Download Set 6A: Frequency Response (Part A)

6. Frequency Response
Reading: Sedra & Smith: Chapter 1.6, Chapter 3.6
and Chapter 9 (MOS portions),
ECE 102, Winter 2011, F. Najmabadi
Typical Frequency response of an Amplifier
 Up to now we have ignored the capacitors. To include the capacitors, we
need to solve the circuit in the frequency domain (or use Phasors).
o Lower cut-off frequency: fL
o Upper cut-off frequency: fH
o Band-width: B = fH − fL
Classification of amplifiers based on the
frequency response
AC amplifier (capacitively-coupled)
DC amplifier (directly-coupled)
fL = 0
Tuned or Band-pass amplifier (High Q)
How to find which capacitors contribute
to the lower cut-off frequency
 Consider each capacitor individually. Let f = 0 (capacitor is open circuit):
o If vo (or AM) does not change, capacitor does NOT contribute to fL
o If vo (or AM) → 0 or reduced substantially, capacitor contributes to fL
Example:
Cc1
vi = 0 → v o = 0
Contributes to fL
CL
No change in vo
Does NOT contribute to fL
How to find which capacitors contribute
to the higher cut-off frequency
 Consider each capacitor individually. Let f → ∞ (capacitor is short circuit):
o If vo (or AM) does not change, capacitor does NOT contribute to fH
o If vo (or AM) → 0 or reduced substantially, capacitor contributes to fH
Example:
Cc1
No change in vo
Does NOT contribute to fH
CL
vo = 0
Contributes to fH
How to find “mid-frequency” circuit
 All capacitors that contribute to low-frequency response should be
short circuit.
 All capacitors that contribute to high-frequency response should
be open circuit.
Example:
Cc1 contributes to fL → short circuit
CL contributes to fH → open circuit
Low-Frequency Response
Low-frequency response of a CS amplifier
 Each capacitors gives a pole.
 All poles contribute to fL (exact value of fL from simulation)
 If one pole is at least two octave (factor of 4) higher than others (e.g., fp2 in the
above figure), fL is approximately equal to that pole (e.g., fL = fp2 in above)
 A good approximation for design & hand calculations:
fL = fp1 + fp2 + fp3 + …
Low-frequency response of a CS amplifier
All capacitors contribute to fL (vo is reduced when f → 0 or caps open circuit)
Cc1 open:
vi = 0 → v o = 0
Cc2 open:
vo = 0
Cs open:
Gain is reduced substantially
(from CS amp. To CS amp. With RS)
See S&S pp689-692 for detailed calculations (S&S assumes ro → ∞ and RS → ∞ )
Vo
s
s
s
= AM x
x
x
Vsig
s + ω p1 s + ω p 2 s + ω p 3
AM = −
RG
g m (ro || RD || RL )
RG + Rsig
ω p1 =
1
1
, ω p3 =
Cc1 ( RG + Rsig )
Cc 2 ( RD || ro + RL )
ω p2 ≈
1
,
Cs [ RS || (1 / g m + RD || RL / ro g m )]
Finding poles by inspection
1. Set vsig = 0
2. Consider each capacitor separately (assume others are short
circuit!), e.g., Cn
3. Find the total resistance seen between the terminals of the
capacitor, e.g., Rn (treat ground as a regular “node”).
4. The pole associated with that capacitor is
f pn =
1
2πRn Cn
5. Lower-cut-off frequency can be found from
fL = fp1 + fp2 + fp3 + …
* Although we are calculating frequency response in frequency domain, we will use time-domain
