Engineering How Does Reversing the Diode Affect a Half-Wave Rectifier Circuit?

AI Thread Summary
Reversing the diode in a half-wave rectifier circuit alters the output voltage behavior, resulting in Vo being zero for positive input and following the negative input. The average output voltage can be calculated using the formula Vo,avg = (1/pi)Vs + (Vd/2), where Vs is the peak voltage and Vd is the diode forward voltage drop. The Peak Inverse Voltage (PIV) is determined to be equal to -Vs, as the rectifier only processes the negative half of the input sinusoid. The discussion highlights confusion regarding the derivation of the average voltage formula and seeks clarification on the calculations. Understanding these concepts is crucial for accurately analyzing rectifier circuits.
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Homework Statement


Given a Half wave rectifier circuit with the diode reversed . Vs is a sinusoid with 12 V peak , R = 1.5 ohms Vd = 0.7 V
find the average Value of Vo , and the PIV of the Diode .

Homework Equations


average Value of Vo = (1/pi)Vs - (Vd/2)


The Attempt at a Solution



i calculated Vo for circuit and got Vo = (0.7 + Vs) for Vs < -0.7 and Vo = 0 for Vs > -0.7
now i have no idea how to calculate the average value, the equation stated above was just given in the textbook with no explanation as to how they got it, but i did take an educated guess and said Vo,avg = (1/pi)Vs + (Vd/2), also i said that PIV = -Vs since this rectifier rectifies only the negative part of the input sinusoid. am i on the right track ? thanks in advance for your help
 
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