Question

Mヴ CA. Center Tap Rectifier D/ l00 Vo Turns, Vin lIOVAC 60H2i0 Turns io Turns D2 2.7 t, m s 2 2. Vo t, m s ) Vin P-k , Vomın , ,,Vo ripple D e termine : -T/2, т, т, 27 OP-kぅ Draw Vot) curve, indlicate values Show your work in detaite

Answer 1

The solution is given in following 8 pages
"Let us think about the transformer part first As V_1/V-2 = N_1/N_2 Now the wave form of V_in V_in = 110 v (rms) V_in pk = 110 times squareroot 2 = 155.56 v \r\n This is a 60 Hz system so, T = 1/60 = 0.0166 sec = 16.66 ms Ultimately waveform of V_in T/2 8.33 ms T = 16.66 ms 3 T/2 24.99 ms 2 T 33.32 ms same for V_0 \r\n Now about output voltage Think separately V_C = V_D when the instantaneous voltage of secondary winding V_in' or V_n"" is higher than V_c (or V_0) D_1 or D_2 conducts. When instantaneous voltage falls below V_c (capacitor voltage) diodes are reversed biased and capacitor are discharged through resistor (R). \r\n Resulting capacitor voltage V_c varies between maximum value of V_m (V_m = 11 times squareroot 2 = 15.55 v) and minimum value of V_m - V_r(pp) V_rpp is peak to peak ripple voltage In practice V_r (pp) is "