Frequency Response

Summary: connect a capacitor and a resistor in series, provide a AC source with various frequency, measure the rms voltage of both components.

Voltage measurements:

 

 

Summary: at low frequency, the capacitor produces a high reactance and has a high voltage. At low frequency, it acts like a short wire.

Online Tutorial Exercise

Capacitor Charging/Discharging

Capacitor Charging/Discharging

Summary: Set up a circuit for charging and discharging a capacitor. Use corresponding resistors to control the charging/discharging time. Monitor the voltage change with a oscilloscope and determine if the actual charge/discharge time is close to the desire value.

Schematic.

Source voltage=10V, charging interval is 20s, discharging interval is 2s. The energy stored is 2.5mJ. Time constant is 1/5 of time interval. Calculate the proper resistance and capacitance, verify if the values meet actual components' power restriction.

 

Connection diagram and some measured value of devices.

Final set up.

Voltage change curve caught by oscilloscope. Final voltage after charging is 9.24V. Both charging and discharging interval are close to designed value.

Charging

Discharging

Consider the leak resistance of the capacitor. Compare the observed result with the calculated result.

 

 

Notes: The overall process meet the designed expectation; however, the Thevenin resistance has great difference. Maybe the graph of voltages-time is not very accurate.

 

Week 8, Practical Signal Conditioning

Week 8, Practical Signal Conditioning

Summary: The out put of a LM35 is 10mV/Celsius degree. Design an op amp circuit which converts the output of LM35 to 10mV/Fahrenheit degree.

Test the LM35 and make sure it has a good output voltage.

The overall design is changing the LM35 output with a difference amplifier. Notice i0 should be less than 10mA according to the data sheet.
i0=Vref/R1.

Design calculation.

Vref=-0.4V is provided by a voltage divider. The complete schematic is:

Final setup. Before connect the circuit, both power supply were set to 10V, and the 10k pot was adjusted to give a -0.4V.

Results: the output of LM35 was 220mV, the output of the op amp was 720mV. 1.8*220+320=716mV. The circuit is giving the expected result. The multi-meter said the room temperature was 24 degree(240mV-752mV). The percent of difference is about 8%.

Notes:
1. R2/R1=0.8181 rather than 0.8 because the availability of resistors.
2. The output relationship between LM35 and the op amp is almost the same as calculation result; therefore, the circuit design was good. The percent of error completely depends on the performance of the LM35.

Week 5, Operational Amplifiers I

Week 5, Operational Amplifiers I

Summary: Simulate a signal amplification circuit with an op amp. The adjustable signal voltage is replaced by a resistor voltage divider. The output of the op amp is the amplified signal output. The gain is designed to be 10 and all components should satisfy their power limit.

Schematic. The 5V power supply is replaced by one of the 12V.

Calculation for devising the circuit.

 

Components used for the circuit.

 

 

Final setup.

Collected data. The gain is close to 10.

 

 

Measure the power supply current to check if they satisfy the designed current limit. Because I(v1) support both the op amp and the circuit, it is much large than I(v2).

Follow up calculation.

Notes:

1. Because only 1/8W resistors are available, all resistor are designed to satisfy 1/8W limit. Also, because 1k resistors were not available, the Ri and Rf were 2.2k/22k rather than 1k/10k.
2. I measured the total current of power supply one, while I was supposed to measure the current going into the op amp. The KCL was not able to be verify.
3. The gain is about 5% to 10% different to 10.
4. If Rx and Ry are smaller, the result should be better because the amplification circuit has less effect on the series voltage divider.

Week 4, Thevenin Equivalents

Week 4, Thevenin Equivalents

Summary: To compare the output of a linear circuit and its Thevenin equivalent circuit.
1. Build the Thevenin equivalent circuit and measure the voltage cross the terminals of a load resistor.
2. Build the original linear circuit and measure the same voltage. Also, change the load with a factor of R(Th) and verify the load.

The schematic of the original circuit.

The schematic of the Thevenin equivalent circuit.

Calculate the theoretical open circuit voltage, expected voltage with  minimum load(8 V), V(TH), R(TH), for devising the equivalent circuit and comparing the measured values and theoretical values.

Set up the Thevenin equivalent circuit and measure voltage with different load.

Set up the original circuit and do the same measurement. Calculate the power of different load.

Notes:
1. Comparing the Thevenin equivalent circuit and the original circuit, both the open circuit voltage and the voltage with load are very closed, indicating that they are actually equivalent.
2. When R(L)=R(TH), the power of the load is maximum. The result from the last chart agrees this theorem.

Week 3, Transistor Switching

Week 3, Transistor Switching

Summary:
1. Setup a simple circuit with a transistor, switching base current on and off and check the result on the emitter side. Then, use finger tip as the base resistor and try again.
2. Collect current reading from base current and emitter current. The result is expected to be a linear relationship.

Schematic:

Final setup. R1 is the resistor box.

Switching the base current on and off by touching wires. Then replace R1 with finger tip.

Schematic for finding the relationship between base current and emitter current.

Setup:

Data.

Graph. The last point is not used because it is far away from the line.

Notes:
1. The base is the "switch" for the emitter. If there is no current go in the base, the "switch" is off.
2. If the base current if within a certain range, the emitter current is in proportional with the base current. The graph above shows this property; however, the received beta value is far away from what it is supposed to be.

Week 3, Nodal Analysis

Week 3, Nodal Analysis

Summary: Use nodal analysis to find the voltage of a circuit. Construct the circuit, measure voltage and currents, and compare the measured values with the theoretical values.

Schematic
         

Equations for nodal analysis. The theoretical values for V2 and V3 are at the bottom.

Other theoretical calculation for the circuit.

 The actual values of the resistors and DC sources.

Final setup. The positions of all components are the same as they are in the schematic.

Measured values. Except for I(Bat2), all other values are very close to their theoretical values. See notes for the big percent error of I2.

If V2=V3=9V, plug in the equations above one gets V(Bat1)=9.9V, V(Bat2)=10.98V. Using 9.9V and 11.0V, the measured results are very close to the calculated results.

Notes: The reasons for the big difference of I(Bat2) may include:

1. The actual values are slightly different from the desired values. The calculated value will be 2.06mA if one uses all the actual values. The percent of error will be within a reasonable range.
2. The 1.5mA is a very small number. Usually meters are not as accurate in small scale as in large scale.