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Small Signal CE Amplifiers
P
Equipment:
Procedure: (A) No Load
To demonstrate the ac operation of the common emitter CE amplifier.
To demonstrate the effect of the bypass capacitor on amplifier gain (A )
To calculate the loaded and unloaded gain of the amplifier.
1 dc Power Supply
1 Digital Multimeter (DMM)
1 Variable ac Signal Generator
1 Dual Trace Oscilloscope
1 2N3904 npn transistor
6 Resistors 1 - 330 1 - 10 k 1 - 1 k 1 - 3.3 k 1 - 2 k
The input signal to a CE amplifier is applied across the emitter- base junction of our transistor. The output signal is taken from the collector terminal of the transistor.
The CE amplifier is one of the most commonly used BJT amplifier configurations. This amplifier has a relatively high voltage gain and a voltage phase shift of 180 from base to collector. (input to output).
In this lab, we will observe the ac operation of the CE amplifier. We will also observe the effects of the emitter bypass capacitor. Remember that the purpose of this capacitor is to hold the emitter of the transistor at ac ground. This action increases the voltage gain of the CE amplifier.
Construct the circuit shown in Figure 1. Note that R is not in the circuit.
Turn on the power supply. This circuit be midpoint biased. Is it.? (yes/no) ___________ What is the value of V ___________
Hook up the signal generator to the input of the circuit. Set the generator to produce a 1 kHz, 40 mV signal.
Measure and record the peak-to-peak input & output voltage of the amplifier
= ________
- Using and , calculate the ordinary voltage gain of the amplifier. ___________
urpose:
Pre-Lab Preparation See Page 5
Discussion:
V
L
CE
p-p
(40 mV output)
Be sure to connect the signal generator to the circuit before you set the output level to 40 mV The circuit is the load for the generator and it will slightly reduce the output level when it is connected to the generator. Setting the output level after the circuit is connected is the more accurate way to do this.
p-p
p-p
W W
W W
W
- 2 10 F Capacitors
- 1 100 F Capacitor
Calculate the “unloaded”or “no load” voltage gain for Figure 1 Calculate the “loaded” voltage gain for Figure 2 Calculate the “loaded” voltage gain with the bypass capacitor removed in Figure 3
= ________
m m
O
should
Vin V
V V A
out
in out V Show your calculation
h (^) FE= h (^) fe=
330 W
10 mF
10 mF
100 mF
2 kW
10 kW 1 kW
C (^1)
C (^2)
C (^3)
V (^) CC 12 V
R (^2) R (^) E
R 1 R^ C
2N
3904 2N
E B C
Figure 1 - Unloaded Amplifier
1 kHz 40m Vp-p
Procedure (B) 3.3 k W load
Add the 3.3 k load resistor as shown in Fig. 2.
The coupling capacitors in an amplifier are used to block dc while allowing the ac signal to pass. Look at C. Using the oscilloscope, measured the ac peak-to-peak voltage on both sides of the capacitor and record it below. ______________ on the load side of C. ______________ on the transistor side of C.
Now using the DMM, measured the dc voltage on both sides of C and record it below.
V ______________ on the load side of C V ______________ on the transistor side of C.
- Do your measurements support the statement that the coupling capacitor passes ac while blocking dc?
- We know that the CE amplifier produces a voltage phase shift of 180 from base to collector
W
Reset the input signal level to 1kHz at 40 mVp-p
- Measure and record the peak-to-peak output voltage at the collector terminal of the transistor.
= ________ = ________
- Using and , calculate the ordinary voltage gain of the amplifier. ___________
V V
V V
A
in out
in out V Show your calculation
Coupling Capacitors
Phase Shift
2
2 2
2,
dc 2 dc 2
V
V
p-p p-p
O (input to output).
-Using the oscilloscope, connect to the base and to the collector of the transistor
-Adjust the vertical sensitivity of each channel so that each signal amplitude fills less than half of the screen’s total height.
-Adjust the vertical position of the signals so that the input signal is in the upper half of the screen, while the output signal is in lower half of the screen.
-Adjust the time-base so that approximately two complete cycles are shown on the screen.
