2/21: More with op amps

in which Julia operates additional op amps

H&H Lab 8-2: Inverting amplifier

The inverting amplifier op amp circuit (from Electronics pdf).  In our circuit, R1 = 1kΩ and R2 = 10 kΩ.

A few notes on the inverting amplifier op amp circuit:
-Made by connecting Vout to V-
-Vout/Vin = -R2/R1 (from op amp golden rules and Ohm's Law)
-This means, if R2 > R1, Vout > Vin: the signal is amplified
-This gain is independent of the op amp's differential gain (Gdiff must be large though)
-Output is inverted

Driving the circuit with a 1kHz square wave of amplitude 1.00 Vpp made an output with an amplitude of 10 V, which means the gain was 10 for this circuit.

The maximum output swing was 21.6 V, which seems reasonable--it is slightly less than 24 V, as expected when powering the op amp with +/- 12 V.

A triangular input wave generates a triangular output (which is linear even at small timescales).
A sine wave input generates a sine output--but at some high frequency (around 1 MHz), the amplitude of the output wave becomes much less than the input amplitude. [We later learned this is a consequence of the non-ideal op amp property of slew.]

Next we measured the input impedance by adding a 1kΩ resistor in series with input.
This caused the output voltage amplitude to decrease from 10 to 5.4 V.

When we measure from the beginning of second resistor (right before the op amp), we measured a voltage of only 0.504 V.

Output voltage was 5.4 V with a 1kΩ resistor in series with the input.

Now it's a voltage divider, with Vout = 0.504 V and R1= 1 kΩ:

R2/(R1+R2) = Vout/Vin = 0.504/1.00

0.504(1 kΩ + R2) = R2

R2 = 1 kΩ

These calculations agree with the band colors on the resistor.

To attempt to measure the op amp's output impedance, we measured the voltage with an oscilloscope over a small varying resistor; theoretically, at a resistance comparable to the output impedance of the op amp, the voltage over the external resistor would be half the resistor-free output voltage.  However, a 1 Ω resistor would require a current of 1 A--and an op amp can only output about 25 mA of current.  Therefore we must use a very small voltage to test the output impedance--which causes the output signal to be very small.  We were unable to identify the output impedance of the op amp due to these complications.

Lab 8-3: Non-inverting amplifier

The non-inverting amplifier circuit (from Electronics pdf).

The output voltage was 11 V for an input wave of 1.00 V amplitude, so the average gain in this circuit was 11, contrary to the 10 gain observed in the inverting amplifier.

Putting the input signal through a 1 MΩ resistor causes no observable change in the output voltage (after dividing by 11 to negate the gain)--which suggests that the input impedance is huge (greater than 100 MΩ)!  The 411 op amp has an input impedance of 10^12 Ω, or 1000000 MΩ (according to the specifications sheet from Texas Instruments, http://www.ti.com.cn/cn/lit/ds/symlink/lf411-n.pdf), which would be in accordance with our observed results.

At an input frequency of 25 kHz, we get about 7.70 V output amplitude--so the 3dB point occurs at 25 kHz.

Output amplitude dropped 70% (11 V to 7.7 V) when driven by a 25 kHz wave.

ω(3dB) = 1/(R*C)

ω = 2*pi*(25 x 10^3) = 157079.633 = 1/(R*C)

R =1MΩ

C = 1/[(1 x 10^6)*(157080)] = 6.4 x 10^-12 = 6.4 pF

Without such a tiny Cin, Vout would be invisible.

Like in the inverting amplifier, the non-inverting amplifier has a low output impedance, because the circuit is essentially the same here--the output impedance depends only on the op amp, not on the external resistors.

Lab 8-4: Op amp follower

Op amp follower diagram (Electronics pdf).

An op amp follower circuit is simply a special case of the non-inverting amplifier, where R2 = 0 so the circuit has a gain of unity.  Here, Zin is huge because it is an op amp input; Zout is too small to measure for the same reason as the non-inverting amplifier.