Rectifier Circuit Operation (Cont’d)What voltage waveform would we observe across the diode vD?
Voltage Transfer Characteristic of the Rectifier Circuit
The source vS is a sinusoid with a 24 Vpeak amplitude. The diode is used to chargea 12 V battery and the 100 resistor limits the current.We want to determine the fraction of each cycle during which the diodeconducts. We also wish to find the peak-value of the diode current and themaximum reverse-bias voltage that appears across the diode.
SolutionWe reason that the diode will not begin to conduct until vS exceeds 12 V. Likewise, thediode will stop conducting once vS again reaches 12 V.This defines the conduction angle, 2 (i.e., the fraction of the sinusoidal cycle over whichthe diode conducts). Hence, we reason that the diode will conduct provided that 24V sin 12 V. Or 30o ≤ ≤ 150o (or 1/3 of a cycle). 24V 12VThe peak value of the diode current is I D 0.12 Amps 100
Solution (Cont’d)The maximum reverse-bias voltage that appears across the diode is simply: 24 V + 12 V = 36 V
Let’s begin by assuming that diodes D1 and D2 are conducting: 10 0 I D2 1mA 10k KCL (node B): 0 ( 10) I 1mA I 1mA 5kFor this second circuit the same assumption gives nonsensical results (verify this) We assume that D1 is OFF and D2 is ON. Consequently, VB = 3.3 V (verify).