequencies, poles and zeros
interact to determine L, lets assumes that FL(s) has two poles and two zeros.
s + ω )( s + ω )
(
Z1
Z2
A (s ) = A
F (s ) ⇒ A (s ) = A
vL
mid s + ω
vL
mid L
( P1 )(s + ωP 2 )
(s + ωZ 1 )(s + ωZ 2 )
A (s ) = A
F (s ) = A
vL
mid L
mid s + ω
( P1 )(s + ωP 2 )
For physical frequencies "s" may be replaced by j , at
∴
⇒
1
=
2
L
A ( j ωL ) =
A mid
2
(ω + ω )(ω + ω )
(ω + ω )(ω + ω )
(ω + ω ) + (ω ω )
1+
2
2
2
2
2
Z2
L
2
P1
L
1+
2
Z1
L
2
Z1
1
=
2
=
L
2
2
Z1
Z2
ωL 2
(ωP 12 + ωP 22 )
2
P2
+
2
Z2
ωL 4
(ωP 12ωP 22 )
ωL 2
ωL 4
Since the pole L is greater than all other pole and zero frequencies we may neglect
the last term in both the denominator and the numerator.
∴
ωL ≅
ωP 12 + ωP 2 2 − 2ωZ 12 − 2ωZ 2 2
In general, for n poles and n zeros,
ωL ≅
2
2
∑ ωPn − 2∑ ωZn
n
n
Example 1:
⎛ s
⎞
s⎜
+ 1⎟
⎝ 100 ⎠
Find the midband gain, FL(s) and fL for A (s ) = 2000
vL
( 0.1s + 1)( s + 1000 )
Solution:
Rearrange the given transfer function to make it in the standard form discussed
earlier,
∴
s ( s + 100 )
A (s ) = 200
vL
( s + 10 )( s + 1000 )
F (s )
Compare this expression with A (s ) = A
vL
mid L
∴ Amid = 200
s (s + 100)
And F (s ) =
L
(s + 10)(s + 1000)
Zeros are at s=0, and s =-100. Poles are at s= -10, and s=-1000.
∴
∴
f
L
f
1
102 + 10002 − 2(02 + 1002 )
2π
= 158Hz
=
L
All pole and zero frequencies are low and separated by at least a decade.
Dominant pole is at =1000 rad/sec, and fL =1000/2 = 159 Hz.
For frequencies greater than a few rad/s the voltage gain transfer function may be
approximated as:
s
A (s ) = 200
vL
( s + 1000 )
Amplifier High-Frequency Response
The high cutoff frequency may be calculated using FH(s).
If FH(s) can be approximated to the following form
1
F (s ) ≅
H
1 + (s / ω P 3 )
∴
ωH
≅ ωP 3
Where the pole P3 is called the dominant high-frequency pole (< all other poles) and
zeros are at frequencies high enough to not affect H.
If there is no dominant pole at high frequencies, poles and zeros interact to determine
H.
To explain this assumes that FH(s) has two poles and two zeros.
A (s ) = A
F (s )
vH
mid H
(1 + (s / ωZ 1 ) )(1 + (s / ωZ 2 ) )
A (s ) = A
vH
mid (1 + (s / ωP 1 ) )(1 + (s / ωP 2 ) )
For physical frequencies "s" may be replaced by j , at
⇒
H
A mid
A ( j ωH ) =
1
=
2
=
2
(1 + (ω
(1 + (ω
)(
) )(1 + (ω
)
))
2
/ ωZ 12 ) 1 + (ωH 2 / ωZ 2 2 )
2
/ ωP 1
H
H
2
2
H
/ ωP 2
2
ωH 2 ωH 2
ωH 4
+
+
ωZ 12 ωZ 2 2 ωZ 12ωZ 2 2
1
=
ω 2 ω 2
ωH 4
2
1+ H 2 + H 2 +
ωP 1 ωP 2 ωP 12ωP 2 2
1+
Since the pole H is less than all other pole and zero frequencies we may neglect the
last term in both the denominator and the numerator.
1
ωH ≅
1
ω
2
P1
+
1
ωP 2
2
−
2
ω
2
Z1
−
2
ωZ 2 2
In general, for n poles and n zeros,
1
ωH ≅
∑
n
1
ωPn
2
− 2∑
n
1
ωZn 2
Example 2:
Find the midband gain, FH(s) and fH for
⎛ s
⎞
9
⎜ 8 + 1⎟ s + 10
10
⎠
A (s ) = 0.3 ⎝
vH
⎛ s
⎞
6
⎜ 7 + 1⎟ s + 10
⎝ 10
⎠
Solution:
(
)
(
)
Rearrange the given transfer function to make it in the standard form discussed
earlier,
⎞
⎛ s
⎞⎛ s
⎜ 8 + 1⎟⎜ 9 + 1⎟
10
⎠
⎠⎝ 10
A (s ) = 300 ⎝
vH
⎞
⎛ s
⎞⎛ s
⎜ 7 + 1⎟⎜ 6 + 1⎟
⎠
⎝ 10
⎠⎝ 10
∴
F (s )
Compare this expression with A (s ) = A
vH
mid H
∴
A
mid
= 300
⎞
⎛ s
⎞⎛ s
⎜ 8 + 1⎟ ⎜ 9 + 1⎟
10
⎠
⎠ ⎝ 10
And F (s ) = ⎝
H
⎞
⎛ s
⎞⎛ s
⎜ 7 + 1⎟⎜ 6 + 1⎟
⎠
⎝ 10
⎠⎝ 10
Zeros are at s = -108, and s = -109. Poles are at s = -107 , and s =-106.
∴
∴
f
f
H
H
=
1
2π
1
10
−14
+ 10
−12
− 2(10−16 + 100−18 )
= 158.4kHz
All pole and zero frequencies are low and separated by at least a decade.
106
= 159 kHz .
Dominant pole is at = 106 rad/sec, and f L =
2π
The voltage gain transfer function may be approximated as:
1
A (s ) = 300
vH
⎛ s
⎞
⎜ 6 + 1⎟
⎝ 10
⎠
Frequency Response of Capacitively Coupled Transistor Amplifiers
•
•
•
•
•
•
The frequency response of the capacitively coupled transistor amplifiers is
shown in Figure 2.
At low frequency band the coupling and bypass capacitors are in effect.
Since impedance of a capacitor increases with decreasing frequency, coupling
and bypass capacitors reduce amplifier gain at low frequencies.
For dc amplifiers, the absence of the coupling and bypass capacitors cause
fL(s) = 1 and fL = 0, thus the midband gain extends to zero frequency.
At high frequency band the transistor internal capacitances are in effect.
Since impedance of a capacitor decrease with increasing frequency, transistor
internal capacitances reduce amplifier gain at high frequencies (refer to last
lecture).
Figure 2: The three frequency bands that characterize the frequency response of
capacitively coupled amplifiers
Lecture 4b
Determination of Poles & Zeros of CS Amplifiers (1)
Objectives
„ To calculate different poles and zeros which determine the frequency
response of CS amplifiers
„ To calculate the dominant poles and the amplifier bandwidth
Direct Determination of Low-Frequency Poles and Zeros:
The low frequency poles and zeros may be determined by direct
analysis.
However, the impedance of the coupling and bypass capacitors
should be considered.
The following example will illustrate this method for common
source amplifier.
Example 1:
Drive an expression for the low frequency response AvL(s) of the
common source amplifier shown in Figure 1. Hence, determine the
midband gain Amid, low frequency poles and zeros, and the lower
cutoff frequency.
Figure 1: Common source amplifier
Solution:
The CS amplifier is redrawn for small signal analysis as shown in
Figure 2 by grounding the DC voltage source VDD and replacing
the transistor by its small signal model.
The biasing resistors R1 and R2 are combined into RG.
∴
RG = R1 || R 2
Figure 2: The CS amplifier in Figure 1 redrawn for small signal analysis with the
transistor replaced by its small signal model
v (s ) = i (s )R L
o
o
RD
v (s ) = − g mV gs (s )
RL
o
R D + (1/ sC 2 ) + R L
Using current division to calculate io(s)
RD
R D + (1/ sC 2 ) + R L
RD
v (s ) = − g mV gs (s )
RL
o
R D + (1/ sC 2 ) + R L
s
Vgs (s)
= − g m (R L R D )
1
s+
C 2 (R D + R L )
RD
v (s ) = i (s )R L = − g mV gs (s )
RL
o
o
R D + (1/ sC 2 ) + R L
i o (s ) = − g mV gs (s )
∴
Using current division to calculate i o (s)
i o (s ) = − g mV gs (s )
∴
RD
R D + (1/ sC 2 ) + R L
RD
v (s ) = − g mV gs (s )
RL
o
R D + (1/ sC 2 ) + R L
= − g m (R L R D )
s
s+
1
C 2 (R D + R L )
Vgs (s)
Using voltage divider at the amplifier input to determine Vg(s)
RG
Vg (s) =
∴
∴
V (s)
1 sig
R sig + RG +
sC 1
sC 1R G
Vg (s) =
Vsig (s)
sC 1 ( R sig + R G ) + 1
sRG
(RG + R sig )
Vg (s) =
Vsig (s)
s+ 1
C 1 (R sig + RG )
Using voltage division at the amplifier input to determine vg (s)
Vg (s) =
∴
∴
RG
R sig
1
+ RG +
sC 1
Vg (s) =
sC 1RG
Vsig (s)
sC 1 (R sig + RG ) + 1
sRG
Vg (s) =
Vsig (s)
(RG + R sig )
s+ 1
C 1 (R sig + RG )
Vsig (s)
Vgs (s) = Vg - Vs
Vgs (s) = V g − g mV gs
∴
V gs (s ) =
RS
sC s
RS +
1
sC s
s + (1/ C S R S )
Vg (s)
1
s+
C S ⎡⎣(1/ g m ) R S ⎤⎦
Vgs (s) = Vg - Vs = V g − g mV gs
∴V gs (s ) =
RS
sC s
RS +
1
sC s
s + (1/ C S R S )
Vg (s)
1
s+
C S ⎡⎣(1/ g m ) R S ⎤⎦
Vo (s)
A (s) =
v
Vsig (s)
⇒
Vo (s) Vgs (s) Vg (s)
A (s) =
.
.
v
Vgs (s) Vg (s) Vsig (s)
∴
Av
∵
A (s) = A mid F (s)
v
L
∴
A mid
= − g m (R L R D ).
RG
.
RG + R sig s +
= − g m (R L R D )
s + (1/ C S R S )
s
.
1
1
1
s+
s+
C 2 (R D + R L )
C 1 (R sig + RG )
C S ⎡⎣(1/ g m ) R S ⎤⎦
s
.
RG
R G + R sig
s 2 ( s + (1/ C S R S ) )
F (s) =
L
⎞⎛
⎛
⎞⎛
⎞
1
1
1
⎟⎜s +
⎜⎜ s +
⎟⎟ ⎜ s +
⎟
⎜
⎟
⎝ C 1 (R sig + RG ) ⎠ ⎝ C S ⎡⎣(1/ g m ) R S ⎤⎦ ⎠ ⎝ C 2 (R D + R L ) ⎠
The three zero locations are: s = 0, 0, -1/(RS CS).
