Fall 2022
EMT 1255 – BJT Amplifiers
Floyd
One of the primary uses of a transistor is to
amplify ac signals. This could be an audio
signal or perhaps some high frequency
radio signal. It has to be able to do this
without distorting the original input.
Introduction
3
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EMT 1255 – BJT Amplifiers
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What we are most interested in is the ac signal itself. Since
the dc part of the overall signal is filtered out in most cases,
we can view a transistor circuit in terms of just its ac
component.
Recall from the previous chapter that the purpose of dc
biasing was to establish the Q-point for operation. The
collector curves and load lines help us to relate the Q-point
and its proximity to cutoff and saturation. The Q-point is best
established where the signal variations do not cause the
transistor to go into saturation or cutoff.
Amplifier Operation
EMT 1255 – BJT Amplifiers
¾ Troubleshooting amplifier circuits
November 2, 2022
Professor Jang
¾ Multistage amplifiers
¾ Common-emitter, common-base, and
common-collector amplifiers
¾ Internal transistor parameters
¾ The concept of amplifiers
Outlines
Dept of Computer Engineering Technology
BJT Amplifiers
Chapter 3-3
EMT1255
4
2
Floyd
5
Graphical operation of the amplifier showing the variation of the base current, collector current,
and collector-to-emitter voltage about their dc Q-point values. Ib and Ic are on different scales.
The operation can be illustrated graphically on the collector-characteristic
curves, as shown in Figure. The sinusoidal voltage at the base produces a
base current that varies above and below the Q-point on the ac load line.
The Linear Amplifier – A Graphical Picture
EMT 1255 – BJT Amplifiers
For the analysis of transistor circuits from both dc and ac perspectives
the ac subscripts are lower case and italicized. Instantaneous values
use both italicized lower-case letters and subscripts.
Amplifier Operation – AC Quantities
EMT 1255 – BJT Amplifiers
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Projections on the graph of
Figure show the collector
current varying from 6 mA
to 4 mA for a peak-to-peak
value of 2 mA and the
collector-to-emitter Voltage
varying from 1 V to 2 V for
a peak-to-peak value of 1V.
and below the Q-SRLQWEDVHFXUUHQWYDOXHRIȝ$DVVKRZQLQ)LJXUH
Determine the resulting peak-to-peak values of collector current and collector-toemitter voltage from the graph.
Ex 3-14 The ac load line operation of a certain amplifier extends 10 ȝ$DERYH
6
The boundary between cutoff and saturation is called the linear region.
A transistor which operates in the linear region is called a linear amplifier.
Note that only the ac component reaches the load because of the capacitive
coupling and that the output is 180º out of phase with input.
The coupling capacitors shown in Figure block dc and thus prevent the
internal source resistance, Rs, and the load resistance, RL, from changing
the dc bias voltages at the base and collector.
Amplifier Operation - The Linear Amplifier
ac collector
resistance
ac base resistance
ac emitter
resistance
ac beta (Ic / Ib)
ac alpha (Ic / Ie)
Description
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EMT 1255 – BJT Amplifiers
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The two graphs best illustrate the difference between EDC and Eac. If
you pick a Q-point on a curve and cause the base current to vary an
amount ǻIB, then the collector current will vary an amount ǻIC. The
two only differ slightly.
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Comparison of the AC Beta (ǃac) to the DC Beta (ǃDC)
Transistor AC Equivalent Circuits –
EMT 1255 – BJT Amplifiers
rºb
rºc
Įac
ȕac
rºe
r Parameter
rec #
25mV
IE
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25mV
2mA
12.5:
EMT 1255 – BJT Amplifiers
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The common-emitter amplifier exhibits high voltage and current gain.
The output signal is 180º out of phase with the input. The input
signal, Vin, is capacitively coupled into base, and the output signal,
Vout, is capacitively coupled from collector. Now lets use our dc and
ac analysis methods to view this type of transistor circuit.
The Common-Emitter Amplifier
EMT 1255 – BJT Amplifiers
current of 2 mA.
Ex 3-15 Determine the rºe of a transistor that is operating with a dc emitter
r´e = 25 mV/IE
You can determine r´e from this simplified equation.
