Lab 6
Differential Amplifier and Current Source
I Differential Amplifier Properties
Differential amplifiers are commonly used as the input stage of laboratory instruments or in other applications where weak signals may be contaminated
by noise. The differential amplifier has two inputs and the output signal is a
multiple of the difference between the two input signals. In the ideal differential amplifier the output is independent of the magnitudes of the individual
input values, only the difference matters. A differential change at the inputs is
called the “normal mode”. If the inputs are identically changed this is called
the “common mode”. The common mode rejection ratio (CMRR) is defined
by CMMR = (normal mode response)/(common mode response) for the same
amplitude of input signals. A circuit for the differential amplifier is shown in
Figure 1. The small parameter equivalent is shown in Figure 2. This circuit
can be analyzed using Kirchoff’s laws in the usual fashion. The loop equations
are:
v1 − hie1 ib1 − ie Re = 0
(1)
v2 − hie2 ib2 − ie Re = 0
(2)
v1o + ic1R1 = 0
(3)
v2o + ic2R2 = 0
(4)
vo − v1o + v2o = 0
(5)
The node equation at point A is
ie = ie1 + ie2
at E1
(6)
ie1 = ib1(1 + hf e1) = g1 ib1 (7)
at E2
ie2 = ib2(1 + hf e2) = g2 ib2 (8)
If we let
dj = hiej + gj Re , (j = 1, 2) (9)
then we can rewrite (1) and (2) as
d1ib1 + Re g2ib2 = v1 (10)
d2ib2 + Re g1ib1 = v2 (11)
Solving (10) and (11) for the base currents yields
ib1 = (d2v1 − Re g2 v2)/D (12)
ib2 = (d1v2 − Re g1 v1)/D (13)
D = d1d2 − g1 g2Re2 (14)
From the values of the base currents in (12) and (13) we could predict the behavior of the amplifier for general conditions. Usually for differential amplifiers
we try to make the two transistors as alike as possible. This means setting
hie1 = hie2, so then g1 = g2 = g, and d1 = d2 = d. The current change in Re is
then
ie = (v1 + v2)/(2Re + hie/g) = (v1 + v2)/2Re (15)
In (15) we drop the term hie /g which is much smaller than 2Re . Perhaps more
relevant is the comparison of ie to ie1 or ie2. Using the above equations we can
show that
ie1 = v1/2Re + g(v1 − v2 )/2hie (16)
From (16) the second term can be large in the normal mode case where the
difference in the inputs is non zero. If ie were exactly zero then the potential at
point A would be constant. This in turn implies a constant current Ie crosses
resistor Re for all values of inputs signals. In fact, we see that ie is not zero,
but it is very small so the resistor RE acts as a not so great current source. The
output signal can be found as
vo = −(hf e /hie)RL(v1 − v2)
(17)
where we set R1 = R2 = RL. The ratio of h parameters can actually be found
in a way that is relatively inpdependent of the transistor types. Use the expressions in (3)-(6) of Lab 4 for the h parameters and obtain
hf e /hie = ∂Ic/∂Vbe = (e/kT )IC
(18)
where in (18) we used the Ebers-Moll model to calculate the derivative.
It turns out that the amplification depends on the total current drawn by Q1
and Q2. The currents depend on the operating point and the fact that Q1 and
Q2 are identical so 2Ie1 = Ie . (In all this we are assuming the IC = IE .)Finally
in terms of the current IE
vo = −(e/kT )IeRL (v1 − v2)/2
(19).
When v1 = v2 the output is zero. We can rarely get two transistors and all the
other resistor to be indentical. In any real amplifier there will be an output signal even if the input signals are equal. The condition that we want (Q1 = Q2)
can be approached by the use of a dual monolithic transistor. This is a transistor which contains two bases, two collectors and one common emitter. Because
of its small size there can never be a large thermal gradient across the transis-
tor so the h parameters change together. The discussion below assumes such a
monolithic dual transistor, although as far as constructing the circuit goes two
discrete transistors can also be used. A second improvement on the circuit of
Figure 1 is to replace the resistor Re by a current source. If we had a rock solid
current source below point A then
ie = ie1 + ie2 = 0
(20)
If (20) were to hold then we could be freer in our selection of transistors because
we no longer require exact equivalence of the transistors. The current source
forces the emitter currents to change oppositely. An example of a current source
differential amplifier is in the text by Brophy ( fig. 7-24). We will construct
a current source based differential amplifier using a monolithic dual transistor.
Our first task will be to build and study the limitations of a current source.
II
Current Source
A transistor is a simple and quickly available current source. Better ones can
be made, but the current source in Figure 3 will suit our needs. Construct the
circuit in Figure 3. The voltage divider Rx3 , Ry3 provides the base bias. It is easy
to compute the expected value of Vb3 so then we have an approximate value for
Ve3 = Vb3 −0.6V . Knowing Ve3 we can approximate Ie by Ie = (Ve3 −Vee )/Re. We
will determine the range of collector resistance over which a constant current Ic
can be delivered. Within that range the current is programmed by the choice
of bias network and emitter resistance. The transistor can provide constant
current provided it is not saturated. Saturation occurs over an approximate
range 0V < Vce < 0.05V.
