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Chapter 5

Represents the AC characteristics of the
transistor.
 re model
 Hybrid  model
 Hybrid equivalent
model
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
3
Uses circuit elements that approximate the
behavior of the transistor.
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1.
2.
3.
4.
Replace all DC sources with short circuits
Replace all capacitors with open circuits
Remove all components bypassed by 1 & 2
Redraw final circuit
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

5
BJTs are basically current-controlled
devices; therefore the re model uses a diode
and a current source to duplicate the
behavior of the transistor.
One disadvantage to this model is its
sensitivity to the DC level. This model is
designed for specific circuit conditions.
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The diode (B-E) can
be replaced by the
resistor re.
re is resistor equivalent, not
to be confused with RE, or
resistance of the emitter
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Input impedance:
Output impedance:
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Voltage gain:
Current gain:
Current:
Phase Shift:
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Input impedance:
Output impedance:
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Voltage gain:
Current gain:
Current:
Phase shift:
ECET 257 Consumer Power Electronics, PNC

Find
◦
◦
◦
◦
Zi
Vo
Av
Zo
11
Vi=10 mV∠0
IE=0.5 mA ∠0
α=0.980
RL = 1.2 kΩ
Ro =∞Ω
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Input impedance:
Output impedance:
Voltage gain:
Current gain:
Phase Shift:
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The input is applied to the base
The output is taken from the
collector
High input impedance
Low output impedance
High voltage and current gain
Phase shift between input and
output is 180
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AC equivalent
re,model
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Input
impedance:
Output
impedance:
Current gain:
Voltage gain:
Current gain
from voltage gain:
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
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Find
◦
◦
◦
◦
re
Zi
Zo
Av
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re model requires you to
determine , re, and ro.
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Input impedance
Voltage gain
Output impedance
19
Current gain
Current gain from Av
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
Find
◦
◦
◦
◦
re
Zi
Zo
Av
Vcc = 16 V
R1 =39 kΩ
R2= 4.7 k Ω
RC=3.9 k Ω
RE=1.2 k Ω
β=100
Ro = 50 k Ω
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No capacitor parallel to RE.
If capacitor is present, RE is
bypassed and normal CE
format is used
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Input impedance:
Output impedance:
ro can be ignored if
≥ 10( + )
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Voltage gain:
Current gain:
Current gain from Av:
ECET 257 Consumer Power Electronics, PNC

Find
◦
◦
◦
◦
re
Zi
Zo
Av
25
Vcc = 10 V
RB= 390 kΩ
RC= 2.2 k Ω
RE=1.2 k Ω
β=140
ro= 100 k Ω
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 Base input (high Z), Emitter output (low Z)
 Input and output are in phase
 Opposite Z configuration of FB, commonly used for Z matching
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Input impedance:
Output impedance:
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Voltage gain:
Current gain:
Current gain from voltage gain:
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




re
βre
Zi
Zo
Av
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Vcc = 10V
RB=300 kΩ
RE=3 kΩ
β=150
RO=100 k Ω
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The input is applied to the emitter
The output is taken from the
collector
Low input impedance.
High output impedance
Current gain less than unity
Very high voltage gain
No phase shift between input
and output
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Input impedance:
Output impedance:
Voltage gain:
Current gain:
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•
•
•
•
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A variation of the common-emitter fixed-bias configuration
Input is applied to the base
Output is taken from the collector
There is a 180 phase shift between the input and output
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Input impedance:
Output impedance:
Voltage gain:
Current gain:
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
Unloaded amp will
always have the highest
gain
◦

35
>
>
Ideal amp
◦ Low input Z
◦ High output Z
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
Amplitude of the applied signal that reaches
the input of the amplifier

Internal source impedance effect on gain
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
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Must have load impedance
Formula applies to all configurations
=−
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
Find AVL for
◦ RL = 4.7K
◦ RL = 0.5K
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β = 120
Av = - 3.1

Find Vcc
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When using the chart on Pages 292-295,
ensure the circuit meets the requirements for
the equation when ro and re are involved.
Requirements are noted in the sections
covering the configurations.
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44
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
Useful when individual measurements are
impossible or impractical
With Vi set to 0 V:
The voltage across
the open terminals:
where AvNL is the noload voltage gain

45
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No-load
◦

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=
To solve
◦ Find RT using II
◦ Find Vi
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
Vs = 20 mV
Rs= 30 Ω
II = 50 uA
Vo = 3.6 V
Find AVNL
ECET 257 Consumer Power Electronics, PNC

Given Amp PCB specs, find
47
RS=0.2 kΩ
AVNL=-480
Zi= 4 kΩ
Zo = 2 kΩ
◦ AVL with RL = 5.6 kΩ
◦ Ai with RL = 5.6 kΩ
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1.
2.
3.
4.
5.


The output of one amplifier is the input to
the next amplifier
Overall voltage gain is determined by the
product of gains of the individual stages
The DC bias circuits are isolated from each
other by the coupling capacitors
The DC calculations are independent of the
cascading
The AC calculations for gain and impedance
are interdependent
=
×
×
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⋯
=−
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Voltage gain:
Input impedance,
first stage:
Output impedance,
second stage:
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This example is a CE–CB
combination. This arrangement
provides high input impedance
but a low voltage gain.
The low voltage gain of the
input stage reduces the input
capacitance, making this
combination suitable for highfrequency applications.
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The Darlington circuit provides
very high current gain, equal to the
product of the individual current
gains:
D = 1 2
The practical significance is that
the circuit provides a very high
input impedance.
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Base current:
Emitter current:
Emitter voltage:
Base voltage:
AC analysis based on circuit
configuration
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
Find DC Bias voltages
and currents
◦
◦
◦
◦
◦
◦
IB1
IC2
VC1
VE2
VB1
VCE2
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This is a two-transistor circuit that operates like a Darlington
pair, but it is not a Darlington pair.
It has similar characteristics:
• High current gain
• Voltage gain near unity
• Low output impedance
• High input impedance
The difference is that a Darlington uses a pair of like
transistors, whereas the feedback-pair configuration uses
complementary transistors (PNP and NPN).
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
Q. 18, 38
◦ Use exact analysis

Q. 44b
◦ Rewrite as “Find the total AVL and AVS for the
system”
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