notation instead of phasor form (i.e., vsig instead of Vsig ) to avoid confusion with the bias values.
Example: Low-frequency response of a CS amplifier
 Examination of circuit shows that ALL capacitors contribute to the low-frequency
response.
 In the following slides with compute poles introduced by each capacitor (compare
with the detailed calculations and note that we exactly get the same poles).
 Then fL = fp1 + fp2 + fp3
Example: Low-frequency response of a CS amplifier
f p1 =
1. Consider Cc1 :
1
2π Cc1 ( RG + Rsig )
Terminals of Cc1
∞
2. Find resistance between
Capacitor terminals
Example: Low-frequency response of a CS amplifier
f p2 =
1
2π C S [ RS || (1 / g m + RD || RL / ro g m )]
1/ gm +
( RD || RL ) / ro g m
1. Consider CS :
11//ggmm++
(RDD || RLL )/ /rorgo gmm
1/ gm +
( RD || RL ) / ro g m
Terminals of CS
2. Find resistance between
Capacitor terminals
Example: Low-frequency response of a CS amplifier
f p3 =
1. Consider Cc2 :
1
2π Cc 2 ( RL + RD || ro )
Terminals of Cc2
2. Find resistance between
Capacitor terminals
High-Frequency Response
 Amplifier gain falls off due to the internal
capacitive effects of transistors
Capacitive Effects in pn Junction
Majority Carriers
 Charge stored is a function of applied
voltage.
 We can define a “small-signal”
capacitance, Cj
Cj =
dQJ
dVR
VR =VQ
 In reverse-bias region, analysis show
(see S&S pp154-156):
C j0
Cj =
(1 + VR / V0 ) m
 V0 : Junction built-in voltage
 Cj0 : Capacitance at zero reversed-bias
voltage.
 m : grading coefficient (1/2 to 1/3).
 For forward-bias region: Cj ≈ 2Cj0
Capacitive Effects in pn Junction
Minority Carriers
 Excess minority carriers are stored in p and n sides of the
junction.
 The charge depends on the minority carrier “life-time” (i.e.,
how long it would take for them to diffuse through the
junction and recombine.
 Gives Diffusion Capacitance, Cd
 Cd is proportional to current (Cd = 0 for reverse-bias)
Cd =
τT ⋅ ID
VT
Small Signal Model for a diode
Reverse Bias
Cj + Cd
C j ≈ 2 ⋅ C j0
Cd =
τT ⋅ ID
VT
Forward Bias
Cj =
C j0
(1 + VR / V0 ) m
Cd = 0
rD
 Junction capacitances are small and are given in femto-Farad (fF)
1 fF = 10-12 F
Capacitive Effects in MOS
1. Capacitance between
Gate and channel
(Parallel-plate capacitor)
3. Junction capacitance
between Source and Body
(Reverse-bias junction)
2. Capacitance between
Gate & Source and Gate &
Drain due to the overlap of
gate electrode
(Parallel-plate capacitor)
4. Junction capacitance
between Drain and Body
(Reverse-bias junction)
Capacitive Effects in MOS
“Parallel-Plate” capacitances (depends on the channel shape)
Define:
C gate = W ⋅ L ⋅ Cox
Cov = W ⋅ Lov ⋅ Cox
Triode
Saturation
1
⋅ C gate + Cov
2
1
= ⋅ C gate + Cov
2
C gs =
C gd
C gs =
Cut-off
2
⋅ C gate + Cov
3
C gd = Cov
C gb = C gate
Pinched Channel
No Channel
“Junction” capacitances
Csb =
Csb 0
(1 + VSB / V0 ) m
C gs = C gd = Cov
Cdb =
Cdb 0
(1 + VDB / V0 ) m
Small signal for MOS in high-frequencies
For source connected to body
Saturation
C gs =
2
⋅ C gate + Cov
3
C gd = Cov
Cdb =
Cdb 0
(1 + VDB / V0 ) m