We can view transistor circuits by use of resistance or r parameters for
better understanding. Since the base resistance, r´b is small it is normally is
not considered and since the collector resistance, r´c is fairly high we
consider it as an open. The emitter resistance, r´e is the main parameter
that is viewed. You can determine r´e from this simplified equation.
r´e = 25 mV/IE
Determining r ’ e by a Formula
Transistor AC Equivalent Circuits –
Transistor AC Equivalent Circuits – r Parameter
12
10
R2
)VCC
R1 R2
(
6.8k:
)12V
28.8k:
VE
RE
2.13 V
560 :
3.80 mA
EMT 1255 – BJT Amplifiers
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Vb
15
§ Rin ( tot ) ·
¨
¸Vs
¨R R
¸
(
)
s
in
tot
©
¹
If the ac source has a nonzero internal resistance, then three factors
must be taken into account in determining the actual signal voltage
at the base. These are the source (Rs), the bias resistance (R1 || R2),
and input resistance (Rin(base)) at the base.
Signal (AC) Voltage at the Base
AC Equivalent Circuit
13
VCE = VC – VE
= (VCC – ICRC) – VE
= 12 V – 3.86 mA(1.0 kȍ – 2.13 V = 6.07 V
IC ؆ IE = 3.80 mA
Common Emitter Amplifier –
EMT 1255 – BJT Amplifiers
IE
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2.83V
VE = VB – VBE = 2.83 V – 0.7 V = 2.13 V
VB # (
RIN(base) ؆ ȕDCRE = (150)(560ȍ Nȍ
The methods for dc analysis are just are the
same as dealing with a voltage-divider circuit.
The dc component of the circuit “sees” only
the part of the circuit that is within the
boundaries of C1, C2, and C3 as the dc will not
pass through these components. The
equivalent circuit for dc analysis is shown.
The Common Emitter Amplifier - DC Analysis
14
EMT 1255 – BJT Amplifiers
Rout = RC || rºc §RC
The output
resistance is for all
practical purposes the
value of RC.
Rin(tot) = R1 || R2 || Rin(base)
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Rin(base) = Vb/Ib = Ie rºe /(Ie/ȕac) = Eac r´e
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The input resistance as seen by the input voltage can be illustrated by
the r parameter equivalent circuit. The simplified formula below is used.
AC Equivalent Circuit – Input & Output Resistance
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Common Emitter Amplifier
EMT 1255 – BJT Amplifiers
The ac equivalent circuit basically replaces the capacitors with shorts,
being that ac passes through easily through them. The power supplies
are also effectively shorts to ground for ac analysis.
AC Ground
AC Equivalent Circuit
Common Emitter Amplifier –
25mV
6.58:
3.80mA
EMT 1255 – BJT Amplifiers
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A´v = (Vb / Vs )Av or A´v =( Rin(total) / (Rs + Rin(total) ))Av
19
EMT 1255 – BJT Amplifiers
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EMT 1255 – BJT Amplifiers
10 X C d RE
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The XC(bypass) should be about ten times less than RE.
20
The emitter bypass capacitor helps increase the gain by allowing
the ac signal to pass more easily.
17
where Vc = Įac Ie RC §,e RC and Vb = Ie rºe
Av = RC /r´e
Voltage gain can also be
determined by the simplified
formula below.
Av = Vout / Vin = Vc / Vb
Voltage gain can be easily
determined by dividing the
ac output voltage by the ac
input voltage.
Taking the attenuation from the ac supply internal resistance and
input resistance into consideration is included in the overall gain.
§ Rin (tot ) ·
¸Vs
¨
¸
¨R R
(
)
s
in
tot
¹
©
§ 873: ·
¸10mVrms
¨
© 1173: ¹
7.44mVrms
The Common-Emitter Amplifier –
Effect of the Emitter Bypass Capacitor on Voltage Gain
Floyd
Vb
Common Emitter Amplifier –
AC Equivalent Circuit – Voltage gain
Common Emitter Amplifier –
AC Equivalent Circuit – Voltage gain
EMT 1255 – BJT Amplifiers
1
1
1
1
22 k: 6.8 k: 1.05 k:
873 :
Rin(tot) = R1 || R2 || Rin(base)
Rin(base) = ȕacrºe = 160(6.58 ȍ) = 1.05 kȍ
25mV
r 'e #
IE
First, determine the ac emitter resistance,
circuit is the ac equivalent of the amplifier in previous Figure with a 10 mVrms,
300 ȍ signal source, ȕac = 160. IE was previously found to be 3.80 mA.
