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What does input and output impedance (I/O) in electronics mean? This is in my book but I didn't understand it. Please explain it to me

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  • \$\begingroup\$ ratio of input voltage/current and output voltage/current \$\endgroup\$
    – tobalt
    Dec 29, 2021 at 9:05
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    \$\begingroup\$ Does this answer your question? What is input and output impedance of an opamp? \$\endgroup\$
    – JYelton
    Dec 29, 2021 at 17:50
  • \$\begingroup\$ @JYelton thanks a lot \$\endgroup\$
    – user312605
    Dec 29, 2021 at 18:40

1 Answer 1

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Simple simple explanation without diving into too technical terms.

Suppose you have a DC "output", with a given impedance.

If you leave it open (no load) and measure its voltage, it will have, say, +5 volts.

When you connect that output to "something", those 5 volts will reduce. If that "something" has a high impedance (impedance is similar to resistance), the signal will decrease little; if that "something" has a lower impedance it will decrease more. If that "something" is a short-circuit it will have zero impedance and the signal will became zero.

The input impedance is the load the input imposes on the signal you send to it: lower the impedance, bigger will be the load and the signal will decrease more.

That happens if the signal is generated by an output with a certain impedance. If the output has zero impedance, it will always win, and the signal will not decrease.

One could think that an output having 0 impedance is the best, but this is not possible. And one may think that an input with very high impedance is the best, and this is in fact quite possible, but not always desirable: an input with very high impedance is very sensitive to noise.

Given that both input and output will have some impedance, the optimum generally is that they match.

You can think at an output as a voltage generator with a series resistance. That resistance is the impedance. When no current flows (no load), the output signal will have the full voltage of the generator because no voltage drops on the resistor. But if you put a load, current will flow and the resistor will drop some voltage (and dissipate energy).

You can think at an input as an infinite-impedance input with a parallel resistance. That resistance is the impedance: if you apply voltage on that input, some current will flow in the resistor and, if the generator of the signal has some internal impedance (resistor), voltage will drop there and consequently will not reach the input.

I talked about resistance which, I think, it is simpler. Impedance is about the same but more general and takes into account AC signals.

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  • \$\begingroup\$ @tobalt thank you so much \$\endgroup\$
    – user312605
    Dec 29, 2021 at 18:38

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