Semiconductor Diodes: Calculating Rf and Why VD=0

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The discussion centers on calculating the resistance (Rf) of semiconductor diodes in forward bias, where Rf is derived from the diode's forward voltage and current. It highlights the assumption of an ideal diode, which simplifies calculations by considering it as a short circuit with a forward voltage drop (VD) of zero. However, in practical applications, such as rectifying 30 volts AC, a small voltage drop of 0.6 volts can be negligible. The conversation also touches on the importance of measuring actual forward voltages when not assuming ideal conditions. Understanding these concepts is crucial for accurate diode performance analysis.
Ghassan99
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Semiconductor dioedes !

when proving that the ideal diode is short circuit in forward biasing we calculate the diode Resistance Rf which is equal to the diode forward voltage upon it's forward current ...but why we put VD=0 why?
 
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These are two different things.

An ideal diode is one that nobody has invented yet, but we assume we have one just to make the other calculations easier. If you are using a diode to rectify 30 volts AC then a voltage drop of 0.6 volts may be a trivial error which can be ignored.

If we don't want to regard the diode as ideal, then we can measure the forward voltages in the way you described.
 


Thank you :)
 
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