The three pole locations are:
s
= −
1
C 1 (R sig + RG )
,−
1
1
,−
C S ⎡⎣(1/ g m ) R S ⎤⎦ C 2 (R D + R L )
Note:
Each independent capacitor in the circuit contributes one pole and
one zero.
Series capacitors C1 and C2 contribute the two zeros at s = 0 (dc),
blocking propagation of dc signals through the amplifier.
Third zero due to parallel combination of CS and RS occurs at
frequency where signal current propagation through MOSFET is
blocked (output voltage is zero).
Example 2:
Calculate the midband gain as well as all low frequency poles and
zeros for the common source amplifier in Figure 1.
Assume: Rsig = 1kΩ, R1 = 430kΩ, R2 = 560kΩ, RD = 4.3kΩ, Rs =
1.3kΩ, RL = 100kΩ, C1 = C2 = 0.1μF, and CS = 10μF. Also assume
that the transistor is biased such that gm = 1mA/V.
Solution:
Refer to Example 1 solution.
RG = R1 || R 2 = 560k || 430k
RG = 243.2k Ω
A mid
= − g m (R L R D )
Amid = − 4.1
RG
R G + R sig
V /V
The zero locations are : s = 0, s = 0, and s = -77.
The pole locations are: s = -41, s = -177, and s = -96.
Short-Circuit Time Constant Method to Determine ωL
• Lower cutoff frequency for a network with n coupling and
bypass capacitors is given by:
1
i =1 R C
iS i
n
ωL ≅ ∑
where RiS is the resistance at terminals of the ith capacitor Ci with
all other capacitors replaced by short circuits. Product RiS Ci is the
short-circuit time constant associated with Ci.
Example 3:
Derive an expression for the low cutoff frequency for the circuit in
Example 1 using the SCTC method.
Solution:
Using SCTC method
For C1 the resistance seen across the capacitor terminals may be
calculated from Figure 3.
R1S = R sig + (RG R CS
in )
R1S = RS + RG
So the time constant associated with C1 equals C1 · (Rs + RG)
R1S
CS
Rin
RG
RD || RL
Rsig
Figure 3: Circuit used to calculate the short-circuit time constant associated with C1
For C2 the resistance seen across the capacitor terminals may be
calculated from Figure 4.
R 2S = R L + (R D R CS
out )
R 2S = R L + (R D ro )
R 2S ≅ R L + R D
R 2S
= R L + (R D R CS
out ) = R L + ( R D ro )
≅ RL + RD
So the time constant associated with C2 equals C2 · (RL+RD)
Figure 4: Circuit used to calculate the short-circuit time constant associated with C2
For CS the resistance seen across the capacitor terminals may be
calculated from Figure 5.
R SS = R S R CG
out
R SS
= RS
1
gm
So the time constant associated with CS equals CS · (RS||1/gm)
Figure 5: Circuit used to calculate the short-circuit time constant associated with CS
The low cutoff frequency is calculated using the SCTC as:
1
i =1 R C
iS i
1
1
1
≅
+
+
C 1 (R S + RG ) C 2 (R L + R D ) C (R || 1 )
S
S
gm
3
1
1
1
1
≅ ∑
=
+
+
i =1 R iS C i
C 1 (R S + RG ) C 2 (R L + R D ) C ( R || 1 )
S
S
gm
3
ωL ≅ ∑
ωL
ωL
Example 4:
Calculate the low cutoff frequency for the common source amplifier
in Figure 1 using the SCTC method. Assume: Rsig = 1kΩ, R1 =
430kΩ, R2 = 560kΩ, RD = 4.3kΩ, Rs = 1.3kΩ, RL = 100kΩ, C1 =
C2= 0.1μF, and CS = 10μF. Also assume that the transistor is
biased such that gm = 1mA/V.
Solution:
Refer to Example 3 solution.
ωL ≅
1
1
1
+
+
C 1 (R S + RG ) C 2 (R L + R D ) C (R || 1 )
S
S
gm
ωL ≅ 313.7rad/sec
ωL ≅
1
1
1
+
+
= 313.7rad/sec
C 1 (R S + RG ) C 2 (R L + R D ) C (R || 1 )
S
S
gm
Coupling and Bypass Capacitors Design Considerations
• We can choose the capacitors' values to set the lower cutoff
frequency of the amplifier at any desired value.
• Pole associated with a capacitor occurs at the frequency at
which the capacitive reactance is equal to the resistance at
the capacitor terminals.
• In the discussed amplifiers, there are several poles at low
frequencies which result in a bandwidth shrinkage.
• A transfer function with n identical poles at ωo is given by
T ( j ω ) = A mid
T ( j ωL ) =
T ( j ωL ) =
ω 2 + ωo 2
⇒
A mid
2
T ( j ω ) = A mid
(
ωn
(
A mid
2
ωn
ω 2 + ωo 2
→
)
ωL =
)
n
ωo
21/ n − 1
n
ωL =
ωo
21/ n − 1
• From the last equation in the preceding slide, we can
conclude that lower cutoff frequency is higher than the
frequency corresponding to the individual poles.
• Instead of having the lower cutoff frequency set by the
interaction of several poles, it can be set by the pole
associated with just one of the capacitors. The other
capacitors can be chosen to have their pole frequencies
much below fL.
• Capacitor associated with emitter or source part of the circuit
tends to be the largest due to low resistance presented by
emitter or source terminal of transistor and is commonly used
to set fL.
• Values of other capacitors are increased by a factor of 10 to
push their corresponding poles to much lower frequencies.
Lecture 4c
Determination of Poles & Zeros of CS Amplifiers (2)
Objectives
„ To calculate different poles and zeros which
determine the frequency response of CS
amplifiers
„ To calculate the dominant poles and the
amplifier bandwidth
Direct Determination of High-Frequency Poles and Zeros:
The high frequency poles and zeros may be determined by direct
analysis.
In this case all the coupling and bypass capacitors may be
considered short circuit.
However, the high frequency model of the transistor should be
used since the transistor internal capacitances can't considered
open circuit at high frequencies.
The following example will illustrate this method for common
source amplifier.
Example 1:
Drive an expression for the high frequency response AvH(s) of the
common source amplifier shown in Figure 1. Hence, determine the
midband gain Amid, high frequency poles and zeros, and the high
cutoff frequency.
Figure 1: Common source amplifier
Solution:
The CS amplifier is redrawn for small signal analysis as shown in
Figure 2 by grounding the DC voltage source VDD and replacing
the transistor by its high frequency small signal model.
The biasing resistors R1 and R2 are combined into RG.
∴
RG
= R1 || R 2
All the coupling and bypass capacitors are considered short circuit
at high frequency.
Figure 2: The CS amplifier in Figure 1 redrawn for small signal analysis with the
transistor replaced by its high frequency small signal model
The small-signal model can be simplified by using Thevinin
Theorem as shown in Figure 3.
v th = v sig
RG
R sig + RG
and Rth =
R sig RG
R sig + RG
Also, R L′ = R L || R D
Figure 3: Simplified circuit to calculate the frequency response of the CS amplifier
Writing the two nodal equations at nodes A and B in frequency
domain,
vo
+ g mv gs + (v o −v gs )sC GD = 0
R L′
v gs sC GD +
v gs −v th
R th
+ (v gs −v o )sC GD = 0
Solving the two nodal equations found in the preceding slide by
eliminating vgs yields,
V o (s) =
v th (s) (sC GD - g m )
Rth
Δ
⎛C
⎛
1
1 ⎞⎞
1
Δ = s 2 (C GS C GD ) + s ⎜ GS + C GD ⎜ g m +
+
⎟ ⎟⎟ +
⎜ R′
R th R L′ ⎠ ⎠ R L′ .R th
⎝
⎝ L
∴
AvH (s ) =
v o (s )
v sig (s )
∴
AvH (s ) =
1 (sC GD - g m )
R sig
Δ
Let us define
CT
⎛
R′ ⎞
= C GS + C GD ⎜ 1 + g m R L′ + L ⎟
R th ⎠
⎝
∴
Δ = s 2C GS C GD +
∴
AvH (s ) =
sCT
1
+
R L′ R ′.R th
1
R sig C GS .C GD s 2 +
(sC GD - g m )
sC T
1
+
R L′ .C GS .C GD R ′.R th .C GS .C GD
vo
(s) =
v th (s) (sC GD - g m )
Δ
R th
⎛C
⎛
1
1
Δ = s 2 (C GS C GD ) + s ⎜ GS + C GD ⎜ g m +
+
⎜ R′
R th R L′
⎝
⎝ L
v o (s )
1 (sC GD - g m )
AvH (s ) =
=
v sig (s )
R sig
Δ
∴
⎞⎞
1
⎟ ⎟⎟ +
⎠ ⎠ R L′ .R th
⎛
R′ ⎞
= C GS + C GD ⎜1 + g m R L′ + L ⎟
R th ⎠
⎝
sC
1
Δ = s 2C GS C GD + T +
R L′ R ′.R th
Let us define C T
∴
∴
1
AvH (s ) =
R sig C GS .C GD s 2 +
(sC GD - g m )
sC T
1
+
R L′ .C GS .C GD R ′.R th .C GS .C GD
Amid may be found from the last expression by assuming s → 0 .
∴
A mid
=
- g m R L′ .RG
R sig + RG
High-frequency response is given by 2 poles, one finite zero and
one zero at infinity. Finite right-half plane zero, ωZ = + gm/CGD > ωT
can easily be neglected.
For a polynomial s2 + sA1 + A0 with roots a and b, if one root is
much larger than the other one we can assume that a = A1 and b =
A0/A1.
∴
ωP 1 =
ωP 1 ≅
1
R thC T
A0
A1
Assume that the midband gain is very large
⎛ R′ ⎞
( i.e. g m R L′ ⎜1 + L ⎟ , and g m R L′C GD C GS )
⎝ Rth ⎠
gm ⎛
1
1 ⎞
+
ωP 2 ≅
⎜1 +
⎟
C GS ⎝ g m Rth g m R L′ ⎠
gm
ωP 2 ≅
C GS
∴
∴
∴
ωP 1 =
A0
1
≅
A1
R thC T
⎛ R′
Assume that the midband gain is very large ( i.e. g m R L′ ⎜ 1 + L
⎝ R th
gm ⎛
gm
1
1 ⎞
ωP 2 ≅
+
⎜1 +
⎟ ≅
C GS ⎝ g m R th g m R L′ ⎠
C GS
⎞
⎟ , and
⎠
g m R L′C GD C GS )
∴
Smallest root that gives first pole limits frequency response and
determines ωH. Second pole is important in frequency
compensation as it can degrade phase margin of feedback
amplifiers.
Assuming ωp1 ωp2
(ωp1 is the dominanat pole)
∴
Assuming ωp1
∴
1
RthCT
ωp2 (ωp1 is the dominanat pole)
ωH ≅ ωP 1 =
ωH ≅ ωP 1 =
1
R thC T
Dominant pole model at high frequencies for CS amplifier is shown
in Figure 4.
Figure 4: Equivalent circuit assuming dominant pole model
Miller’s Theorem
Another way to find the high frequency poles is by calculating the
equivalent capacitance for CGD at the amplifier input. This may be
done using Miller's Theorem as shown next.
