Day 7: Amplifiers!

OpAmps are very cool and we can make some useful circuits with them!

First, an open loop test circuit

When the pot is turned, the Vout should go from -12V to +12V basically instantaneously.

Picture by Hannah

The max gain cannot be realized, since the max Vout is (+/-) 12 V. So we cannot verify the gain, but we can show that it is consistent with a gain >> Vout, since the change in Vout is instantaneous.

Now inverting amplifiers

Second picture by Hannah

By driving with a Vin of 1 Vpp and 1 kHz sine wave (not pictured), we get a Vout of 5 Vp.
G=5/0.5 = 10. (Recall that V peak to peak is 2 * V peak)
This is consistent with our prediction of R2/R1 = 10
The out put swing is Vpp = 10 V.

Something cool is going to the rail: this is when you get the max output from the opamp, so the sine wave has cut off tops and bottoms.

Picture by Hannah

 The maximum output of the swing when it is at the rails is 22 V (2 less than the input into the amplifier).

A circuit is linear if Vout is in the same shape as Vin (ignoring the 180 degree phase shift). Since the output of the amplifier is in the same shape, it is therefore linear.

At high frequency the amplitude of Vout actually decreases! So it acts as a low pass filter.

With a 1 kOhm resistor in series with Vin, as above, we can see that Vin = 2 Vpoint, indicating that the input impedance, Rth, is the same as the 1 kOhm resistor.

Similarly, when a resistor is in series with Vout, we see that the output impedance must be very small compared to 1 kOhm, since Vpoint is the same as Vout.

Now for the non inverting amplifier!

With a 1 Vpp 1 kHz sine wave, we get a Vout of 5.5 V that is in phase with Vin.
This indicates a gain of 11, which is consistent with our prediction of 1+R2/R1.

Our input impedance can be determined in a similar way to above:

With a DC input, Vpoint = Vin.
This indicates Rth >> 1 MOhm

Using an AC input, we expect the capacitor's impedance to dominate. (It will be much less than the resistor's very high impedance.)
The frequency is 1 kHz
Vpoint = 1/2 Vin, so impedance is 1 MOhm
Impedance = 1 /(frequency * capacitance)
Capacitance = 1/(frequency * impedance) = 1000 pF

The output impedance should still be really small.