Lab 5 Op-amps III, The Op-amp Integrator
Authored
by Jeremy Garrod
3/7/2017
garrod@unlv.nevada.edu
Pre-lab work
Lab Work
- Calculate the frequency response of the following circuit. Ensure you show your clear hand calculations.
- What can you neglect to simplify the calculation?
In
order to simplify the calculation, the 100k resistor can be neglected.
It is only there to make sure that an offset does not cause the output
to saturate at the rail voltage. It is also quite a bit larger than the
other resistor, so its affect is minimal.
- Does the circuit work if you remove the 100k? Why or why not?
In the case of an ideal Op-amp, the circuit will work perfectly fine
without the 100k resistor. However, once you use a real Op-amp that has
an offset it will not work without the resistor. The offset will cause
the voltage to just sit at the rails.
- Does the 100k have much of an effect on the frequency response?
Since
the A/C current can flow through the capacitor, which has a much lower
impedance than the resistor, it effectively acts like a short to the
resistor. The majority of the current will flow through the capactior,
causing the resistor to have little impact.
- Verify your calculations with experimental results.
- Is the phase shift between the input and the output what you expect? Why or why not?
The
wave was starting to clip on the lower half in my original experiments.
I found that my voltage divider did not ouput exactly 2.5V, instead it
output 2.48V. Once I set the offset of the function generator to match
that of the voltage divider, that issue went away.
The
amplitude is the same between the input and output with a 90 degree
phase shift at 160Hz. This is what was calculated and expected.
The
lower the frequency, the higher the gain of the Op-amp. This can be
seen in the images below, which are at 75Hz, 125Hz, 190Hz, and 250Hz.
At 75Hz, a gain of roughly 2 was measured.
At 125Hz, a gain of roughly 1.25 was measured.
At 190Hz, a gain of roughly .83 was measured.
At 250Hz, a gain of about .63 was measured.
- Next, design, simulate, and build a square-wave to triangle wave generation circuit.
- Assume the input/output frequency is 10 kHz and the output ramp must swing from 1 to 4 V centered around 2.5 V.
- Show all calculations and discuss the trade-offs (capacitor and resistor values, input peak, min, and average, etc.)
The hand calculations below were used to determine the required capacitor and resistor values. A capacitor value was chosen based on what was availble in the lab and the resistor value was calculated from that. My input was a 10kHz square wave that had a peak-peak voltage of 5V.
I
had initially used a capacitor value of 10nF which resulted in a
resistor value of 4.166k. However, the output triangle was heavily
distorted. I went to 1.5nF instead, which had a much better output. I
did forget to take a picture of this though. I chose a 3MEG resistor
since I wanted a couple of orders of magnitude larger than the 27.77k
resistor.
Schematic for square-wave to triangle-wave generation circuit
Simulation of the above schematic
Schematic
and simulation of my original design. It can be seen that it works the
same when dealing with ideal models. However, once it was impleneted in
the lab, this design did not work very well.
Oscilloscope output of square-wave to triange-wave generation circuit
The
results of this experiment did not exactly match my simulation. I had
to use a 27k resistor instead of the value 12.8k that I had calculated.
My capacitor was also not exactly 1.5nF, but I was not able to get an
accurate value for it, so all I have to go off of is the value that was
listed on it. Having those values not be perfect could have easily have
influenced the output that I had recieved
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