# Deriving AC resistance from Shockley's equation

Discussion in 'Homework Help' started by Protic, Oct 19, 2012.

1. ### Protic Thread Starter New Member

Oct 19, 2012
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I'm having problem while deriving equation for AC resistance from Shockley's equation. Can anybody please tell me how the equation d/dVD (ID)= d/dVD [Is (eVD/nVT-1)]

becomes this equation

dID/dVD=(1/nVT) (ID + IE) ?

2. ### WBahn Moderator

Mar 31, 2012
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4,699
Is this for a diode or a transistor? You are throughing around ID and IE. Please clarify.

It really shouldn't be referred to as AC resistance, but rather the small signal resistance.

You are used to Ohm's Law, in which

R = V/I

So R tells you how much V changes in response to a change in I. This is 'large signal" resistance because it doesn't matter how big V and I are.

But for a diode, it does matter. So we use the notion that we can break out signal into two pieces, a "bias" signal and an "incremental" signal (different names are used, as well) with the total actual signal being the sum of the two. The bias voltage is almost always considered to be constant (hence is generally referred to as the DC component) and the incremental signal are the deviations in the real signal from the DC signal. This is why it is so common to call it the AC component. But "incremental" or "small" are much better names because they emphasize that, which using this whole notion, it is important that the deviations be relatively small. How small? A lot of people will say something like, "Small relative to the DC signal", but this is completely false. The real answer is something like, "Small enough such that the deviation of the small signal model from the real behavior is acceptable."

But what does this mean? Well, the idea behind the whole approach is that, given any V-I characteristic, no matter how nonlinear (as long as we stay away from outright discountinuities), if we zoom in tight enough the curve looks like a straight line and, if we only operate along that portion of the curve, then the thing looks linear, meaning that we can use the slope (the derivative) as the incremental, or small signal, resistance:

v(t) = V_DC + i(t)*r

where V_DC is the DC voltage, i(t) is the small signal current (the difference between the total current and the DC current, which in turn is the curre(t)nt that corresponds to V_DC) and r is the incremental resistance given by:

r = dv/di

Protic likes this.

Oct 19, 2012
2
0
Thanks buddy