# Lab 7 Operational Amplifiers and Negative Feedback Operational ```Figure 1: Symbolic representation for an op-amp. Note that the power supplies +Vcc and -Vcc
may not always be shown on the circuit diagram but they must always be present in the real
circuit.
Lab 7 Operational Amplifiers and Negative Feedback
Operational amplifiers are high gain differential amplifiers. The gain A can be
as large as 106. Other noteworthy characteristics are very large input impedance
(Zin &gt; 106Ω) and very small output impedance (Zout &lt; 100Ω). A symbol for
the op-amp is shown in Figure 1. The + and - signs on the inputs do not refer
to the polarity of the input signals but rather to the effect on the ouput signal
given an input signal. The output is inverted with respect to the input if the
input signal is applied to the inverting(-) input. The output is in phase with
the input if the signal is applied to the noninverting(+) input. By virtue of
these characteristics the workings of op-amps in circuits are fairly simple to
understand. It turns out we can ignore the internal circuitry of the op-amp.
The function of the op-amp is then largely determined by the feedback network
associated with it. Because the feedback network is typically, but not exclusively made from resistors or other stable elements the operation of the op-amp
with negative feedback is quite stable against temperature changes. We will
look at three aspects of op-amp use in this experiment; the notion of “virtual
ground”, use as an inverting amplifier, and use as a component in a current
source. Because of the large inherent amplification one finds it is necessary to
offset null the op-amp first. That is, if both inputs are at identical potentials
we expect vo to be zero(It is a difference amplifier.).
Offset Null Procedure
Figure 2: Pinout for the 741 op-amp. Use Vcc = +- 12V from the fixed supplies.
The pinout of the 741 is shown in Figure 2. Connect the circuit as shown in Figure 3. If you have a potentiometer on the experimenter’s board, adjust the pot
until vo is approximately zero. You should be able to get within 5mV of ground.
Virtual Ground
Discussion
The equivalent circuit for the op-amp is shown in Figure 4. Ri and Rc are the
input resistances, which are both very large. The output impedance Ro is very
small. The feedback netwrok is represented by β. It returns a portion of the
output, βvo, to the input terminal. We wish to determine the difference v2 − v3.
For Kirchoff’s loop going through the generator we obtain
vo = a(v3 − v2) + io Ro
(1)
but
v2 = βvo
(2)
βvo = βa(v3 − v2) + βio Ro
v2 = βa(v3 − v2) + βio Ro
(3)
(4)
Figure 3: Offset null procedure.
Figure 4: Equivalent circuit for the op-amp.
Ry
(Ω)
.
.
.
Rw
(Ω)
.
.
.
Vo
(V)
.
.
.
V2
(V)
.
.
.
V3
(V)
.
.
.
V2 /V3
.
.
.
Table 1: Op-amp golden rules
(1 + aβ)v2 = aβv3 + βio Ro
(5)
we have β &lt; 1 but still keep aβ &gt;&gt; 1 so
aβ(v2 − v3) = βio Ro
(6)
or
v2 − v3 = io Ro /a ≈ 0
(7)
So it is a good approximation that the output adjusts in order that the feedback
network makes v2 = v3.
Circuit
Build the circuit in Figure 5. Use the nominal values for the resistors. Check
the relation v2 = v3 by filling in Table 1. Find the average ratio v2/v3 and the
standard deviation. Notice if we set v3 = 0 by attaching terminal 3 to ground,
for example, then v2 = 0 from the above results but no, or very little, current
actually goes to ground through the op-amp because the input impedance is so
high. This is the origin of the term virtual ground. Terminal 2 is at ground
potential but no current goes to ground through the terminal 2 input.
Inverting Amplifier
Using the the approximation v2 = v3 we can easily compute the gain G of an
op-amp with feedback. Consider Figure 6. Because v3 = v2 we also have v2 = 0.
Also ia ≈ 0, hence ii + if = 0. Now vi = ii Ri and vo = if Rf and since G = vo /vi
then G = if Rf /iiRi so G = −Rf /Ri . Build the circuit in Figure 6 and feed vi
with a 100Ω/10KΩ voltage divider on the sine wave signal. Typical values for
Figure 5: Circuit to check the effect of negative feedback on the relation between v2 and v3 .
fnominal
(Hz)
.
.
T
(S)
.
.
tp
(S)
.
.
phase
.
.
Vi
(V)
.
.
Vo
(V)
.
.
gain
.
.
Table 2: Inverting Amplifier
Rf are around 10KΩ and Ri around 1KΩ. Actually you can choose any set.
For four frequencies covering the range of the signal generator measure the gain
G and the phase shift between the input and output as in Table 2
Current Source
A very good current source capable of sinking a current through a load to
ground can be built with an op-amp. The simplest arrangement is in Figure 7.
See Horowitz and Hill for other circuits. The circuit in Figure 7 can be understood as follows:
(a) v3 is determined by the Rx /Ry voltage divider.
(b) v3 = v2 = Ve .
(c) Ie = (Vcc − Ve )/Re
(d) Ie = Ic + Ib
(e) Ic ≈ Ie for very small Ib
(f) steps (e) and (d) work if Vce is bigger than the saturation voltage of a few
tenths of a volt.
Build the circuit in Figure 7. First calculate the approximate value of the
maximum load resistor RLmax so that the transistor Vce is not saturated. Next
choose five values of RL such that RL = RLmax , 2RLmax and three values of RL &lt;
Figure 6: Inverting amplifier.
RL
(Ω)
.
.
Vcc
(V)
.
.
V3
(V)
.
.
V2
(V)
.
.
Vb
(V)
.
.
Vc
(V)
.
.
IL
(A)
.
.
Table 3: Current source
RLmax and fill in Table 3. Find IL = Ic over the flat range for your source. How
does it compare to the predicted value?
Figure 7: Op-amp based current source.
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