Solution to Exercise 55:
Since this is an ideal op-amp, we know that:
Rf
Vout
=−
Vin
Rs
(1)
This tells us that the voltage at output of the op-amp is:
−Vs
Rf
Rs
= −4mVp−p
(2)
12K
1.2K
= −40mVp−p
(3)
(4)
Now we have to deal with the potentiometer. A potentiometer is essentially
a voltage divider:
R1
Vin
R2 Vout
We are told that the wiper of the potentiometer is one quarter of the way
up from the ground connection; in other words:
R1 = 3R2
(5)
So the voltage divider (alone) gives us
Vout = Vin
= Vin
R2
R1 + R2
R2
3R2 + R2
=
1
Vin
4
(6)
(7)
(8)
Now we can combine equation (4) with equation (8) to get our answer, the
level of the output voltage:
1
Vout =
1
(−40mVp−p )
4
= −10mVp−p
(9)
(10)
An ideal op-amp doesn't change the frequency or the shape, so the output
wave is still a sine wave at 440 Hz, but due to the negative sign it is inverted
i.e. when the input sine wave rises, the output sine wave falls.
2