Chuck,
On page 16 of? ?
there's an equation for the capacitance of a reverse biased diode
? Cj =? ?A/2 * ((2*q*e/(V0-V))*(Nd*Na/(Nd+Na)))**(1/2)
From that, it appears that the junction capacitance is inversely proportional
to the sqrt of the voltage applied (voltage measured relative to V0)
Which to my eye, agrees with those curves.
If you happen to be exceptionally curious, the derivation is in the pages prior.
And if you're lucky, he defines some of the variables and constants in lecture 26.?
Extra credit:?
So, as the voltage across the diode decreases, the change in capacitance due to some delta-V (voltage noise) increases.
But since we are using a resistive divider to create the voltage across the diode,
any noise in the power supply voltage is reduced proportional to that resistive divider.
Which one wins?
Jerry
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On Thu, Dec 19, 2019 at 12:12 PM, Chuck Carpenter wrote:
Jerry, the shape of the curve in fig 3 is typical of?junctions of non?"purpose built" diodes. (like that term).
I've attached a set of curves from K7QO.? The curve at the bottom is a power diode.? Note that there is not much change from about 5 V so higher voltages don't help much.
--
Chuck, W5USJ (ex K2OFN)
Point, Rains Co, TX? EM22cv
