Hi,

Ive read the ebook on diodes but was wondering how the resistance goes so high so fast when reverse biased, so I measured some diodes on the meter for capacitance and got readings.

I applied the formula Xc = 1 / (2 pi f C ) f --0 = ∞ { DC = zero freq }

The formula was shown to me by one of the senior members.

IN4007 = 27 pF , Xc = 5,894,627,600 Ω

UF4007 = 42 pF , Xc = 3,789,403,400 Ω

This capacitance could only be because of the separation between the P type material and the N type material and the depletion region expanding and narrowing.

1 Does this explain the momentary open circuit condition when reverse bias ?

2 Is this what is contributing to stray capacitance ??

 

The method you used is a little unclear. Perhaps you could elaborate. Your results suggest you made your observations at a frequency of 1Hz???

I believe the modelling of diode capacitances is a rather more complex than solely attributing it to the reverse biased condition. There is capacitance in the forward bias case as well - such as that related to minority carrier diffusion. There's also things like the physical construction of the capacitor to consider.

Google might provide some enlightenment.

 

What momentary open circuit condition?
Diodes actually remain closed (on) for some time after applying reverse bias causing a lot of headaches in switching circuits.
http://archive.chipcenter.com/eexpert/akruger/akruger004.html