High-frequency response of a CS amplifier
1) MOS “internal” capacitors are shown “outside” of the transistor to see their impact.
2) All MOS capacitors contribute to fH (vo is reduced when f → ∞ or caps short circuit)
3) For f → ∞ , all coupling (Cc1 and Cc2 ) and by-pass capacitors are short circuit
Cgd short:
Input is connected to output
Gain is reduced to 1
For f → ∞ Cgs short:
vi = 0 → v o = 0
Cdb short:
vo = 0
High-frequency response of a CS amplifier
In General:
1) One internal capacitors shorts input to the ground (Cgs here)
2) One internal capacitors shorts output to the ground (Cdb here)
3) One internal capacitors shorts input to output (Cgd here)
Cgd appears in parallel to Cdb
See S&S pp712-714 for detailed calculations (S&S assumes Cdb → 0)
Vo
1
= AM x
Vsig
1 + s / ω p1
AM = −
RG
g m (ro || RD || RL )
RG + Rsig
ω p1 =
1
Cin ( RG || Rsig )
, ω p2 =
(C gd
1
,
+ Cdb )(ro || RD || RL )
Cin = C gs + C gd (1 + g m ro || RD || RL )
Cgd appears in parallel to Cgs
(with a much larger value)
High-frequency-relevant capacitors
 High-frequency-relevant capacitors appear between
o
o
o
input & ground,
output & ground, and
input & output.
 Capacitors that are connected between input & output
provide feedback. In the case of CS amplifier, we saw that
they appeared in the transfer function as capacitors in
parallel to input & ground and output & ground capacitors.
 We can use Miller’s Theorem to replace capacitors connected
between input & output and simplify the analysis.
Miller’s Theorem
 Consider an amplifier with a gain A with an impedance Z attached
between input and output
 V1 and V2 “feel” the impedance of Z only through I1 and I2
 We can replace Z with any circuit as long as a current I1 flows out of
V1 and a current I2 flows out of V2.
V2 = A ⋅ V1
I1 =
V1 − V2 (1 − A) ⋅ V1
=
Z
Z
I2 =
V2 − V1 ( A − 1)V1 ( A − 1)V2
=
=
Z
Z
Z⋅A
I1 =
V1
V
= 1,
Z /(1 − A) Z1
I2 =
V2
V
Z
= 2 , Z2 =
Z ⋅ A /( A − 1) Z 2
1−1/ A
Z1 =
Z
(1 − A)
Miller’s Theorem
 If an impedance Z is attached between input and output an
amplifier with a gain A , it can be replaced with two
impedances between input & ground and output & ground
Other parts of the circuit
V2 = A ⋅ V1
V2 = A ⋅ V1
Z
Z1 =
1− A
Z2 =
Z
1−
1
A
Example of Miller’s Theorem:
Inverting amplifier
Solution using Miller’s theorem:
Recall from ECE 100, if A0 is large
Z
1− A
Rf
Rf
≈
Rf1 =
1 + A0 A0
Z1 =
Rf
vo
=−
vi
R1
vo = A0 ⋅ (v p − vn ) = − A0 ⋅ vn
Z2 =
Rf 2 =
Z
1−1/ A
Rf
1 + 1 / A0
≈ Rf
Rf 1
vn
=
vi R1 + R f 1
− A0 R f 1 − A0 ( R f / A0 )
− Rf
− Rf
vo
v
= − A0 n =
=
=
≈
vi
vi R1 + R f 1 R1 + ( R f / A0 ) R1 + ( R f / A0 )
R1
Finding fH by inspection
1. Set vsig = 0
2. Use Miller’s Theorem to replace capacitor between input &
output with two capacitors at the input and output.
3. Consider each capacitor separately (assume others are open
circuit!), e.g., Cn
4. Find the total resistance seen between the terminals of the
capacitor, e.g., Rn (treat ground as a regular “node”).
5. Compute the
f pn =
1
2πRn Cn
6. Upper cut-off frequency can be fund from:
1
1
1
1
=
+
+
+ ...
fH
f p1 f p 2 f p 3
Caution:

