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Advance Electronics
Prof. Rajput Sandeep
Assist. Prof., EC Dept.
HCET ,Siddhpur
Chapter 1:
Transistor at High Frequency
Lecture : 1
The hybrid- pi CE Transistor Model
The hybrid π model at low frequencies
Chapter 1:
Transistor at High Frequency
Lecture : 2
The Input Conductance - gb’e

To obtain the expression for the input conductance gb’e , refer to the
low frequency hybrid – π model shown in figure 1.
Figure 1:The hybrid π model at low frequencies

Note that all the capacitor from the high frequency hybrid- π model
have been removed because the capacitances at low frequencies are
of negligible values.
The Input Conductance - gb’e

Figure 2 shows the h parameter model at low frequency, for the same
transistor.
Figure 2:The hybrid model at low frequencies

Now, rb’c > rb’e therefore almost all current Ib will flow through rb’e .
so, V b’e = Ib x rb’e
(1)
The Input Conductance - gb’e

So that, the short circuit collector current in figure 1 is given by ,
Ic = gm Vb’e = gm ( Ib x rb’e )

(2)
The short circuit current gain is defined as,
hfe = Ic / Ib = gm rb’e
or,
rb’e = hfe / gm = (hfe VT) / | Ic |

(3)
Now, input conductance , gb’e = Ib / Vb’e
= 1 / rb’e

By substituting the value of rb’e we can get,
gb’e = gm / hfe

This is the required expression for the input conductance.
(4)
The Input Conductance - gb’e

Conclusion :

By referring the below equation ,
rb’e = hfe / gm = (hfe VT ) / | Ic |
(5)

It can be concluded that if hfe is constant over a certain range of
current, then rb’e is directly proportional to temperature (due to V T )
and it is inversely proportional to current Ic .

Therefore input conductance gb’e will be inversely proportional to
temperature and directly proportional to current if hfe is constant.
The feedback Conductance – gb’c

Let us define the reverse voltage ratio hre using the low frequency
hybrid – pi circuit of figure 1 as follows:
hre = (Vb’e / V ce ) | for Ib = 0

(6)
Let’s see the modified low frequency hybrid – pi model in figure 3.
Figure 3: Modified hybrid – pi model at low frequencies
The feedback Conductance – gb’c

From the figure 3 it is evident that rb’e and rb’ c form a potential divider
across Vce .
So that , Vb’e = { rb’e / (rb’e + r b’c ) } x Vce
(7)
so, hre = Vb’e / V ce
= { rb’e / (rb’e + r b’c ) }
(8)
Rearrange this expression ,
h re rb’e + hre rb’c = rb’e
so,
h re rb’c = rb’e - hre rb’e
= rb’e ( 1 – hre )
(9)
The feedback Conductance – gb’c

But, hre < < 1. Hence (1- hre ) = 1.
so that , hre rb’c = rb’e

The feedback conductance gb’c is defined as,
gb’c = Ib / Vce
But ,
Ib = Vb’e / r b’e
so that gb’c = Vb’e / ( Vce * rb’e )
But, Vb’e / Vce = hre and 1 / rb’e = gb’e

So that,
The feedback conductance = hre * gb’e .
(11)
The Base Spreading Resistance – rb’b

Let us define hie from the low frequency hybrid- pi model.
Figure 4: Modified hybrid – pi model at low frequencies

From the figure 3 it is evident that rb’e and rb’ c from a potential divider
across Vce and rb’c > > r b’e so that,
(rb’e || rb’c ) = rb’e
(12)
The Base Spreading Resistance – rb’b

Hence hie = (Vbe / Ib ) | Vce =0
= {Ib rb’b + Ib * (rb’e || rb’c )}/ Ib
So that , hie = rb’b + (rb’e || rb’c ) = rb’b + rb’e

This is the expression for hie in terms of hybrid – pi parameter.

