# What does impedance seen from input/output mean?

I am seeing it usually like impedance seen from input/output when we need to match impedances between two circuits. What does it even mean? Does impedance change based on where do you look from?

• Input and output are typically different devices at opposite ends of a cable. For example you might have a 50 Ohm function generator output connected to a 1 Meg Ohm oscilloscope input. Commented Jul 12 at 14:43
• substitute the word seen, with the word measured Commented Jul 12 at 14:43

Does impedance change based on where do you look from?

More like, impedance changes where you look into.

Imagine a black box, with several ports on it. Each port can have a different impedance. If you've not heard of ports before, then a port is a pair of terminals. Often, one of those terminals is ground, but not necessarily.

You can measure the impedance of a port by pushing a current into it, and seeing what voltage is generated there as a result. Equivalently, put a voltage across it, and see what current is drawn. The impedance is the ratio of voltage to current. Depending on what's inside the box, one or the other method might be easier to perform, or to conceptualise. You can either measure the absolute output impedance by using the actual voltage and current, or the dynamic / differential / incremental impedance by using the change in voltage and change in current.

If each port had its own isolated resistor behind it, then the impedance of the ports would all be independent, and easy to work by by reading the resistor value. But that would make for a pretty uninteresting black box. Usually the ports will have interconnections and components between them. Then the impedance of one port would also be dependent on what you had connected to the other ports, whether they were open, short-circuited, connected to some finite load, or even connected to each other.

If the black box was an amplifier, with an input port and an output port, then you would estimate the port impedances as above by injecting a signal current into them, and measuring the voltage developed. With a good isolating or buffer amplifier, you would find the impedance of one port would be essentially unaffected by the loading conditions on the other ports. If we regard the power supply to the amplifier as another port, then we would know we had a very flaky amplifier design indeed if the loading conditions on the power supply port affected the signal port impedances.

• Have a look at the Howland current source question I've just posted, where I use precisely this method to measure the output impedance of an output port. Commented Jul 13 at 9:58

Imagine a simple resistor divider which divides input voltage by two, with for example two 10k resistors.

If you apply a 10V battery on input and measure the output with multimeter, you see a 5V output.

The input is just two 10k resistors in series to ground. Impedance of the input is thus 20 kohms. You can disconnect the battery and verify with multimeter that it is 20 kohms.

But the input is driven by about 0 ohms of the battery, assume it is ideal.

So the battery sees it is loaded by 10kohm impedance while the divider input sees it is driven by 0 ohm impedance.

When there is 5V on output, there is 10k to ground and 10k to the ideal battery. The output impedance is thus 5 kohms. You can verify it with multimeter, if you measure short circuit current of the output, you only get 1mA current, as there is only 10kohms to 10V. It matches the case if there was a 5 V source with 5kohm source impdance.

But the output drives nothing, it is open or unterminated. Full 5V voltage with no current into any load.

So the next device would see it being driven by 5kohm impedance, but there is no next device. The output will thus see infine impedance.

Or the output would see zero impedance when it is shorted to ground.

So yes, it does depend from the viewpoint of the terminal or connector. The impedance of port X or Y is different than the impedance port X or port Y sees by the other party.

• This is very confusing without a diagram and I understand what you're trying to say. Be better to edit it and add a diagram. Commented Jul 12 at 17:40