Ex 3-16 Determine the signal voltage at the base of the transistor in Figure. This
RE
10
560:
10
56:
1
2SfX C
1
1.42 PF
2S (2kHz )(56:)
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EMT 1255 – BJT Amplifiers
Rc
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RC RL
RC RL
Av = Rc / rºe
23
When a load, RL, is connected to the output through the coupling capacitor
C3, as shown in Figure, the collector resistance at the signal frequency is
effectively RC in parallel with RL.
Common Emitter Amplifier - AC Equivalent Circuit
– Effect of a Load on the Voltage Gain
EMT 1255 – BJT Amplifiers
C2
The capacitance value is determined at the
Minimum frequency of 2 kHz as follows:
XC
10XC = RE
Since RE = 560 ȍ, XC of the bypass capacitor,
C2, should be no greater than
Ex 3-17 Select a minimum value for the emitter bypass capacitor, C2, in Figure
if the amplifier must operate over a frequency range from 2 kHz to 10 kHz.
RC
r 'e RE
RC
r 'e
1.0k:
152
6.58:
833:
127
6.58:
Floyd
(1.0k:)(5k:)
833:
6k:
The unloaded gain was found to
be 152 in previous example.
Rc
r 'e
Therefore,
RC RL
RC RL
EMT 1255 – BJT Amplifiers
Av
Rc
The ac collector resistance is
Ex 3-19 Calculate the base-to-collector voltage gain of the amplifier in Figure
when a load resistance of 5 kȍ is connected to the output. The emitter is
effectively bypassed and rºe = 6.58 ȍ.
EMT 1255 – BJT Amplifiers
Av
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1.0k:
1.76
566.58:
With C2, the gain is
Av
From Ex 3-15, rºe = 6.58 ȍ for this same amplifier.
Without C2, the gain is
Ex 3-18 Calculate the base-to-collector voltage gain of the amplifier in Figure
both without and with an emitter bypass capacitor if there is no load resistor.
24
22
RC
r 'e RE1
RC
RE1
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s
Vs
Rin ( tot ) Rs
EMT 1255 – BJT Amplifiers
Ap = Av Ai
The power gain is the product of the
overall voltage gain and current gain.
Is
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I c Where I is the total signal current
I s Produced by the source.
The total current signal produced
by the source is
Ai
The total current gain from base to collector is Ic / Ib or ȕac.
The overall current gain of the amplifier is
The Common-Emitter Amplifier –
Current & Power Gain
EMT 1255 – BJT Amplifiers
Av #
If RE1 is at least ten times larger than
rºe, then the effect of rºe is minimized
and the approximate voltage gain for
the swamped amplifier is
Av
The ac voltage gain will be
The bypass capacitor makes the gain unstable since transistor amplifier
becomes more dependent on IE. This effect can be swamped or
somewhat alleviated by adding another emitter resistor(RE1).
The Common-Emitter Amplifier –
Stability of the Voltage Gain
27
25
RC
RE1
3.3k:
10
330:
Floyd
EMT 1255 – BJT Amplifiers
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The common-collector amplifier is usually referred to as the emitter
follower because there is no phase inversion or voltage gain (Av §).
The output is taken from the emitter. The common-collector amplifier’s
main advantages are its high current gain and high input resistance.
The Common-Collector Amplifier
EMT 1255 – BJT Amplifiers
Av #
RE2 is bypassed by C2. RE1 is more than
ten times rºe so the approximate voltage
gain is
Ex 3-20 Determine the voltage gain of the swamped amplifier in Figure.
Assume that the bypass capacitor has a negligible reactance for the
frequency at which the amplifier is operated. Assume rºe = 20 ȍ.