For the circuit shown we can state the following.
Figure 5: Miller's Theorem
v o (s ) = − Av 1 (s )
i s (s ) = sC [v 1 (s ) − v o (s ) ]
i s (s )
= sC (1 + A )
v 1 (s )
v o (s ) = − Av 1 (s )
Y (s ) =
i s (s ) = sC [v 1 (s ) −v o (s ) ]
Y (s ) =
i s (s )
= sC (1 + A )
v 1 (s )
The total input capacitance = C (1+A) because total voltage across
C is vc = vi (1+A) due to the inverting voltage gain of amplifier.
Applying Miller's Theorem to the CS Amplifier
For the common source amplifier referring to Figure 6 we may
write.
CT
CT
= C GS + C GD (1 + A )
= C GS + C GD (1 + g m R L )
CT
= C GS + C GD (1 + A ) = C GS + C GD (1 + g m R L )
Figure 16a: Applying Miller's theorem to the CS amplifier
Figure 6b: Resulting circuit from applying Miller's theorem to the CS amplifier
Open-Circuit Time Constant Method to Determine ωH
At high frequencies, impedances of the coupling and bypass
capacitors are small enough to be considered short circuits.
Open-circuit time constants associated with the impedances of
device capacitances are considered instead.
ωH
≅
1
m
∑ R ioC i
i =1
where Rio is the resistance at the terminals of the ith capacitor Ci
with all other capacitors open-circuited.
Example 2:
Derive an expression for the high cutoff frequency for the circuit in
Example 1 using the OCTC method.
Solution:
Using OCTC method for the circuit in Figure 3.
For CGS the resistance seen across the capacitor terminals may be
calculated from Figure 7.
RGSO = R th
So the time constant associated with CGS equals CGS. Rth
Figure 7: Circuit used to calculate the open-circuit time constant associated with CGS
For CGD the resistance seen across the capacitor terminals may be
calculated from Figure 8.
vx
ix
RGDO =
RGDO = Rth ⋅ (1 + g m R L′ +
R L′
)
Rth
vx
R′
= R th .(1 + g m R L′ + L )
ix
R th
RGDO =
So the time constant associated with CGD equals CGD · Rth (1 +
gmR`L + R`L/Rth)
Figure 8: Circuit used to calculate the open-circuit time constant associated with CGD
The high cutoff frequency is calculated using the SCTC as:
ωH
≅
1
2
∑ R ioC i
i =1
1
+ RGDOC GD
ωH ≅
ωH
ωH
RGSOC GS
1
≅
R thCT
1
1
≅ 2
=
∑ R ioC i RGSO C GS + RGDO C GD
i =1
=
1
R thC T
Lecture 05 a
Common Emitter Amplifier Frequency Response (1)
Objectives
To analyze the CE amplifiers and to calculate different poles
and zeros which determine its frequency response
To calculate the dominant poles and the CE amplifier bandwidth
Introduction
In the last lectures the frequency response of the amplifier circuits
were examined.
Also, the frequency response of the common source amplifier was
calculated and the dominant poles were determined.
In this lecture the frequency response of the common emitter
amplifier will be considered using the SCTC and OCTC techniques
introduced in the last lectures.
Short-Circuit Time Constant Method to Determine ωL
As mentioned in the last lectures midband gain and upper and
lower cutoff frequencies that define bandwidth of the amplifier
are of more interest than complete transfer function.
Lower cutoff frequency for a network with n coupling and
bypass capacitors is given by:
1
i =1 R C
iS i
n
ωL ≅ ∑
where RiS is the resistance at terminals of the ith capacitor Ci with
all other capacitors replaced by short circuits.
Product RiS Ci is the short-circuit time constant associated with Ci.
In the next example the low cutoff frequency of the CE amplifier
will be determined using the SCTC method.
Example 1:
Derive an expression for the low cutoff frequency for the CE
amplifier circuit in Figure 1 using the SCTC method.
Figure 1: Common Emitter Amplifier Circuit
Solution:
The small signal circuit may be obtained by short circuiting the DC
supply as shown in Figure 2.
Figure 2: CE amplifier circuit used for small signal analysis
RB is the equivalent resistance for R1 and R2 given by:
R B = R1 || R 2
Using SCTC method, for C1 the resistance seen across the
capacitor terminals may be calculated from Figure 3.
R1S = R sig + (R B R inCE )
R1S = R sig + (R B rπ )
So the time constant associated with C1 equals C1·(Rsig + RB||rπ )
B
Figure 3: Circuit used to calculate the short-circuit time constant associated with C1
For C2 the resistance seen across the capacitor terminals may be
calculated from Figure 4.
CE
R 2S = R L + (RC Rout
)
R 2S = R L + (RC ro )
R 2S ≅ R L + RC
So the time constant associated with C2 equals C2·(RL + RC)
Figure 4: Circuit used to calculate the short-circuit time constant associated with C2
For CE the resistance seen across the capacitor terminals may be
calculated from Figure 5.
R ES = R E R CC
out
R ES
= RE
rπ + (R sig R B )
βo + 1
So the time constant associated with CE equals CE·RES.
Figure 5: Circuit used to calculate the short-circuit time constant associated with CE
The low cutoff frequency is calculated using the SCTC as:
1
i =1 R C
iS i
3
ωL ≅ ∑
ωL ≅
C 1 (R sig
1
1
+
+
+ (R B rπ )) C 2 (R L + RC )
1
⎛
r + (R sig R B ) ⎞
C E ⎜ RE π
⎟⎟
⎜
βo + 1
⎝
⎠
Please note that due to the finite input resistance of the CE
amplifier compared to the CS amplifier, the lower cutoff frequency
will be higher in the CE amplifier compared to the CS amplifier.
Example 2:
Calculate the low cutoff frequency for the common emitter amplifier
in Figure 1 using the SCTC method. Assume: VCC=12V, Rsig = 1kΩ,
R1 = 10kΩ, R2 = 30kΩ, RC = 4.3kΩ, RE = 1.3kΩ, RL = 100kΩ, C1 =
2μF, C2 = 0.1μF, and CE = 50μF.
Solution:
Refer to Example 1 solution.
ωL ≅
C 1 (R sig
1
1
+
+
+ (R B rπ )) C 2 (R L + RC )
1
⎛
r + (R sig R B ) ⎞
C E ⎜ RE π
⎟⎟
⎜
1
β
+
o
⎝
⎠
To calculate ωL we need to find rπ, this value may be found from
the dc operating point of the amplifier.
The DC circuit of the amplifier may be obtained from the CE
amplifier circuit in Figure 1 by considering all the capacitors are
open circuits and the AC sources are zero.
Performing this on the circuit in Figure 1 results in the circuit given
in Figure 6.
Figure 6: DC circuit for the CE amplifier in Figure 1
The circuit in Figure 6 may be further simplified using Thevinin's
Theorem as shown in Figure 7.
Figure 7: Simplified circuit for DC analysis
Rth = R B = R1 || R 2 = 7.5k Ω
V th = V CC ⋅
R1
10
= 12 ×
= 3V
R1 + R 2
40
Applying KVL in the B-E loop and assuming that the transistor is in
the active mode:
∴
V th = I B R th +V BE + I E R E
∴
V th = V BE + I B (R th + (1 + βo )R E )
∴
IB
=
V th −V BE
(R th + (1 + βo )R E )
∴
IB
=
3 − 0.7
= 16.6 μ A
7.5 + 101× 1.3
I C = βo I B ≅ I E = 16.6mA
Applying KVL at the C-E loop
∴
V CC = I C .RC +V CE + I E .R E
∴
V CE = 12 −1.66 × (1.3 + 4.3) = 2.7V
Since VCE > VCEsat , then the transistor is in active region as
assumed.
Q-point = (16.6mA, 2.7V)
rπ
VT
IB
=
⇒r
π
=
25mV
16.6 μ A
= 1.5k Ω
Substitute in the expression of ωL
ωL ≅ 222 + 96 + 750 = 1068 rad sec
Lecture 5b
Common Emitter Amplifier Frequency Response (2)
Objectives
„ To analyze the CE amplifiers and to calculate different poles and zeros
which determine its frequency response
„ To calculate the dominant poles and the CE amplifier bandwidth
„ To analyze the effect of the emitter resistor on the CE amplifier gain
and bandwidth
Introduction
In the last lectures the frequency response of the amplifier circuits were
examined.
Also, the frequency response of the common source amplifier was calculated
and the dominant poles were determined.
In this lecture the frequency response of the common emitter amplifier will be
considered using the SCTC and OCTC techniques introduced in the last
lecture.
The design trade-off using the emitter resistor will also be explored.
Direct Determination of High-Frequency Poles and Zeros:
Similar to what we did in the last lectures.
The high frequency poles and zeros may be determined by direct analysis.
In this case all the coupling and bypass capacitors may be considered short
circuit.
However, the high frequency model of the transistor should be used since the
transistor internal capacitances can't considered open circuit at high
frequencies.
The following example will illustrate this method for the common emitter
amplifier.
Example 1:
Drive an expression for the high frequency response AvH(s) of the common
emitter amplifier shown in Figure 1. Hence, determine the midband gain Amid,
high frequency poles and zeros, and the high cutoff frequency.
Figure 1: Common Emitter Amplifier Circuit
Solution:
The CE amplifier is redrawn for small signal analysis as shown in Figure 2 by
grounding the DC voltage source VCC and replacing the transistor by its high
small signal model.
The biasing resistors R1 and R2 are combined into RB.
∴
RB
= R1 || R 2
All the coupling and bypass capacitors are considered short circuit at high
frequency.
Figure 2: The CE amplifier in Figure 2 redrawn with the transistor
replaced by its high frequency small signal model
The small-signal model can be simplified by using Thevinin's theorem as
shown in Figure 3.
The simplified circuit is shown in Figure 4.
Figure 3: Using Thevinin's Theorem to simplify the CE amplifier circuit in
Figure 2
v th = v sig
v th 2 =
RB
R sig + R B
and Rth =
v th .rπ
Rth + rx + rπ
R sig R B
R sig + R B
and rth 2 = rπ ( R th + rx )
Also, R L′ = R L || RC || ro
Figure 4: Simplified circuit to calculate the frequency response of the CE
amplifier
Writing the two nodal equations at nodes A and B in the frequency domain,
v be′ ⋅ sC π +
v be′ −v th 2
+ (v be′ −v o )sC μ = 0
Rth 2
vo
+ g mv gs + (v o −v be′ )sC μ = 0
R L′
Solving the last two equations by eliminating v`be yields,
vo
v th2 (s) (sC μ - g m )
R th 2
Δ
(s) =
⎛C
⎛
1
1
Δ = s 2 (C πC μ ) + s ⎜ π + C μ ⎜ g m +
+
⎜ R′
R th 2 R L′
⎝
⎝ L
∴
AvH (s ) =
⎞⎞
1
⎟ ⎟⎟ +
R L′ ⋅ R th 2
⎠⎠
(sC μ - g m )
v o (s )
RB
=
v sig (s )
(R th + rx ) ⋅ ( R sig + R B )
Δ
Let us define
CT
⎛
R′ ⎞
= C π + C μ ⎜1 + g m R L′ + L ⎟
R th 2 ⎠
⎝
∴
Δ = s 2C π C μ +
∴
AvH (s ) =
sC T
R L′
+
1
R L′ ⋅ R th 2
RB
⋅
(R th + rx ) ⋅ (R sig + R B ) ⋅C π ⋅C μ s 2 +
(sC μ - g m )
sC T
1
+
R L′ ⋅C π ⋅C μ R L′ ⋅ R th 2 ⋅C π ⋅C μ
Amid may be found from the last expression by assuming s → 0 .
