ES330 Laboratory Experiment No. 1
NPN Common-Emitter Amplifier
[Reference: Section 7.5.2 of Sedra & Smith (pp. 470-471)]
Objectives:
1. Design the amplifier for voltage gain AV to be at a minimum of -100 (V/V) and choose
resistor values of RC and RE by calculation.
2. Measure the voltage gain of the amplifier to see how it compares with your
calculated voltage gain. Display waveforms on an oscilloscope.
3. Measure the output resistance RO of the amplifier looking into the output port.
4. Replace NPN transistor with a different device and see what changes. [extra-credit]
Materials:
1. Breadboard
2. One NPN transistor – type 2N3904
3. Three large 47 microfarad capacitors
4. Several resistors of various values (two resistors are 10 k in value)
5. Jumper wires for use on breadboard
6. Function generator, digital multimeter and oscilloscope
Circuit Schematic:
1
Circuit Parameters:
The parameters of the circuit are listed in the table below.
Component
Value
CC1, CC2 and CE
47 F each
NPN BJT
2N3904
V+ = (-V_ )
 10 volts (nominal)
Resistors RB and RL
10 k each
Signal Generator Rsig
50  or 600 
Not specified in the table are values for the emitter resistor RE and collector resistor RC.
These you will determine using specified design goals, namely, the collector current and the
voltage gain of the amplifier.
The minimum voltage gain AV of the amplifier is to be at least -100 (V/V) in magnitude –
the minus sign indicates that the common-emitter BJT amplifier is inverting (i.e., produces a 180
degree phase shift in the output waveform). In addition, we want the DC collector current IC
equal to 1 milliampere (1 mA).
AV  -100 (V/V) and IC = 1 mA
Transistor Parameters:
Nominal parameters for the 2N3904 NPN bipolar transistor can either be measured
using the transistor you selected, or estimated from the 2N3904 Data Sheet. The Data Sheet is
reproduced in the Appendix. The DC current gain of the transistor () is designated as hFE on
the Data Sheet. You should use a nominal value for hFE rather than the minimum or maximum
value for hFE for design purposes. We will ignore the transistor’s output resistance ro in the
small-signal model of the transistor because it very large for the 2N3904.
Determining RE and RC:
The value of IC is set by choosing RE because the forward-biased base-emitter junction
voltage is close to VBE = 0.7 volt at IC = 1 mA. The transconductance gm is
gm 
IC
I (mA) 1 mA
 C

 0.0385 (A/V) .
 kT / q  26 mV 26 mV
Remember the base node potential (VB) is close to ground potential (i.e., 0 volts) if the base
current IB flowing in RB is quite small. You should check to see that this is true in your design.
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In the small-signal hybrid-pi model the parameter r is given by
r 
 kT / q  
IB
26 mV
I B (mA)
where hFE 
IC
IB
The voltage gain AV is then
AV   gm  RC RL 
The value of resistor RC controls the voltage gain because it is in parallel with resistor RL.
However, there is another constraint on resistor RC because the collector-emitter voltage VCE
must be large enough to keep the transistor in the forward-active region of operation while
allowing for adequate voltage swing along the load line. In other words, be sure to pay
attention to the Q-point location of the transistor.
Note: There are effectively two load lines – a DC load line which excludes RL and an AC
load line where the load line resistance is RC and RL in parallel.
DC Operating Point Analysis:
Sketch the DC schematic circuit. The three large capacitors are all open circuits for DC
analysis. Also, the signal generator assumes no role in the DC analysis. Begin by determining
the values of IB and IE where IC = IB + IE = 1 mA. The nominal power supply V_ = - 10 volts.
A. What is the base node voltage VB? VB = ______volt
B. Determine the value of RE required to set the collector current at IC = 1 mA.
The resistor RE = _______ ohms
C. Is the value of RE equal to commonly available resistor values? If not, what can you do to
establish IC = 1 mA? Comment on this.
AC Analysis:
We next determine VCE and RC. Sketch the AC small-signal circuit. The three large
capacitors now become short circuits (i.e., the capacitors have negligible impedance at higher
frequencies). Also, the power supply connections are assumed to be AC grounded.
A. What happens with the emitter resistor RE in the AC analysis?
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B. Let vi be the small-signal voltage at the base node of the transistor and vsig is the internal
voltage of the signal generator. You should recognize that the signal generator is shown in its
Thévenin equivalent circuit. Be sure to check the value of resistance Rsig is for the signal
generator (i.e., function generator) you use in this laboratory. Note that we use vi in the
voltage gain expression for the amplifier.
Estimate of vi/vsig = ____________
C. Derive an expression for AV = vo/vi, where vo is the output voltage as defined in the
schematic circuit on page 1. Find a value for resistor RC giving a voltage gain equal to, or a little
larger than, AV = -100 (V/V). Be sure that this value of RC also gives an acceptable value for VCE.
What value of VCE do you calculate? VCE = _______ volts
D. Now that you know the value of resistor RC, calculate the output resistance Ro at the output
node as defined in the schematic circuit.
Ro = _______ ohms
Prototyping:
Now you have reached the point where you can prototype the circuit you designed. It
should look something like the one in the photograph.
Be sure to keep wire lengths and lead lengths short to prevent unwanted oscillations.
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Measurements:
Make the following measurements:
A. Using a digital multimeter, measure the DC voltages at the base node (VB), emitter node (VE)
and collector node (VC) of the transistor.
B. Using a function generator set its sinusoidal “peak-to-peak” amplitude to 10 mVpk-pk with a
frequency of 1,000 Hz (i.e., 1 kHz). This is the small-signal voltage vi driving the amplifier. Now
measure the amplifier’s midband voltage gain AV.
C. Using an oscilloscope generate plots of vo and vi versus time t.
D. Measure the output resistance Ro. You can do this by replacing the 10 k RL load resistor
with say a 1 Meg resistor and again measuring the voltage gain. This gives a maximum
voltage gain value. By adjusting (i.e., lowering) the value of RL we can find a value of RL such
that the voltage gain is one-half the value you found with the 1 Meg resistor. That particular
value of RL is equal to the output resistance Ro of the amplifier.
E. Increase the sinusoidal “peak-to-peak” amplitude of the function generator to a level where
the output sinusoidal waveform distorts. Describe what you see with respect to the distorted
waveform’s shape.
Post-Measurement Exercise:
A. Calculate the values of VBE and VCE at the DC bias from the ones you measured in the lab.
How do they compare with the calculated values?
B. Compare the measured and the calculated voltage gain values. Explain the difference.
C. Compare the measured and calculated output resistance values.
Extra-Credit (NPN Bipolar Transistor Substitution)
Obtain a different NPN bipolar transistor (not a 2N3904) from the instructor and plug it
into the breadboard amplifier. Measure the voltage gain, bias point (VBE and VCE) values, and
output resistance Ro. How does it differ from results you got with the 2N3904?
APPENDIX (2 pages)
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