CONTENTS
CONTENTS
+ 0 ) 2 6 - 4
Learning Objectives
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General
DC Equivalent Circuit
AC Equivalent Circuit
Equivalent Circuit of a
CB Amplifier
Effect of Source Resistance
Rs on Voltage Gain
Equivalent Circuit of a
CE Amplifier
Equivalent Circuit of a
CC Amplifier
Small-signal Low-frequency
Model or Representation
T-Model
Formulas for T-Equivalent of
a CC Circuit
What are h-parameters ?
Input Impedance of a Two
Port Network
Voltage Gain of a Two Port
Network
The h-parameters of an
Ideal CB Transistor
The h-parameters of an
ideal CE Transistor
Approximate Hybrid
Equivalent Circuits
Transistor
Amplifier
Formulae Using
h-parameters
Typical Values of Transistor
h-parameters
Approximate Hybrid
Formulas
Common Emitter
h- parameter Analysis
Common Collector
h-parameter Analysis
Conversion of
h-parameters
CONTENTS
CONTENTS
#'
TRANSISTOR
EQUIVALENT
CIRCUITS AND
MODELS
Ç
Generalised hybrid parameter equivalent
circuit
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Electrical Technology
59.1. General
We will begin by idealizing a transistor with the help of simple approximations that will retain
its essential features while discarding its less important qualities. These approximations will help us
to analyse transistor circuits easily and rapidly.
We will discuss only the small signal equivalent circuits in this chapter. Small signal operation
is that in which the ac input signal voltages and currents are in the order of ± 10 per cent of Q-point
voltages and currents.
There are two prominent schools of thought today regarding the equivalent circuit to be substituted for the transistor. The two approaches make use of
(a) four h-parameters of the transistor and the values of circuit components,
(b) the beta (β) of the transistor and the values of the circuit components.
Since long, industrial and educational institutions have heavily relied on the hybrid parameters
because they produce more accurate results in the analysis of amplifier circuits. In fact, hybrid-parameter equivalent circuit continues to be popular even to-day. But their use is beset with the following difficulties :
1. The values of h-parameters are not so readily or easily available.
2. Their values vary considerably with individual transistors even of the same type number.
3. Their values are limited to a particular set of operating conditions for reasonably accurate
results.
The second method which employs transistor beta and resistance values is gaining more popularity of late. It has the following advantages :
1. The required values are easily available;
2. The procedure followed is simple and easy to understand;
3. The results obtained are quite accurate for the study of amplifier circuit characteristics.
To begin with, we will consider the second method first.
59.2. DC Equivalent Circuit
(a) CB Circuit
In an ideal transistor, α = 1 which
means that IC = IE.
The emitter diode acts like any
forward-biased ideal diode. However,
due to transistor action, collector diode acts as a current source. In other
words, for the purpose of drawing dc
equivalent circuit, we can view an ideal
Fig. 59.1
transistor as nothing more than a rectifier diode in emitter and a current source in collector. In the dc equivalent circuit of Fig. 59.1 (b),
current arrow always points in the direction of conventional current.
As per the polarities of transistor terminals (Art. 59.3) shown in Fig. 59.1 (a), emitter current flow from E to B and collector current from B to C.
The dc equivalent circuit shown in Fig.
59.2 for an NPN transistor is exactly similar
except that direction of current flow is
opposite.
Fig. 59.2
Transistor Equivalent Circuits and Models
(b) CE Circuit
Fig. 59.3 shows the dc equivalent circuit of an NPN transistor when connected
in the CE configuration. Direction of current flow can be easily found by remembering the transistor polarity rule given in
Art 59.3. In an ideal CE transistor, we disregard leakage current and take a.c beta
equal to dc beta.
2239
Fig. 59.3
59.3. AC Equivalent Circuit
(a) CB Circuit
In the case of small input ac signals, the emitter diode does not rectify, instead it offers resistance called ac resistance. As usual, collector diode acts as a current source.
Fig. 59.4 shows the ac equivalent circuit of a transistor connected in the CB
configuration.* Here, ac resistance offered
by the emitter diode is
rac = junction resistance (rj) + base-spreading resistance (rB)**
= rj
=
Fig. 59.4
rB is negligible
25 mV
IE
— where IE is dc emitter current in mA
It is written as re´ signifying junction resistance of the emitter i.e. a.c resistance looking into the
emitter.
25 mV
re´
IE
Hence, the a.c equivalent circuit of a
CB circuit becomes as shown in Fig. 59.5.
Since changes in collector current are
almost equal to changes in emitter current,
∆ic = ∆ ie.
(b) CE Circuit
Fig. 59.6 (a) shows the equivalent circuit when an NPN transistor has been connected in the CE configuration.
The a.c resistance looking into the
base is
rac
*
**
Fig. 59.5
25 mV
— d.c current is IB not IE
IB
This circuit is valid both for PNP as well as NPN transistors because difference in the direction of ac
current does not matter.
Also called bulk-resistance.
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Electrical Technology
25mV
IC /
,
,
25mV
IE
25mV
IC
re´
Strictly speaking,
rac= (1 + β) r´ ≅ β re´ — Art 7.24
The a.c collector current is β times
the base current i.e., ic = β ib.
Fig. 59.6
59.4. Equivalent Circuit of a CB Amplifier
In Fig. 59.7 (a) is shown the circuit of a common-base amplifier. As seen, emitter is forwardbiased by –VEE and collector is reverse-biased by + VCC. The a.c signal source voltage υs drives the
+VCC
-VEE
C1
Vs
RE
A
RC
NPN
Vin
B
+VCC
-VEE
RC
RE
C2
RL
Vout
IE
IC
VCB
(b)
(a)
Fig. 59.7
emitter. It produces small fluctuations in transistor voltage and currents in the output circuit. It will be
seen that ac output voltage is amplified because it is more than Vs.
(a) DC Equivalent Circuit
For drawing dc equivalent circuit, following procedure should be adopted :
(i) short all ac sources i.e. reduce them to zero,
(ii) open all capacitors because they block dc.
If we do this, then as seen from Fig. 59.7 (a), neither emitter current can pass through C1 nor
collector current can pass through C2. These currents are confined to their respective resistances RE
and RC (earlier we had been designating it as resistance RL).
Here, IC ≅ IE
and
VCB = VCC – ICRL
Hence, the dc equivalent circuit becomes as shown in Fig. 59.7 (b).
(b) AC Equivalent Circuit
For drawing ac equivalent circuit, following procedure is adopted :
(i) all dc sources are shorted i.e. they are treated as ac ground,
(ii) all coupling capacitors like C1 and C2 in Fig. 59.7 (a) are shorted and
(iii) emitter diode is replaced by its a.c resistance rac = re´
re '
25mV
IE
where IE is dc emitter current in mA.
As seen by the input a.c signal, it has to feed RE and re´ in parallel [Fig. 59.8 (a)]. As looked
Transistor Equivalent Circuits and Models
2241
from point A in Fig. 59.7 (a), RE is
grounded through V EE which has
been shorted and re´ is grounded via
the base.
Similarly, collector has to feed
RC and RL which are connected in parallel across it i.e
at point B in Fig. 59.7 (a). The ac
signal in collector sees an output load
resistance of rL = RC || RL.
Fig. 59.8
Hence, ac equivalent circuit is as shown in Fig. 59.8 (b). Here, collector diode itself has been
shown as a current source.
Following two points are worth noting :
(i) changes in collector signal current are very nearly equal to changes in emitter signal current.
Hence, ∆ic ≅ ∆ie.,
(ii) directions of ac currents shown in the circuit diagram are those which correspond to positive
half-cycle of the a.c input voltage. That is why i c is shown flowing upwards in
Fig. 59.8 (b).
(c) Principal Operating Characteristics
1. Input Resistance
As seen from Fig. 59.8 (a), the input resistance of the circuit (or stage) is given by
rin
RE || re´
re´ RE
RE
re´
In practice, RE is always much greater than re´ so that parallel combination RE || re´ ≅ re´
∴ rin ≅ = re´ input resistance of the emitter diode
2. AC Load Resistance
The collector load as seen by output ac signal consists of a parallel combination of RL and RC.
This has already been designated as rL.
∴
rL=RL || RC
It should be carefully noted that it is the output resistance as seen by collector and not the ac
output resistance when looking into the collector.
Note. In case, RL has not been connected, then rL = RC (Ex. 59.1)
3. Current Gain
It is given by the ratio Ai
ic
ie
4. Voltage Gain
It is given by the ratio
Vout
Av
Vin
Now, Vin* = ie rin = (1 + β) ib . rin
Vout
*
Fig. 59.9
= ic rL = βib rL
Here, Vin equals υs because there is no internal resistance of the source. In case it is there, the two will not
be equal (Art 59.5).
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Electrical Technology
∴ Av
(1
ib rL
) ib rin
Taking the ratio β / (1 + β) as unity, AV
rL
rin
rL
r´e
5. Power Gain
Ap =
Av. Ai
When expressed in decibels, it is written as Gp = 10 log10 Ap dB.
When expressed in terms of rin and rL, the a.c equivalent circuit becomes as shown in Fig. 59.9.
Example 59.1. For the single-stage CB amplifier shown in Fig. 59.10 (a), find rin , rL, Ai, Aυ and Ap.
What would be the rms value of the signal voltage across the load if vs has an rms value of
1.5 mV? Assume silicon material and transistor α = 0.98.
(Electronics-II, Madras Univ.)
Fig. 59.10
Solution. For finding , let us first find IE.
10 0.7
VEE VBE
1.86 mA
IE
5K
RE
re´
25 mV
I E mA
25
1.86
14
(i) rin= re´ || RE = 14 Ω || 5 K ≅ 14 Ω
(ii) rL = RC = 4 K
The ac equivalent circuit becomes as shown in Fig. 59.10 (b).
rL
rL
4K
14
rin
re´
Gp= 10 log10 280 = 24.5 dB
(iii) Ai = α = 0.98
(iv)
Av
(v) Ap = Ai Av= 0.98 × 286 = 280
Output ac voltage = Av × input voltage
∴
Vout
= Av × Vin= 286 × 1.5 = 429 mV = 0.429 V
Example 59.2. For the CB amplifier circuit shown in Fig. 59.11 (a), find
(i) stage rin
(ii)
rL
(iii) a.c output voltage Vout (iv)
Take VBE= 0.7 V.
