﻿ Load Voltage In A Half-Wave Rectifier

Load Voltage In A Half-wave Rectifier

Strings (SiPjAjk) = S7P5A51     Base Sequence = 12735     String Sequence = 12735 - 5 - 51

Expressions Of Pj Problems
Load Voltage In A Half-Wave Rectifier
Math AC retification (change of AC signal to DC signal) is one of the important diode applications.

Circuit 121.8 is a half-wave rectifier. Determine the DC value of the rectified waveform if the AC voltage source, vs = 52cosωt V. Assume ideal diode.

The strings: S7P5A51 (Physical Change).

The math:
Pj Problem of Interest is of type change (physical change). The half-wave rectifier is a change device. The positive half of an AC signal is changed to a DC signal. Diode of a half-wave rectifier conducts only during the positive half cycle of the AC signal
So, vD ≥0 only when AC source voltage is positive.
Diode acts an open circuit during the negative half cycle of AC signal.
So, vD <0 when AC source voltage is negative.
iD = vs/RL when vD >0
So, vL = iDRL = vs
Average value of load voltage (DC load voltage) is obtained by integrating load voltage over one period, T and dividing by the period, T.

So,DC load voltage vL = (1/T)∫0 T 52cosωt dt
Where T = 2π/ω

So, vL = (ω/2π)2[∫0 π/2ω 52cosωt dt]----- since negative AC half cycle is zero.

So, vL = (ω/π)[(1/ω)[52sinωt]0 π/2ω
So, DC load voltage vL = 52/π =16.55 V. The point . is a mathematical abstraction. It has negligible size and a great sense of position. Consequently, it is front and center in abstract existential reasoning.
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