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3 answers

The capacitance causes leakage as the other answer says.

This capacitane creates problem in your calculations. Say, if there were only resistance in junction, then you can determine the performance of the junction at different operating voltages easily [hold temperature at fixed value].

But, capacitance causes the frequency dependance. So, the junction impedance becomes non-linear with the operating frequency. So, calculations are complicated.

2006-11-17 20:03:56 · answer #1 · answered by Maverick from the sky 2 · 0 0

Same as the problem caused by any stray capacitance in a structure: the capacitor acts in parallel to the junction. If the diode is used to rectify AC current, the current can bypass the junction through the capacitor during reverse bias. The cutoff frequency, at which the bypass current equals .707 times the reverse current of the diode is 2πRC, where R is the reverse resistance of the diode. Since R is very high (tens of megohms) it doesn't take much C to cause current leakage at modest frequencies.

2006-11-17 20:01:48 · answer #2 · answered by gp4rts 7 · 0 0

Unwanted high frequency leakage.

2006-11-17 18:34:06 · answer #3 · answered by LeAnne 7 · 0 0

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