Solution. IE = (20 – 0.7)/30 K= 0.64 mA ; re´ = 25/0.64 = 39 Ω
(i)
(ii)
stage
rin
= re´ || RE´ = 39Ω || 30 K ≅ 39 Ω
rL
= RL || RC= 30 K || 15 K=10 K
286
voltage gain Av.
2243
Transistor Equivalent Circuits and Models
The amplifier circuit alongwith its ac equivalent circuit is shown in Fig. 59.11.
(iii) We will first find the value of nout as a drop across rL = RC || RL. For that purpose, we will
employ rms values.
Now,
ie
Vin
rin
Vs
rin
1.6
3.9
*
0.04 mA
Fig. 59.11
ic= ie = 0.04 mA ; Vout=ic rL= 0.04 × 10 K = 0.4 V
(iv)
rL
rL rL 10 K
250
rin
re´ re´ 39
The rms value of ac output voltage may be found with the help of Av.
Av
Now,
Av
Vout
Vin
Vout
Av .Vin 250
1.6
400 mV
0.4 V
59.5. Effect of Source Resistance Rs on Voltage Gain
The voltage gain for the CB amplifier circuit
has so far been calculated on the assumption that
the resistance RS of the ac signal source is negligible. The voltage gain will decrease as RS increases
because more and more of VS will drop on RS rather
than on rin and so will not appear in the output.
The basic circuit is shown in Fig. 59.12 where
RS is the internal resistance of the ac signal source.
The ac equivalent circuit is shown in Fig.
59.13 (a). Here, RS is in series with (re´ || RE). The
input ac signal voltage v S drops on these two
series resistors. Obviously, Vin is the drop across
re´ || RE and is less than VS. Using Proportional Voltage Formula,
*
+VCC
-VEE
RE
RS C1
Vs
RC
Vin
C2
RL
Vout
Fig. 59.12
Strictly speaking, it is not ie but is . a small part of the ac source current goes through RE = 30 K and
the balance (major part) goes through parallel resistance re´. However, if we neglect current through 30
K, then ie = is
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Electrical Technology
re´ || RE
Vin Vs
Rs re´ || RE
In practice, RE >> re´, so that re´ || RE ≅ re´
Fig. 59.13
Vin Vs
∴
re´
Rs
re´
Vs
R
1 s
re´
As RS increases in comparison to re´, the ratio RS / re´ increases thereby making vin increasingly
less than Vs.
Moreover, as seen from the source, input resistance is
RS + re´ // RE ≅ RS + Re´
Vout
Vs
∴
rL
( RS
re´ )
rL
RS
— if RS >> re´
Making RS much larger than re´ is called swamping out the emitter diode. This makes voltage
gain independent of re´ and hence the particular transistor is used. The above gain is different from the
voltage gain considered from emitter to collector which is
Vout
Vin
rL
rin
rL
r ´e
Example. 59.3. In the CB amplifier circuit of Fig. 59.14, find
(a) voltage gain from source to output
(b) voltage gain from emitter to output
(c) approximate value of Vin
(Electronics-I, Patna Univ.)
Solution. (a) The voltage gain from source to output is given by
Vout
VS
rL
( RS
re´
-25 V
)
rL = RC || RL = 10 K || 2M
25 K
≅ 10 K ; IE = 25/25 K = 1 mA
re´= 25/1=25 Ω
Vout
Vs
(1K
10 K
26
10
...(i)
(b) The voltage gain from emitter to output is
Vout
Vs
rL
rin
10 K
1K
Vs
)
+20 V
RS
2 mV
(rms)
Vin
Fig. 59.14
2M
Vout
Transistor Equivalent Circuits and Models
∴
Now, rin = RE // re´ = 25 K || 25 Ω ≅ 25 Ω
Vout
VS
10 K
25
400
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...(ii)
(c) The value of Vin may be found by the following two ways :
(i) As seen from Eq. (ii) above, Vin=
Vout
, Now, from Eq. (i) above, we have
400
Fig. 59.15
20 103
µV
50µ
400
(ii) As seen from the ac equivalent circuit of the amplifier (Fig. 59.15), Vs is dropped proportionally on the two series resistances RS and re´ (strictly speaking re´ || RE). The drop across rin is called
Vin.
re´
25
2 mV
µV
∴
50µ
Vin Vs
´
1000
25
Rs re
∴
Vout = 10 Vs = 10 × 2 = 20 mV
Vin
59.6. Equivalent Circuit of a CE Amplifier
Consider the simple CE amplifier circuit of Fig. 59.16 (a) in which base bias has been employed.
IC
IB
RC
RB
iS
C1
B
IC
IB
RL
RC
VCC
C2
ib
Vs
VCC
E
Vin
RL
Vout
IE
(b)
(a)
Fig. 59.16
(a) DC Equivalent Circuit
Fog drawing the dc equivalent circuit, same procedure is adopted as given in Art. 59.2. It is
shown in Fig. 59.16 (b).
As seen, I B
VCC
VBE
RB
VCC
; I = β . IB ; IE ≅ IC ≅ βIB
RB C
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Electrical Technology
(b) AC Equivalent Circuit
Let us now analyse the ac equivalent circuit given in Fig. 59.17.
As proved in Art. 59.3, the ac resistance as seen by the input signal when looking into the base
is = β re´. It may be called rin(base) to distinguish it from rin which is resistance of the CE stage as
shown in Fig. 59.17 (a).
Fig. 59.17
It may be noted that in the absence of ac voltage source resistance RS , whole of Vs acts across RB
as well as β re´ because the two are connected in parallel across it.
Another point worth noting is that major part of source current is passes through β re´ and an
extremely small part ( = vs /RB) passes through RS . Since RB is usually very large, current passing
through it can be easily neglected. Hence, as shown in Fig. 59.17 (b), ib = is.
(c) Principal Operating Characteristics
1. Input Resistance
As seen from Fig. 59.17, input resistance of the stage is
rin
= RB // β re´ ≅ β re´
RB >> β re´
= input resistance of the base
rin(stage) = rin(base)
2. AC Load Resistance. rL = RC || RL
ic
3. Current Gain. Ai
ib
4. Voltage gain. The voltage gain of the stage or circuit is Av
Now,
vin ib re´
∴
Av
ib rL
ib re´
and
vout
ic . rL
vout
vin
ib rL
rL
re´
— if RB >> β rs´
5. Power Gain. Ap= Ai . Av and Gp = 10 log10 Av dB
Example. 59.4. If in the CE circuit of Fig. 59.16 (a), VCC = 20 V, RC = 10 K, RB = 1 M, RL = 1
M, vs = 2mV and β = 50, find (i) ib and ic (ii) rin
(iii) rL
(iv) An
(v) Ap and Gp.
(Basic Electronics, Bombay Univ. 1991)
Solution. IB = 20/1 M = 20 µA ;
IC = β IB = 50 × 20 mA = 1 mA
re´= 25 mV/1 mA = 25 Ω ;
rin(base) = βre´ = 50 × 20 = 1250 Ω
(i) As seen from Fig. 59.17 (a), ac base current is given by
2 mV
µ A ; ic = βib = 50 ×1.6 = 80µ
µA
1.6µ
1250
rin (base )
(ii) stage rin = RB // βre´ = 1 M || 1250 Ω ≅ 1250 Ω
(iii)
rL = RL || RC = 1 M || 10 K ≅ 10 K
ib
vs
Transistor Equivalent Circuits and Models
(iv)
Av
(v)
Ap =
rv
re´
rL
r´
Ai . Ave=
2247
10 K
400
25
50 × 400 = 20,000 ; Gp = 10 log10 20,000 = 43 dB
Example 59.5. In the CE amplifier circuit of Fig. 59.18 employing emitter feedback, find :
(i) rin (ii) rL (iii) Av (iv) Ap and (v) Gp
Take transistor β = 100. How will these values change if emitter bypass capacitor is removed ?
(Electronics—II, Madras Univ.)
Solution. It should be carefully noted that emitter bypass capacitor C provides ac ground to the signal i.e. it shorts out RE to ground so far as ac signal is
concerned. However, it plays its normal role so far
as dc quantities are concerned.
Hence, the ac equivalent circuit of the amplifier
becomes as shown in Fig. 59.19.
Now, I B
RB
VCC
RE
30
2M 100 10K
= 10 µA ; IC = β IB
= 100 × 10 mA = 1 mA
IE ≅ IC = 1 mA, re´ = 25/1=25 Ω ;
β re´ = 100 × 25 = 2500 Ω
Fig. 59.18
(i) As seen from Fig. 59.19
rin = RB || b re´
Ω
= 2M || 2500 Ω ≅ 2500Ω
(ii) rL = RC || RL = 10 K || 20 K
= 6.67 K
(iii) AV
6.67 K
25
Ap = Ai. Av = 100 × 267 = 26,700 ; Gp = 10 log10 26,700 = 44.3 dB
Fig. 59.19
(iv)
Vout
Vin
rL
re´
267
When Emitter Bypass Capacitor is Removed
When bypass capacitor is removed, ac ground is removed. Now, the ac signal will have to pass
through RE also. According to β-rule of Art. 57.24, the total emitter resistance referred to base will
become (1 + β) (re´ + RE) ≅ β
(re´ + RE) because re´ is in series with RE as shown in Fig.
59.20 (a).
Now, the ac equivalent
circuit becomes as shown in
Fig. 59.20 (b).
Stage rin = RB || β
(r´e + RE) ≅ RB || β RE
It is so because R E is
much greater than re´.