The base spreading resistance is given by,
rb’b = hie – rb’e
In the expression for hie, Substitute the expression of

(13)
rb’e = (hfe VT ) /|Ic | to get,
hie = rb’b + rb’e = rb’b +{ hfe VT / |Ic |}
(14)
Chapter 1:
Transistor at High Frequency
Lecture : 3
The output conductance – gce

The output conductance hoe can be obtained from the low frequency
hybrid – pi equivalent circuit by open circuiting the input terminal, i.e.
Ib = 0.
Figure 5 : Modified hybrid – pi model at low frequencies
From figure 5 , hoe = ( Ic /Vce ) | Ib = 0 ,
(16)
But Ic = I1 + I2 + I3
Now, I1 = Vce / r ce , I2 = gm V b’e = gm hre Vce and I3 = Vce / (rb’e + rb’c)
The output conductance – gce

Now Substituting (1 / r ce ) = gce and (1/rb’c ) = gb’c to get,
hoe = gce +gm hre + gb’c
(17)
Hence output conductance, gce = hoe – gm hre – gb’c
Substitute gm = hfe gb’e and hre = gb’c / gb’e to get,
gce = hoe – [ hfe gb’e * (gb’c / gb’e ) ] – gb’c
(18)
so that , gce = hoe – hfe gb’c – gb’c
gce = hoe – ( 1 + hfe ) gb’c
This is the required expression for the output conductance.
(19)
The Hybrid – pi Capacitance

The Basic high frequency hybrid – pi model includes two
capacitance.

Cb’c or Cc and Ce.

Capacitance Cb’c or Cc :

Cb’c or Cc is the collector junction capacitance. It is measured CB
capacitance between collector ( C ) and base ( B ) with the input (E)
open.

Cb’c or Cc is usually specified by the manufacturer as Cob.

In the active region of the transistor operation , the CB junction is
reverse biased. Hence Cc represents the transition capacitance CT.

Cc is inversely propositional to VCE, because as VCE is increased, the
width of the depletion region also increases and Cc decreases.
The Hybrid – pi Capacitance

Capacitance Ce : Ce capacitance appears between base and emitter,
in the hybrid – pi model.

Ce represents the sum of emitter diffusion capacitance (CDe ) and the
emitter junction capacitance CTe.

But the emitter junction is forward biased and the diffusion
capacitance of a forward biased junction is much higher than its
transition capacitance.
i.e. CDe > > CTe
Therefore, Ce = (CDe + CTe ) = CDe

(20)
Ce or CDe is propositional to the emitter bias current Ic and it is
almost independent of temperature.
The Hybrid – pi Capacitance

The diffusion capacitance CDe is mathematically expressed as,
CDe = gm (W2 / 2DB )
(21)

Where DB = Diffusion constant for minority carriers in the base.

Experimentally Ce or CDe is determined from the measurement of fT ,
a frequency at which the CE short circuit current gain reduces to 1.

The value of Ce in terms of fT is given by, Ce = gm / 2 π fT
But, Ce = CDe + Cte
(22)
(23)
Therefore , (CDe + CTe ) = gm / 2 πfT
So that fT = gm / 2 π (CDe + CTe ) .
(24)
Simplified Hybrid – pi Model
Figure 6 : Simplified hybrid – pi model

If we neglect the resistance rb’b because it is too small and rb’c
because it is too high, Then we get simplified hybrid model as shown
in figure 6.
Validity of Hybrid – pi Model

Consider the CDe and Ce equations,
CDe = gm (W2 / 2DB )
(25)
= (W2 / 2DB r‘e)
= (W2 IE/ 2DB VT )

And g m = α ∂IE / ∂ VE ,

In both the equations we have assumed that VBE varies slowly. This is
done to maintain the minority carrier distribution in the base region
triangular as shown in figure 7.
(26)
Validity of Hybrid – pi Model

If the distribution remains triangular under the condition of varying
VBE then the slope at x =0 and x = W are equal. Hence the emitter and
collectors currents will also be equal.
Figure 7 : Modified- carrier charge distribution in the base region

If the rate of change of VBE is small such that the base incremental
current IB is small in comparison to the collector incremental current
IC then under such dynamic condition the hybrid – pi model is valid.
Validity of Hybrid – pi Model

