Modern Electronics (ETIN70) – lab 2
Laboratory exercise 2 - Feedback Amplifiers Using a BJT
In this lab you will study a transconductance (voltage to current) and a transresistance (current to
voltage) amplifier with a negative feedback. You will learn how the transistor operating point affects
its parameters and how it affects the closed loop gain At. You will also study the effect of including
feedback on the circuit properties such as gain.
The BJT has three different (forward) modes of operation seen below. Consider these as you change
the biasing conditions (with potentiometer or by changing resistors).
Needed: BJT BC547B, resistors R1 = R2 = 4.7 kΩ, capacitance C1=10 µF, potentiometer Rpot=10 kΩ,
power supply (VCC=15 V), signal generator, multimeter. The gain of the transistor is assumed to be
βf=200.
Closed loop gain with feedback = At (A in lectures)
Open loop gain without feedback = a = id/v1
Feedback gain = f
Loop gain T=af
Closed loop gain when a -> ∞ = At∞
Preparation tasks
The following tasks should be solved before the lab session and discussed with the lab supervisor.
Read through all lab instructions before solving the tasks.
1. Use nullors to draw the feedback circuit for the transconductance amplifier. Determine the
polarity and confirm that the feedback is negative.
2. The closed loop gain for a->∞ should be At∞=-1/4700 A/V. Choose a suitable feedback
resistor Rf.
3. Draw the small signal model for the transconductance amplifier with a BJT as the active
element. Use a load resistor RL=2.7 kΩ.
4. Write down the expression for the loop gain T=af. Is the feedback negative?
5. The closed-loop gain should be at least -110. Calculate the operating point IC that fulfills this
condition.
6. What is the required VB (in figure 2) to obtain the operating point from question 5? If you did
not manage to solve question 5, use VBE = 0.6 V and IC =1.3 mA.
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Modern Electronics (ETIN70) – lab 2
7. Use nullors to draw the feedback circuit for the transresistance amplifier. The closed-loop
gain should be At∞=-1 V/mA. Determine the feedback resistor Rf that should be used.
8. In the transresistance amplifier in figure 3 the input is a current (ig) but in the lab there is only
a signal generator available which is a voltage source. Design a simple circuit with a resistor
so that the voltage generator (vg) can be used as a current generator (ig). As the voltage is
changed between 0 and 10 V the current should change between 0 to 1 mA. Hint: a
transresistance amplifier has a low input impedance.
Part 1 – Transconductance amplifier
Biasing
To operate the transistor as an amplifier it has to be biased in the forward active mode (need large
high gain and linear amplification). The large-signal model below describes how biasing should be
performed.
Figure 1: Biasing model for the transconductance amplifier.
In order to reach the forward active mode the base-emitter junction have to be forward biased and
the base-collector junction reverse biased. This implies a positive VCE and a positive IC flowing from
the emitter into the collector. An applied VBE and a base current IB is also needed. Remember that the
collector current is related to the base current through the gain as Ic=βf*IB. A suitable circuit to
achieve these conditions is shown in figure 2.
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Modern Electronics (ETIN70) – lab 2
Figure 2: Transconductance amplifier with a biased BJT.
The BJT is biased by the power supply (VCC), the resistor RE is the feedback resistor and RL is a load
resistor. Since RE also affects the biasing condition of the BJT in addition to the feedback, the bias
current (IC) and the gain can not be set independent of each other which is a drawback for this circuit
configuration. To determine the output current, an oscilloscope is used to measure the voltage drop
over the load resistor (i= v / RL). The inclusion of a sufficiently small load resistor should not affect the
biasing since IC is controlled by RE and VB+vg in the forward active mode i.e. changes in VCE should not
have any effect if the Early effect is disregarded and as long as the base-collector junction is reverse
biased.
The base voltage (VB) is set by two resistors (R1 and R2) connected in series with a potentiometer
(Rpot) which allows for straightforward variation of VB. In this lab you will perform both DC and AC
measurements and apply a DC bias (VB) plus a small AC signal (vg). Since vg is symmetric around 0 V
(no DC offset) it have to be isolated from the DC bias (VB). This is achieved using a capacitor (C1)
which will act as a short for AC signals and as an open circuit for DC signals.
DC measurements
First, the circuit has to be checked so that the connections are correct and the operating point (IC)
can be varied. Perform these measurements with a multimeter and without the signal generator.
