# Op-Amp Examples - ```Op-Amp Examples -- Solutions
Q1
(a) (i) Voltage gain is infinite, so v+ = v- (approx) = 0 V
This means that no current can enter at the I.I. (Ohm’s law 0/R in = 0)
and the current in the 10-k = the current in the 100-k.
So by Ohm’s Law vin/10 = (0 - vout)/100 and vout/vin = -10.
(b) (I) Again v+ = v- and by voltage division v- = [10/(100 + 10)] x vout,
so vout = 11 v-.
But v- = v+ = vin in this circuit, so vout = 11 vin and the voltage gain = 11.
(a) (ii) We now have that vout = 10000(v+ - v-) = -10000v-, so v- = -10-4vout.
The currents in the 10-k and the 100-k must be equal again, so
(vin - (-10-4 vout))/10 = (-10-4 vout -vout)/100
and by multiplying out and collecting terms
0.1 vin = vout(-10-5 - 10-6 - 0.01)
and vout/vin = -0.1/(0.010011) = -9.99.
(b) (ii) We now have vout = 10000(v+ - v-) = 10000[vin - 10/(100 + 10)vout]
so vout(1 + 100000/110) = 10000vin
and vout/vin = 10000 / 909.1 = 10.988 (NB approximated to 10.99 on sheet)
Input resistance for (a) -- approximately 0 V at the I.I., so vin faces a
resistance of 10 k down to 0 V and so Rin = 10 k. It will be much higher for
circuit (b) because Rin is very high anyway and much less than the full vin
appears across it because of the feedback. (You will probably need to discuss
this point with them).
For the voltage gain of (b) with the full equivalent circuit, we first need to draw
the full equivalent circuit:
Ro = 150 ohm
vin =
v+
vout
Rin =
10 k
+
10000(v+ - v-)
--
100 k
v10 k
The studenti are not necessarily expected to work the analysis through to
completion but please discuss the approach with them.
We start off by examining the junction of the 100-k and 10-k resistors and
noting that (a) the currents into the junction sum to zero, and (b) we can
obtain expressions for the currents by Ohm’s law.
Current in Rin + Current in 10-k resistor + current in 100-k resistor = 0
(vin - v-)/10 + (0 - v-)/10 +(vo - v-)/100 = 0
If we set vin = 1 V, our value of vo will be equal to the voltage gain.
So 0.1 - 0.21v- + 0.01 vo = 0, or 0.1 = 0.21v- - 0.01vo (Eqn 1).
If we assume the output to be open-circuit, we can also obtain another
equation by noting that the current in Ro is the same as that in the 100-k
resistor:
[(10000(vin - v-) - vo]/0.15 = (vo - v-)/100, or again with vin = 1 V:
10000/0.15 - 10000v-/0.15 - vo/0.15 = 0.01vo - 0.01v66666.67 - 66666.67v- - 6.6667vo = 0.01vo - 0.01v66666.67 = 66666.57v- + 6.6567vo
Solving these by your favourite method (I used MATLAB) gives:
v- = 0.9989 V, vo = 10.977 V (technically the answer on the sheet should be
rounded to 10.98 V, not 10.97 V). So the voltage gain is 10.977 ... very near
the nominal 11 in spite of a very poor op-amp !
Q2. We assume an ideal op-amp again, so v+ = v-, and Rin is infinite.
By voltage division, v+ = 1.0 x 22/(22 + 10) = 0.6875 V.
This is also the value of v-, so by Ohm’s law
(0.5 - 0.6875)/10 = (0.6875 - vo)/22
so vo = 0.6875 - 22/10*(0.5-0.6875) = 1.1 V.
(The ‘magic formula’ for the voltage gain of this circuit is ‘Rf/R1 x (v2 - v1)’, or
2.2 x (1 - 0.5) = 1.1 V.)
Q3
Please remind (or tell) the group that the gain of the inverting amp is -Rf/R1 !
The derivation using this relation is that Zc = 1/(Cs), so using the Z instead of
the R (the deadly Generalised Impedance) we get - (1/(Cs))/R = -1/(Rcs), so it
works out to -1/(10000 x 0.22 x 10-6 x s) = -454/s. For a.c., s = jw, so that is
where the j and w come from ...
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