Question

In the circuit shown, a center-tapped transformer is used to create a full- wave center-tapped configuration. Note that half of the secondary voltage appears across each half winding. 120V : 7.8V 120 Vrms) 60 Hz of 7 R } Vout a. Find the maximum value of Vout. Note that half of the total secondary voltage appears across each half of the secondary. b. Suppose the load resistance is 10 ohms. Find the required filter capacitance so that the peak-to-peak ripple voltage will be 0.1V. What is the RLCF product in milliseconds? c. Now suppose the load resistance is changed to 20 ohms. What is the new value of filter capacitance to obtain a peak-to-peak ripple voltage of 0.1V? What is the associated RCF product and how does it compare to the value from part b? d. Finally, suppose that the load resistance is kept at 20 ohms and the filter capacitor is changed to obtain a peak-to-peak ripple of 0.2V. How will the filter capacitance and RLCF product change compared to part c?

Answer 1

SOLUTION:

solution @ Total Rms Secondary voltage = 7.8V peak Secondary voltage -sqrt (2) X 7.= 11.0V Peak voltage across half the secondary winding = 11.0/2=5.5v de voltage after the diode drop ? Therefore maximum 5.5-0.7=4.80 © Vycpp) téc x[Vpeat -0.7) where fa ripple frequency , RL = load resistance 6c - filter capacitance.

Alote that ripple frequency is twice the line frequency in case of a felll - wave rectifieu. if Uripp = 0.1, then os Taxbbyiotice X (5.5-07) OF = 4.8 / 120X1080.1 = 4.8 10.04 fasad 126 =40,000 micro forad and RiXCE = o.oux 10-0.4 second . @ the new Ef value is a cf = 0.81(120x20x0.1) = 20,000 micro fasad and REXEC 20.024 2020.4 sec @ 0.2=4.8 | (120 X 200CF) cf = 4.8 120X20X0.2.) – = 0.01 farad Hence CF X RC = 0.01 X20 = 0.2 sec.

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