Lab 5 - EE 420L
reedj35@unlv.nevada.edu
For
the following questions and experiments assume VCC+ = +5V and VCC- = 0V.
- Calculate the
frequency response of the following circuit. Ensure you show your clear hand
calculations.
Figure 1: Integrator Circuit
Since we are looking at the frequency response of this circuit, we
are only looking at the AC components of this circuit.
Therefore, the non-inverting terminal of the op-amp is at 0V (AC).
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Frequency
Response Calculations:
The
unity frequency of an op-amp is when the gain of the op-amp is equal to 1.
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- What
can you neglect to simplify the calculation?
- For
this calculation, the large resistor in parallel with the capacitor, R2
can be neglected. This is because
it is much larger than R1
and the real term in the denominator goes to zero.
- Does
the circuit work if you remove the 100k? Why or why not?
- Since
op-amps in real life have an offset voltage and are not ideal, the
circuit does not work if you remove the 100k resistor.
The offset voltage is a DC
voltage and if there is no resistor in the feedback loop, then there is only
the capacitor. The capacitor
acts as an open in a DC
circuit and so there will only be a small difference in the inverting input and
non-inverting input voltages.
We know that 
and so 
which will be large and thus, the output
voltage will go to the power rails
of either +5V or 0V in our
case.
- Does
the 100k have much of an effect on the frequency response?
o
In this circuit, the 100k resistor does not have much of an effect
on the frequency response. This is because when an AC signal is introduced,
the capacitor will start to
play a role in the circuit. The 100k resistor has a much larger impedance than
the capacitor and so most of the
current flows through the
capacitor.
- Show,
at the unity-gain frequency of the integrator, that the input and the
output have the same peak values.
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Figure 2: Integrator Implemenation at unity-gain frequency |
Figure 3: Phase difference between
input and output |
- Is
the phase shift between the input and the output what you expect? Why or
why not?
We can see that there is a time
delay of 1.6ms between the input and output waveforms. Knowing the equation to
calculate the delay:
and then rearranging: 
The
calculated value matches the phase shift I was expecting above.
- Next, design, simulate, and
build a square-wave to triangle wave generation circuit.
o
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.)
Referring to Figure 1 above and ignoring the big resistor, we can
see the current flowing through R1 is equal to the current flowing
through C1. Therefore, we can say that the potential across the
capacitor is just Vout:
Solving the integral:
Since we want the output ramp to swing from 1V to 4V, 
.
Since the output needs to be centered about 2.5V, I will have an
input square wave from 0V to 5V with a frequency of 10kHz;
and my Vcm will be 2.5V.
For this design, I will be choosing the value for C1 that
will keep the value for my resistor relatively small and so that I
may find a capacitor that is readily available in the lab. The capacitor
value I will be choosing is 
With a frequency of 10kHz, the period is 100µs.
Solving for R1:
The closest resistor value I could find in the lab is 4.3kΩ.
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Figure 4: Integrator Design with
values used |
Figure 5: Simulated integrator
waveforms |
Figure 6: Experimental integrator waveforms
As said above, the value
used for the capacitor help set what resistor I can choose. I chose the
capacitor based off of lab availability, and then this
helped the resistor be large
enough for the circuit
to operate. As can be seen in the experimental waveforms, the change in the
output voltage is 2.96V and it is approximately centered about 2.5V. The
DC offset needed to be
changed only slightly to help the waveform look like what was calculated. This
is because the voltage at the inverting input slightly differed from the
voltage at the
non-inverting input.