🚀 go-pugleaf

jon_sree@world.std.com (Jon Sreekanth) writes:

>A rather curious query : what is the audio frequency impedance of 
>a reverse biased diode ? 

>Off-hand, one would say, it's high, but just how high ? 

>Specifically, for a small signal diode like 1N4148, between 0 - 70 C,
>with about 2.5 V or more of reverse bias, is the small signal audio
>impedance at least > 1 Meg ? 10 Meg ? I don't need an exact number,
>just a minimum, to see if it interferes with other circuitry. 

You can calculate it from the diode equation by measuring the
saturation current for the diode in question--that will get you the
resistive component.  I think you will find that the most significant
component at audio frequencies, however, is the capacitance, which
will be in the low-megohms region above a kilohertz.

Diode equation:

    I=Isat * (e^(qV/kT)-1)

Differentiate with respect to V to find dI/dV; invert for small
signal resistive component.  Isat you can measure as essentially the
current at about half the reverse breakdown voltage, but as you can
see from the equation, the exact voltage you measure it at isn't
critical.  (V is negative at that point, and will be many times
larger than kT/q)  This should be fine for most signal and power
diodes, but watch your step for things like tunnel or PIN diodes...

In practice, resistive leakage paths will far outweigh the contribution
of the diode equation, also, at the voltages you mention.  V of -2.6
volts is about 100 times larger than kT/q at room temp, so you get
an e^-100 which is pretty small...  It may well come down to how clean
your construction is, as far as resistive components go.