We successfully built the speed controller whose speed varied as we turned the potentiometer:
Wednesday, May 22, 2013
MOSFET Controlled Electric Motor
The experiment was to use a MOSFET transistor connected to a potentiometer to make a speed controller.
The design looked like the following:
We successfully built the speed controller whose speed varied as we turned the potentiometer:
We successfully built the speed controller whose speed varied as we turned the potentiometer:
FreeMat 2
The second FreeMat assignment dealt with the use of phase angles in sinusoidal waves.
Adding Sinusoids
The result of adding two sinusoids
Adding Sinusoids
The result of adding two sinusoids
Assignment 1
1. Two circuits with the output 2e^(-t/τ) with two different time constants. The circuit with the smaller time constant will have the lowest output sooner.
2. The output is now 2(1-e^(-t/τ)).
Assignment 2
1. Determine the output when adding 3 sin(2t + 10) and 5 cos(2t-30)
2. For a large number of frequencies.
Tuesday, May 21, 2013
Operational Amplifiers II
The second part of the op-amp experiment is to observe the changing of input and feedback resistors. We began by setting up the inverting amplifier
If the fixed input resistor is 10 kΩ, then we need a 100 kΩ resistor to achieve a gain of -10.
If Vsen = 1 V, the current leaving the op-amp is IOP = 0.1 mA.
We constructed the following setup and made measurements
Vin
Desired
|
Vin
Actual
|
Vout
Measured
|
VRF
Measured
|
IOP
Calculated
|
ICC
Measured
|
IEE
Measured
|
0.25 V
|
0.248 V
|
-2.48 V
|
2.48 V
|
0.0248 mA
|
|
|
0.5 V
|
0.5 V
|
-4.81 V
|
4.83 V
|
0.05 mA
|
|
|
1.0 V
|
1 V
|
-10.03 V
|
9.95 V
|
0.1 mA
|
0.889 mA
|
-0.982 mA
|
Kirchhoff's Current Law is confirmed when we add ICC and IEE to yield IOP.
The power supplied by the 12 V supplies were
PCC = (12 V) (ICC) = (12 V) (0.889 mA) = 10.67 mW
PEE = (12 V) (IEE) = (12 V) (0.889 mA) = 11.78 mW
We added a 1 kΩ resistor to the output shown below
and measured the following
Vin
Desired
|
Vout
Measured
|
VRF
Measured
|
IOP
Calculated
|
ICC
Measured
|
IEE
Measured
|
1.0 V
|
-9.97 V
|
9.75 V
|
0.1 mA
|
0.887 mA
|
-0.987 mA
|
We can see that KCL is obeyed when we sum ICC and IEE.
Again, we calculated the power:
PCC = 10.64 mW
PEE = 11.84 mW
Extra Credit
In order to achieve a gain of -5 in our inverting amplifier, our value for the feedback resistor must be 50 kΩ.
Our measurements were:
Vin
Desired
|
Vout
Measured
|
VRF
Measured
|
IOP
Calculated
|
ICC
Measured
|
IEE
Measured
|
1.0 V
|
-5.04 V
|
4.98 V
|
0.1 mA
|
0.885 mA
|
-0.985 mA
|
The behavior of the circuit is exactly the same as the two previous measurements made in the experiment.
Second Order Circuit Tutorial
This serves as acknowledgement that we completed the second order circuit assignment. The results of each question is posted below:
Question 1
Question 2
Question 3
Question 4
Question 5
Question 6
Question 7
Question 8
Question 9
Question 10
Question 11
Question 12
Question 13
Question 14
Question 15
Question 16
Question 17
Capacitor Charging/Discharging
The experiment was to design a charge/discharge network using capacitors and resistors in series. The design required us to find a way to charge the capacitor in 20s, then abruptly discharging it in 2s. With a few calculations, our design was built in the following manner:
We determined the values of the capacitor, charge resistor, and discharge resistor to be 47 µF, 85 kΩ, and 8.5 kΩ, respectively. We ran the circuit through a 9 V power supply, charged the capacitor by closing the left half of the circuit and discharged it by immediately closing the right half. and recorded the data on Logger Pro. Our results were:
= 85 kΩ * 992.5 kΩ / (85 kΩ + 992.5 kΩ)
= 78.3 kΩ
2. During discharging,
RTh = Rdischarge * Rleak / (Rdischarge + Rleak)
= 8.5 kΩ * 992.5 kΩ / (8.5 kΩ + 992.5 kΩ)
= 8.4 kΩ
3. When V = 0.6321 * (final voltage), τc = 4.0 s. Thus, R = τc / C = 85.11 kΩ
The value is indeed close to the value in question 1.