∴
A mid =
∴
A mid =
A mid
=
- g m R L′ ⋅ R B ⋅ R th 2
( R th + rx ) ⋅ ( R sig + R B )
- g m R L′ ⋅ R B ⋅ rπ
( rπ + R th + rx ) ⋅ ( R sig + R B )
-βo R L′ ⋅ R B
( rπ + R th + rx ) ⋅ ( R sig + R B )
High-frequency response is given by 2 poles, one finite zero and one zero at
infinity.
Finite right-half plane zero, ωZ = + gm/Cμ > ωT can easily be neglected.
For a polynomial s2 + sA1 + A0 with roots a and b, if one root is much larger
than the other one, we can assume that a =A1 and b=A0/A1.
∴
ωP 1 =
A0
A1
∴
ωP 1 ≅
1
Rth 2CT
Assume that the midband gain is very large
⎛
R′ ⎞
( i.e. g m R L′ ⎜ 1 + L ⎟ , and
⎝ R th 2 ⎠
g m R L′C μ C π )
∴
ωP 2 ≅
gm ⎛
1
1 ⎞
+
⎜1 +
⎟
C π ⎝ g m R th 2 g m R L′ ⎠
∴
ωP 2 ≅
gm
Cπ
Smallest root that gives first pole limits frequency response and determines
ωH.
Second pole is important in frequency compensation as it can degrade phase
margin of feedback amplifiers.
Assuming ωp1 ωp2 (ωp1 is the dominanat pole)
∴
ωH ≅ ωP 1 =
1
Rth 2CT
Dominant pole model at high frequencies for CE amplifier is shown in Figure
5.
Figure 5: Equivalent circuit assuming dominant pole model
Example 2:
Find the midband gain, high frequency poles and zeros, ωH, and the
bandwidth for the CE amplifier in Example 1. Given the following transistor
parameters fT = 500 MHz, βo =100, Cμ = 0.5 pF, rx =250 Ω, and VA = 10V.
Solution:
Referring to example 2 solution in Lecture 5a, the Q-point is (1.66 mA, 2.7 V)
IC
VT
∴
gm =
∴
g m = 40 ⋅ I C
rπ
= 66.4mS
= 1.5k Ω
ro =
Cπ =
VA
IC
=
gm
2π f T
100
= 60.2k Ω
1.66
− C μ = 20.6pF
R L′ = R L || RC || ro
R L′ = 100 || 4.3 || 60.2 = 3.86kΩ
Rth = R B R sig = 7.5 1 = 0.88kΩ
R th 2 = rπ ( R th + rx ) = 0.64 k Ω
CT
⎛
R′ ⎞
= C π + C μ ⎜ 1 + g m R L′ + L ⎟
R th 2 ⎠
⎝
CT
= 152.3pF
ωP 1 =
1
Rth 2CT
= 10.26M rad/sec
ωP 2 ≅
gm ⎛
1
1 ⎞
+
⎜1 +
⎟
C π ⎝ g m R th 2 g m R L′ ⎠
ωP 2 ≅ 3.31 G rad/sec
ωz =
gm
Cμ
= 132.8 G rad/sec
It is clear that ωp1 is much lower than ωp2 and ωz which means that ωp1 is the
dominant pole.
∴
ωH ≅ ωP 1 = 10.26 M rad/sec
fH
=
ωH
= 1.63 MHz
2π
BW
.
= f H −f L
BW
.
= 1.63MHz − 170 Hz
A
A
≅ 1.63MHz
- βo R L′ ⋅ R B
( rπ + R th + rx ) ⋅ ( R sig + R B )
mid
=
mid
= − 129.5 V/V
Gain-Bandwidth Product Limitations of the C-E Amplifier
As we can see from the expressions of the midband gain and the CE amplifier
bandwidth, Rth is appearing in both expressions.
If Rth is reduced to zero in order to increase the bandwidth, then Rth2 would not
be zero but would be limited to approximately rx.
GBW = Av ωH
⎛
⎞⎛ 1 ⎞
βo R L′
≤ ⎜
⎟⎜
⎟
⎝ R th + rx + rπ ⎠ ⎝ R th 2C T ⎠
If Rth = 0, rx << rπ so that rx = Rth2 and CT
= C μ ( g m R L′ )
∴
GBW ≤
1
rx C μ
If we used the same values in Example 2
⇒
GBW ≤ 8G rad/sec
The Actual GBW in Example 2 is 1.334G rad/sec.
Open-Circuit Time Constant Method to Determine ωH
As mentioned in the last lectures the open-circuit time constants
associated with the transistor capacitances may be used to simplify the
determination of the high cutoff frequency of the amplifier.
In the next example the high cutoff frequency of the CE amplifier will be
determined using the OCTC method.
Example 3:
Derive an expression for the high cutoff frequency for the circuit in Example 2
using the OCTC method.
Solution:
Using OCTC method for the circuit in Figure 4
For Cπ the resistance seen across the capacitor terminals may be calculated
from Figure 6.
R πO = R th 2
So the time constant associated with Cπ equals Cπ· Rth2
Figure 6: Circuit used to calculate the open-circuit time constant
associated with Cπ
For Cμ the resistance seen across the capacitor terminals may be calculated
from Figure 7.
R μo =
vx
R′
= Rth 2 ⋅ (1 + g m R L′ + L )
ix
Rth 2
So the time constant associated with Cμ equals Cμ· Rth2 (1 + gmR`L + R`L/ Rth)
Figure 7: Circuit used to calculate the open-circuit time constant
associated with Cμ
The high cutoff frequency is calculated using the OCTC as:
ωH
≅
1
2
∑ R ioC i
=
1
Rπo C π + R μ o C μ
=
1
rth 2 C T
i =1
Which is the same expression found before using direct calculations in
Example 3 of lecture 5a.
Gain-Bandwidth Trade-off Using Emitter Resistor
The emitter resistance of the common emitter amplifier may be used to trade
off between the gain and the amplifier bandwidth.
In this case no bypass capacitor is used.
The small signal circuit is shown in Figure 8.
Where Rth and vth are the same as calculated in Example 3 of lecture 5a.
v th = v sig
RB
R sig + R B
and Rth =
R sig R B
R sig + R B
Figure 8: Small signal CE Amplifier circuit at high frequency with no
bypass capacitor
The midband gain may be calculated from Figure 8 by assuming the transistor
capacitors to be open circuit.
A mid
for
= −
βo R L′ .R B
( R th + rx + rπ + ( βo + 1) R E ) ⋅ ( R sig + R B )
rπ >> R th + rx , rB >> R sig and g m ⋅ R E
A mid ≅ −
>> 1
R L′
RE
The midband voltage gain decreases as the emitter resistance increases and
the bandwidth of CE amplifier will correspondingly increase.
To find the effect of RE on the amplifier bandwidth the high cutoff frequency
may be calculated as follows:
Using OCTC method for the circuit in Figure 8
For Cπ the resistance seen across the capacitor terminals may be calculated
as rπ || Req.
Where Req is calculated from Figure 9.
Req =
Rth + rx + R E
vx
=
ix
1+ g m RE
R π O = rπ Req ≅
Rth + rx + R E
1+ g m RE
So the time constant associated with Cπ equals Cπ · Rπo
Figure 9: Circuit used to calculate the open-circuit time constant
associated with Cπ
For Cμ the resistance seen across the capacitor terminals may be calculated
from Figure 10.
Figure 10: Circuit used to calculate the open-circuit time constant
associated with Cμ
To simplify the calculation, the test source ix is first split into two equivalent
sources as shown in Figure 11 and then superposition is used to find vx = (vb
- vc).
Figure 11: Modified circuit used to calculate the open-circuit time
constant associated with Cμ
Assuming that βo >>1 and ( R th + rx
R μo =
)
<<
( rπ + ( βo + 1)R E )
vx
ix
⎛
g m R L′
R L′ ⎞
R μ o = (R th + rx ) ⎜ 1 +
+
⎟
⎝ 1 + g m R E R th + rx ⎠
So the time constant associated with Cμ equals Cμ · Rμo
The high cutoff frequency is calculated using the OCTC as:
ωH
≅
1
2
∑ R ioC i
i =1
ωH ≅
1
⎛ Cπ
⎛
⎛
R L′
RE ⎞
g m R L′
(R th + rx ) ⎜⎜
+
⎜1 +
⎟ + C μ ⎜1 +
⎝ 1 + g m R E R th + rx
⎝ 1 + g m R E ⎝ R th + rx ⎠
⎞⎞
⎟ ⎟⎟
⎠⎠
If we used the same amplifier discussed in Example 2 with CE = 0, we will
obtain:
A mid ≅ −
ωH
ωH
≅
R L′
RE
=
−3.86
= − 2.97 V/V
1.3
1
⎛ Cπ
(R th + rx ) ⎜⎜
⎝ 1+ g m RE
⎞
⎛
g m R L′
R L′ ⎞ ⎞
+
⎟ + C μ ⎜1 +
⎟ ⎟⎟
⎠
⎝ 1 + g m R E R th + rx ⎠ ⎠
1
=
⎛ 20.6 pF ⎛
1.3
66.4 × 3.86
3.86 ⎞ ⎞
⎞
⎛
(0.88 + 0.25)k Ω ⎜
+
⎜1 +
⎟ + 0.5 pF ⎜ 1 +
⎟⎟
⎝ 1 + 66.4 * 1.3 0.88 + 0.25 ⎠ ⎠
⎝ 1 + 66.4 ×1.3 ⎝ 0.88 + 0.25 ⎠
⎛
RE
⎜1 +
⎝ R th + rx
ωH = 211 M rad/sec
Note that this is a higher bandwidth on the expense of a lower midband gain.
Lecture 6
More Single Stage Amplifiers Frequency Response
Objectives
„ To calculate the dominant poles and the amplifier bandwidth for the remaining
BJT single stage amplifier configurations namely the common-base and the
common-collector amplifiers.
„ To calculate the dominant poles and the amplifier bandwidth for the remaining
MOSFET single stage amplifier configurations namely the common-gate and
the common-drain amplifiers.
„ To compare the bandwidth of the different single stage amplifier
configurations.
Introduction
In the last two lectures the frequency response of the amplifier circuits were
examined. Also, the frequency responses of the common-source and common-emitter
amplifiers were calculated and the dominant poles were determined. In this lecture the
frequency response of the other single stage amplifier configurations will be
considered using the SCTC and the OCTC techniques. A comparison between the
different configurations from the point of view of the amplifier bandwidth will be
considered.
