Download Lab03 - Weber State University

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Lab 03: Differential Amplifiers (BJT) (20 points)
NOTE:
1) Please use the basic current mirror from Lab01 for the second part of the lab (Fig. 3).
2) You can use the same chip as the basic current mirror; make necessary connections for
the diff pair. Avoid using Q5 on CA3046.
3) The procedure is same for both types of amplifiers (common-emitter or commonsource).
Objectives
The purpose of this lab is build and analyze Differential amplifiers based on NPN transistors
(or NMOS transistors).
In this lab, we will build the differential amplifier and determine Common Mode Rejection
Ratio (CMRR) for two types of configuration for the biasing the emitter: one with bias resistor
and the other with current mirror. A major benefit of using a differential amplifier is to get rid of
noise or interference signal present in the input signal.
You can use any NPN transistors that meet the objective (recommendation: 2N3904, 2N4401, or
CA3046 (NPN Array)).
For 2N3904, VA=90 V (measured), β=80 (datasheet).
CA3046 parameters are very similar: VA=96 V (measured), β=100 (measured).
Fig. 1: Differential amplifier with bias resistor (iC ≈ iE)
Weber State University
EE3120 Microelectronics II
Suketu Naik
2
1.0 Differential Amplifier with Bias Resistor
Fig. 1 shows the differential amplifier which uses the resistor REE as the bias resistor.
Here VCC=VEE=10 V, Ic=1mA (IE≈1mA). Q1 and Q2 should be a matched pair of NPN
transistors (you will have to find the two that are closely matched by trial and error).
As explained the class, the differential amplifier amplifies the differential input signal, Vid =
(Vin+) - (Vin-), where the differential signals are applied at the bases of the transistors as shown
in Fig. 1. The differential output can be taken across the collectors as, Vod=(Vo+)-(Vo-). Note that
the differential input (and output) signals have phase shift of 180 deg between them (180 deg
phase shift corresponds to the opposite polarity). An ideal differential amplifier should have
differential input signals with identical amplitudes.
First determine the values of RC and REE. REE can be found by finding the DC common emitter
voltage (VE) based on the base-emitter voltage drop (see the datasheet for the transistor that you
will be using). Remember that REE = [VE-(-VEE)]/IE. Select RC such that the transistor operates
deep in the active-mode. For example, RC=10 kΩ will make the dc voltage at the collector (VC)
~5V. Collector current in each of the branches will be about half of the bias current (iC/2). It is
best to make sure that the collector voltage is less than VCMmax in eq. 8.66.
L1: What are the values of RC and REE?
In order to determine the common-mode rejection ratio (CMRR), we need to calculate the
common-mode gain, Acm and differential-mode gain, Ad.
1.0.1 Common-mode Gain Acm
The amplifier exhibits the common-mode gain when both the input signals are same (common).
Ideally the amplifier should completely reject the common-mode signal. However, the amplifier
will not completely reject the common-mode signal due to the mismatches in the scale currents,
the bias resistors, the collector resistors, the finite impedance of the current source at the
collector and so on. Consequently, the common-mode gain represents the ability of the amplifier
to amplify noise. So the lower the common-mode gain, the better the amplifier will be at
rejecting common-mode interference or noise.
Let’s apply a 1Vdc (or 1-kHz sinusoidal signal of 2V peak-to-peak) to both inputs. Calculate
common-mode gain based on these input signals (see eq. 8.99, assume 1% mismatch between the
resistors).
L2: What is the calculated common-mode gain Acm of the differential amplifier?
1.0.2 Differential Voltage Gain Ad
The amplifier will exhibit differential voltage gain when you apply differential input signals
(equal amplitude and opposite polarity). The differential voltage gain is the ratio of output
differential voltage over the input differential voltage. This is the central theme of a differential
amplifier.
Weber State University
EE3120 Microelectronics II
Suketu Naik
3
For the purposes of calculation and simulation, we will assume that the two input signals have
equal amplitudes and are 180 deg out of phase with each other (opposite polarity). Calculate the
differential voltage gain for input signals of 20 mV peak-to-peak at 1 kHz (see eq. 8.95).
L3: What is the calculated differential voltage gain Ad of the differential amplifier?
1.0.3 Common Mode Rejection Ratio (CMRR)
The CMRR is a measure of the effectiveness of the differential pair in amplifying the differential
signal while rejecting common-mode interference. It is simply the ratio of the magnitude
differential voltage gain to the magnitude of common-mode gain (see eq. 8.100 and 8.50b).
Calculate the CMRR in dB (20*log10(x)). Note: We will assume that the main contribution to
CMRR is the mismatch between the collector resistors.
L4: What is the calculated common-mode rejection ratio CMRR (dB) of the differential
amplifier?
1.1 Simulation
Simulate the circuit shown in Fig. 1 and determine the CMRR (in dB) of the amplifier.
