Inverting Differentiator

Inverting Differentiator

In this lab, we use an op amp (OP 27) and a capacitor to create an inverting differentiator. This means that the output voltage V_o is (a multiple of) the negative derivative of the input voltage V_in, or V_o = -RC*(dv_i/dt), where R is the feedback resistance, C is the capacitor capacitance, and v_i is the input voltage. If the input voltage v_i = Acos(ωt), where A is the input amplitude and ω is the angular frequency, then its derivative is (dv_i/dt) = -Aωcos(ωt). Thus the output voltage is V_o = RCAωsin(ωt).
We wire the circuit according to the diagram above using a 470 nF capacitor, and a 470 Ω (464 Ω measured) feedback resistor.
We will be applying three sinusoidal waves for the input. Their amplitudes are all 100 mV and are centered at zero (offset = 0). The frequencies will vary from 1 kHz, 2 kHz, and 500 Hz.
First we solve for the expected output voltages V_o1, V_o2, and V_o3, which correspond to the 1 kHz, 2 kHz, and 500 Hz waves, respectively. Earlier we saw that the output of the voltage is V_o = RCAωsin(ωt). Since the input frequency is given in hertz, we will need to convert angular frequency to linear frequency using ω = 2πf. Therefore the output voltage can be rewritten as V_o = 2πfRCAsin(2πft).
The numbers were plugged in for the three inputs and we obtained the three theoretical time-varying output voltages.

Here is a picture of our circuit, ready to be analyzed. The long orange wires measure the input voltage V_i on the oscilloscope and are on channel 1 (C1) on the WaveForms software. The yellow wire is connected to the input voltage V_i. The two top green wires are connected to the output pin and ground, respectively to measure the output voltage V_o which is on channel 2 (C2) on the software. The white and brown wires provide the -/+ 5 V rail voltages. The bottom green wires are both connected to ground (the second is redundant).

Sinusoidal Wave - Input: 100 mV @ 2 kHz

Here we see the input (orange) and output (blue) voltages. The input voltage is 100 mV at 2 kHz. We see that the blue wave is the negative derivative of the input multiplied by some constant. However, from examining the waves, we see there are two errors. The input voltage is actually not a cosine wave but a sine wave. Second, the amplitude is much greater than we expected. The first problem can be fixed by simply changing the phase angle of the input wave but this does not concern us much. We are more interested in seeing the input become differentiated. However, the amplitude does not match what we expected. We are supposed to be getting an output amplitude of around 0.2776 V but we are observing an amplitude of about 2.5 V!

After double checking everything, we notice the capacitor we used was not 470 nF but 4.7 µF (4.24 µF measured). This means that the capacitance was off by a factor of 10. We must take this into consideration.
Looking more closely at the graph, we see the input amplitude is 100 mV with a little offset. The output amplitude is obtained by taking the average of the wave peaks. We obtain an output of (2.35+2.4)/2 = 2.375 V. Recalculating what the output amplitude should be, A_output = 2πfRCA = 2π(2000)(464)(4.24E-6)(0.100) = 2.4723 V. Comparing this to our theoretical value 2.375 V, we get a percent error of 3.9 %. This makes our predictions accurate, and we now regain trust in our inverting differentiator.

Sinusoidal Wave - Input: 100 mV @ 1 kHz

Now we repeat the experiment for a sinusoidal voltage input of 100 mV with a 1 kHz frequency. We calculate the output amplitude once again as done previously as A_output = 2πfRCA = 2π(1000)(464)(4.24E-6)(0.100) = 1.2361 V. Looking at the graph, we see the input has an amplitude of 100 mV (with a small offset). Taking the average of the top and bottom peaks we find the amplitude is (1.1+1.25)/2 = 1.175 V. Compared to our theoretical value, we have a 4.9 % error, which again is acceptable.

Sinusoidal Wave - Input: 100 mV @ 50 Hz

Finally, we repeat the experiment with an input wave of amplitude 100 mV at 500 Hz. The theoretical value is once again calculated. A_output = 2πfRCA = 2π(500)(464)(4.24E-6)(0.100) = 0.6181 V. Looking at the graph, we see the same input wave with its same offset. We look at the blue wave and take an average of the peaks again to find the amplitude of the output. The calculations give our output amplitude to be (0.7+0.55)/2 = 0.625 V. Comparing to our theoretical value, we obtain a 1.1 % error.

Data Chart

This chart shows the data obtained from the three graphs above. The percent error (% difference above) is below 5%, so our predictions had good accuracy. The graphs all show a phase difference of roughly a quarter of a period, as expected.