Practical Questions
1. C = 2E / V2
= 2 * 160 MJ / (15 kV)2
= 1.42 F
2. If the capacitance is arranged in the schematic, the value of each capacitor, C, is 0.71 F
We determined the values of the capacitor, charge resistor, and discharge resistor to be 47 µF, 85 kΩ, and 8.5 kΩ, respectively. We ran the circuit through a 9 V power supply, charged the capacitor by closing the left half of the circuit and discharged it by immediately closing the right half. and recorded the data on Logger Pro. Our results were:
As expected, the capacitor charges at 20 s and discharges in 2 s.
Follow-up Questions
1. During charging,
RTh = Rcharge * Rleak
/ (Rcharge + Rleak)
= 85 kΩ * 992.5 kΩ / (85 kΩ + 992.5 kΩ)
= 78.3 kΩ
2. During discharging,
RTh = Rdischarge * Rleak / (Rdischarge + Rleak)
= 8.5 kΩ * 992.5 kΩ / (8.5 kΩ + 992.5 kΩ)
= 8.4 kΩ
3. When V = 0.6321 * (final voltage), τc = 4.0 s. Thus, R = τc / C = 85.11 kΩ
The value is indeed close to the value in question 1.
Practical Questions
1. C = 2E / V2
= 2 * 160 MJ / (15 kV)2
= 1.42 F
2. If the capacitance is arranged in the schematic, the value of each capacitor, C, is 0.71 F
Monday, May 20, 2013
Oscilloscope 101
This exercise was to familiarize ourselves with the use of the oscilloscope and to study and adjust different waveforms on the frequency generator.
Exercise 1
We began by displaying a simple sinusoid on the oscilloscope with the following setup:
We measured the DC and AC voltage with the DMM.
VDC = 5.35 V
VAC = 2.45 V
Exercise 3
We measured the DC and AC voltages.
VDC = 5.2 V
VAC = 5.13 V
The expected VAC value for a square wave is 5 V, which is similar to the value we measured with the DMM.
Mystery Signals
We were given an unknown signal generated by the frequency generator and attempted to trigger on the signal. However, the limitations of the oscilloscope prevented us from fully triggering on the signal. The output is as follows:
Exercise 1
We began by displaying a simple sinusoid on the oscilloscope with the following setup:
According to the scope,
The period was at 208 microseconds.
Peak-to-peak amplitude was 10 V.
Zero-to-peak amplitude was 5 V.
Anticipated RMS value was 3.54 V.
We measured the VDC and VAC RMS values and got the following:
VDC = 6.2 mV
VAC = 2.34 V
The VAC value was close to our anticipated RMS value.
Exercise 2
The next exercise was to add a 5 V offset to the existing waveform.
VDC = 5.35 V
VAC = 2.45 V
Exercise 3
We measured the DC and AC voltages.
VDC = 5.2 V
VAC = 5.13 V
The expected VAC value for a square wave is 5 V, which is similar to the value we measured with the DMM.
Mystery Signals
We were given an unknown signal generated by the frequency generator and attempted to trigger on the signal. However, the limitations of the oscilloscope prevented us from fully triggering on the signal. The output is as follows:
The signal was a triangle pulse with the following:
DC voltage: 0.5 V
Frequency: 7 Hz
Pk-pk Amp: 1 V
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