Scientist Giacollecto has shown that the network elements of hybrid –
pi model are frequency independent provided that,
2πf W2 / 6DB < < 1
(27)
CDe = gm (W2 / 2DB )
(28)
Now we know that,
= (W2 IE / 2DB VT)
And
Ce = gm / 2 π fT
Therefore,
W2 /6DB = Ce / 3gm = 1/ 6πfT
So that f < < 6πfT / 2π = 3fT

(29)
Thus the hybrid π model is valid for frequencies approximately up to
fT / 3.
Chapter 1:
Transistor at High Frequency
Lecture : 4
The CE Short Circuit Current Gain

Consider the single stage CE transistor amplifier or the last stage of
cascade configuration with load RL on this stage.

The hybrid – pi equivalent circuit for a single transistor with a
resistive load RL is show in figure 8.
Figure 8 : The hybrid – pi equivalent circuit for a single transistor with resistive load RL
The CE Short Circuit Current Gain

In this as we want to obtain the short circuit current gain. So we will
short the output terminals, C and E. Therefore RL = 0.

The approximate equivalent circuit with RL = 0 is as shown in figure 9.
Figure 9 : Equivalent circuit with RL = 0

Note that the resistance rb’c has been neglected as it is of very high
value, and rce is short circuited.

The capacitance Ce and Cc will appear in parallel with each other.
The CE Short Circuit Current Gain

The short circuit load current IL is given by,
IL = - gmVb’e

(30)
Now let us calculate Vb’e from figure 9. Let the parallel combination
of rb’e and Cc+Ce be represented by Z.
Z = { rb’e * 1/(jw ( Ce + Cc ) )} / { rb’e + 1/(jw ( Ce + Cc )) } (31)
= rb’e / {rb’e * (jw ( Ce + Cc ) + 1)}
= rb’e / {1 + rb’e * (jw ( Ce + Cc ))}

(32)
Including this term into the equivalent circuit we et the simplified
hybrid – pi equivalent circuit as shown in figure 10
The CE Short Circuit Current Gain
Figure 10 : Equivalent circuit with Z included

From figure 10 we can write that,
Vb’e = Ii x Z = (Ii * rb’e ) / (1+ jw rb’e (Cc + Ce))
Substituting this value into equation (31) we get,
IL = - gm rb’e Ii /{1 + rb’e * (jw ( Ce + Cc ))}
(33)
The CE Short Circuit Current Gain

Therefore, the current gain is given by,
AI = IL / Ii = - gm rb’e /{1 + rb’e * (jw ( Ce + Cc ))}

But hfe = gm rb’e
So that, AI = - hfe /{1 + rb’e * (jw ( Ce + Cc ))} where ,w = 2πf
Figure 11: The short circuit CE current gain Vs frequency (plotted on the Log- Log scale
The CE Short Circuit Current Gain

That means the current gain is inversely propositional to frequency.

At very low frequencies, the term jw rb’e (Cc + Ce ) is very small as
compared to 1.Thereforecurrent gain AI ~ hfe.

But as the frequency increases the, current gain goes on decreasing.

Figure 11 shows the variation in the current gain AI expressed in
decibels (i.e. 20log|AI |) against frequency on a logarithmic frequency
scale.

Note that AI (dB) = 0 dB at frequency f = fT and for f > > fB the
graph is a straight line having a slope of 6 dB / octave or 20
dB/decade.