Analysis
Fill in Table 1 for your range of Rc .
Table 1 Current Source Compliance Table
Rc
.
.
.
Vb3
.
.
.
Vc3
.
.
.
Ve3
.
.
.
Vee
.
.
.
Ic
.
.
.
Ie
.
.
.
From your table compute the mean current < Ic > and the standard deviation
SI . Consider carefully what values of current Ic you should use in computing
the mean!
Plot Ic vs Rc and on the same graph plot Vce vs Rc .
What is the approximate saturation voltage for your transistor?
III Differential Amplifier
Build the circuit in Figure 4. Leave your current source from part II intact.Notice that the resistance Rc is replaced by the differential amplifier. We
will first determine tha amplification for d.c.. This is done by changing the
vales of resistors Rv1 and Rv2 . Different values for these resistor will produce
different values of the input signals Vb1 and Vb2 . An efficient way to proceed
is first to leave Rv2 open (i.e. infinite resistance) and adjust the values of Rv1 .
Then go back and leave Rv1 open and vary Rv2 . In this manner you will cover
both the positive and negative axis of (Vb1 − Vb2). Record the values in Table
2. Also record the value of the power supply because it tends to drift. Typical
values are mV for the base voltage Vb1 and Vb2 and mA for the currents.
Table 2
Differential Amplifier
Rv1 Vb1 Vb2 Vc1 Vc2 VA Ve3 Vcc Vee Ic1 Ic2 (Ic1+Ic2 ) Ie3 (Ic1 +Ic2 )/Ie3
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
Make a third table, Table 3, derived from the data in Table 2 and plot the
Table 3 data. From your plot obtain the amplification, vo /vi. How well does
the amplificaton agree with the prediction in (19)?
Table 3
Input/Output Curve Differential Amplifier
(Vb1−Vb2) (Vc1−Vc2)
(mV)
(V)
.
.
.
.
Common Mode Rejection
We will now investigate the ability of this amplifier to reject common signals.
First we will record the response to a normal mode signal. Leave Rv2 open and
set Rv1 =330K except connect Rv1 to a sine wave generator instead of Vcc (See
Figure 5). With the circuit input as in Figure 5 make a reasonably accurate
plot of the magnitudes and relative phases of vb1, vb2, vc1, vv2. You can also
obtain the amplification here by using the oscilloscope to subtract vb1 from vb2
and vc1 from vc2. Do this by putting channel 1 to Q1 parameters, channel 2 to
Q2 parameters, set channel 2 to invert and use the add function on the display
mode.
How does (vc1 − vc2 )/(vb1 − vb2) compare to the d.c. value?
Now take an additional connection from vb1 to vb2, i.e., a single wire connects
the bases of Q1 and Q2. This puts the same signal into both inputs. Make the
same free hand sketch as above for the input and output wave forms.
Is there a visible output now?
Notice that the individual signals vc1 and vc2 are also considerably smaller than
in the normal mode. The constant current source only requires that ie1 +ie2 = 0,
and not that ie1 or ie2 are separately zero.
Explain why ie1 and ie2 are also much suppressed in the common mode compared to the normal mode.
IV
Noise reduction
The suppression of common signals on the inputs is of great utility in suppressing noise. A weak signal, say from a distant instrument, may suffer pick up
noise. Noise can come from many sources, 60Hz being a ubiquitous signal, for
example. If a parallel wire which does not carry the signal but which is in close
proximity to the signal carrying wire is also fed into the differential amplifier
the noise component can be greatly suppressed. This assumes that noise source
affects both wires in the same way.
A differential amplifier may have been set up in the lab to illustrate noise reduction possibility due to the common mode rejection behavior of the amplifier.
In Figure 6 the amplifier is represented by a triangle. The signal plus noise is
going into Q1 and the noise is going into Q2. Noise has been introduce to the
two lines by attaching antennas to them(loops of wire). With a square wave
pulse as a signal record the width of the noise band in the output signal with
Figure 1: A differential amplifier circuit from the text by Brophy(fig. 6-21).
the noise pick up line both attached and not attached to the base of Q2 . There
should be a noticeable reduction in the noise band. Noise can not be totally
suppressed because there may also be uncorrelated noise on the lines. This can
not be subtracted away.
Reduction of noise and improvement of the signal/noise ratio is an unending
task of the experimenter. If we had, for example, two signals on the line in
Figure 6 which were close together in amplitude, our ability to resolve them
would depend on the noise present. With low noise our experimental resolution
is probably good. A sketch of this effect is shown in Figure 7.
Figure 2: Small signal equivalent circuit of the differential amplifier.
Figure 3: Transistor Current Source.Re = 1K, Rx3 = 2.7K, Ry3 = 13K. RC , 5 values between
0Ω and 6KΩ, then three values between 6KΩ and 8KΩ and 2 values above 8KΩ. Vee =-12V.
Figure 4: Current source based differential amplifier using a dual transistor.
Figure 5: Introducing a small signal to the amplifier. Figure 6: Noise reduction using a
differential amplifier. Figure 7: Sketch of resolving peaks with good and bad resolution.