# 9. For the op-amp circuit with a resistive divider in the input network ```9. For the op-amp circuit with a resistive divider in the input network shown in Fig.2
answer the following questions (Rs = 5kΩ, Rf = 1MΩ, R1 = 10kΩ):
Rf
I in
Rs
-
V in
V ou
R1
+
Figure 2.
a) (3 points) Is this op-amp inverting or not inverting?
Inverting
b) (4 points) Write out the equation for Vout as a function of Vin.
Vout/Vin = -Rf/Rs = -1000/5 = -200
(Note that R1 does not enter the final answer because of virtual ground
assumption: the voltages on + and – terminals of op amp are zero therefore there
is no current through R1)
a) (3 points) What are the input (Vin/Iin) and output (Vout/Iout) impedances of an ideal
op-amp?
Input – infinity, output – zero.
c) (3 points) What is the input impedance (Vin/Iin) of this circuit?
Vin/Iin = Rs = 5kΩ
(Readers) The questions in b) and d) are much simpler than last quarter op amp questions.
If student just write down the correct answers in b) and d) without derivations or
explanation why R1 is not important, take out one point.
10. For the speaker circuit shown in Fig. 3,
a) (3 points) Explain why do we need the transistor TR4 in the circuit (what is its
function)?
Transistor acts as a switch or gate for the oscillator circuit
b) (3 points) Why do we connect the input to the speaker to Vout from the second
inverter (as shown) instead of the third inverter?
The 2nd inverter has a zero output when oscillator is turned off and we do not
want to use battery power to drive DC current trough the speaker.
c) (3 points) What elements will you replace to change a frequency of the gated
oscillator?
Either the resistor R12 or the capacitor C9.
d) (3 points) Suppose that the output voltage Vout of the gated oscillator goes from
5V to 2.5 V in 1 second. How long will it take if we triple the value of R12?
It will take 3 seconds.
R 12 68 k Ω
R 13
C 9 3300 pF
1kΩ
+
R 19 1.8 k Ω
+
V in
+
TR 4
5V
V out
R 20
120 k Ω
-
Figure 3.
-
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