Wednesday, April 30, 2014
Sunday, April 27, 2014
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.
Consider the leak resistance of the capacitor. Compare the observed result with the calculated result.
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.
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.
Saturday, April 12, 2014
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.
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.
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