-Be sure to set the
-Neatly draw the waveforms shown on the scope on the grid.
-Label the input and the output.
-Fill in Time/Div and V/Div in the space provided.
Channel 1 Channel 2
vertical mode switch to chop
Time/Div:
Volts/Div: Channel 1
Volts/Div: Channel 2
h (^) FE= h (^) fe=
330 W
10 mF
10 mF
100 mF
2 kW
10 kW 1 kW
C (^1)
C (^2)
C (^3)
V (^) CC 12 V
R (^2) R (^) E
R 1 R (^) C
2N
3904 2N
E B C
3.3 kW
R (^) L
Figure 2 - Loaded amplifier with 3.3 k Wload
R is addedL
1 kHz 40m Vp-p
Procedure: (B) 1k Load (cont)
Coupling Capacitors
Phase Shift
Procedure: (C) Remove Emitter Capacitor
W
- In the procedure (B) section , you measured and compared the that appears on both sides of capacitor C.
(i) Does the capacitor appear to or the ac signal? _________________
- In the procedure (B) section , you measured and compared the that appears on both sides of capacitor C.
(i) Does the capacitor appear to or the dc signal? _________________
Why are capacitors needed in the circuit?
In the procedure (B) section , you sketched the ac signal that appears at the input and output of the amplifier. The output should be 180 out of phase with the input?
(i) Why is the output 180 out of phase with the input?
- In the pre-lab section, you calculated the loaded and swamped gain for the circuit shown in Figure 3. In procedure (C), you calculated the voltage gain from the measured values. Compare these two values of.
__________ ( Calculated from the schematic values ) __________ ( Calculated from the measured values )
As before, the is ( Calculated from the measured values ) value above. The two values above should be close to the same value.
(i) Give the main reason why these values above are near the same value
ac signal
dc signal
A
A
most accurate voltage gain
2
2
pass block
pass block
O
O
AV
V V
(pre-lab) (procedure (C)
(procedure (C)
Questions (cont.)
h (^) FE= h (^) fe=
330 W
10 mF
10 mF
100 mF
2 kW
10 kW 1 kW
C (^1)
C (^2)
C (^3)
V (^) CC 12 V
R (^2) R (^) E
R 1 R^ C
2N
3904 2N
E B C
Figure 1 - Unloaded Amplifier
1 kHz 40m Vp-p
In Figure 1, to the right, the amplifier is operating “unloaded” since there is no load resistor on the amplifier.
The unloaded gain is found as
Calculate the unloaded gain for Figure 1. Be sure to include R (^) basein your calculation.
Pre-Lab (Complete all of this page)
A V
R C
r’ (^) e
In Figure 2, to the right, the amplifier is operating “loaded” since there is a load resistor on the amplifier.
The loaded gain is found as
Calculate the loaded gain for Figure 2. .
In Figure 3, the amplifier is operating “loaded” and swamped. The bypass capacitor has been removed, which exposes the 330 resistor
(R ) to the ac signal. This resistance must now be included in the
gain equation. We now call it since it now an ac value.
The swamped gain is found as
Calculate the swamped gain for Figure 3.
W
E
rE
A (^) V where r = R ||RC C L
rC r’ (^) e
= (^) r
A V
r’ (^) e
r (^) C where r = R ||RC C L
h (^) FE= h (^) fe=
330 W
10 mF
10 mF
100 mF
2 kW
10 kW 1 kW
C (^1)
C (^2)
C (^3)
V (^) CC 12 V
R (^2) R E
R 1 R^ C
2N
3904 2N E B C
3.3 kW
R (^) L
Figure 2 - Loaded amplifier with 3.3k Wload
1 kHz 40m Vp-p
h (^) FE= h (^) fe=
330 W
10 mF
10 mF
2 kW
10 kW 1 kW
C (^1)
C (^2)
V (^) CC 12 V
R (^2) R (^) E
R 1 R^ C
2N
3904 2N E B C
3.3 kW
R (^) L
Figure 3 - Loaded amplifier with bypass capacitor removed
1 kHz 40m Vp-p
A V =
A V =
A V =