Common-Base amplifier low cut-off frequency ωL estimation
•
In the next example the low cutoff frequency of the CB amplifier will be
determined using the SCTC method.
Example 1: Derive an expression for the low cutoff frequency of the CB amplifier
circuit in Figure 1 using the SCTC method. Calculate the numerical value
assuming βo =100, VCC=VEE=5V.
Figure 1 Common Base Amplifier Circuit
Solution:
The small signal circuit may be obtained by short circuit the DC supply as shown in
Figure 2.
Figure 2 CB amplifier circuit used for small signal analysis
Using SCTC method. For C1 the resistance seen across the capacitor terminals may be
calculated from Figure 3.
(
)
R1S = R sig + R E R inCB = R sig + ( R E re )
So the time constant associated with C1 equals C1.(Rsig+RE||re )
Figure 3 Circuit used to calculate the short-circuit time constant associated with C1
For C2 the resistance seen across the capacitor terminals may be calculated from
Figure 4.
CE
R 2S = R L + (RC R out
) ≅ R L + (RC ro ) Using Miller's Thereom and assuming high gain
≅ R L + RC
So the time constant associated with C2 equals C2.(RL+RC)
Figure 4 Circuit used to calculate the short-circuit time constant associated with C2
The low cutoff frequency is calculated using the SCTC as:
1
1
1
=
+
i =1 R iS C i
C 1 ( R sig + ( R E re )) C 2 ( R L + RC )
2
ωL ≅ ∑
To calculate ωL we need to find re, this value may be found from the dc operating
point of the amplifier.
Figure 5 DC circuit for the CB amplifier in Figure 1
The DC circuit of the amplifier may be obtained from the CE amplifier circuit in
Figure 1 by considering all the capacitors are open circuits and the AC sources are
zero. The DC circuit is shown in Figure 5.
Applying KVL in the B-E loop and assuming that the transistor is in the active mode:
∴V BE + I E R E −V EE = 0
V EE −V BE 5 − 0.7
=
= 0.1mA
43
RE
I C = αo I E ≅ 0.1m A
∴I E =
Applying KVL at the C-E loop
∴V CC = I C .RC +V CE + I E .R E +V EE
∴V CE = 5 − 0.1*(22 + 43) + 5 = 3.5V
Since VCE > VCEsat , then the transistor is in active region as assumed.
Q-point = (0.1mA, 3.5V)
V T 25mV
=
= 250Ω
I E 0.1mA
Substitute in the expression of ωL
ωL ≅ 610 + 10 = 620rad / sec
re =
Common-Gate amplifier low cut-off frequency ωL estimation
•
In the next example the low cutoff frequency of the CG amplifier will be
determined using the SCTC method.
Example 2: Derive an expression for the low cutoff frequency of the CG amplifier
circuit in Figure 6 using the SCTC method. Calculate the numerical
value assuming that the MOSFET is biased such that gm=1mA/V.
Figure 6 Common Gate Amplifier Circuit
Solution:
Using SCTC method. For C1 the resistance seen across the capacitor terminals may be
calculated from Figure 7.
⎛
1 ⎞
R1S = R sig + R S R inCG = R sig + ⎜⎜ R S
⎟
g m ⎟⎠
⎝
So the time constant associated with C1 equals C1.(Rsig+RS||(1/gm) )
(
)
Figure 7 Circuit used to calculate the short-circuit time constant associated with C1
For C2 the resistance seen across the capacitor terminals may be calculated from
Figure 8.
CG
R 2S = R L + (R D Rout
) ≅ R L + (R D ro ) Using Miller's Thereom and assuming high gain
≅ RL + RD
So the time constant associated with C2 equals C2.(RL+RD)
Figure 8 Circuit used to calculate the short-circuit time constant associated with C2
The low cutoff frequency is calculated using the SCTC as:
2
1
1
1
ωL ≅ ∑
=
+
i =1 R C
C 2 (R L + R D )
1
iS i
))
C 1 (R sig + ( R S
gm
ωL ≅ 197.5 + 10 = 207.5rad / sec
Please note that since the input resistance of the CG amplifier is higher than that of
the CB amplifier (since the transconductance of the MOSFET is less than that of the
BJT), the low cutoff frequency will be lower in the CG amplifier compared to the CB
amplifier for the same capacitances' and resistances' values.
Common-Collector amplifier low cut-off frequency ωL estimation
•
In the next example the low cutoff frequency of the CC amplifier will be
determined using the SCTC method.
Example 3: Derive an expression for the low cutoff frequency of the CC amplifier
circuit in Figure 9 using the SCTC method. Calculate the numerical value
assuming βo =100, VCC=VEE=5V.
Figure 9 Common Collector Amplifier Circuit
Solution:
The small signal circuit may be obtained by short circuit the DC supply as shown in
Figure 10.
Figure 10 CC amplifier circuit used for small signal analysis
Using SCTC method. For C1 the resistance seen across the capacitor terminals may be
calculated from Figure 11.
(
( (r
)
R 1S = R sig + R B R inCC = R sig + R B
π
+ ( βo + 1) ( R E R L )
))
So the time constant associated with C1 equals C1.R1S
Figure 11 Circuit used to calculate the short-circuit time constant associated with C1
For C2 the resistance seen across the capacitor terminals may be calculated from
Figure 12.
⎛
r + R th ⎞
CC
R 2S = R L + (RC R out
) = RL + ⎜ RE π
⎟
⎜
βo + 1 ⎟⎠
⎝
R th = R sig R B
So the time constant associated with C2 equals C2.R2S
Figure 12 Circuit used to calculate the short-circuit time constant associated with C2
The low cutoff frequency is calculated using the SCTC as:
1
=
i =1 R C
iS i
C 1 R sig + R B
2
ωL ≅ ∑
(
( (r
π
1
+ ( βo + 1) ( R E R L )
)))
+
1
⎛
⎛
r + R th
C 2 ⎜ R L + ⎜⎜ R E π
⎜
βo + 1
⎝
⎝
To calculate ωL we need to find rπ, this value may be found from the dc operating
point of the amplifier.
⎞⎞
⎟⎟ ⎟⎟
⎠⎠
Figure 13 DC circuit for the CB amplifier in Figure 1
The DC circuit of the amplifier may be obtained from the CC amplifier circuit in
Figure 9 by considering all the capacitors as open circuits and the AC sources equal to
zero. The DC circuit is shown in Figure 13.
Applying KVL in the B-E loop and assuming that the transistor is in the active mode:
∴ I B R B +V BE + I E R E −V EE = 0
⎛ R
⎞
∴ I E ⎜ B + R E ⎟ +V BE −V EE = 0
⎝ 1 + βo
⎠
V −V BE
5 − 0.7
∴ I E = EE
=
= 1.08mA
100
RB
+3
+ RE
101
1 + βo
I C = αo I E ≅ 1.08m A
Applying KVL at the C-E loop
∴V CC =V CE + I E .R E +V EE
∴V CE = 5 − 1.08*3 + 5 = 6.76V
Since VCE > VCEsat , then the transistor is in the active region as assumed.
Q-point = (1.08mA, 6.76V)
βV
. T
100* 25mV
= 2315Ω
IC
1.08mA
100
Rth = R sig R B = 1 100 =
= 0.99kΩ
101
Substitute in the expression of ωL
ωL ≅ 133 + 0.21 = 133.2rad / sec
rπ =
=
Common-Drain amplifier low cut-off frequency ωL estimation
•
In the next example the low cutoff frequency of the CD amplifier will be
determined using the SCTC method.
Example 4: Derive an expression for the low cutoff frequency of the CD amplifier
circuit in Figure 14 using the SCTC method. Calculate the numerical value assuming
that the MOSFET is biased such that gm=1mA/V.
Figure 14 Common Drain Amplifier Circuit
Solution:
Using the SCTC method. For C1 the resistance seen across the capacitor terminals
may be calculated from Figure 15.
(
)
R1S = R sig + R S R inCD = R sig + RG
So the time constant associated with C1 equals C1.R1S
Figure 15 Circuit used to calculate the short-circuit time constant associated with C1
For C2 the resistance seen across the capacitor terminals may be calculated from
Figure 16.
⎛
1 ⎞
CG
R 2S = R L + (R S Rout
) ≅ R L + ⎜⎜ R S
⎟
g m ⎟⎠
⎝
So the time constant associated with C2 equals C2. R2S
Figure 16 Circuit used to calculate the short-circuit time constant associated with C2
The low cutoff frequency is calculated using the SCTC as:
2
1
1
1
ωL ≅ ∑
=
+
i =1 R C
⎛
C 1 ( R sig + RG )
⎛
1 ⎞⎞
iS i
C 2 ⎜ R L + ⎜⎜ R S
⎟⎟
⎜
g m ⎟⎠ ⎟⎠
⎝
⎝
ωL ≅ 41 + 0.9 = 41.9rad / sec
Common-Base amplifier high cut-off frequency ωH estimation
•
In the next example the high cutoff frequency of the CB amplifier will be
determined using the OCTC method.
Example 5: Derive an expression for the high cutoff frequency for the circuit in
Example 1 using the OCTC method. Calculate the amplifier bandwidth assuming that
VCC=VEE=5V. Given the following transistor parameters fT =500 MHz, βo =100, Cμ
=0.5 pF,and rx =250Ω, VA= ∞ V.
Solution:
The CB amplifier is redrawn for small signal analysis as shown in Figure 17 by
replacing the transistor in Figure 2 by its high frequency small signal model. Also, all
the coupling and bypass capacitors are considered short circuit at high frequency.
Figure 17 The CB amplifier in Figure 2 redrawn with the transistor replaced by its high
frequency small signal model
The small-signal model can be simplified by using Thevinin's theorem. The simplified
circuit is shown in Figure 18.
v th = v I
R sig R E
RE
and R th =
R sig + R E
R sig + R E
Also, R L′ = R L || RC
Figure 18 Simplified circuit to calculate the high frequency response of the CB amplifier
Using the OCTC method for the circuit in Figure 18. For Cπ the resistance seen across
the capacitor terminals may be calculated from Figure 19-a. The calculation may be
simplified by considering the circuit in Figure 19-b.
R πO = rπ
vx
ix
v x = v b ′ −v e = i x rx − ( g m v x − i x )Rth
v x (1 + g m R th ) = i x (Rth + rx )
R +r
vx
= th x
i x 1 + g m R th
Rth + rx
R π o = rπ
1 + g m Rth
Assuming that βo >>1 and rx << rπ
R +r
R πo ≅ th x
1 + g m Rth
So the time constant associated with Cπ equals Cπ. Rπo
e
rπ
Rth
b′
c
vbe
+
gm vbe
Rπo
R′L
rx
b
(a)
(b)
Figure 19 Circuit used to calculate the open-circuit time constant associated with Cπ
For Cμ the resistance seen across the capacitor terminals may be calculated from
Figure 20. Note that split-source transformation is used to ease the calculation.
Assuming that βo >>1 and rx << rπ
These assumptions will imply that ib=0 or ie=ic=gmvb`e as well as all the current
sourced from ix which shunt rx will pass in rx only.
∴v b ′ ≅ i x rx
Using KVL in the loop containing rx, rπ, and Rth.