Method in previous slide is called the time-constant approximation to fH
(see S&S page 724).
1
fH =
(S&S Eq. 9.73)
2π Σ Rn Cn
n

Since f pn
1
=
, the above formula give
2πRn Cn
1
1
1
1
= 2π Σ Rn Cn =
+
+
+ ...
n
fH
f p1 f p 2 f p 3
This is the correct formula to find fH

However, S&S gives a different formula in page 722 (contradicting
formulas of pp724). Ignore this formula (S&S Eq. 9.68)
1
=
fH
1
1
1
+
+
+ ...
2
2
2
f p1 f p 2 f p 3
Applying Miller’s Theorem to Capacitors
V2 = A ⋅ V1
Z1 =
Z
1− A
Z2 =
Z
1−
1
A
1
j ωC
Z
Z1 =
⇒ C1 = (1 − A)C
1− A
Z
Z1 =
⇒ C2 = (1 − 1 / A)C
1−1/ A
Z=
Example: High-frequency response of a CS amplifier
o Circuit includes CL which is often used to set the “dominate pole”.
o The first step is to identify which capacitors are relevant to highfrequency response (Cgs ,Cdb , Cgd , and CL ). The other capacitors,
Cc1 and Cc2 are relevant to low-frequency response.
o At high frequency, Cc1 and Cc2 will be short.
Example: High-frequency response of a CS amplifier
Use Miller’s Theorem to replace
capacitor between input & output
(Cgd ) with two capacitors at the
input and output.
A=
vd
= − g m (ro || RD || RL ) ≡ − g m RL′
vg
C gd ,i = C gd (1 − A) = C gd (1 + g m R′) ≈ g m R′C gd
C gd ,o = C gd (1 − 1 / A) = C gd (1 + 1 / g m RL′ ) ≈ C gd
* Assuming gmR’L >> 1
Cin = C gs + C gd ,i
C L′ = Cdb + C gd ,o + C L
Example: High-frequency response of a CS amplifier
f p1 =
1
2π Cin ( RG || Rsig )
1. Consider Cin :
∞
Terminals of Cin
2. Find resistance between
Capacitor terminals
Example: High-frequency response of a CS amplifier
f p2 =
1
2π C L′ (ro || RD || RL )
1. Consider C’L :
Terminals of C’L
2. Find resistance between
Capacitor terminals
High-frequency response of a CS amplifier
AM = −
RG
g m (ro || RD || RL )
RG + Rsig
1/f p1 = 2π Cin ( RG || Rsig ), 1/f p 2 = 2π C L′ (ro || RD || RL )
Cin = C gs + C gd (1 + g m ro || RD || RL )
C L′ = C gd + Cdb + C L
1/f H = 1/f p1 + 1/f p 2
Miller’s Theorem vs Miller’s Approximation

For Miller Theorem to work, ratio of V2/V1 (amplifier gain) should be
independent of feedback impedance Z.

This was correct for OpAmp example where the gain of the chip, A0 ,
remains constant when Rf is attached (output resistance of the chip is
small).

However, the capacitor that connect the input and output changes the
frequency response of the amplifier (i.e., its gain) and so we cannot
“strictly” apply Miller’s Theorem.

In our analysis, we used mid-band gain of the transistor and ignored
changes in the frequency response due to the feedback capacitor. This is
called “Miller’s Approximation.”

Miller’s Approximation only gives “approximate” values of the poles and
the higher cut-off frequency.

More importantly, Miller’s Approximation “misses” the zero introduced
by the feedback resistor (which can cause “unstable” operation).
Example: High-frequency response of a CG amplifier
C L′ = C gd + C L (+Cdb )
o Cdb is ignored in the above. Including body effect, one sees Cdb
actually appears between drain and ground (parallel to CL in the
above circuit) and is “absorbed” in CL.
o Note that Cgd is also between the drain and the ground and is in
parallel to CL.
Example: High-frequency response of a CG amplifier
f p1 =
2π C gs [ Rsig
1
|| (ro +RD || RL ) / g m ro ]
Terminals of Cgs
1. Consider Cgs :
ro +RD || RL
g m ro
ro +RD || RL
g m ro
≈
ro +RD || RL
g m ro
2. Find resistance between
Capacitor terminals
Example: High-frequency response of a CG amplifier
f p2 =
1
2π C L′ [ RD || RL || ro (1 + g m Rsig )]
1. Consider C’L :
≈ ro (1 +g m Rsig )
ro (1 +g m Rsig )
2. Find resistance between
Capacitor terminals
High-frequency response of a CG amplifier
AM = +
Ri
g m (ro || RD || RL )
Ri + Rsig
Ri = (ro +RD || RL ) / g m ro
1/f p1 = 2π C gs ( Rsig ||Ri ) ,
1/f p 2 = 2π C L′ [ RD || RL || ro (1 + g m Rsig )]
C L′ = C gd + C L (+Cdb )
1/f H = 1/f p1 + 1/f p 2