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Electronic Measurement and Testing Assignment

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Added on  2020-05-11

Electronic Measurement and Testing Assignment

   Added on 2020-05-11

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Running head: Electronics measure and test1Electronics measure and testAuthorCourse: ElectronicsDate
Electronic Measurement and Testing Assignment_1
Electronics measure and test2Task 1AAt 1khz frequency, input and output voltages as shown in the images below are;Input voltage = 5 volt/div*2div=10volts peak to peakOutput voltage= 2 volts/div*4div=8volts peak to peak Cut off frequency calculationsFc= 12πCR= 12π∗0.1μ∗1k= 1591.55HzWhen function generator frequency was set to cutoff (1591.55 Hz) the input and output voltages are;Input voltage = 5 volt/div*2div=10volts peak to peakOutput voltage= 2 volts/div*3.3div=6.6volts peak to peak
Electronic Measurement and Testing Assignment_2
Electronics measure and test3From the results of the two cases above, it was noticed that there is variation in the amplitude of the capacitor voltage, as frequency of the signal is changed. This is attributed to change in capacitive reactance, and capacitive reactance varies inversely with frequency. The resistor value will remain constant because it’s frequency independent. Capacitive reactance is greater than resistive impedance at low frequencies thus the capacitor voltage will be greater than the resistor voltage drop, which from our results was found to be 8 volts pp. At cut off frequency which is the highest frequency for this circuit( frequency above cutoff is attenuated) the capacitive reactance is small compared to restive value of the resistor, and the capacitor voltage is expected to be small which from the results was found to be 6.4 volts peak to peak.
Electronic Measurement and Testing Assignment_3
Electronics measure and test4Task 1B1)Time constant (τ) = RCChosen resistor value is 10Kohms and capacitor value is 470uFτ =10000*470*10^-6 =4.7Time at which capacitor voltage will be 90% of maximum value is-t=4.7{exp(1-(0.9*12)/12)}=10.8secsTime at which capacitor voltage will be 90% of maximum value is-t=4.7{exp(1-(0.1*12)/12)}=1.72secs2)On switching the SW1 to connect 12volts supply the above image was obtain. And the capacitor voltage was found to be , 2 volts/div * 4 div = 8 voltsCalculated value for maximum capacitor voltage, Vc=12[1-exp(-3.49/2.2)]=9.5 voltsRise time is the time taken for the signal to cross minimum voltage (10%) by maximum voltage (90%)
Electronic Measurement and Testing Assignment_4

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