Fig. 59.20
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Electrical Technology
re´ = 25 Ω
β(re´ + RE) ≅ β RE = 100 × 10 K = 1 M
rin = RB || β RE = 2 M || 1M
2
= M = 666.7 K
3
rL = 6.67 K
Vin = ib β (re´ + RE);
(i)
∴
∴
(ii)
(iii)
Av
∴
Vout
Vin
ib rL
ib . ( re´ RE )
— as before
— much greater than before
— it remains unchanged
Vout= ic. rL = β ib rL
rL
(re´ RE )
rL
RE
6.67 K
0.667
—reduced drastically
10 K
(iv)
Ap = Ai Av=100 × 0.667 = 66.7
— reduced considerably
It is seen that by removing bypass capacitor, excessive degeneration has occurred in the amplifier circuit.
Example 59.6. For the circuit shown in Fig. 59.21, compute (i) rin (base) (ii) Vout (iii) Av (iv) rin.
Neglect VBE and take β = 200.
Solution. It may be noted that voltage divider
bias has been used in the circuit.
V2
15
15
45
30 7.5 V
VE = V2 – VBE ≅ V2 = 7.5 V
IE = 7.5/7.5 K = 1 mA
re´ = 25 mV/1 mA = 25Ω
The ac equivalent circuit is shown in Fig. 59.22.
Since dc souce is shorted, 45 K, resistor is ac
grounded. On the input side, three resistance
Fig. 59.21
become paralleled across vs i.e. (i) 15 K (ii) 45 K and
(iii) β re´ or rin(base).
Capacitor C2 ac grounds the emitter resistance RE, so does C2 to 5 K and VCC to 10 K.
(i) rin(base) = β re´ = 200 × 25 = 5 K
(ii)
ib = 5 mV/5K = 1 µA
(Obviously, ib is not the current which leaves the source but that part of the source current is
which enters the base). Now, collector load resistance is ic = β ib = 200 × 1 = 200 µA = 0.2 A
rL = 10 K || 5 K = 10/3 K = 3.33 K, Vout = ic rL = 0.2 × 3.33 = 0.667 V
(iii) Av
or
Av
Vout
Vin
rL
re´
0.667 V
5mV
133
3.33 K
133
25
(iv) rin means the input ac resistance as seen from the source i.e. from
point A in Fig. 59.22. It is different
from rin(base) . Obviously, rin is equal
to the equivalent resistance of three resistances connected in parallel. rin = 15 K || 45 K || 5 K = 3.54 K.
Fig. 59.22
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Transistor Equivalent Circuits and Models
Example 59.7. For the CE amplifier circuit of Fig. 59.23, find out (i) rin (ii) rL (iii) Av (iv) Ap
and (v) Vout. Take transistor β = 50 and RS = 0.
(Applied Electronics-II, Punjab Univ. 1992)
Solution.
IE
VEE
RE
20V
20 0.5 mA
40
20 K
re´ = 25/0.5 = 50 Ω
βre´ = 50 × 50 = 2500 Ω
(i) rin = 10 K || 2500 Ω = 2000 Ω
(ii) rL = 20 K || 20 K = 10 K
(iii)
Av
10 K
50
rL
re´
RS
20 K
Vs
10 K
Fig. 59.23
(iv) Ap = Av . Ai
= 200 × 50 =10,000
Gp = 10 log10 10000 = 40 dB
(v) Vout = Av × Vin=20 × 2 = 40 mV (r.m.s.)
Since RS is zero, whole of vs appears across the diode base.
Example. 59.8. For the single-stage CE amplifier circuit of Fig. 59.24, find approximate value of
(i) rin (ii) rL (iii) Av (iv) Ap and Gp. Take transistor β = 100.
Use re´ = 50 mV/IE. (Electronics-II, Gujarat Univ. 1991)
Solution.
V2 20
5
2 V;
5 45
20 V
R1 45 K RC
C1
Vs
R2
5K
V2
2 2 mA
1
RE
re´ = 50/2 =25 Ω ;
βre´ = 25 × 100 = 2.5 K
= R1 || R2 || βre´ = 45 || 5 || 2.5 = 1.6 K
5K
C2
+
IE
(i) rin
Vout
C3
40 K
2mV
(rms)
200
C2
C1
2
1 K RE
_
RL
5K
C3
Fig. 59.24
rL
re
2.5 K
25
(ii) rL = 5 K || 5 K = 2.5 K
(iii) Av
100
(iv) AP = Av . Ai = 100 × 100 = 10,000
(v) Gp = 10 log10 10,000 = 40 dB
Note. RE did not come into picture because it was ac grounded by the bypass capacitor C3.
Example 59.9. Find the approximate values of the quantities of Ex. 59.8 in case bypass capacitor C3 in Fig. 59.24 is removed.
Solution. (i) In this case,
rin = R1 || R2 || β (re´ + RE) = 45 K || 5 K || 100 (25 + 1 K)
≅ 45 K || 5 K || 100 K = 5 K
rL
rL
rE
2.5 K
1K
(ii) rL = 2.5 K — as before
(iii) Av
(iv) Ap = 100 × 2.5 = 250
(Please note the reduction)
(v) Gp = 10 log10250 = 24 dB
( re´
RE )
2.5
2250
Electrical Technology
59.7. Effect of Source Resistance Rs
Greater the internal resistance of the ac signal
source, greater the internal voltage drop and hence lesser
the value of Vin because
Vin= Vs
– drop across RS.
Consider the CE circuit shown in Fig. 59.25 whose
ac equivalent circuit is shown in Fig. 59.26.
As seen from Fig. 59.26, on the input side, RS is in
series with RB || β rE´. Hence, VS is divided between them
in the direct ratio of their resistances.
If RS is much less than RB || βre´, then
Vin ≅ Vs
VCC
RC
RS
Vs
RB
C
RE
RL
Fig. 59.25
Vout rL
The voltage gain from base to output is still given by V
re´
in
If, in any question, Vs is given, we will first find out how much of it appears across the base as
vin. Then, for determining Vout, we will simply multipy this Vin by the voltage gain.
Example 59.10. In the CE amplifier circuit of Fig. 59.27, find the rms signal output voltage when RS is (a) 100 Ω
and (b) 3 K. Take b = 50.
Fig. 59.26
Solution. This circuit is similar to that shown if
Fig. 59.23 except for the addition of RS.
As found in Ex. 9.8, re´ = 50 Ω,
βre´ = 2500 Ω.
∴
RB || βre´ =10 K || 22500 Ω = 2000 Ω
(a) When RS = 100 Ω
In this case, 2 mV are dropped across a series
combination of 100 Ω and 2000 Ω. Drop over 100 Ω is
negligible as compared to that on 2000 Ω. Hence, it
can be presumed that Vs = Vin = 2mV. We have already found that Av = 200.
∴
Vout = Av × Vin= 200 × 2 = 400 mV (rms)
(b) When RS = 3 K
In this case, drop across RB || βre´ = 2 K is Vin Vin
∴
Vout = 200 × 0.8 = 160 mV
Obviously, as RS is increased, Vout is decreased.
RB || r e´
Rs
( RB ||
re´ )
2
Fig. 59.27
3
2
2
0.8 mV
59.8. Equivalent Circuit of a CC Amplifier
We will consider the two-supply emitter-bias circuit shown in Fig. 59.28 (a) in which the
collector is placed at ac (not dc) ground. The ac input signal is coupled into the base and output
signal is taken out of the emitter. This circuit is also called emitter follower circuit because the
Transistor Equivalent Circuits and Models
2251
emitter signal follows the signal at the base both in magnitude and phase.
(a) DC Equivalent Circuit
It is drawn in the usual way by opening all the capacitors and shorting all ac sources. It is
shown in Fig. 59.28 (b).
VEE VBE VBE
IE
RE RB /
RE
(b) AC Equivalent Circuit
It is obtained by shorting all ac sources and capacitors and is shown in Fig. 59.29.
(c) Principal Operating Characteristics
1. Input Resistance
The input resistance of the CC
stage is given by the parallel combination of RS and rin(base). Now, rin(base)
is the input resistance looking into the
base. It is found to be equal to (1 + β)
(re´ + rL) ≅ β rL i.e. β times the ac load
seen by the emitter.
∴ rin or rin(stage) = RB || rin(base)
= RB || β (re´ + rL) ≅ β (re´ + rL)
—when RB is very large
≅ βrL — when re´ << rL
= βRE — if RL = 0
Fig. 59.28
2. AC Load Resistance : rL = RE || RL
It is the ac output resistance as seen by the emitter
(and not the one when looking into the emitter).
3.
Current Gain.
4.
Voltage Gain.
Av
Vout
Vin
ie
(1 )
ib
Vin = ib . rin = ib . β (re´ + rL) :
Vout= i.e . rL= βib . rL
Ai
ib rL
ib .
( re´
rL
re´
rL )
rL
— if re´ << rL
rL
rL
1
Fig. 59.29
It means that output signal from emitter has the same
magnitude as the input signal at the base.
5. Power Gain
Ap = Av . Ai = 1 × (1 + β) = (1 + β) ≅ β
Gp= 10 log10Ap dB
It would be noted from above that main usefulness of the emitter follower is to step up the
impedance level i.e. it transforms load impedance to a much higher value. It does not increase the
signal voltage. Hence, primary application of emitter follower or CC stage is as an impedance matching device. It offers a higher impedance at the input terminals i.e. rin and a low output impedance (rL)
– something opposite of typical basic transistor amplifier.
Another point worth keeping in mind is that the above formulas are not exact because we have
used ideal transistor approximations. However, these formulas do help in rapidly grasping the essential features of an emitter follower. Moreover, they are adequate for preliminary analysis and design.
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Electrical Technology
Example. 59.11. For the emitter follower shown in Fig. 59.30, find
(i) rin or rin(stage)
(ii) rL
(iii) Av
(iv) Ap. Take transistor β = 50.
(Electronics, Delhi Univ.)
Solution. IE = 20/20 = 1 mA
+20
re´ = 25/1=25 Ω
rL = RE || RL = 20 || 5 = 4 K
rin(base) = β(re´ + rL) ≅ β rL = 50 × 4 = 200 K
∴
(re´ of 25 Ω has been neglected as compared to
rL of 4 K)
(i) rin(stage) or rin = RB || rin(base)
rin
Vs
RB 400 K
20 K RE RL 5 K Vout
= 400 K || 200 K
-20
= 133.3 K
than 1.
