Why is input and output impedance important when analyzing electronic circuits?

Hello


I want to Know the reason for being interested in the input and output impedence of the circuits we analyse

The input and output impedances are important because of the reason of voltage division. Voltage gets broken up in a circuit depending on the impedances present in the circuit. For voltage to fall on a device, the device needs high impedance.

For full detail, check out:

http://www.learningaboutelectronics.com/Artic...gh-input-impedance-and-a-low-output-impedance

-David
www.learningaboutelectronics.com
www.allthingsdiscussed.com

There's a number of important reasons we are interested in the input and output impedance of a circuit. The impedance of a circuit needs to be known to match associated and connecting circuits. In the case of audio or high frequency signals being input into a circuit, if the impedance of the driver into the circuit isn't matched with the input impedance of the circuit, then you will get reflections and loss in power (which could result in lower volume in audio applications).

Another reason to know the impedance is if you have a number of the same circuit, all being driven by the one signal, then you will need to know if that signal is being driven strong enough to see clean transitions on each individual circuit. As the number of devices that the signal is being driven into increase, so will the power needed to drive this signal.

The same applies for the output impedance. Hope that helps you.

the amount of signal transmitted from one stage to another stage depend on input and output impedance.

A distinction needs to be made between maximum voltage transfer and maximum power transfer and even maximum current transfer.

David is talking about maximum voltage transfer.

Mike is talking about maximum power transfer (and the resultant lack of reflected power).

Then there is maximum current transfer: high impedance output into low impedance input (e.g. current source/sink).

I'm going to try an analogy that may be a stretch, and other commenters may follow on and say "wrong, Wrong, WRONG!", but here goes:

If you've ever sloshed water in a bathtub, you know it hits the flat end under the faucet and splashes right back.
The other end usually slants a bit. But what if it really, really slants, like 10 feet or so?
Can you picture the same size wave that would bounce back from the flat end will instead slide way up the slant, and barely come back as a ripple?

It's all about energy transfer. The flat end shows a high impedance to its wave and reflects most of the energy back,
while the slanted end dissipates its wave's energy, in this case turning it into a tiny amount of heat.

When electronic waves reach the end of a wire, cable, whatever, the circuit at the endpoint either accepts their energy efficiently or bounces it back to some degree or another, depending on how well the cable impedance matches the receiver circuit.

Unlike water in a bathtub, this matching is very dependent on frequency - the analogy isn't perfect!

It's an excellent analogy. The flat end reflects "voltage" waves--i.e. waves with peaks and valleys that represent a variance in pressure, whereas the slopped end reflects "current" waves--i.e. waves that slosh in the horizontal direction with lots of horizontal velocity (or _current_). Either of these extremes results in a rapid dissipation of the energy. I wager that there is an optimal slope that would match the impedance of water and thus lead to the longest period of wave activity in the tub.

That would be the optimal transfer of power (i.e. the power in the water wave would be optimally transferred to traveling up the ramp and then the energy of the water traveling down the ramp would be optimally transferred back to the tub.

So, if the objective is to move a float up and down with the greatest degree of vertical travel, then the flat termination (or high impedance) solution would be optimal. If the objective is to generate rapid horizontal flow, then the highly sloped termination (or low impedance) solution would be best. And if you want the water to slosh around in the tub for a long time, then the medium slope solution would be prime.

Thanks for the props, Steve.

The whole voltage / current thing makes my ones and zeros conditioned mind want to run and hide.

Once in a physics class, the prof put this huge 2 or 3 foot coil in series (parallel?) with house current and a 100 watt light bulb. Then he gradually inserted an iron bar to intensify the coil's magnetic field. The light dimmed and finally went out.

"The current is now totally out of phase with the voltage," he explained. Say what?!? Give me an if-else statement every time ;^)

Gotta ask: I'm pretty sure that 2 perfectly vertically flat tub ends would reflect the water 'voltage' wave back and forth a really long time, so I don't see a very rapid dissipation of the energy in that case.

What am I missing? Are you sure about the medium slope sloshing the longest? Have we stretched the analogy past its breaking point? Can we commission a bathtub manufacturer to test this for us?

Nope, I'm not sure ;)

I would love to see an experiment--beyond my current resources, though.

Maybe we can get the mythbusters to do it!