1
\$\begingroup\$

It is said that the input impedance of a buffer must be high enough so that it can isolate its input voltage and carry it to its output for driving purposes.

What I do not understand is why we treat all the independent sources as shorts while calculating input impedance. I know that input impedance is the impedance seen by input voltage present at buffers input.

There is, however, a huge difference between treating independent sources as shorts while calculating input impedance, and not doing so. Someone on this website said you can try finding input impedance without treating independent sources as shorts and get the same result as if you did.

I do not believe that. Can you enlighten me on the use of input impedance and the logic behind its calculation?

Input impedance found by treating sources as shorts cannot be used as an impedance replicate of a circuit connected to some other circuit which we try to solve without writing lots of node equations and therefore has no practical use. Also I looked at posts regarding it on this website and none answered my question.

\$\endgroup\$
3
  • 2
    \$\begingroup\$ During calculation of any input impedance I do not "kill" anything. However, I take into account the fact that the internal resistance of a DC source is zero. That´s all. \$\endgroup\$
    – LvW
    Jan 14 at 8:08
  • 2
    \$\begingroup\$ "No DC sources were killed in the making of these calculations." \$\endgroup\$
    – ocrdu
    Jan 14 at 13:05
  • 1
    \$\begingroup\$ I have always considered it as an application of the principle of superposition: you consider the effect of the 'probing signal' applied to a port, when all other (independent) sources are off. i.e. voltage sources are shorted and current sources are opened. \$\endgroup\$ Jan 14 at 13:15
1
\$\begingroup\$

What I do not understand is why we treat all the independent sources as shorts while calculating input impedance.

Hopefully, this answer will make the reason clearer.

Someone on this website said you can try finding input impedance without treating independent sources as shorts and get the same result as if you did.

They are correct.

When we refer to the input impedance of a circuit, we are (almost always) referring to the small signal input impedance. That is, the input impedance gives us the ratio between how much the current will change if we make a small change in input voltage or vice versa.

$$Z_{in} = \frac{\Delta V_{in}}{\Delta I_{in}}$$

or, using the terms of calculus

$$Z_{in} = \frac{dV_{in}}{dI_{in}}$$

If we are given a circuit, such as this:

schematic

simulate this circuit – Schematic created using CircuitLab

We can calculate the relationship between the input voltage and the input current as follows.

$$I_{in} = \frac{V_{in} - V_{internal}}{R_{in}}$$

or

$$V_{in} = V_{internal} + R_{in}I_{in}$$

Differentiation gives us

$$Z_{in} = \frac{dV_{in}}{dI_{in}} = \frac{dV_{internal}}{dI_{in}} + \frac{d(R_{in}I_{in})}{dI_{in}} = \frac{dV_{internal}}{dI_{in}} + R_{in}$$

But since \$V_{internal}\$ does not depend upon \$I_{in}\$,

$$ \frac{dV_{internal}}{dI_{in}} = 0$$

So

$$Z_{in} = R_{in}$$

Which is exactly the impedance we would get if we simply treated the voltage source \$V_{internal}\$ as a short circuit.

If you understand how this works with one internal resistance and one internal voltage source, then it shouldn't be difficult to see that the same principal applies when there are a network of internal resistances and voltage sources.

The results we get "the long way" is equivalent to simply treating the internal voltage sources as shorts. Since it is also generally quicker to just do the latter, in practice, that is what we generally do. We treat internal voltage sources as shorts when calculating input impedance.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.