Fig. 59.30
(ii) rL = 20 K || 5 K = 4 K
(iii) Av ≅ 1 since re´ is negligible as compared to rL. If it is not, then Av could be even less
(iv) Ap = Ai . Av β × 1 = β = 50 ; Gp = 10 log1050 = 17 dB
Example 59.12. For the beta-stabilized emitter follower circuit of Fig. 59.31, find
(i) rin(base)
(ii) rin or rin(stage)
(iii) rL and
(iv) Av. Take transistor β = 100.
(Electronics Technology, Mysore Univ.)
Solution. V2 VCC
V2
RE
IE
R1
R2
R2
5
5
20 10 5 V
40
20V
VCC
R1 30 K
1mA
re´ = 25/1=25 Ω
rL = 5 || 5 = 2.5 K
(i) rin(base) = β (re´ + rL)
+
Vs
R2 10 K V2
= 100 (25 + 2500) = 250 K
(ii) rin(base) = R1 || R2 ||
RE 5 K RL 5 K Vout
_
Fig. 59.31
rin(base) =30K || 10 K|| 250 K ≅ 8 K
(iii) rL = 5 K || 5 K = 2.5 K
(iv) Ap = Ai . Av ≅ 1 × β = 100 ; Gp = 100 log10100 = 20dB
Example 59.13. For the CE circuit of Fig. 59.32, find
(i) rin or rin(stage) (ii) Av (iii) Ap. Take transistor β = 200
Solution . IE = 20/20 = 1 mA re´ = 25/1 = 25 Ω
rL = 20 K || 50 Ω = 50 Ω
rin(base) = β (re´ + rL) =200 (25 + 50) =15 K;
(i) rin or rin(stage) = RB || rin(base)
= 100 K || 15 K = 13 K
(ii) Since re´ is not negligible as compared to rL,
we will use the expression
Av
rL
re ´ rL
50
0.667
25 50
Fig. 59.32
Transistor Equivalent Circuits and Models
2253
(iii) Ap = Av . Ai = 0.667 × 200 = 133.3
59.9. Small-signal Low-frequency Model or Representation
Although many transistor representations or models have been suggested and widely used, the
equivalent T-model is the easiest to understand because, in this representation, component parts retain their identity in all configurations leading to rapid appreciation of a given network.
59.10. T-Model
Such models are not in common use today because they do not take into account any gain
between input and output.
(a) CB Circuit
In Fig. 59.33 is shown the low-frequency T-equivalent circuit of a transistor connected in CB
configuration. It utilizes T- or r-parameters. All these parameters are ac parameters and are measured
under open-circuit conditions. Here, r e represents the ac resistance of the forward-biased
emitter-base junction. Its value is
25 mV
re
— for Ge
IE
50 mV
— for Si
IE
This resistance is fairly small and depends on IE. Also, re represents the ac resistance of the
reverse-biased C/B junction. It is of the order of a few MΩ. Finally, rb represents the resistance of the
base region which is common to both junctions. Its value depends on the degree of doping. Usually,
rb is larger than re but much smaller than rc.
E
re
rc
rb
I/P
C
O/P
E
re
rc
rb
I/P
(a)
Fig. 59.33
O/P
E
re
re
rc
ie
C
rb ib
B
B
B
C
ie
(b)
Fig. 59.34
However, circuit shown in Fig. 59.34 (a) is not complete because it does not illustrate the
forward current transfer ratio. Since current in the output of a transistor depends on the current at the
input, a dependent current-generator in parallel with rc must be included as shown in Fig. 59.34 (b).
As it is the usual practice, all currents are shown flowing inwards even though some of them may
actually be flowing in the opposite direction.
The current generator may be
replaced by a voltage generator with
the help of Thevenin’s theorem as
shown in Fig. 59.35 (a). In that case,
the T-equivalent circuit becomes as
shown in Fig. 59.35 (b). The generator has a voltage of αie rc = rin ie
Fig. 59.35
where rin = αre.
Typical values of different parameters are :
re = 25 to 50 Ω; rb = 100 to 1000 Ω; rc = 1 MΩ
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Electrical Technology
(b) CE Circuit
The T-equivalent circuit for such a configuration is shown in Fig. 59.36. Whereas circuit shown
in Fig. 59.36 contains a parallel current-generator that shown in Fig. 59.37 contains a series-voltage
generator.
(c) CC Circuit
The T-equivalent circuit for such a configuration is shown in Fig. 59.38.
Fig. 59.36
Fig. 59.37
Fig. 59.38
59.11. Formulas for T-Equivalent of a CB Circuit
1. rin
rb (1
re
re
(1
rb
ro rc
3.
Ai = α
4.
Av
vout
vin
Av
Vout
Vin
RL
rb (1
re
( re
Vs
)
re
E
Vin
ie
r in
rc
ic
ib
rin ie
_+
ro
rb
C
Vout
B
Transistor
Fig. 59.39
re
rbrc
re rb RS
2.
5.
)
RS
rL=RL
In Fig. 59.39 is shown a small-signal
low-frequency T-equivalent circuit for CB
configuration. The ac input signal source
has a resistance of RS and voltage of Vs.
Of course, dc biasing circuit has been
omitted and only ac equivalent shown. The
approximate expressions for input and output resistance and voltage and current
gains as derived by applying KVL to the
input and output loops are given below
without derivation :
rb
rc (1
rc
) RL
rL
)
RL
Rs ) rb (1
)
RL
(re Rs )
AP = Ai . Av
2
RL
(re rb (1
)
— no source resistance
2
( re
RL
)
Rs rb (1 )
Gp = 10 log10Ap
— with source resistance
Transistor Equivalent Circuits and Models
2255
59.12. Formulas for T-Equivalent of a CE Circuit
1.
rin
rb
RS
rc
B rb
ib
Vs
Vin
re
( (1 )
+
rmie
C
Vout
re
ie
E
rin
ic
_
ro
rL=RL
The low-frequency small-signal
T-equivalent circuit for such a configuration is shown in Fig. 59.40.
The approximate expressions for
various resistances and gains as found by
applying KVL to the input and output loops
are given below :
Fig. 59.40
= rb + (1 + β)re
2.
4.
ro re (1
Vout
Av
Vin
)
rb
RL
rb (1
re
5. Ap = Av.Ai
re re
re RS
(1
3.
)
2
)[( re
RL
rb (1
)]
Ai = β
Vout
Avs
Vs
(1
(1
) [( re
)
re
2
( rb
RL
Rs ) (1
RL
( rb RS ) (1
)
)]
59.13. Formulas for T-Equivalent of a CC Circuit
The low-frequency T-equivalent circuit for such a configuration is shown in Fig. 59.41.
The approximate expressions for various resistances and gains are given below :
RL
1
(1
) RL
1.
rin
2.
rv = re + (1 – α) (rb + RS)
(rb RS )
re
(1 )
3.
Ai
4.
(1
An = 1
1
)
(1
)
Fig. 59.41
1
(1 ) Gp = 10Ap dB
(1 )
Example 59.14. A junction transistor has re = 50 Ω, rb = 1 K, rc = 1M and α = 0.98. It is used
in common-base circuit with a load resistance of 10 K. Calculate the current , voltage and power
gains and the input resistance.
(Applied Electronics, Punjab Univ. 1993)
Solution. (i) Ai = α = 0.98
5.
Ap
0.98 10, 000
RL
140
(iii) Ap= Aν . Ai = 140 × 0.98 = 137
re rb (1 ) 50 1000 (1– 0.98)
(iv) rin = re + rb (1 – α) = 50 + 1000 (1 – 0.98) = 70 Ω
Example 59.15. A P-N-P junction transistor is used as a voltage amplifier in the groundedbase circuit with a load resistance being 300 K and the internal resistance of the original source
being 200 Ω.
Derive an expression for the voltage gain of the amplifier and calculate its magnitude if the
transistor T-network parameters are : re = 18 Ω, rb= 700 Ω , rc = 1 M and rm = 976 K.
(Applied Electronics and Circuits, Grad. I.E.T.E.)
(ii) Av
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Electrical Technology
Solution. Here, rm= α (rc + rb) ≅ αrc
Av
Avg
V2
V1
V2
Vg
rc
RL
r (1
0.976 300 103
) 18 700 (1 – 0.976)
RL
rc rb (1
RG
∴ α = rm / rc= 976 K / 1000 K = 0.976
)
8,410
0.976 300 103
1247
200 18 700 (1– 0.976)
Example 59.16. Calculate the input and output resistance, overall current, voltage and power
gains for a CE connected transistor having following r-parameters :
rb= 30 Ω, re = 400 Ω , rc = 0.75 M, α = 0.95, RL = 10 K and RS = 400 Ω
Also, calculate the power gain in decibels.
(Electronics Technology, Bangalore Univ. 1999)
Solution. (i) rin
(ii) ro
re (1
re
30
(1 – )
rb
)
rb
1
8030 Ω
rc re
rc RS
750, 000 (1 – 0.95)
(iii) Ai
400
(1 – 0.95)
0.95 750, 000 400
380, 870 0.38 M
(30 400 400)
0.95
19
(1 0.95)
RL
0.95×10×103
= 22.5
re ( rb Rs ) (1 ) 400 + (30 + 400) × 0.05
(v) Ap = Aν . Ai = 22.5 × 19 = 427.5
Power gain in decibels, Gp = 10 log10427.5= 26.3 dB
(iv) A
59.14. What are h-parameters ?
These are four constants which describe the behaviour of a two-port linear network. A linear
network is one in which resistance, inductances and capacitances remain fixed when voltage across
them is changed.
Consider an unknown linear network contained
in a black box as shown in Fig. 59.42. As a matter of
convention, currents flowing into the box are taken
positive whereas those flowing out of it are considered
negative.
Fig. 59.42
Similarly, voltages are positive from the upper to
the lower terminals and negative the other way around.
