AC small signal model

So why is resistance of diode is infinite if it's reverse biased by a voltage source? The way I understand it, is that holes (positive) and electrons (negative) will be further apart in the diode, which will make the capacitance less ... ? So because of that resistance would be higher (infinite)?

Here's the context, for those who want to know more.


1 Answer 1


The resistance being "infinite" (its not actually, but its certainly very large) is a result of the DC behavior of the diode.

When you reverse bias a diode, the DC current flowing through it drops to basically zero. In the case of a PN diode, this is due to the lack of carriers flowing across the increased depletion region (until you reach breakdown). This is represented in the model by the resistor (a DC component) becoming infinite. However, as you said, the capacitance is still there and gets smaller because the depletion is wider (the AC component).

The AC impedance of the diode model never truly reaches infinity because of the capacitor, so to an AC signal there is still some current flowing in to the diode when it is reverse biased. If you think about it a bit, the current during reverse bias is produced by the depletion region expanding, a process that requires additional electrons or holes, but that current only occurs so long as the depletion region is actually changing. As it reaches its size for the applied voltage, the current will dwindle to near zero...similar to a capacitor charging. Hence, it can be represented by a capacitor in the model, whose capacitance is strong function of the applied voltage (which is very useful in oscillators since you can change the value of a component (the capacitor) simply by changing the voltage applied).

Normal ceramic capacitors also have a similar voltage-dependent effect, but I believe it's far weaker (requires larger voltage changes).


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