1.1.1 Common-mode Gain
step1: Make sure to change either of collector resistor to +/- 1% of the other resistor value (e.g.
10 kΩ and 9.9kΩ represent the mismatch).
step2: Apply both the input voltages with same phase (use 2V peak-to-peak at 1 kHz). This
will make the two input signals common-mode. Run the simulation.
step3: Plot (vo+) - (vo-). Note the peak value and calculate the ratio of this value to input
amplitude. This is the common-mode gain.
L5: What is the simulated common-mode gain Acm of the differential amplifier?
1.1.2 Differential Voltage Gain
step1: Set the phase of one of the input voltages to 180 deg and amplitudes to small values (e.g.
10 mVamp). Run simulation.
step2: Plot (vin+)-(vin-) and (vout+)-(vout-) on the same plot, use markers and obtain the ratio of
output differential voltage to input differential voltage.
L6: What is the simulated differential voltage gain Ad of the differential amplifier?
Based on the above numbers, find the CMRR in dB (20*log10(x)).
L7: What is the simulated CMRR (dB) of the differential amplifier?
Please see the next page for the experimental set up.
Weber State University
EE3120 Microelectronics II
Suketu Naik
4
1.2 Experiment
Don’t forget to check the multi-meter mode before you measure a current or voltage, if
it is set incorrectly you will blow a fuse!
HIGH-Z MODE
Make sure you put the function generator in High-Z mode. Otherwise
your signal amplitudes will be off by factor two.
Repeat 1.1 to determine the CMRR of the amplifier.
1.2.1 Common-mode Gain
Before you begin, first check the input signal, vin (same input applied at both the bases) and the
output signal, vo+ or vo- (output signal at either of the collectors) on the oscilloscope. Ensure that
there is no distortion. If you observe noticeable distortion, reduce the input signal amplitude until
the distortion disappears. If you cannot lower the amplitude enough to get rid of the distortion
completely, build a resistive voltage divider to attenuate the input signal further. If you use a
voltage divider to attenuate your input signal, don’t forget to factor the attenuation into your gain
calculations. You can also put resistors at the emitters to improve the linearity (see Fig. 8.18 in
the text).
Apply a 1-kHz sinusoidal signal of 0.1V peak-to-peak to both inputs. Measure the amplitude of
the differential output signal, vod=vo+ - vo-. Calculate the common-mode gain (=vod / vicm). Is the
common mode gain higher or lower than your calculated and simulated values? Why?
L8: What is the measured common-mode gain Acm of the differential amplifier?
1.2.2 Differential Voltage Gain
To measure the differential mode gain, you will need two equal amplitude signals with opposite
polarity. We can supply the needed signals with the circuit shown in Fig. 2. You can use any opamp
that accomplishes this task (alternatively you can use any other scheme that produces differential
input signals). You can use a potentiometer for R2 which will be useful for fine adjustment. Make
R2=R1=1kΩ. Connect a 1-kHz, 0.1Vamp sinusoidal signal and measure Vin+ and Vin- on the
oscilloscope (you should reduce the amplitude later after you get the op-amp working). Set the scope
to sum the two channels (use math function). The sum should be as close to zero as possible (since
Vin+ and Vin- have opposite polarity). Adjust R2 until you achieve zero sum. Note that Vin+ and Vinmay need to pass through the blocking caps CB before connecting them to the transistors if you have
dc offset from the op-amp.
Fig. 2: Circuit with op-amp for creating differential input signals
Weber State University
EE3120 Microelectronics II
Suketu Naik
5
Now reduce the amplitude of the input signal Vin until the differential output signal, Vod is
sinusoidal (for no distortion, you may need to build the voltage divider to attenuate the input signal
further). Measure both Vid and Vod. Remember that Vid=(Vin+)-(Vin-) and Vod=(Vo+)-(Vo-); use math
function. Please make note of where the two output signals are measured in Fig. 1. Calculate the
differential voltage gain as the ratio, Vod/Vid.
L9: What is the measured differential voltage gain Ad of the differential amplifier?
Based on the above numbers, find the CMRR in dB (20*log10(x)).
L10: What is the measured CMRR (dB) of the differential amplifier?
Create a table with calculated, simulated and measured values of Acm, Ad, and CMRR in
your report. Comment on the discrepancies.
2.0 Differential Amplifier with Current Mirror
Fig. 3: Differential amplifier with current mirror
Build the circuit shown in Fig.3 by using a current mirror at the common emitter to improve the
CMRR. You can use the basic current mirror you constructed in lab1 to accomplish the task (select
the appropriate value of Rp to provide the bias current of 1 mA). Is CMRR higher than the previous
circuit? If so, why?
L11: What is the measured CMRR (dB) of the differential amplifier with current mirror?
Weber State University
EE3120 Microelectronics II
Suketu Naik