FYS3220 - LAB B H23
University of Oslo
Daniel Tran
10. November 2023
LAB B H23 - FYS3220
UiO
Daniel Tran
Task 1A
Figure 1: Plot of current going into the resistor
From the plot you can see 3 different circuits, with their own damping factor. The first circuit is
overdamped, the second is critically damped and the third is underdamped. The different damping
factor is because of the value of the resistor. The higher the value of the resistor, the higher the
damping factor. The reason for why the value of the resistor affects the damping factor is because
of the voltage drop over the resistor. Less current will flow through the resistor if the resistor is
higher, this will affect the capacitor and the inductor. The capacitor will charge slower and the
inductor will have less current flowing through it. And the magnetic field from the inductor will be
smaller.
R1 Circuit
R2 Circuit
R3 Circuit
Zero
0
0
0
α [kHz]
100
10
1
β [kHz]
√
30 ∗ 11
0
9.9i
p1 [kHz]
−0.5
−10
−1 + 9.9i
Table 1: Pole and alpha/beta values
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p2 [kHz]
−199.5
−10
−1 − 9.9i
LAB B H23 - FYS3220
UiO
Daniel Tran
Task 1B
Figure 2: Bode Plot of the three circuits
From the bode plot you can see that circuit 3 has a more pointy slope than the rest. The reason
for this is because of the imaginary part of the pole.
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LAB B H23 - FYS3220
UiO
Daniel Tran
Task 2A
Figure 3: Bode plot showing all four operational amplifier circuits
An ideal OpAmp has infinite gain and infinite bandwidth. By looking at the blue line from
figure 3 you can see a simulation of an ideal OpAmp. But in the real world, the OpAmp is a
non-ideal one, has a gain and a bandwidth. The gain is the ratio between the output voltage and
the input voltage. The bandwidth is the range of frequencies that the OpAmp can amplify. You
can see from three other lines in the figure 3 that the higher the gain, the lower the bandwidth.
By looking at the table 2 you can see that Out 4 has the lowest gain, but the highest bandwidth.
Even though the three circuits have different gain, they all are a lowpass filter.
Are the calculated and
simulated gains comparable at f = 100 Hz?
Are the calculated and
simulated gains comparable at f = 100 kHz?
What is the simulated BW?
Out 1
Out 2
Out 3
Out 4
Yes
Yes
Yes
Yes
Yes
No
No
Yes
∞
9.78kHz
89.13kHz
501kHz
Table 2: Gain and BW Values
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LAB B H23 - FYS3220
UiO
Daniel Tran
Task 2B
(a) Circuit
(b) A subfigure
Figure 4: Bode plot of circuit 3 and 4
From the table 3 below you can see that the gain is the same for both circuits, but the bandwidth
is different. The reason for this is the new circuit 4 is a second order lowpass filter. By begin a
second order filter it has a higher bandwidth than the first order filter, but also a higher roll off.
Mentioned earlier, the correlation between gain and bandwidth is that the higher the gain, the
lower the bandwidth.
Gain (dB)
BW
Out 3
20
91.2kHz
New "Out 4"
20
154.9kHz
Table 3: Gain and BW Values
Task 3
We have the following equation:
Vout = aku(t), where k =
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Lf 1
Ri1
(1)
LAB B H23 - FYS3220
UiO
Daniel Tran
With Lf 1 = 700mH and Ri1 = 1.2kΩ we get that:
Vout


if t < 0ms
0V,
1000V 700mH
=
·
· u(t) = 0.583V, if 0ms < t < 1ms

s
1.2kΩ

0V,
if t > 1ms
(2)
Figure 5: Plot of the voltage for V_out and V_in
By looking at the plot in figure 5 you can see that the voltage for Vout is 0.583V
when 0ms < t < 1ms. This is the same value as the calculated value.
If we were to change the linear function present in the voltage source with a sinus signal, we could
expect the output to be a cosine signal. But if you were to integrate a triangle signal, you would
get a "pseudo-sine" wave.
Task 4
By using KCL we get the following equation:
Vin (t)
C · dVout (t)
=
Rf 22
dt
=> Vout (t) = −
1
Rf 22 C22
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Z
(3)
t
Vin (t)dt
−∞
(4)
LAB B H23 - FYS3220
UiO
By using the values R = 10kΩ and C = 1µF , we get the following equation:
Z t
1
Vin (t)dt
Vout (t) = −
1kΩ · 1nF −∞
=> Vout (t) = 3.23V
Daniel Tran
(5)
(6)
The reason for why we changed Rf 22 to such an odd value is because it is our scaling factor and it
gives us a nice amplitude for the output voltage.
Figure 6: Plot of the output voltage for the three circuits
In this task we have 3 different circuits, where 2 of them is a integrating circuit. For the first
circuit we integrate over a triangle signal, and for the second circuit we integrate over a cosine
signal. Figure 6 shows the output voltage for the three circuits. Even though they almost match
perfectly, they are not the same. As mentioned earlier, if you integrate a triangle signal you get a
"pseudo-sine" wave. This means that we do not get a sine signal, but a signal that looks like a sine
signal.
Task 5a
It is given that:
Q=
1
(3 − G)
(7)
1
2Q
(8)
ζ=
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LAB B H23 - FYS3220
UiO
Daniel Tran
and
1
RC
With G = 2, R6 = 20kΩ and C = 10nF we get that the quality factor is,
ω0 =
Q=
(9)
1
=1
(3 − 2)
(10)
1
1
=
2Q
2
(11)
the damping factor is,
ζ=
and the resonant frequency is,
ω0 =
1
1
=
= 5kHz
R6 C
20kΩ · 10nF
(12)
The relation between Q and ζ values and the shape of the bode plot is that the Q determines the
Figure 7: Bode plot of the Wien Bridge Filter circuit
width of the peak and the ζ determines the slope of the peak. By changing the value of R6 = 29kΩ
we get a higher Q factor. From the bode plot we get a narrower and pointy peak compared to plot
from figure 7.
The reason for the different spikes in the FFT plot from figure 8 is because of the system will
start oscillating. And we can see it oscillates at the resonant frequency.
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LAB B H23 - FYS3220
UiO
Daniel Tran
Figure 8: FFT of the Wien Bridge Filter circuit with R6 = 32kΩ
Task 5B
The reason for why the FFT plot from figure 9 is more choppy from the FFT plot from figure 8 is
because of the system is less damped. And the reason for why the system is less damped is because
of the higher value of R6 .
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LAB B H23 - FYS3220
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Figure 9: FFT the Wien Bridge Filter circuit with R6 = 35kΩ
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Daniel Tran
LAB B H23 - FYS3220
UiO
Daniel Tran
Task 5C
The output of the OpAmp is fed back to both the inputs of the OpAmp. But before the output is
fed back to the inputs, it is treated differently. The first part is a resistor divider network. And it
is this part that determines the gain of the circuit. The gain of the circuit must be 3 ≥ G for the
circuit to oscillate, and the gain is determined by the value of the resistors in the resistor divider
network. The other part is a combination of a resistor and a capacitor in series and parallel. This
forms the feedback network, and it is this positive feedback that gives rise to the oscillation.
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