Let us now define various frequencies as fα , fβ and fT .
Chapter 5:
Operational Amplifiers
Lecture : 5
Operational Amplifiers

“operational” was used as a descriptor early-on because this
form of amplifier can perform operations of
•
adding signals
•
subtracting signals
•
integrating signals,
 x(t )dt
 The
applications of operational amplifiers ( shortened to op amp )
have grown beyond those listed above.
 At
this level of study we will be concerned with how to use the op
amp as a device.
 The
internal configuration (design) is beyond basic circuit theory and
will be studied in later electronic courses. The complexity is
illustrated in the following circuit.
Operational Amplifiers
 The
op amp is built using VLSI techniques. The circuit diagram of an
LM 741 from National Semiconductor is shown below.
Internal circuitry of LM741
Operational Amplifiers
Fortunately, we do not have to sweat a circuit with 22 transistors and
twelve resistors in order to use the op amp.
The circuit in the previous slide is usually encapsulated into a dual inline pack (DIP). For a single LM741, the pin connections for the chip
are shown below.
Pin connection, LM741.
Operational Amplifiers
The basic op amp with supply voltage included is shown
in the diagram below.
V-
inverting input
output
noninverting input
V+
Basic op am diagram with supply voltage.
Operational Amplifiers
In most cases only the two inputs and the output are
shown for the op amp. However, one should keep in
mind that supply voltage is required, and a ground.
The basic op am without a ground is shown below.
Outer op am diagram.
Operational Amplifiers
A model of the op amp, with respect to the symbol, is
shown below.
V1
_
Vd
+
V2
Ro
Ri
AVd
Op Amp Model
Vo
Operational Amplifiers
The previous model is usually shown as follows:
Ri
V1
+
_
Vd
Ri
AVd
Vo
+
V2
_
Working circuit diagram of op amp.
Operational Amplifiers
Application: As an application of the previous model,
consider the following configuration. Find Vo as a
function of Vin and the resistors R1 and R2.
R2
R1
_
+
+
Vin
_
+
Vo
_
Op amp functional circuit.
Operational Amplifiers
In terms of the circuit model we have the following:
R2
Ri
R1
a
+
_
Vin
Vi
_
b
+
Ri
AVi
Vo
+
_
Total op amp schematic for voltage gain configuration.
Operational Amplifiers
R2
Ri
R1
+
_
Vin
Vi
_
b
a
+
Ri
AVi
Vo
+
Circuit values are:
R1 = 10 k
A = 100,000
_
R2 = 40 k
Ri = 1 meg 
Operational Amplifiers
We can write the following equations for nodes a and b.

( Vin  Vi )
 Vi
Vi  Vo 


10 k
1 meg
40 k
  ( Vin  Vo ) 
Vo  50
 AVi

40 k


Operational Amplifiers
Equation simplifies to;
 25Vo 126Vi 100Vin
simplifies to;
4.005 x10 Vo  4 x10 Vi  0
5
9
Operational Amplifiers
From Equations we find;
Vo   3.99Vin
Eq 8.5
This is an expected answer.
Fortunately, we are not required to do elaborate circuit
analysis, as above, to find the relationship between the
output and input of an op amp. Simplifying the analysis
is our next consideration.
Operational Amplifiers
For most all operational amplifiers, Ri is 1 meg  or
larger and Ro is around 50  or less. The open-loop gain, A,
is greater than 100,000.
Ideal Op Amp:
The following assumptions are made for the ideal op amp.
1. Infinite open  loop gain; A  
2. Zero output ohms ; Ro  0
3. Infinite input ohms ; Ri  
Operational Amplifiers
Ideal Op Amp:
i1 = 0
+
V1
_
i2 = 0
+
_ _
Vi
+ +
V2 = V1
_
+
Vo
_
Figure 8.9: Ideal op amp.
(a) i1 = i2 = 0: Due to infinite input resistance.
(b) Vi is negligibly small; V1 = V2.
Operational Amplifiers
Ideal Op Amp:
Find Vo in terms of Vin for the following configuration.
R2
R1
_
+
+
Vin
_
+
Vo
_
Figure 8.10: Gain amplifier op amp set-up.
Operational Amplifiers
Ideal Op Amp:
R2
R1
a _
+
Vi
Vin
+
+
_
Vo
_
Writing a nodal equation at (a) gives;

(Vin  Vi )
Vi  Vo 

R1
R2
Eq 8.6
Operational Amplifiers
Ideal Op Amp:
(Vin  Vi )  Vi  Vo 

R1
R2
With Vi = 0 we have;
V0 
Eq 8.7
 R2
Vin
R1
With R2 = 4 k and R1 = 1 k, we have
Vo   4Vin
Earlier
we got
Vo   3.99Vin
Operational Amplifiers
Ideal Op Amp:
When Vi = 0 in Eq 8.7 and we apply the Laplace Transform;
V0 ( s )
 R2