The electrical behaviour of such a circuit can be described with the help of four hybrid parameters or
constants designated as h11, h12, h21, h22. In this type of double-number subscripts, it is implied that
the first variable is always divided by the other. The subscript 1 refers to quantities on the input side
and 2 to the quantities on the output side. The letter ‘h’ has come from the word hybrid which means
mixture of distinctly different items. These constants are hybrid because they have different units.
Out of the four h-parameters, two are found by short-circuiting the output terminals 2-2 and the
other two by open-circuiting the input terminals 1-1 of the circuit.
(a) Finding h11 and h21 from Short-Circuit Test
As shown in Fig. 59.43, the output terminals have been shorted so that v2 = 0, because no
voltage can exist on a short. The linear circuit within the box is driven by an input voltage v1. It
Transistor Equivalent Circuits and Models
2257
produces an input current i1 whose magnitude depends
on the type of circuit within the box.
h 11
V1
i2
— output shorted
h21
i2
i1
— output shorted
Fig. 59.43
These two constants are known as forward parameters.
The constant h11 represents input impedance with output shorted and has the unit of ohm. The
constant h21 represents current gain of the circuit with output shorted and has no unit since it is the
ratio of two similar quantities.
The voltages and currents of such a two-port network are related by the following sets of equations or V/I relations.
V1= h11 i1 + h12 V2
...(i)
i2 = h21 i1 + h22 V2
... (ii)
Here, the hs are constants for a given circuit but change if the circuit is changed. Knowlege of
parameters enables us to find the voltages and currents with the help of the above two equations.
(b) Finding h12 and h22 from Open-circuit Test
As shown in Fig. 59.44, the input terminals are open so that i1 = 0 but there does appear a
voltage v1 across them. The output terminals
are driven by an ac voltage v2 which sets up
current i2.
h12
V1
V2
— input open
h22
i2
V2
— input open
Fig. 59.44
As seen, h12 represents voltage gain (not forward gain which is v2 / v1). Hence, it has no units.
The constant h22 represents admittance (which is reverse of resistance) and has the unit of mho or
Siemens, S. It is actually the admittance looking into the output terminals with input terminals open.
Generally, these two constants are also referred to as reverse parameters.
Summary of h-parameters
h11 = input impedance
with output shorted
h21 = forward current gain
h12 = reverse voltage gain
with input open
h22 = output admittance
59.15. Input Impedance of a Two Port Network
Consider the two-port linear network shown in
Fig. 59.45 which has a load resistance rL across its output terminals. The voltage source V1 on the input side
drives the circuit and sets up current i1. As seen, Zin =
V1 / i1. Substituting the value of V1 from Eq. (i) of Art.
59.14, we get
V h 11i 1 h 12 v2
Z in 1
i1
i1
h 11
h 12V2
i1
Fig. 59.45
... (i)
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Electrical Technology
As seen from Fig. 59.45 above i2 = – ν2 / RL*
Substituting this value of l2 in Eq. (ii) of Art 59.14, we have
– v2
RL
h 22 i1 h 22 v2
or
v2
i1
h21
h22 1 / RL
Substituting this value in Eq. (i) above, we have
Z in
h11
h 12 . h 21
h 22 1 / RL
59.16. Voltage Gain of a Two Port Network
The voltage gain of such a circuit (Fig. 59.45) is Av = ν2 / ν1. Now ν1 = i1. Zin. Hence,
Av = ν2 / i1.Zin.
Substituting the value of ν2 / i1 as found earlier in Art. 59.15, we get
Av
h 21
Z in ( h 22 1 / R L )
Example. 59.17. Find the h-parameters of the circuit shown in Fig. 59.46 (a).
Solution. First of all, let us find the forward parameters hu and h21. For that purpose, a short is
put across the output terminals as shown in Fig. 59.46 (b).
Fig. 59.46
(i) The input impedance of the network as viewed from input terminals is
h11 = 2 + 4 || 4 = 4 Ω
(ii) As seen from Fig. 9.46 (b), input current i1
divides into two equal parts at point A. The
output current i2 = – i1/2 (negative sign has
been taken because actually it is flowing out
of the box).
∴ h21
–i1 / 2
i1
i2
i1
1 – 0.5
2
Fig. 59.47
(iii) Now, for finding reverse parameters, we will
keep input terminals open and apply v2 across output terminals as shown in Fig. 59.47. It will
produce a current i2 which will produce equal drops across the two 4 Ω resistors. The voltage
which appears across input terminals as v1 is the drop across the vertical 4 Ω resistor connected
at point A. Hence, v1 = v2 / 2.
∴
h12
v1
v2
v 2/ 2
0.5
v2
* The negative sign is used because actual load current is opposite to that shown in figure.
Transistor Equivalent Circuits and Models
2259
The input impedance of the network when viewed from output terminals with input terminals
open is = 4 + 4 = 8 Ω.
∴
h22 = 1/8 = 0.125 Siemens (i.e. mho)
Hence, for the network shown in Fig. 59.46 (a), the h-parameters are as under :
h11 = 4 Ω
h21= – 0.5
h12=0.5 and
h22= 0.125 S
59.17. The h-parameter Notation for Transistors
While using h-parameters for transistor circuits, their numerical subscripts are replaced by the
first letters for defining them.
h11 = hi = input impedance
output shorted
h21= hf = forward current gain
h12 = hr = reverse voltage gain
h22 = h0 = output admittance
input open
A second subscript is added to the above parameters to indicate the particular configuration.
For example, for CE connection, the four parameters are written as :
hie
hfe
hre
and
hoe
Similarly, for CB connection, these are written as hib, hfb, hrb, hop and for CC connection as hic,
hfc, hrc and hoc.
59.18. The h-parameters of an Ideal Transistor
As stated earlier, every linear circuit has a set of parameters associated with it, which fully
describe its behaviour. When small ac signals are involved, a transistor behaves like a linear device
because its output ac signal varies directly as the input signal. Hence, for small ac signals, each
transistor has its own characteristic set of h-parameters or constants.
The h-parameters depend on a number of factors such as
1. transistor type 2. configuration
3. operating point
4. temperature 5. frequency
These h-parameters can be found experimentally or graphically. The parameters hi and hr are
determined from input characteristics of the CE transistor whereas hf and h0 are found from output
characteristics.
59.19. The h-parameters of an Ideal CB Transistor
In Fig. 59.48 (a), a CB-connected transistor has been shown connected in a black box. Fig.,
59.48 (b) gives its equivalent circuit. It should be noted that no external biassing resistors or any
signal source has been shown connected to the transistor.
(i) Forward Parameters
The two forward h-parameters can be found from the circuit of Fig. 59.49 (a) where a short has
been put across the output. The input impedance is simply re.
∴
hib = re
The output current equals the input
current i.e. Since it flows out of the box, it
is taken as negative. The forward current
gain is
–ie
1
h fb
ie
Fig. 59.48
(It also called the ac α of the CB circuit.)
(ii) Reverse Parameters
The two reverse parameters can be found from the circuit diagram of Fig. 59.49 (b). When input
terminals are open, there can be no ac emitter current. It means that ac current source (inside the box)
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Electrical Technology
has a value of zero and so appears as
an ‘open’. Because of this open, no
voltage can appear across input terminals, however, large v2 may be. Hence,
ν1 = 0.
∴
hrb
v1
v2
0
v2
0
Similarly, the impedance, looking
into the output terminals is infinite. Consequently, its
admittance (= 1/∞) is zero.
∴
hob = 0
Fig. 59.49
E
Summary
The four h-parameters of an ideal transistor connected in CB configuration are
hib = re ;
hfb = –1, hrb = 0 ; hob= 0
The equivalent hybrid circuit is shown in Fig. 59.50.
ic
ie
h ib
B
hfb ie
C
B
Fig. 59.50
Note. In an actual transistor, hrb and hob are not zero but have some finite value though extremely small (ranging
–
–6
from 10 to 10 ).
In reality, output impedance is not infinity but very high so that hob is extremely small. Similarly,
there is some amount of feedback between the output and the input circuits (even when open) though
it is very small. Hence, hrb is very small.
59.20. The h-parameters of an ideal CE Transistor
Fig. 59.51 (a) shows a CE-connected
ideal transistor contained in a black box
whereas 59.51 (b) shows its ac equivalent
circuit in terms of its β and resistance
values.
(a) Forward Parameters
The two forward h-parameters can be
Fig. 59.51
found from the circuit of Fig. 59.52 (a)
where output has been shorted. Obviously, the input impedance is simply βre.
∴
hie=βre
The forward current gain is given by
ib
i
h fe 2
i1 ib
(It is also called the ac beta of
the CE circuit)
(b) Reverse Parameters
These can be found by reference
to the circuit of Fig. 59.52 (b) where
input terminals are open but output
Fig. 59.52
terminals are driven by an are voltage source v2. With input terminals
open, there can be no base current so that ib = 0. If ib = 0, then collector current source has zero value
and looks like an open. Hence, there can be no v1 due to this open.
Transistor Equivalent Circuits and Models
2261
v1 0
0
v2 v2
Again, the impedance looking into the output terminals is
infinite so that conductance is zero.
∴
hoe = 0
Hence, the four h-parameters of an ideal transistor conFig. 59.53
nected in CE configuration are :
hie=βre ; hfe= β, hre= 0 ; hoe = 0.
The hybrid equivalent circuit of such a transistor is shown in Fig. 59.53.
Note. In practice, hre and hoe are not exactly zero but quite small for the same reasons as given in the
Note to Art 59.19.
∴
hre
59.21. Approximate Hybrid Equivalent Circuits
So far, we did not take into account the following two factors which exist in an actual transistor
(as opposed to an ideal one).
(i) because of the transistor’s non-unilateral behaviour, there is a ‘feedback’ of the output
voltage into the input voltage. This feedback is represented by a voltage-controlled generator hrv2 as shown in Fig. 59.54 and 59.55.
By definition, an ideal amplifier is one which responds only to signals applied to its input
terminals. It should not do the reverse i.e. reproduce at the input any portion of the ac signal
applied at the output. Such an ideal one-way device is called a unilateral device. A real
transistor cannot be unilateral because of unaviodable interaction between its input and
output circuits (after all, it consists of a single piece of a crystal). Therefore, not only its
output responds to its input but, to a lesser degree, its input also responds to its output.
