What is the behavior of ideal voltage/current sources in small signal analysis?

[jinyong]

Can someone please explain why ideal voltage/current sources are short circuit for voltage source and open circuit for open circuit in small signal analysis? Any mathematical proof to this?

 

[berkeman]

I would think of it in terms of $$Z = \frac{dv}{di}$$ (although there may be other ways).

For a good voltage source, the output voltage is very stiff (doesn't change much) as the output current changes, so

$$Z = \frac{dv}{di} = \frac{0}{di} = 0$$

But for a good current source, you get very little change in the output current over a wide range of output voltages, so

$$Z = \frac{dv}{di} = \frac{dv}{0} = infinity$$


EDIT -- okay, I give up. How do you make the little infinity symbol in LaTex? "\inf" didn't work.

 

[waht]

Berkman is correct.

Voltage source can provide infinite amount of current. And a current source can provide an infinite amount of voltage.

Z = V/I

infty
$$\infty$$

 

[berkeman]

infty
$$\infty$$

Thanks what!