Connect the transconductance amplifier according to figure 2. Set VCC = 15 V and VB = 6.7 V by
adjusting the potentiometer (use the multimeter to measure VB). Measure the voltage over the load
resistor (RL = 2.7 kΩ), calculate IC and compare it to what you obtained in preparation task 5.
The operating point (IC) is adjusted by changing VB. Try to increase and decrease VB to e.g. 7.7 V and
5.5 V. How does the operating point change? How would this affect the closed-loop gain?
AC measurements (small-signal)
In this task you will apply a small signal (vg) on the input and study the gain of the amplifier. Use an
oscilloscope to measure the collector voltage vc (see figure 2). Set the oscilloscope to “AC” which
removes any DC offset. Measure the collector current (IC) using the multimeter.
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Modern Electronics (ETIN70) – lab 2
1. Set VB=6.7 V and connect the signal generator to vg. Use a sinewave with 2 kHz and 1 V
amplitude. Measure the closed-loop gain At and compare to the given closed loop gain when
a-> ∞ : At∞ = 1/4700 A/V.
2. Determine the influence of the load resistor (RL) on the output signal ic. Change to RL=1.0 kΩ
and measure vc (calculate ic). Repeat with RL=4.7 kΩ. How is the closed-loop gain At affected?
How is IC affected?
3. Change back to RL=2.7 kΩ and change RE=3.3 kΩ. How is the closed-loop gain At affected?
How is IC affected?
4. Change back to RE=4.7 kΩ. Change the operating point IC by varying VB. How is the closedloop gain At affected? What does this imply about the loop-gain?
AC measurements (large-signal)
(optional if you have time)
1. Increase VB (and possibly also increase in the input vg) until the BJT operates in the saturation
mode. Sketch the oscilloscope signal in the left part of the figure below.
2. Decrease VB (and possibly also increase in the input vg) until the BJT operates in the cut-off
mode. Sketch the oscilloscope signal in the right part of the figure below.
3. Set VB=6.7 V and the input signal from the signal generator to 1V. Connect a 47 nF capacitor
in parallel to RE. Vary the frequency on the function generator. How is At affected? How is IC
affected? Why?
Part 2 – Transresistance amplifier
In this part you will study a transresistance amplifier (current to voltage). In this circuit the feedback
resistor (Rf) is connected between the collector and base (see figure 3) and a load resistor (RL = 4.7
kΩ) is connected between the power supply and collector. To be able to control IC a resistor RE is
introduced. Since IC should not pass though the feedback resistor (Rf) a 10 µF capacitor is connected
in series to prevent any DC current from passing. To avoid that the resistor RE influences the smallsignal gain, a 0.47 µF capacitor acting as a short for AC signals is connected in parallel to it.
To obtain a loop gain of T=-75 an IC=0.7 mA should be used (calculated similarly as in preparation task
5). For the transconductance amplifier the feedback resistor (RE) was dependent of the bias. In this
circuit RE can be chosen more freely depending on the desired output voltage swing. For a fixed
operating point IC, there will be a constant voltage drop over RE. The maximum peak-to-peak output
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Modern Electronics (ETIN70) – lab 2
voltage swing is then VCC – “voltage drop over RE”. In this setup choose RE=4.7 kΩ which gives VB = VBE
+ RE*IC ≈ 4V. The DC bias VB is done using a voltage divider with two resistors (R1 and R2) connected in
series with a potentiometer (Rpot) as for the transconductance amplifier. The small-signal input
current (ig) passes through a capacitor to avoid drawing a large DC current from the supply.
Figure 3: Transresistance amplifier circuit with bias.
Measurements
In this part you will perform both DC and AC measurements simultaneously. Make sure that you
understand the differences between these. The DC component of IC can be determined by measuring
the voltage drop over RL using a multimeter and the AC measurements are done between vc and
ground.
1. Use VB = 4 V and check so that IC = 0.7 mA.
2. Set the signal generator to a current ig = 0.5 mA at 2 kHz (no DC offset). Check if the
measured small-signal gain At corresponds to the calculated At∞.
3. Vary VB and study how this affects both the small-signal gain At and the operating point IC.
4. Set VB = 4 V and reduce the feedback resistance Rf by 50% through a parallel connection with
two identical resistors. How does this affect At and IC? What is the difference compared to
the transconductance amplifier?
5. Change back to the previous Rf and change the load resistor to RL=3.3 kΩ. How does this
affect At and IC. What conclusions can be made concerning the loop gain T?
6. Change back to RL = 4.7 kΩ. Reduce RE by 50%, measure At and IC and explain what is
happening?
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