28
26
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29
500:
505.8:
Re
r 'e Re
Av
EMT 1255 – BJT Amplifiers
0.989 Ai
I in
Ie
I in
2 mA rms
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16.3
1 Vrms
123 PA rms
8.16 k:
2 mA rms
123 PA rms
Vin
Rin(tot)
AvVb 1Vrms
#
500 :
Re
25mV
5.8:
4.3mA
Ve
Re
25mV
IE
Ie
r 'e #
4.3mA
VE
RE
IE
4.3V
1.0k:
§ R2 ·
¸¸VCC VBE = (0.5)(10 V) – 0.7 V = 4.3 V
¨¨
© R1 R2 ¹
VE
Rin(tot) = R1 || R2 || Rin(base)
= 18 kȍ || 18 kȍ || 87.5 kȍ = 8.16 kȍ
Rin(base) # EacRe = (175)(500 ȍ) = 87.5 kȍ
Re = RE || RL = 1.0kȍ|| 1.0kȍ= 500 ȍ
31
Ap(load)=Ap/2
= 16.3/2
= 8.15
Ap§$i=16.3
Ex 3-21 Determine the total input resistance of the emitter-follower in Figure. Also find the voltage
gain, current gain, and power gain in terms of power delivered to the load, RL. Assume ȕac = 175 and that
the capacitive reactances are negligible at the frequency of operation.
§R ·
Rout # ¨¨ s ¸¸ || RE
© E ac ¹
EMT 1255
Where Rs is the resistance of the input source.
If Re >> rºe
Rin(tot) = R1 || R2 || Rin(base)
Rin(base) # EacRe
Rin(base) # Eac(r´e + Re)
Because of its high input resistance the
common-collector amplifier used as a
buffer to reduce the loading effect of
low impedance loads. The input
resistance can be determined by the
simplified formula below.
The Common-Collector Amplifier –
Input & Output Resistance
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30
EMT 1255 – BJT Amplifiers
Rin = Eac1Eac2RE
The darlington pair is used to
boost the input impedance to
reduce loading of high output
impedance circuits. The
collectors are joined together
and the emitter of the input
transistor is connected to the
base of the output transistor. The
input impedance can be
determined the formula below.
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32
The Common-Collector Amplifier – The Darlington Pair
EMT 1255 – BJT Amplifiers
The power gain is approximately equal to the
current gain (Ap §Ai).
The voltage gain is approximately 1.
The current gain(Ai) is approximately Eac.
The output resistance is very low. This makes it
useful for driving low impedance loads.
The Common-Collector Amplifier
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§ 12 k: ·
¨¨
¸¸ 1.76 V
© 68 k: ¹
VE
RE
Av
Rc
r 'e
1.8 k:
23.6 :
76.3
Also, Ai §and Ap §$v = 76.3
35
EMT 1255 – BJT Amplifiers
Phase relationship
between Vin and Vout
Output Resistance
23.6 :
Current Gain
Voltage Gain
25 mV
IE
25 mV
1.06 mA
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Û
Û
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Not inverted
Very Low
High
Inverted
Rout § 5s/ȕac)//RE
High
Medium
Rout = RC
Rin(tot) = ȕacRE1
High
High
Rin(tot) = ȕacr´e
Ai = ȕac
Not inverted
Û
High
RC
Very Low
Rin(tot) = r´e
Low
Ai §1
High
Low
High
Ai = ȕac
Av = Rc /r´e
CB
Av §
CC
Av = - Rc /r´e
CE
Table 3-1 Comparison of amplifier ac parameters. Voltage – divider bias is assumed
for all amplifiers with an unbypassed emitter resistor in the CE and CB configurations.
EMT 1255 – BJT Amplifiers
Input Resistance
33
The output resistance is approximately equal to RC.
The input resistance is approximately equal to r´e.
The power gain is approximately equal to the voltage gain.
The current gain is approximately 1.
The common-base voltage gain(Av) is approximately equal
to Rc / r´e
The Common-Base Amplifier
1.06 V
1.06 mA
1.0 k:
Rc = RC || RL = 2.2 kȍ || 10 kȍ = 1.8 kȍ
Rin # r 'e
IE
VE = VB – 0.7 V
= 1.76 V – 0.7 V = 1.06 V
§ R2 ·
¸¸VCC
VB # ¨¨
© R1 R2 ¹
First, find IE so that you can determine rºe.
Then Rin §Uºe. Since ȕDCRE >> R2, then
Ex 3-22 Find the input resistance, voltage gain, current gain, and power gain
for the amplifier in Figure. ȕDC = 250.
EMT 1255 – BJT Amplifiers
The common-base amplifier has high voltage gain with a current
gain no higher than 1. It has a low input resistance making it ideal
for low impedance input sources. The ac signal is applied to the emitter
and the output is taken from the collector.
The Common-Base Amplifier
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