EE 420L Engineering Electronics II Lab – Lab 5
Op-amps III, the op-amp integrator.
Authored by
Shadden Abdalla
Email: abdals1@unlv.nevada.edu
March 13,
2019
PRELAB WORK:
Watch the
video op_amps III, review notes and simulate
circuits.
LAB WORK:
This lab
uses the LM324 op-amp and assumes that VCC+ is 5V and VCC- is 0V.
Calculate
the frequency response of the following circuit.
The calculations above
show that the frequency is about 160Hz and the phase shift should be 90
degrees.
a. What can you neglect to simplify the
calculation?
You can neglect the Rbig
resistor, 100k. It will only slightly change the output. In the calculation, it
will only slightly change the result because of the ratio of R2 to R1. The huge
resistor is there as the feedback path for the DC current. Without the
resistor, the circuit will rail.
b. Does the circuit work if you remove the 100k? Why or why not?
It does not work in a non-ideal
circuit because the offset will cause it to rail. The huge
resistor is there as the feedback path for the DC current. Without the
resistor, the circuit will rail.
c. Does the 100k have much of an
effect on the frequency response?
It does not affect the frequency
response significantly at all because the AC current affects the capacitor a
lot more than it affects the resistor.
PART ONE
Show, at the unity-gain frequency of the integrator, that the
input and the output have the same peak values.
Below is the frequency generator input of 160Hz which
is what we calculated above with a 100mVpp amplitude and an offset of 2.5VDC.
To the right is a photo of the circuit pictured above on the breadboard.
Below are the results on the
oscilloscope. The yellow signal is the input and it measured a delta of 204mV.
To the right is the measurement of the blue signal, the output of the circuit,
which measured a delta of 148mV.
This shows that the
circuit worked properly because were were trying to find the unity gain value.
In order to do that, you divide the input delta by the square root of two to
get the value of the output.
We changed the frequency
until we got a value that is similar to the one we calculated, 144.25mV. We
reached 140mV at a frequency of 220Hz.
Is the phase shift between the input and the output what you
expect? Why or why not?
The phase shift is what
we expect, there is a 90 degree phase shift between the
two signals. You can see from the calculation above that the experimental
reading matches the calculation. You can see the difference in the oscilloscope
reading above. The output leads the input by 90 degrees.
PART TWO:
- 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.)
We picked a
capacitor in the lab which measured to be 0.427nF. From that value, we
calculated the resistor value that we needed, which was about 100k. We
calculated the resistor value based on the capacitor value to ensure the best
results and measured the capacitor to create a more accurate oscilloscope
reading. The tradeoff of using a smaller capacitor is that we need a larger
resistor value, but we had both of those values, so it was not a huge problem.
The calculations above show
how we choose to build the circuit. We found the period by using the frequency
and solved for the resistor value by using the measured capacitor value. Below
is the input signal from the frequency generator, with the circuit next to it
on the breadboard. The output shows that the input, a square wave, produced a
triangle wave. It was a slightly uneven circuit, but it is a triangle.
Conclusion and tradeoffs:
The
LTSpice simulation confirms that our values work. It is centered about 2.5V and
goes in between 1V to 4V. The LTSpice simulation is not exactly what we want
because it is simulating the ideal case that I produced using the breadboard.
That is why the values do not go exactly to 4V or 2.5V but they are very close
because it is a non-ideal case. The tradeoffs regarding the average values
leveled out and overall, the output was what we intended it to be. For my input
peak, I used a square wave that went from 0 to 5 in order to make it easier to
get from 1 to 4. I also choose those values in order to get closer to VCM and
have an average that is very close to VCM to reduce offset.