∴v b ′ = i x rx ≅ v b ′e + g mv b ′e R th = v b ′e (1 + g m Rth )
i x rx
(1 + g m R th )
Writing the nodal equation at the collector node
⎛
⎞
g m rx
∴v c = −( g mv b ′e + i x ).R L′ = −i x . ⎜ 1 +
⎟ .R L′
g
R
(1
)
+
m th ⎠
⎝
∴v b ′e =
⎛
g m R L′ ⎞
⎜1 +
⎟ + R L′
⎝ 1 + g m R th ⎠
So the time constant associated with Cμ equals Cμ. Rμo
R μo =
v x v b′ − v c
=
= rx
ix
ix
Figure 20 Circuit used to calculate the open-circuit time constant associated with Cμ
The high cutoff frequency is calculated using the OCTC as:
1
1
ωH ≅ 2
=
=
⎛ Rth
∑ R ioC i Rπ oC π + R μoC μ r ⎛ C π
x ⎜
i =1
⎜ 1 + g R ⎜1 + r
m th ⎝
x
⎝
1
⎛
⎞
g m R L′
⎟ + C μ ⎜1 +
⎠
⎝ 1 + g m R th
⎞⎞
⎟ ⎟⎟ + C μ R L′
⎠⎠
Neglecting the first term which is of order of 1/ ωT and hence the last term will be
dominant.
∴ω
H
≅
1
C μ R L′
R L′ = R L R C = 75 22 = 17kΩ
ωH =
1
= 117.6M rad/sec
C μ R L′
BW =
ωH − ωL ωH
≅
= 18.72MHz
2π
2π
By comparing the B.W of the CB amplifier to that of the CE amplifier we find the
B.W of the CB is much higher. This may be explained by the absence of the Miller's
capacitance associated with the common emitter configuration.
Common-Gate amplifier high cut-off frequency ωH estimation
•
In the next example the high cutoff frequency of the CG amplifier will be
determined using the OCTC method.
Example 6: Derive an expression for the high cutoff frequency and the bandwidth for
the CG amplifier in Example 2 using the OCTC method.
Solution:
The CG amplifier is redrawn for small signal analysis as shown in Figure 21 by
replacing the transistor in Figure 6 by its high frequency small signal model. Also, all
the coupling and bypass capacitors are considered short circuit at high frequency.
Figure 21 The CG amplifier in Figure 6 redrawn with the transistor replaced by its high
frequency small signal model
The small-signal circuit can be simplified by using Thevinin's theorem as shown in
Figure 22.
v th = v I
R sig R S
RS
and Rth =
R sig + R S
R sig + R S
Also, R L′ = R L || R D
Figure 22 Simplified circuit to calculate the high frequency response of the CG amplifier
Using the OCTC method for the circuit in Figure 22. For CGS the resistance seen
across the capacitor terminals may be calculated from Figure 23.
Writing KCL at the source node for the circuit in Figure 23-b
v
i x = g m .v x + x
Rth
Rth
v
RGSo = x =
i x 1 + g m Rth
So the time constant associated with CGS equals CGS. RGSo
(a)
(b)
Figure 23 Circuit used to calculate the open-circuit time constant associated with CGS
For CGD the resistance seen across the capacitor terminals may be calculated from
Figure 24.
Applying KVL
∴ −v gs = g mv gs R th
∴v gs (1 + g m R th ) = 0 → v gs = 0
vx
= R L′
ix
So the time constant associated with CGD equals CGD. R'L
RGDo =
Figure 24 Circuit used to calculate the open-circuit time constant associated with CGD
The high cutoff frequency is calculated using the OCTC as:
1
1
1
ωH ≅ 2
=
=
∑ R ioC i RGSo CGS + RGDo CGD C ⎛ R th
GS ⎜
i =1
⎝ 1 + g m R th
BW =
⎞
⎟ + C GD R L′
⎠
ωH − ωL ωH
≅
2π
2π
By comparing the B.W of the CG amplifier to that of the CS amplifier we find the
B.W of the CG is much higher. This may be explained by the absence of the Miller's
capacitance associated with the common source configuration.
Common-Collector amplifier high cut-off frequency ωH estimation
•
In the next example the high cutoff frequency of the CC amplifier will be
determined using the OCTC method.
Example 7: Derive an expression for the high cutoff frequency for the circuit in
Example 3 using the OCTC method. Calculate the amplifier bandwidth assuming that
VCC=VEE=5V. Given the following transistor parameters fT =500 MHz, βo =100, Cμ
=0.5 pF,and rx =250Ω, VA= ∞ V.
Solution:
The CC amplifier is redrawn for small signal analysis as shown in Figure 25 by
replacing the transistor in Figure 10 by its high frequency small signal model. Also,
all the coupling and bypass capacitors are considered short circuit at high frequency.
Figure 25 The CC amplifier in Figure 10 redrawn with the transistor replaced by its high
frequency small signal model
The small-signal model can be simplified by using Thevinin's theorem as shown in
Figure 26.
v th = v I
R sig R B
RB
and R th =
R sig + R B
R sig + R B
Also, R L′ = R L || R E
Figure 26 Simplified circuit to calculate the high frequency response of the CC amplifier
Using the OCTC method for the circuit in Figure 26. For Cπ the resistance seen across
the capacitor terminals may be calculated from Figure 27-a. The calculation may be
simplified by considering the circuit in Figure 27-b.
R πO = rπ
vx
ix
v x = v b ′ −v e = i x (Rth + rx ) − ( g m v x − i x )R L′
v x (1 + g m R L′ ) = i x (Rth + rx + R L′ )
v x Rth + rx + R L′
=
1 + g m R L′
ix
R + r + R L′
R π o = rπ th x
1 + g m R L′
R th + rx + R L′
1 + g m R L′
Rπo ≅
So the time constant associated with Cπ equals Cπ. Rπo
(a)
(b)
Figure 27 Circuit used to calculate the open-circuit time constant associated with Cπ
For Cμ the resistance seen across the capacitor terminals may be calculated from
Figure 28.
R μo = (R th + rx ) R CC
in = ( R th + rx ) ( rπ + ( βo + 1) R L′ )
≅ (Rth + rx )
So the time constant associated with Cμ equals Cμ. Rμo
Figure 28 Circuit used to calculate the open-circuit time constant associated with Cμ
The high cutoff frequency is calculated using the OCTC as:
1
1
1
ωH ≅ 2
=
=
C
π
∑ R ioC i Rπ oC π + R μoC μ (R th + rx + R L′ )
+ (Rth + rx )C μ
i =1
1 + g m R L′
A better estimate is obtained if we assumed that R'L << (Rth+rx) in the expression of
Rπo.
1
∴ωH ≅
⎛ Cπ
⎞
+C μ ⎟
( R th + rx ) ⎜
⎝ 1 + g m R L′
⎠
Referring to example 2 solution, the Q-point is (1.08mA, 6.76V)
IC
= 40.I C = 43.2mS
VT
βV
. T 100* 25mV
rπ =
=
= 2315Ω
IC
1.08mA
100
Rth = R sig R B = 1 100 =
= 0.99kΩ
101
gm =
Cπ =
gm
− C μ = 13.25pF
2π f T
R L′ = R L R C = 47 3 = 2.82kΩ
ωH =
1
Cπ
(R th + rx + R L′ )
+ (R th + rx )C μ
1 + g m R L′
1
= 945M rad/sec
(438 + 620) *10−12
ω − ωL ωH
BW = H
≅
= 150MHz
2π
2π
By comparing the B.W of the CC amplifier to that of the CE amplifier we find the
B.W of the CC is much higher. This may be explained by the absence of the Miller's
capacitance associated with the common emitter configuration.
=
Common-Drain amplifier high cut-off frequency ωH estimation
•
In the next example the high cutoff frequency of the CD amplifier will be
determined using the OCTC method.
Example 8: Derive an expression for the high cutoff frequency and the bandwidth for
the CD amplifier in Example 4 using the OCTC method.
Solution:
The CG amplifier is redrawn for small signal analysis as shown in Figure 29 by
replacing the transistor in Figure 6 by its high frequency small signal model. Also, all
the coupling and bypass capacitors are considered short circuit at high frequency.
Note that since the drain is grounded CGD can be redrawn as in Figure 29.
Figure 29 The CD amplifier in Figure 14 redrawn with the transistor replaced by its high
frequency small signal model
The small-signal circuit can be simplified by using Thevinin's theorem as shown in
Figure 30.
v th = v I
R sig RG
RG
and R th =
R sig + RG
R sig + R G
Also, R L′ = R L || R S
Figure 30 Simplified circuit to calculate the high frequency response of the CG amplifier
Using the OCTC method for the circuit in Figure 30. For CGS the resistance seen
across the capacitor terminals may be calculated from Figure 31.
vx
ix
v x = v g −v s = i x .Rth − ( g m v x − i x )R L′
RGSO =
v x (1 + g m R L′ ) = i x (Rth + R L′ )
∴ RGSO =
R + R L′
vx
= th
i x 1 + g m R L′
So the time constant associated with CGS equals CGS. RGSo
G
Rth
D
+
vgs R
GSo
s
R′L
(a)
G
ix
Rth
D
+
vx
s
R′L
(b)
Figure 31 Circuit used to calculate the open-circuit time constant associated with CGS
For CGD the resistance seen across the capacitor terminals may be calculated from
Figure 32.
R GDo = R th R CD
in = R th
So the time constant associated with CGD equals CGD. Rth
s
RGDo
Rth
G
CD
Rin
gm vgs
+
vgs
D
R′L
Figure 32 Circuit used to calculate the open-circuit time constant associated with CGD
The high cutoff frequency is calculated using the OCTC as:
1
1
1
ωH ≅ 2
=
=
∑ R ioC i RGSo CGS + RGDo CGD C ⎛ R th + R L′
GS ⎜
i =1
⎝ 1 + g m R L′
⎞
⎟ + C GD R th
⎠
Please note that this expression may be obtained from that for the common collector
by setting rπ as infinite and rx as zero in the final expression.
BW =
ωH − ωL ωH
≅
2π
2π
By comparing the B.W of the CD amplifier to that of the CS amplifier we find the
B.W of the CG is much higher. This may be explained by the absence of the Miller's
capacitance associated with the common source configuration.
Lecture 07
Differential Amplifier Frequency Response
Objectives
To calculate the dominant poles and the amplifier bandwidth for differential amplifier.
To calculate the frequency response of the CMRR of the differential amplifier.
To calculate the dominant poles and the amplifier bandwidth for the CC-CB amplifier.
Introduction
In the last lectures the frequency response of the single stage amplifier
circuits were examined.
Also, the dependence of the amplifier bandwidth on the amplifier
configuration as well as the amplifier parameters was examined.
In this lecture the frequency response of the differential amplifier will be
considered.
The analysis will be based on the CE and CS analysis done in the previous
lectures.
To accept the proposed methodology, we will start by reviewing the half
circuit analysis technique of the differential amplifier.
Finally, the CC-CB two stage amplifier configuration will be examined and
the similarities with the differential amplifier will be considered.
Analysis of Differential Amplifiers Using Half-Circuits
Half-circuits are constructed by first drawing the differential amplifier in a
fully symmetrical form.