(ii) even when input circuits is open, there is some effective value of conductanece when looking into the transistor from its output terminals. It is represented by h0.
We will now draw low-frequency small-signal hybrid equivalent circuits after taking into
account the ‘feedback’ voltage generator and output admittance.
(a) Hybrid CB Circuit
In Fig. 59.54 (a) is shown an NPN transistor connected in CB configuration. Its ac equivalent
circuit employing h-parameters is
shown in Fig. 59.54 (b). The V/I relationships are given by the following
two equations.
veb = hib ie + hrb vcb
ie = hfb ie + hob veb
These equations are self-evident
because applied voltage across input
terminals must equal the drop over hib
Fig. 59.54
and the generator voltage. Similarly,
current ic in the output terminals must equal the sum of two branch currents.
As per current convention stated earlier (Art. 59.14), collector ie is shown flowing inwards
though actually this current flows outwards as shown by the arrow inside the ac current source.
Similarly, ac voltage polarities have been taken by considering upper terminal positive and lower
one as negative (please remember that the dc biasing rule of Art. 59.12 does not apply here).
It may be noted that no external dc biasing resistor or ac voltage sources have been connected
to the equivalent circuit as yet.
Incidentally, it may be noted that the ac equivalent circuit contains a Thevenin's circuit in the
input and a Norton’s circuit in the output. It is all the more a reason to call it a hybrid equivalent
circuit.
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Electrical Technology
(b) Hybrid CE Circuit
The hybrid equivalent of the
transistor alone when connected in CE
configuration is shown in Fig. 59.55
(b). Its V/I characteristics are described
by the following two equations.
νbe = hie ib + hre νec
ie = hfb ib + hoe νec
Fig. 59.55
We may connect signal input
source across its input terminals and load resistance across output terminals (Art. 59.22).
(c) Hybrid CC Circuit
The hybrid equivalent of a transistor alone when connected in CC configuration is shown in Fig. 59.56 (b). Its V/I
characteristics are defined by the following two equations :
νbe = hie ib + hre νec
ie = hfe ib + hoc νce
We may connect signal input source
Fig. 59.56
across ioutput terminals BC and load resistance across output terminals EC to get a CC amplifier.
59.22. Transistor Amplifier Formulae Using h-parameters
As shown in Fig. 59.57, if we add a signal source across input terminals 1-1 of a transistor and
a load resistor across its output terminals
i2
RS 1 e i1 h i
2
2–2, we get a small-signal, low-frequency
e
iL hybrid model of a transistor amplifier. It is
ie
v2
v1
valid for all the three configurations and
+
rin
he
re rL holds good for all types of load whether a
_
vS
hf i1
hrv2
resistance of an impedance. We will now
find expressions for its gains and impedb
ances.
1
2
Before undertaking the above derivaFig. 59.57
tions, let us consider different components
in the hybrid model of Fig. 59.57. The input resistance looks like a resistance (hi) in series with a
voltage generator (hr ν2). This generator represents the voltage feed-back from the output to the
input circuit. It is known as voltage-controlled generator because its value is determined by v2 (as hr
is a dimensionless constant). The output circuit also has two components (i) h0 component which
represents the conductance as seen from output terminals and (ii) the current-controlled generator
(hf i1) which simulates the transistor’s ability to amplify. The parameter hf is a dimensionless constant.
The above model can be described mathematically by using the following two equations :
Input Circuit
ν1 = sum of voltage drops from a to b
= hi i 1 + h r ν 2
...(i)
Output Circuit
i2 = sum of currents leaving junction c
= hf i1 + h0 ν2
...(ii)
Transistor Equivalent Circuits and Models
2263
Now, ν2 = –i2 rL. Substituting this value in Eq. (ii) above, we have
i2 = hf i1 – h0 i2 rL
Eq. (i) and (iii) can now be used to find various gains of a transistor.
(i) Current Gain
It is given by Ai = i2/i1
Dividing both sides of Eq. (iii) by i1, we get
i2
i1
hf
h0
i2
r
i1 L
or
... (iii)
Ai = hf – h0 Ai rL
hf
1 h 0 rL
If
rL = 0 or h0 rL << 1, then Ai = hf
Current Gain Taking RS into Account
The source current is not the transistor input current because i1 partly flows along RS and partly
along rin.
To illustrate this point, consider the Norton's
equivalent of the source (Fig. 59.58). The overall current gain Ais is given by
∴
Ai
i2
is
Ais
i2
i1
i1
is
Ai
i1
is
As seen from Fig. 9.58
is Rs
rin Rs
i1
or
i1
is
Rs
rin Rs
Fig. 59.58
∴
Ais = Ai . RS / (rin + RS)
(ii) Input Impedance
It is defined as the resistance when looking into the amplifier from its input terminals. Hence,
rin = v1/i1.
From Eq. (i) above, we have
rin
h i i1 h r
i1
1
i1
2
hi
hr .
2
i1
Substituting the value of v2 = – i2 rL = –Ai i1 rL, we get
rin h i
hi
1
h i Ai r L h i
hrL
h0 rL
h f hr r L
1 h 0rL
hi
h f hr
h0 1/ rL
where
∆h = hi h0 – hf hr
≅ hi
– if hr or rL is very small.
It is seen that rin depends on rL i.e. ac resistance of the load across output terminals of the
transistor.
(iii) Voltage Gain
Av = v2 / v1. It is also known as the internal voltage gain of the transistor. It is different from
Avs = v2 / vs which is the gain from the source to the output terminals and is known as stage gain or
overall gain.
As seen from above, v2 = –Ai i2 rL and v1 = i1 rin
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Electrical Technology
∴
i 2 A irL
iL rin
2
Av
1
Ai
hf
1 h 0rL
rL
rin
r L (1 h 0 . rL )
hi
h . rL
h f rL
h i (1 h0rL ) h f h r rL
hi
hf .
h f .rL
h . rL
rL
hi
Overall voltage gain is
v2
Avs
v2 vi
v1 vs
s
Av
vi
vs
Now, vs drops over series combination of RS and rin.
Drop across rin constitutes v1. Hence, v1 = vs × rin/(RS + Rin)
v1
vs
∴
( RS
rin
Avs Av
or
rin )
rin
( R5 rin )
As seen, if RS = 0, Avs = Aν
Value of Avs may also be obtained by adding RS to hi in the expression for Av.
(iv) Output Impedance
v2
|
i2 vs
It is defined as r0
g0
or
0
i2
v2
Dividing both sides of Eq. (ii) by v2, we get
g0
hf .
i2
v2
h0
..... (iv)
Taking vs = 0 and then applying KVL to the input circuit in Fig. 59.57, we get
–i1(hi + RS) – hr v2 = 0
or
i1/v2 = –hr / (hi + RS)
Substituting this value in Eq. (iv) above, we have
g0
h0 .
h f hr
(hi RS )
ho RS
h
(hi RS )
∴
ro
1
go
hi RS
ho RS h
It is seen that rin depends on rL whereas ro depends on RS.
If RS is very large (i.e. circuit is driven by a current source) or hr is negligible, then ro ≅ 1/ho.
(v) Power Gain
Ap
P2 v2 i2
P1 v1i1
Av Ai
Ai2
rL
rin
The above formulae are summarized below :
(i)
Ai
Ai
hf
1 horL
hf
h f RS
(1 horL )( rin RS )
ho RS
( rin RS )
(ii)
rin hi
h f hr rL
1 ho rL
hi
(iii)
Av
h f hr
h . rL
h f rL
; Avs
hi
hi
h f rL
(hi RS )
Transistor Equivalent Circuits and Models
ro
(iv)
1 hi / RS
ho h / RS
hi RS
ho RS
h
2265
1
ho
59.23. Typical Values of Transistor h-parameters
In the table below are given typical values for each parameter for the broad range of transistors
available today in each of the three configurations.
Parameter
CB
hi
25 Ω
3 × 10
hr
CE
CC
1K
–4
hf
–0.98
ho
0.5 × 10 S
–6
1K
2.5 × 10
–4
≅1
50
–50
–6
25 × 10 S
–6
25 × 10 S
59.24. Approximate Hybrid Formulas
The approximate hybrid formulas for the three connections are listed below. These are applicable when ho and hr are very small and RS is very large. The given values refer to transistor terminals. The values of rin(stage) or rin´ and ro(stage) will depend on biasing resistors and load resistance
respectively.
Item
CE
CB
CC
rin
hie
hib
hic + hfeRL
ro
1
hoc
1
hoB
hie
h fc
Ai
hfe = β
–hfb ≅ 1
–hfe ≅ β
Av
hie RC
his
f fb
R
hib C
1
59.25. Common Emitter h-parameter Analysis
The h-parameter equivalent of the CE circuit of Fig. 59.59 (a) is shown in Fig. 59.59 (b). In
Fig. 59.59 (a), no emitter resistor has been connected.
Fig. 59.59
However, Fig. 59.60 shows the CE circuit with an emitter resistor RE.
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Electrical Technology
Fig. 59.60
We will now derive expressions for voltage and current gains for both these circuits.
1. Input Impedance
When looking into the base-emitter terminals of the transistor, hie is in series with hre no.
For a CE circuit, hre is very small so that hre v0 is negligible as compared to the drop over
hie. Hence, rin ≅ hie.
Now, consider the circuit of Fig. 59.60. Again ignoring hrevo, we have
v1 = hieib + ieRE = hie ib + (ib + i) RE
= hieib + ib RE + hfe ib RE
= ib [hie + RE (1 + hfe)]
v
((
ic h feib )
v
1
1
h (1 h fe ) RE *
rin= rin(base) = i
ib ie
1
rin or rin(base) = R1 || R2 || rin(base)
∴
2. Output Impedance
Looking back into the collector and emitter terminals of the transistor in Fig. 59.59 (b),
ro ≅ 1/hoe.
