The circuit shown in Fig. P7.20 is a battery charger. The current flows when the

Question

The circuit shown in Fig. P7.20 is a battery charger. The current flows when the instantaneous voltage out of the transformer exceeds the battery voltage plus the diode threshold voltage. As the battery is charged, its voltage increases until the current stops flowing. For this problem, however, consider the battery voltage constant at 12.6 V. Consider that it takes 0.7 V to turn ON the diode. Find the peak value of the secondary voltage Vp. such that the peak current in the battery is 10 A. With this value for Vp, what percent of time is the diode conducting? What is the peak inverse voltage that the diode must withstand? figure P7.20

Explanation / Answer

a)Apply KVL in the secondary loop. The secondary winding voltage must be equal to the voltage drop in the diode, resistor and the battery. V(peak)= V(diode,on) +R*i(peak) +V( battery) V(peak)= 0.7 +0.25*10 +12.6 =15.8 V b) With this value of V(peak), the voltage at the secondary will oscillate between +15.8 to -15.8 sinusoidally. And the diode id conducting when the secondary voltage is greater than V(diode) +V(battery), i.e secondary voltage >=13.3V Hence 15.8cos(wt) = 13.3 hence wt = cos-1(13.3/15.8) wt = 32.67 deg, 327.32 deg; When the phase changes from 327.32 to 360 i.e (for 32.67 deg) and then again from 360 ( equal to 0 deg) to 32.67 deg, the diode is conducting. Hence total conduction occurs for 32.67*2 = 65.34 deg out of total 360 deg of phase in a cycle. so percentage of time the diode is in conduction = (65.34/360)*100 =18.15% c)When the voltage in secondary reaches the negative peak value then the diode is switched off and it experiences a maximum inverse voltage across it. This is peak inverse voltage given by: v( diode) = V (secondary)- V(battery) = -15.8-12.6 = -28.4V ( peak inverse voltage) The negative sign indicates that the voltage at cathode of diode is higher than the voltage at anode.

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