As shown in Figure 1, power supplies are split into two equal halves in
parallel and emitter resistor is separated into two equal resistors in parallel.
a) BJT Differential Amplifier
b) MOSFET Differential Amplifier
Figure 1: Redrawn Differential Amplifier Circuit for Half Circuit Technique
None of the currents or voltages in the circuit is changed.
For differential mode signals, points on the line of symmetry are virtual
grounds connected to ground for AC analysis.
For common-mode signals, points on line of symmetry are replaced by
open circuits.
Example 1:
Using the half circuit technique, calculate the single ended voltage gain,
differential output voltage gain, and the differential input resistance for the
circuit in Figure 1-a assuming that v1 = -v2 = vid/2.
Solution:
Applying the given rules for drawing the half-circuits, the two power supply
lines and emitter become AC grounds.
The half-circuit represents a C-E amplifier stage as shown in Figure 2.
RC
RC
vC1
ib
vC2
ib
+ vod
vid
2
vid
2
Figure 2: BJT Differential Amplifier Half Circuit for Differential Input Calculations
Direct analysis of the half-circuits yield:
vc1 = − g m RC
vid
2
vc2 = + g m RC
vid
2
∴
A dSE
=
v c1
v id
=
g m RC
2
vod = vc1 − vc2 = − g m RC vid
∴
R id
A dd
=
=
v od
vid
vid
= 2rπ
ib
= g m RC
Example 2:
Using the half circuit technique, calculate the single ended voltage gain,
differential output voltage gain, and the common-mode input resistance for
the circuit in Figure 1-a assuming that v1 = v2 = vic. Write an expression for
the common-mode rejection ratio (CMRR).
Solution:
Applying the given rules for drawing the half-circuits, the half-circuit for
small signal analysis will represent a C-E amplifier stage with emitter
resistance 2 REE as shown in Figure 3.
Figure 3: BJT Differential Amplifier Half Circuit for Coomon Mode Input Calculations
Direct analysis of the half-circuits yield:
vc1 = vc2 =
−α RC
v ic
re + 2R EE
vc1 = vc2 ≅
− RC
v ic
2R EE
∴
A cSE
=
v c1
vid
=
RC
2R EE
vod = v c1 − vc2 = 0
∴
A cd
=
v od
v id
= 0
R ic =
vic
r
= e + R EE ≅ R EE
2i b
2
CMRR =
AdSE
AcSE
= g m R EE
Differential Amplifier Frequency Response
The Differential amplifier is a direct coupled amplifier then ωL = 0
Figure 4 represents the BJT differential amplifier circuit used for frequency
response analysis; CEE is the total capacitance at the emitter node of the
differential pair.
Figure 4: Differential Amplifier Circuit Used for Frequency Response Analysis
By referring to Figure 5, differential mode half-circuit is similar to a C-E
stage.
Figure 5: Differential-mode Half Circuit at High Frequency
Bandwidth is determined by the R π o CT product (refer to lecture (9&10), As
the emitter is a virtual ground, CEE has no effect on differential-mode
signals.
B .W = ωH ≅
B .W = ωH ≅
1
R C
πo T
1
⎛
⎛
( rπ r ) ⎜⎜C π + C μ ⎜⎜1 + g
x
⎝
⎝
m
RC +
RC
rπ rx
⎞⎞
⎟⎟ ⎟⎟
⎠⎠
Figure 6 represents the frequency response of the differential-mode gain
for the differential amplifier.
⎛g R ⎞
20 log ⎜ m C ⎟
⎝ 2 ⎠
Figure 6: Differential gain frequency response
The half circuit used for common-mode high frequency response analysis
is shown in Figure 7.
RC
vC1
Q1
vcm
C EE
2
2REE
Figure 7: Common-mode Half Circuit at High Frequency
Transmission zero due to CEE is
s = − ωZ
= −
1
C EE R EE
Common-mode half-circuit is similar to a C-E stage with emitter resistor
2REE.
OCTC for Cπ and Cμ is similar to the C-E stage (refer to lecture (9&10),
OCTC for CEE / 2 is:
= 2R EE
R EEO
rπ + rx
1
≅
gm
βo + 1
The high frequency pole may be determined using the OCTC (refer to
lecture (9&10), as follows:
ω
≅
ω
≅
P
P
1
3
∑ R C
io i
i =1
⎛
C
π
(r + r ) ⎜
π x ⎜ 1 + 2g R
m EE
⎝
1
⎛
2R
EE
⎜1 +
⎜ r +r
π x
⎝
⎞
⎛
g R
R
m C
C
⎟ + C ⎜1 +
+
μ
⎟
⎜ 1 + 2g R
r +r
π x
m EE
⎠
⎝
⎞⎞ C
⎟ ⎟ + EE
⎟ ⎟ 2g
m
⎠⎠
As REE is usually designed to be large,
∴
ω
P
≅
∴
ω
P
≅
1
2C π + C EE
+ C μ (RC + rπ + rx )
2g m
1
C μ (RC + rπ + rx )
The frequency response of the common-mode gain may be plotted by
noting that at very low frequencies,
AcSE (0) ≅
RC
<< 1 (refer to Example 2)
2R EE
And the calculated transmission zero and high frequency pole above.
Figure 8 represents the frequency response of the common-mode gain for
the differential amplifier.
⎛ R ⎞
20 log ⎜ C ⎟
⎝ 2R EE ⎠
Figure 8: Common-mode Gain Frequency Response
The CMRR may be obtained from Figure 6 and 8 and noting that fp is much
higher than fH.
The CMRR frequency response is shown in Figure 9.
20 log ( g m R EE )
Figure 9: CMRR Frequency Response
Please note that by increasing REE the CMRR will increase.
However, the bandwidth for the CMRR will decrease by the same ratio.
Common Collector – Common Base Cascade Frequency Response
From the last lectures we found that both the CC and CB have much wider
bandwidth compared to the CE.
However, if we want to have a voltage amplifier neither of them will be an
attractive choice.
The voltage gain of the CC is almost unity while the CB has small input
resistance and its voltage gain highly depends on the input source
resistance.
The CC-CB cascade is used to overcome the low input impedance of the
CB amplifier which will eliminate the dependence on the input source
resistance as well.
This configuration entails a high gain voltage amplifier with wide
bandwidth.
The CC-CB amplifier may be considered as a differential amplifier with one
input grounded. The small signal circuit is shown in Figure 10.
Figure 10: CC-CB Cascade Amplifier
REE is assumed to be large and may be neglected.
This will enable us to redraw the circuit in Figure 10 as shown in Figure 11.
Figure 11: CC-CB Amplifier redrawn with REE neglected
Direct analysis of Figure 11 will result in:
CC 1
=
rπ 1 + rx 1
1
≅
g m1
βo 1 + 1
CB 2
=
rπ 2 + rx 2
1
≅
gm2
βo 2 + 1
R out
R in
Sum of the OCTC of Q1 is (refer to lecture 6 (11&12)
⎛
⎞
⎜
⎟
Cπ1
⎛C
⎞
rx 1 ⎜
+ C μ1 ⎟ = rx 1 ⎜ π 1 + C μ1 ⎟
1
⎜ 1+ g
⎟
⎝ 2
⎠
m1
⎜
⎟
g
m2
⎝
⎠
Sum of the OCTC of Q2 is (refer to lecture 6 (11&12)
⎛
⎛
⎜
⎜
Cπ 2
g m RC
+ C μ 2 ⎜1 +
rx 2 ⎜
1
1
⎜ 1+ g
⎜ 1+ g
m2
m2
⎜
⎜
g m1
g m1
⎝
⎝
⎞⎞
⎟⎟
⎟ ⎟ + C μ 2 RC
⎟⎟
⎟⎟
⎠⎠
⎛C
⎛ g R
R ⎞⎞
≅ rx 2 ⎜ π 2 + C μ 2 ⎜1 + m C + C ⎟ ⎟
⎜
rx 2 ⎠ ⎟⎠
2
⎝
⎝ 2
Combining the OCTC for Q1 and Q2, and assuming that the transistors are
matched,
∴
B .W
= ωH
=
1
⎛
⎛
g R
R ⎞⎞
rx ⎜⎜C π + C μ ⎜ 2 + m C + C ⎟ ⎟⎟
rx ⎠ ⎠
2
⎝
⎝
Lecture 08
Multistage Amplifier Frequency Response
Objectives
„ To understand the principle of operation and the advantage of the cascode amplifier.
„ To explore the advantage of the CC-CE cascade amplifier and compare it with the
cascode amplifier.
„ To calculate the dominant poles and the amplifier bandwidth for a multistage
amplifier.
Introduction
In the last lecture the frequency response of the differential amplifier as well as the
common collector-common base (CC-CB) cascade amplifier were examined. As we saw in
the last lecture the obtained high bandwidth in the CC-CB amplifier was obtained by
avoiding the use of the common emitter amplifier which is know with its poor frequency
response. In this lecture the frequency response of the common emitter amplifier will be
enhanced by means of cascading it with other configurations. The analysis will be based on
the single amplifier configurations analysis done in the previous lectures. At the end a
complete cascade amplifier will be considered.
The cascode amplifier
(a) The BJT Cascode
(b) The MOSFET Cascode
Figure 1: The Cascade Amplifier
•
•
•
•
Figure 1 shows the cascode amplifier for both BJT and MOSFET transistors.
The cascode amplifier was initially invented for the triode vacuum tubes amplifiers
to overcome the Miller's capacitance effect.
The name cascode is a shortened version of "cascaded cathode" since the anode of
the first triode was feeding the cathode of the next triode.
The basic idea of the cascode is to combine the high input resistance and large
transconductance of the common emitter amplifier with the current buffer and
excellent frequency performance of the common base amplifier.
•
The cascode amplifier may be designed either to have the same DC gain as CE (CS)
amplifier with a wider bandwidth or higher DC gain with same gain-bandwidth
product. The first possibility will be examined by the following example.
Example 1:
Calculate the high 3dB frequency of the BJT cascode amplifier in Figure 1-a assuming a
load resistor RL is connected at vo.
Solution:
The small signal circuit of the cascode amplifier in Figure 1-a may be drawn as in Figure 2.
From which it is clear that the cascode amplifier may be treated as a common emittercommon base amplifier.
Figure 2: BJT cascode amplifier small signal circuit
To simplify the analysis we may treat the amplifier as a cascade of CE amplifier with an
output resistance = re2 =1/ gm2 and a common base amplifier with source resistance equals
the output resistance of Q1 or ro1.