(
As seen, ro´ or ro(stage) = ro || RL = (1/ho) || rL
Since 1/hoe is typically 1 M or so and RL is usually much smaller, ro´ ≅ RL = rL
3. Voltage Gain
Av
Now,
∴
v2
v1
vo
vin
– Fig. 9.59 (b)
vo= – ic RL and
Av
is RL
ib hie
rL RL )
ic RL
.
ib hie
νin ≅ ib hie
h f RL
hie
Now, consider Fig. 59.60 (b). Ignoring hre vo, we have from the input loop of the circuit
vin = ib [hie + RE (1 + hfe)]
–proved above
∴
Av
vo
vin
RL
RE
ic RL
ib [ hie RE (1 h fe )]
h fe RL
h fe (1 h fe ) RE
— if (1 + hfe) RE >> hie
* The above result could also be obtained by applying β-rule (Art. 9..24)
Transistor Equivalent Circuits and Models
2267
In general,
4. Current Gain
Ai
Ais
i2
i1
h fe
1 hoe rL
h fe
— if hoe rL << 1
h fe . R1 || R2
rin R1 || R2
5. Power Gain
Ap = Aν × Ai
59.26. Common Collector h-parameter Analysis
The CC transistor circuit and its h-parameter equivalent are shown in Fig. 59.61.
One can make quick approximations of CC gains and impedance if one remembers that
hre = 1 i.e. all of νo is fed back to the input (Art. 59.23).
1. Input Impedance
vin = ib hic + hrc νo = ibhic + νo = ib hic + ie RL
= ibhic + hfe ib RL = ib (hic + hfe RL)
ic = hfe ib
Fig. 59.61
rin
∴
vin
ib
hic h fe RL
As seen,
rin(stage) = rin(base) || R1 || R2 = rin(base) || RB where RB = R1 || R2
2. Output Impedance
ro
v2
|
i2 vs
vo
|
ic vs
0
0
Now,
ie ≅ ic = hfe ib= hfc i1
Since
vs = 0, ib is produced by hrc vo = vo
Hence, considering the input circuit loop, we get
ib
v0
v0
hic ( RS || R1 || R2 ) hic RS || RB
ic h fc ib
h fc vo
hic
( RS || RB )
where RB = R1 || R2
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Electrical Technology
Vo
ie
ro
∴
hic ( RS || R1 || R2 )
h fe
Also, ro´ or ro(stage) = ro || RL
3. Voltage Gain
v2
v1
Av
Now,
vo = ie RL = hfe ib RL and ib = (νin – νo) / hic
vo
∴
vo
vin
h fe RL
(vin vo )
hie
h fe RL / hic
1 h fe RL / hic
vo
vin
Av
vo 1
or
h fe RL
hic
h fe RLvin
hic
1
4. Current Gain
Ai
i2
i1
ie
ib
h fe ; Ais
h fe RB
rin RB
where RB = R1 || R2
59.27. Conversion of h-parameters
Transistor data sheets generally specify the transistor in terms of its h-parameters for CB connection i.e. hib, hfb, hrb and hob. If we want to use the transistor in CE or CC configuration we will
have to convert the given set of parameters into a set of CE or CC parameters. Approximate conversion formulae are tabulated over leaf :
Table No. 59.2
From CB to CE
From CE to CB
From CE to CC
hie
hib
1 h fb
hie
hie
1 h fe
hic = hic
hoe
hob
1 h fb
hob
hoe
1 h fe
hoc= hoe
h fe
hre
1
h fb
h fb
hib hob
1 h fb
h fb
hrb
hrb
1
h fe
h fe
hie hoe
1 h fe
hfe = –(1 + hfe)
hre
hre = 1 – hre ≅ 1
Example. 59.18. A transistor used in CB circuit has the following set of parameters.
–4
–6
hib= 36 Ω, hfb = 0.98, hrb = 5 × 10 ,hob= 10 Siemens
With RS = 2 K and RC = 10 K, calculate (i) rin(base) (ii) rout (iii) Ai and (iv) Aν.
(Applied Electronics-I; Punjab Univ. 1991)
Solution. Approximate Values
(i) rin = hib = 36 Ω
1
(ii) ro h
ob
(iii) Ai = hfb = – 0.98
(iv) Av
h fb
hib
1
10 6
1M
RC
0.98
36
10K
272
Transistor Equivalent Circuits and Models
2269
More Accurate Values
(i) rin(base)
hrb h fb
hob 1/ rL
hib
–Art 9.22
0.98 5 10 4
10 6 1/10 103
= 36
(∴
rL = RC since there is no RL)
= 36 + 4.9 = 40.9 Ω
It is the input resistance at transistor terminals.
Fig. 59.62
(ii) ro
ho ( hib
hib Rs
RS ) h fb . hrb
36
10
6
(36
2000
2000)
( 0.98) 5 10
4
0.8 M
It is the output resistance at transistor terminals.
h fb
1 hob rL
(iii) Ai
–0.97
h fb rL
hib (1 hob rL ) h fb hrb . rL
Av
(iv)
0.98
1 10 6 104
(
Here, ac load rL = RL = 10 K = 104 Ω
∴ Av
36(1 10
6
( 0.98) 104
104 ) ( 0.98) 5
10
4
104
249
Example 59.19. A transistor used in CE connection (Fig. 59.63) has the following set of
h-parameters : hir = 1 K, hfe = 100, hre = 5 ×10–4 and hoc
= 2 × 10–5 S. With RS = 2 K and RC = 5 K, determine
(i) rin (ii) ro (iii) Ai and (iv) Av
Solution.
(i) rin
hie
h fb rL
ho 1/ rL
= 1000
5 10
2 10
5
4
100
1/ 5
103
Fig. 59.63
= 723 Ω
(ii) ro
hie Rs
( hie Rs ) hoe h fe . hre
= 30,000Ω = 0.03 M
1000 2000
(1000
2000) 10
5
100 5 10
4
rL
RL )
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Electrical Technology
Fig. 59.64
(iii) Ai
(iv)
Ai
1
h fe
hoe rL
(hie
1
2
100
10 5 5
103
91
h fe rL
Rs ) (1 hoe . rL ) h fe hre rL
100 5 103
–164
(100 2000) (1 2 10 5 5 103 ) 100 5 10 4 5 103
The negative sign indicates that there is 180° phase shift between the input and output ac signals.
Obviously, it is the overall (or circuit) voltage gain and not the voltage gain of the transistor alone.
Example 59.20. In the CE circuit shown in Fig. 59.65, the transistor parameters are :
–4
–5
hie=2 K, hfe=100, hre = 5 × 10 , hoe = 2 × 10 S
Calculate (i) rin(base), (ii) rin(stage), (iii) ro, (iv) ro(stage), (v) Ai and (vi) Av.
(Electronics–1, Karnataka Univ.)
Solution. The hybrid equivalent circuit is shown in Fig.
59.66. We will use the approximate formulas given in Art.
59.24.
(i) rin(base) = hie
= 2K
(ii) rin(stage) = 2 K || 250 K = 1.98 K
–5
(iii) ro = 1 / hoe = 1/2 × 10 = 50 K
It is the output impedance of the transistor only.
(iv) ro(stage) = ro´ = 50 K || 5 K = 4.54 K
The impedance takes into account the collector load.
(v) Ai ≅ hfe = 100
(vi) Av
h fe rL
hie
100 5
2
–250
Fig. 59.66
Fig. 59.65
Transistor Equivalent Circuits and Models
2271
Example 59.21. Determine the various gains of the circuit of Fig. 59.67 if an emitter resistance of 0.5 K is included in the circuit.
(Applied Electronics, Punjab Univ. 1991)
Solution. The CE circuit with RE included is shown in Fig. 59.47. Different performance characteristics of the circuit are as under :
+VCC
(i)
(ii)
(iii)
(iv)
(v)
rin(base)
hie
(1
) RE
250 K RB
= 2 + 101 × 0.5 = 52.5 K
rin(stage) = RB || rin = 250 || 52.5
= 43.3K
ro = 1/hoe = 50 K
ro(stage) = ro´=50 K || 5 K
= 4.54 K
Av = hfe = 100
(vi) Av
h fe rL
hie (1 ) RE
2
100 5
50.5
RC 5 K
C1
vo
C2
=100
RS 1 K
vs
–9.5
RE
0.5 K
Fig. 59.67
The value is reduced from 250 to 9.5.
Example 59.22. The transistor of Fig. 59.68 has the following set of h-parameters :
–4
–6
hie = 2K, hfe = 50, hrs = 4 × 10 , hoe = 25 × 10 Siemens
Determine (i) rin(base) (ii) rin(base) (iii) ro (iv) ro(stage) and (v) Av.
(Electronics-I, Patna Univ. 1991)
Solution. we will use the formula derived in Art.
59.25.
(i)
rin(base)
hie (1 h fe) R
E
(ii)
= 257 K
rin(stage) = rin(base) || R1 || R2 = 257 || 80 || 40
2
(1
50)
5
= 24 K
–6
(iii) ro = 1/hoe = 1/25 × 10 = 40 K
(iv) ro(stage)
= ro || rL where rL = 15 K || 30 K
= 10 K = 40 K || 10 = 8 K
(v)
h fe RL
hie (1 h fe ) RE
Av
50 10
257
Fig. 59.68
–2.9
Example 59.23. Draw a hybrid small-signal model for a transistor in CE configuration.
A single source with an open circuit voltage of 1 mV and internal impedance of 600 W is
connected to the input of transistor AC 125 in CE configuration. The small-signal parameters
measured at VCB = –5 V and IE = 2 mA, are as follows :
–4
hie = 1.7 K, hre = 6.5 × 10 , hfe = 125, hoe = 80 µS.
Calculate the input and output impedances and signal amplification for a load of 5.6 K. Also,
find the value of the signal voltage at the output.
(Electronics-I, Gwalior Univ.)
3
–6
–4
Solution. ∆h = 1.7 × 10 × 80 × 10 – 125 × 6.5 × 10 = 0.055
(i)
rin
hie
h . RL
1 hoe RL
1700 5.6 103 55 10 3
1 80 10 6 5.6 103
–Art. 59.22
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Electrical Technology
(ii)
ro
hie RS
hoe RS
h
(iii)
Av
h fe RL
hie RL . h
1700
80
10
6
600
600
125 5600
1700 308
0.055
2300
0.103
22.3 K
–348.6
The negative sign merely indicates that there is phase reversal of 180° between the output and
input voltages.
