AC Primary Voltage is 120 V with Transformer
Assume full wave rectifier is used to supply DC voltage to a load. Also assume that AC primary voltage is 120 V with transformer turns ratio of 10:1. a) How much DC voltage will be applied to the load?(include the diode drop in your calculations) b) How much DC voltage will be applied to the load if a bridge rectifier is used? (include the diode drop in your calculations)
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see the diagram.
AC voltages are specified in RMS with a transformer. The turns ratio = 10: 1. So the secondary coil voltage will be = 120/10 = 10 V rms. So the peak value will be = 10√2V.
a) two-diode rectifier
There is one diode voltage drop of 0.7V, to be subtracted from the peak value.
Vpk = 10√2 - 0.7 Volts
The V_DC at the load = V_avg = 2 * Vpk / π = 8.56 V DC
b) Bridge rectifier with 4 diodes.
Input Vpk to the rectifier = 10√2V.
There is one diode drop (from the peak Voltage) in each half cycle.
Peak Voltage across load will be : 10 √2 - 0.7 V
V_DC across load = 2 Vpk / π = 8.56 V DC
AC voltages are specified in RMS with a transformer. The turns ratio = 10: 1. So the secondary coil voltage will be = 120/10 = 10 V rms. So the peak value will be = 10√2V.
a) two-diode rectifier
There is one diode voltage drop of 0.7V, to be subtracted from the peak value.
Vpk = 10√2 - 0.7 Volts
The V_DC at the load = V_avg = 2 * Vpk / π = 8.56 V DC
b) Bridge rectifier with 4 diodes.
Input Vpk to the rectifier = 10√2V.
There is one diode drop (from the peak Voltage) in each half cycle.
Peak Voltage across load will be : 10 √2 - 0.7 V
V_DC across load = 2 Vpk / π = 8.56 V DC
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