Vin ( s )
R1
Eq 8.8
In fact, we can replace R2 with Zfb(s) and R1 with Z1(s) and
we have the important expression;
 Z fb ( s )
V0 ( s )

Vin ( s )
Z in ( s )
Eq 8.9
Operational Amplifiers
Ideal Op Amp:
At this point in circuits we are not able to appreciate the
utility of Eq 8.9. We will revisit this at a later point in
circuits but for now we point out that judicious selections
of Zfb(s) and Zin(s) leads to important applications in
• Analog Filters
• Analog Compensators in Control Systems
• Application in Communications
Operational Amplifiers
Ideal Op Amp:
Example 8.1: Consider the op amp configuration below.
Assume Vin = 5 V
6 k
1 k
a
+
_+
Vin
_
+
3V
V0
_
Figure 8.11: Circuit for Example 8.1.
Operational Amplifiers
Example 8.1 cont.
6 k
1 k
a
+
_+
Vin
_
+
3V
V0
_
At node “a” we can write;
( Vin  3 )  3  V0 

1k
6k
From which;
V0 = -51 V
Eq 8.10
Operational Amplifiers
Example 8.2: Summing Amplifier. Given the following:
R fb
R1
a
R2
V1
V0
V2
Figure 8.12: Circuit for Example 8.2.
 V0
V1 V2


R1 R2
R fb
Eq 8.11
Operational Amplifiers
Example 8.2: Summing Amplifier. continued
Equation 8.11 can be expressed as;
 R fb 
 R fb  
V1  
V2 
V0   
 R2  
 R1 
Eq 8.12
If R1 = R2 = Rfb then,
V0   V1 V2 
Eq. 8.13
Therefore, we can add signals with an op amp.
Operational Amplifiers
Example 8.3: Isolation or Voltage Follower.
Applications arise in which we wish to connect one circuit
to another without the first circuit loading the second. This
requires that we connect to a “block” that has infinite input
impedance and zero output impedance. An operational
amplifier does a good job of approximating this. Consider
the following:
+
Circuit 1
Vin
_
The
"Block"
+
Vout
_
Figure 8.13: Illustrating Isolation.
Circuit 2
Operational Amplifiers
Example 8.3: Isolation or Voltage Follower. continued
The Block
Circuit 1
+
+
Vin
V0
_
_
Circuit 2
Figure 8.14: Circuit isolation with an op amp.
It is easy to see that: V0 = Vin
Operational Amplifiers
Example 8.4: Isolation with gain.
20 k
+
_
+
_
+
10 k
Vin
a
+
Vin
V0
10 k
_
_
Figure 8.15: Circuit for Example 8.4:
Writing a nodal equation at point “a” and simplifying gives;
V0  2Vin
Operational Amplifiers
Example 8.5: The noninverting op amp.
Consider the following:
R
fb
R0
a
+
_
V2
+
_
+
V0
_
Figure 8.16: Noninverting op am configuration.
Operational Amplifiers
Example 8.5: The noninverting op amp. Continued
Writing a node equation at “a” gives;
V2 (V2  V0 )

0
R0
R fb
so
1
V0
1 
 V2  

R fb
R
R
 0
fb 

which gives,
 R fb 
V2
V0   1 
R0 

Remember this
Operational Amplifiers
Example 8.6: Noninverting Input.
Find V0 for the following op amp configuration.
2 k
Vx

+
_
+
10 k
+_ 4 V
6 k
a
V0
5 k
_
Figure 8.17: Op amp circuit for example 8.6.
Operational Amplifiers
Example 8.6: Noninverting Input.
The voltage at Vx is found to be 3 V.
Writing a node equation at “a” gives;
V x ( V x  V0 )

0
5k
10k
or
V0  3Vx  9V