Lab 5 - EE 420L
Authored
by Steven Leung
3/1/15
leungs@unlv.nevada.edu
Pre-lab work
Introduction
The main purpose of this lab is to understand how the op-amp integrator
works and designing parameters to design output signals from an integrated signal.
Experiment 1
Schematic for experiment 1(Figure 1)
- Calculate the frequency response of the following circuit. Ensure you show your clear hand calculations.
Frequency response with feedback resistor
(Figure 2)
Frequency responser without
feedback resistor (Figure 3)- What can you neglect to simplify the calculation?
- We can neglect the feedback resistor R2 to make calculations simpler.
The reason why we can neglect the R2 or Rf resistor in calculations is
because the main purpose of that resistor is for a DC feedback path and
the only effect that the resistor has on calculations is a DC gain, but
since we are setting the voltage at Vp to match the DC voltage of the
input, the circuit is designed to have to gain of only 1. In other
words, since there is no DC current flow (no voltage drop across R1)
there will be no DC gain.
- Does the circuit work if you remove the 100k? Why or why not?
- The circuit will not work if the 100K resistor is removed because
that resistor is the only feedback path for DC signals. Without that
resistor, there will be no feedback path for DC signals since a
capacitor is a open in DC. As a result, since the open loop gain of an
op-amp is very high (infinite in ideal case) the output will be equal
to (Vp-Vn)Aol. Since the rails of our op-amp is 0 and 5 volts, any
small difference in Vp and Vn will cause the output to jump to one of
the rails. When the 100k resistor is included, this causes a smaller DC
gain and therefore allowing for a small change between V+ and V- to not
cause the output to rail.
- Does the 100k have much of an effect on the frequency response?
- The
100k resistor does not have a significant effect on the frequency
response. The reason why the 100K resistor has to be large is because
when
we perform AC analysis, we want all the AC current to flow through the
capacitor and not the feedback resistor causing the capacitor to charge
or discharge. The charging and discharging of the capacitor is what
causes the integrating effect (summing up the charge). When there is
little to no AC current flowing through the 100k resistor, it can be
said that that resistor is an AC open. In other words, there is only
one purpose for the 100k resistor and it is to ensure a DC feedback
path so that the integrator will not jump to a rail and saturate.
- Verify your calculations with experimental results.
Magnitude
response at 1K frequency (Figure 4)
Unity gain frequency (Figure 5)- Show, at the unity-gain frequency of the integrator, that the input and the output have the same peak values.
- Is the phase shift between the input and the output what you expect? Why or why not?
- Figure
5 shows the input and output of the circuit at the unity gain
frequency. Ignore the peak to peak measurements on the scope because
there are some spikes in the data and notice that when the time scale
for both input and output are set to the same values, the amplitude of
the output versus the input is the same. The phase shift as shown in
figure 5 is what i expect because in the calculations, the phase change
was to be +90 degrees for all frequencies(see figure 3). This phase
shift can be see from figure 5 because when the output is at the "off"
part of the squire wave, the output should be integrating (increasing in
voltage as input goes up) but it is decreasing. The same is also true
for the "on" part of the input signal.
Experiment 2
- 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.)
As
we can see, the output of a square wave input is similar to a triangle
wave since the integral of a horizontal line a ramp function. Having
the output swing and input swing of the op-amp be a similar value, our
output signal will increases and decrease for at the same rate and the
same total amount causing the signal to increase and decrease similar to
a ramp function resulting in a triangle wave. For example, if the input
signal is centered at 2.5 and the positive swing is .2 Volts and the
negative swing is also .2 Volts, the capacitor in the feedback will
charge and discharge at the same rate causing the output to be
proportional square wave. One thing to take note of when
designing a triangle wave output is the rails of the op-amp. The rails
for our design being 0 and 5 volts, the output cannot exceed 5 volts.
Our integrator circuit will have an offset because of a DC gain caused
by the offset voltage of an op-amp and a mismatch in the VCM voltage
and DC voltage of the input. The offset caused by VCM not matching the
DC voltage of the input will be seen in this lab compared to the
previous ones because since the DC gain is large, a small mismatch will
result in a notable change. One way to resolve this problem is to play
with the offset feature on the signal generator and find the right
offset voltage that will place the triangle signal centered at 2.5
volts because even if the input is set to have a DC voltage of 2.5
volts and VCM is set to be 2.5 volts, the output will not have an
offset of 2.5 volts because of other factors that are amplified such as
offset voltage.(see figure 4)
See below for calculations to designing the specified triangle wave.
Design
values for triangle wave (Figure 6)
Experimental results with
calculated values (Figure 7)
The
most noticeable trade off is the value of the resistor and capacitor.
Notice that they are multiplied together to form a time constant. While
the value of the time constant cannot change, the vales of each
individual part that compose of the time constant can change. The trade
off of having a large resistor and therefore a small capacitor is that
with a large resistor, there will be a larger DC gain and therefore
shifting the signal to saturation. To increase the peak of our output
triangle wave, the input square wave has to have bigger since a larger
voltage (top and bottom of square wave) will result in the capacitor
charging to a hinger value, but the tradeoff to this is again that the
circuit may saturate.
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