Now, 1 mV signal voltage is divided between rin and RS. The input voltage vin is that which
drops over rin = 1 × rin/(rin + RS) = 1 × 1387/(1387 + 600) = 0.698 mV
∴ output voltage = –348.6 × 0.698 = –243.3 mV.
Example 59.24. The transistor of Fig. 59.69 has the following set of h-parameters :
–4
–5
hie = 2 K, hfe = 100, hre = 5 × 10 , hoe = 2.5 × 10 S
Find the voltage gain and the ac impedance of the stage.
(Electronics-II, Bombay Univ. 1992)
Solution. Using somewhat exact formulas given in Art. 59.22, we have
rin(base)
hie
h fe hre
hoe 1/ rL
Now, collector load
rL = 10 K || 30 K = 7.5 K
∴
200
rin(base)
100 5 10 4
2.5 10 5 1/ 7.5 103
=2000 – 316 = 1684 Ω
The ac input impedance of the stage i.e.
impedance when looking into point B is
rin(base) = rin(base) || R1 || R2 = 1.684 || 50 || 25 = 1.53K
∴
Av
h fe
rin(base) ( hoe 1/ rL )
Av
100
0184 ( 2.5 10 5 1/ 7500)
Now,
Fig. 59.69
rL = 10 K || 30 K = 7.5 K
− 375
Obviously, RE does not come into the ac picture because it is ac grounded by the bypass capacitor.
Example 59.25. In the CC circuit of Fig. 59.70, the transistor parameter are hic = 2K and hfc =
100. Calculate the circuit input and output impedance and voltage, current and power gains.
(Electronic Technology, Bangalore Univ.)
Solution.
rin ≅ hic + hfe RL = 2 + 100 × 5 = 502 K
rin(stage) = R1 || R2 || rin
= 10 || 10 || 502 = 4.95 K
ro
hie ( Rs || R1 || R2 )
h fe
2
(1 || 10 || 100)
100
28.3 Ω
Fig. 59.70
Transistor Equivalent Circuits and Models
2273
ro(stage) = ro || RL = 28.3 Ω || 5K
Ω
= 28.1Ω
Aν ≅ 1 and Ai = hfe = 100
Example 59.26. A transistor with hie = 1.5 K and hfe = 75 is used in an emitter follower circuit
where resistances R1 and R2 are used for normal biasing. Calculate (i) Ai (ii) rin (iii) ro and (iv) Av if RE
= 860 Ω, R1 || R2 = 20K and RS = 1K.
(Electronic Engg.; Indore Univ. 1992)
Solution. Let us first convert the given CE values of h-parameters into their equivalent CC
values with the help of Table No. 59.2. It is seen that hic = hie = 1.5 K and hfc = (1 + hfe) = 76.
(i) Ai = hfe= 76
(iii) ro
(ii) rin = hic + hfc RL = 1.5 + 76 × 0.860 = 66.9 K
hic ( Rs || R1 || R2 )
h fe
1.5 1 ||20
76
32.3 Ω
(iv) Av ≅ 1
Tutorial Problems No. 59.1
1.
2.
Using ideal transistor approximations for the single-stage CB amplifier of Fig. 59.71, find
(i) stage rin (ii) rL (iii) Avand (iv) Ap. Take transistor α = –0.99.
[ (i) 25 Ω (ii) 10 K (iii) 400 (iv) 396]
For the single-stage CB amplifier circuit shown in Fig. 59.72, find
(i) rin (ii) rL (iii) Av and (iv) Ap
Fig. 59.71
Fig. 59.72
Transistor α = –0.9. Take Ge transistor material. [(i) 25 Ω (ii) 5 K (iii) 200 (iv) 196]
3.
For the CB amplifier circuit of Fig. 59.73, find approximate
values of
(i)
(ii)
(iii)
(iv)
(v)
(vi)
stage rin
rL
Aν=νout / νin
Aνσ = νout / νs
Ap
Gp
Take transistor α = –0.98
[(i) 25 Ω (ii) 5K (iii) 200 (iv) 22.2 (v) 196 (vi) 22.9 dB]
4.
Fig. 59.73
Find rin, rL, Ai, Av and Gp for the CE amplifier shown in Fig. 9.74.
[(i) 1250 Ω (ii) 5K (iii) 50 (iv) 200 (v) 30 dB]
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Electrical Technology
Fig. 59.74
5.
Fig. 59.75
For the single-stage CE amplifier circuit of Fig. 59.75, find
(i) rin
(ii) rL
(iii) Av
(iv) Ap and
Take β = 100 and use re = 50 mV/IE.
(v) Gp
[(i) 160 K (ii) 5K (iii) 2.5 (iv) 250 (v) 24 dB]
6. For the CE amplifier of Fig. 59.76, find approximate values of
(i) rin
(ii) rL
(iii) Av
(iv) Ap and Gp.
Take transistor β = 50 and use re = 50 mV/IE.
[ (i) 2K (ii) 10K (iii) 200 (iv) 10,000 (v) 40 dB]
7. For the emitter follower circuit shown in Fig. 59.77, find
(ii) rin(stage)
(i) rin(base)
(iii) stage Av
(iv) Ap in decibles
Take β = 100 and use re = 25 mV/IE. [(i) 10K (ii) 9.52K (iii) 0.75 (iv) 18.75 dB]
Fig. 59.76
Fig. 59.77
8. An NPN silicon transistor is connected as a CE amplifier with load RL and a source resistance RS. The hparameters are :
hie = 1.1 K, hre = 2.5 × 10–4, hfe = 50 and hoe = 25 µS
If RL = 10K and RS = 1K, find the various gains and input and output impedances.
Derive the relations used.
[Ai = 40, Ais = 20, rin = 1K, Av = –400, Avs = –200, ro = 52.5 K, Ap = 16,000, Gp = 42 dB]
9. Calculate the gain of a common-emitter transistor amplifier whose hybrid parameters are:
–4
hie = 1100 ohms, hre = 2.5 × 10 ; hfe = 50, hoe = 25 µS, RL = 5K.
Derive any formula used.
(–213) (Electronic Engg., Indore Univ.)
–4
10. A CE amplifier has hie = 1.1 K, hfe = 50, hre = 2.5 × 10 ; hoe = 25 µS, RS = RL = 500 ohm. Calculate the
output impedance and voltage gain.
(Applied Electronics-I, Punjab Univ. Dec.)
–4
11. A junction transistor has the following h-parameters hie = 2000 ohm, hre = 15 × 10 , hfe = 49, hoe = 50 µS.
Determine the current gain, voltage gain, input resistance and output resistance of the CE amplifier if the
Transistor Equivalent Circuits and Models
2275
load resistance is 10K and source resistance is 600 ohm. Derive the expressions used.
(Applied Electronics-1, Punjab May)
–4
12. A CE amplifier has hie = 1.1 K, hfe = 50, hre = 2.5 × 10 , hoe = 25 µS, RS = RL= 1K.
Calculate the output impedance and voltage gain.
(Applied Electronics-1, Punjab Univ. Dec.)
13. A junction transistor has the following h-parameters hie = 1kW, hre = 1, hfe = –50, hoe = 25mA/V. This
transistor is connected to a source of internal resistance 600 ohm and a load of 40 kΩ. Calculate the current
gain, voltage gain, input resistance and output resistance of the amplifier. Derive the expressions used.
(Applied Electronics-I, Punjab Univ. June)
14. A transistor connected in CE configuration has following h-parameters.
hie = 1.1 kΩ
hre = 2.5 × 10–4
hfe = 50
hoe = 25 µ Siemens
and
rs = rL = 1 kΩ
Calculated current gain, input impedance and voltage gain.
(Electronics Engg. Bangalore Univ. 2001)
OBJECTIVE TESTS – 59
1.
2.
3.
4.
5.
6.
In an ac amplifier, larger the internal resistance
of the ac signal source
(a) greater the overall voltage gain
(b) greater the input impedance
(c) smaller the current gain
(d) smaller the circuit voltage gain.
The main use of an emitter follower is as
(a) power amplifier
(b) impedance matching device
(c) low-input impedance circuit
(d) follower of base signal.
An ideal amplifier is one which
(a) has infinite voltage gain
(b) responds only to signals at its input
terminals
(c) has positive feeback
(d) gives uniform frequency response.
The smallest of the four h-parameters of a transistor is
(b) hr (c) ho (d) hf.
(a) hi
The voltage gain of a single-stage CE amplifier is increased when
(a) its ac load is decreased
(b) resistance of signal source is increased
(c) emitter resistance RE is increased.
(d) ac load resistance is increased.
When emitter bypass capacitor in a CE amplifier is removed, its .................... is considerably reduced.
(a) input resistance
(b) output load resistance
7.
8.
9.
10.
11.
(c) emitter current
(d) voltage gain
Unique features of a CC amplifier circuit is
that it
(a) steps up the impedance level
(b) does not increase signal voltage
(c) acts as an impedance matching device
(d) all of the above.
The input impedance h11 of a network with
output shorted is given by the ratio
(b)
ν1 / ν2
(a) ν1/i1
(c) i2 / i1
(d)
i2 / ν2
The h-parameters of a transistor depend on
its
(a) configuration
(b) operating point
(c) temperature
(d) all of the above
The output admittance h0 of an ideal transistor connected in CB configuration is
.................... siemens.
(a) 0
(b) 1/r
(d) –1.
(c) 1/ βre
A transistor has hfe = 100, hie = 5.2 K Ω, and
rbb = 0. At room temperature, VT = 26 mV, the
collector current, | IC | will be.
(a) 10 mA
(b) 5 mA
(c) 1 mA
(d) 0.5 mA
ANSWERS
1. (d)
2. (b)
3. (b) 4. (c)
5. (d)
6. (d)
7. (d)
8. (a)
9. (d)
10. (a)
11. (a)
ROUGH WORK
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