OCTC of Q1 with load resistor 1/ gm2 may be obtained as (refer to lecture 5, for
Authorware this is the lecture of CE high frequency response):
⎛
⎛
⎞⎞
g
1
R π o 1C π 1 + R μo 1C μ1 = R π o 1C T 1 = R π o 1 ⎜ C π 1 + C μ1 ⎜ 1 + m 1 +
⎟⎟
⎟
⎜
⎝ g m 2 g m 2 Rπo1 ⎠ ⎠
⎝
(
Where R π o 1 = rπ 1 ( rx 1 + R s )
)
As IC2 = IC1, gm2 = gm1, gain of the first stage is unity. Assuming gm2 Rπo1 >>1,
∴ R π o 1C T 1 ≅ R π o 1 (C π 1 + 2C μ1 )
OCTC of Q2 with source resistor ro1 may be obtained as (refer to lecture 6, for Authorware
this is the lecture of CB high frequency response):
R
⎛ Cπ 2 ⎛
⎛
r ⎞
g R ⎞⎞
+R
= rx 2 ⎜
C
C
1 + o 1 ⎟ + C μ 2 ⎜1 + m 2 L ⎟ ⎟ + C μ 2 R L
⎜
⎜
⎟
πo 2 π 2
μo 2 μ 2
⎝ 1 + g m 2 ro 1 ⎠ ⎠
⎝ 1 + g m 2 ro 1 ⎝ rx 2 ⎠
Assuming that ro1 >> RL and μf = gmro>>1:
∴R
C
πo 2 π 2
+R
C
μo 2 μ 2
⎛C
⎞
≅ ⎜ π 2 + C μ 2 ( rx 2 + R L ) ⎟
⎝ gm2
⎠
Assuming matched devices,
∴ ωH =
1
(
)
(
))
(
⎛
1 ⎞
⎜ rπ ( rx + R s ) +
⎟C π + rx + R L + 2 rπ ( rx + R s ) C μ
gm ⎠
⎝
To compare the obtained bandwidth of the cascode with the case of CE amplifier let us
assume the same parameters as in Example 4 in lecture 5 (for Authorware this is the lecture
of CE high frequency response)
g m = 66.4mS
rπ = 1.5k Ω
ro = 60.2k Ω
C μ = 0.5pF
rX = 250Ω
C π = 20.6pF
R L = R L′ = 3.86kΩ
R π o 1 = R th 2 = 0.64k Ω
ωH =
1
= 61.77M rad/sec
⎛
1 ⎞
⎜ R πo1 +
⎟C π + ( rx + R L + 2R π o 1 )C μ
gm ⎠
⎝
The high frequency obtained for the common amplifier in the example was 10.26Mrad/sec.
The CD-CS, CC-CE, and CD-CE cascade Amplifier
VCC
VCC
VDD
I2
Q1
I2
Q1
vo
Q2
I1
vo
Q2
I1
Figure 3: (a) CD–CS amplifier. (b) CC–CE amplifier. (c) CD–CE amplifier.
•
•
As discussed in the last section the cascode configuration was invented to reduce
the effect of Miller's capacitance associated with the CE (CS) amplifier to increase
its bandwidth. The same result may be obtained with different configuration by
cascading a CC (CD) amplifier with the CE (CS) amplifier as shown in Figure 3.
Although, both configuration reduce the Miller's capacitance. The difference
between the cascode and the CC-CE or (CD-CS) may be summarized as:
In the cascode we isolate the load resistance from the CE collector by means of
low input resistance of the CB stage
In the CC-CE the large Miller capacitor sees a small resistance from the output
of the CC which isolate the effect of the source resistance
•
•
Example 2:
Calculate the high 3dB frequency of the BJT CC-CE amplifier in Figure 3-b assuming a
source with a source resistor Rsig is connected at the input.
Solution:
The small signal circuit of the CC-CE amplifier in Figure 3-b may be drawn as in Figure 4.
Rsig
Q1
Vsig
vo
Q2
RL
Figure 4: BJT cascade amplifier small signal circuit
The common collector amplifier has a load resistance equals the input resistance of Q2 or
(rπ2 + rx2) , and The CE amplifier has a source resistance equals the output resistance of Q1
or (Rsig+rx1+ rπ1)/β.
OCTC of Q1 with load resistor rπ2 may be obtained as (refer to lecture 6, for Authorware
this is the lecture of CC high frequency response):
R
Cπ1
C +R
C = (R S + rx 1 + rπ 2 + rx 2 )
+ ( ( R sig + rx 1 ) R in 1 )C μ1
π o1 π 1
μo1 μ1
1 + g m 1rπ 2
R in 1 = rπ 1 + (1 + β1 )(rπ 2 + rx 2 )
Assuming β >> 1, Rsig <<Rin1
R C +R
C ≅ (R sig + rx 1 )C μ1
π o1 π 1
μo1 μ1
OCTC of Q2 with source resistor Ro1 may be obtained as (refer to lecture 5, for
Authorware this is the lecture of CE high frequency response):
R
C
πo 2 π 2
Ro1 =
+R
C
μo 2 μ 2
(
= rπ 2 ( rx 2 + R o 1 )
R sig + rx 1 + rπ 1
1+ β
Assuming matched devices
)
⎛
⎛
⎜C + C ⎜1 + g R
m2 L
π2
μ2
⎜
⎜
⎝
⎝
+
(r
π2
RL
( rx 2 + Ro 1 ) )
⎞⎞
⎟⎟
⎟⎟
⎠⎠
∴ ωH =
1
⎛ ⎛
R sig + rx + rπ
⎜⎜ rπ ⎜ rx +
1+ β
⎝ ⎝
⎛
⎛ ⎛
R sig + rx + rπ
⎞⎞
⎟ ⎟⎟C π + C μ ⎜⎜ (1 + g m R L ) ⎜⎜ rπ ⎜ rx +
1+ β
⎠⎠
⎝ ⎝
⎝
⎞⎞
⎟ ⎟ + R L + R sig + rx
⎠ ⎠⎟
To compare the obtained bandwidth of the CC-CE with the case of CE amplifier let us
assume the same parameters as in Example 4 in lecture 5 (for Authorware this is the lecture
of CE high frequency response)
g m = 66.4mS
rπ = 1.5k Ω
ro = 60.2k Ω
C μ = 0.5pF
rX = 250Ω
C π = 20.6pF
R L = R L′ = 3.86kΩ
R sig = 1k Ω
ωH = 26.8M rad/sec
The high frequency obtained for the common amplifier in the example was 10.26Mrad/sec.
Multistage Amplifier
Multistage amplifiers are utilized either to have a high overall voltage gain or to enhance
the amplifier characteristics by increasing the input resistance, decreasing the output
resistance, or increasing the bandwidth as studied in the previous two sections by using
cascode or CC-CE cascade.
As a final example let us consider multistage amplifier using CS amplifier as an input stage
to provide high input impedance and CC as a final stage to provide low output impedance
with most of the gain provided by a CE amplifier.
Example 3:
Use open-circuit and short-circuit time constant methods to estimate upper and lower cutoff
frequencies and bandwidth of the multistage amplifier shown in Figure 5. The transistors'
parameters are as follows:
M1 has fT1 =200 MHz, λ =0.02V-1, Kn=12.5mA/V2, and CGD1 =1.0 pF.
Q2 has fT2 =500 MHz, βo2 =150, Cμ2 =1.0 pF, rx2 =250Ω, and VA2=80V.
Q3 has fT3 =500 MHz, βo3 =80, Cμ3 =1.0 pF,and rx3 =250Ω, VA3=100V.
The transistors are biased at the following operating points:
M1 is biased at (4mA, 8.72).
Q2 is biased at (1.5mA, 2.7).
Q3 is biased at (2mA, 5.4).
⎞
⎟
⎟
⎠
Figure 5: Multistage Amplifier
Solution:
Using the given operating points given, the small signal transistors' parameters may be
calculated a follows:
g m 1 = 2K n I D 1 (1 + λV DS 1 ) = 10.8mS
1
= 12.5k Ω
λI D1
ro 1 =
C GS 1 =
g m1
− C GD 1 = 7.6 pF
2π f T 1
gm2 =
IC 2
= 40.I C 2 = 60mS
VT
rπ 2 =
1 + βo 2
= 2.52k Ω
gm2
ro 2 =
V A 2 80
=
= 53.3k Ω
I C 2 1.5
Cπ 2 =
gm2
− C μ 2 = 18.1pF
2π f T 2
gm3 =
IC 3
= 40.I C 3 = 80mS
VT
rπ 3 =
1 + βo 3
= 1.01k Ω
gm3
ro 3 =
V A 3 100
=
= 50k Ω
IC 3
2
Cπ3 =
gm3
− C μ 3 = 24.5pF
2π f T 3
Coupling and bypass capacitors determine the low-frequency response.
SCTC for each of the six independent coupling and bypass capacitors has to be determined.
R1S = R I + (RG R in 1 ) = 10kΩ + 1MΩ ∞ = 1.01MΩ
R 2S = R S 1
1
1
= 200Ω
= 63.3 Ω
0.0108S
g m1
R 3S = ( R D 1 RO 1 ) + ( R B 2 R in 2 ) = ( R D 1 ro 1 ) + ( R B 2 rπ 2 ) = 2.79kΩ
R th 2 = R B 2 R D 1 ro 1 = 571Ω
R 4S = R E 2
R th 2 + rπ 2
= 20.2Ω
βo 2 + 1
R 5S = ( RC 2 RO 2 ) + ( R B 3 R in 3 )
(
= ( RC 2 ro 2 ) + R B 3 ( rπ 3 + ( βo 3 + 1)(R E 3 R L ) )
)
= 18.66kΩ
R th 3 = R B 3 RC 2 ro 2 = 3.99kΩ
R 6S = R L + R E 3
Rth 3 + rπ 3
= 311Ω
βo 3 + 1
1
= 3154rad/s
i =1 R iS C i
n
ωL ≅ ∑
fL =
ωL
= 502Hz
2π
Device capacitances affect high-frequency response.
At high frequencies, the coupling and bypass capacitors are short circuits. Small signal
model for the multi-stage amplifier is shown in Figure 6.
Figure 6: Equivalent small signal circuit of the multi-stage amplifier at high frequencies
OCTC for each of the two capacitors associated with each transistor has to be determined.
For M1,
R L 1 = R I 12 rπ 2 = 483Ω
⎛
⎛
R ⎞⎞
R thCT 1 = R th ⎜C GS 1 + C GD 1 ⎜1 + g m 1R L 1 + L 1 ⎟ ⎟
⎜
R th ⎠ ⎟⎠
⎝
⎝
= 0.137 μ s
For Q2,
rπ o 2 = rπ 2 (R th 2 + rx 2 ) = 619Ω
(
)
R L 2 = ( R I 23 R in 3 ) = R I 23 ( rπ 3 + ( βo 3 + 1)(R E 3 R L ) ) = 3.54kΩ
⎛
⎛
R ⎞⎞
rπ o 2C T 2 = rπ o 2 ⎜ C π 2 + C μ 2 ⎜ 1 + g m 2 R L 2 + L 2 ⎟ ⎟ = 0.147 μ s
⎜
rπ o 2 ⎠ ⎟⎠
⎝
⎝
For Q3,
R th 3 = R I 23 ro 2 = 3.99kΩ
R EE = R L R E 3 = 232Ω
⎛
(R th 3 + rx 3 + R EE ) ⎞
R π 3O C π 3 + R μ 3O C μ 3 = ⎜ rπ 3
⎟C π 3 + (R th 3 + rx 3 )C μ 3 = 8.81ns
1 + g m 3 R EE
⎝
⎠
1
ωH ≅ m
= 3.42 ×106 rad/s
∑ R ioC i
i =1
fH =
ωH
= 544kHz
2π
B.W = f H − f L ≅ f H
=
544kHz