What are input and ouput impedances?

[PiraxaStudios]

While I do know that instruments and hi Z microphones have a high output impedance (same unit, Ω and resistance), and regular microphones use lower output impedances, this doesn't help me much in the world of outboard gear like compressors, EQs and preamps - all of which use open circuitry. I recognise that in these pieces of gear, there's going to be a low input impedance, and a high output impedance (measured the inputs and outputs of my TL-C1 valve compressor, and sure enough, ≈6kΩ on input, and ≈1MΩ on output).

What I don't understand though, is what part of a "black box" part of a unit's schematic stops being the input, and starts being the output. I have heard that it's down to what part of the circuit is active, but if that's the case then there will be loads of different input and output stages in something like a compressor, due to all the transistors and opamps in the signal path.

I can read schematics, but have a limited understanding of electronics having last done it at A-Level in school 10 years ago (and we never learnt anything about AC, so this will be fun)

 

[Whoops]

While I do know that instruments and hi Z microphones have a high output impedance

It's not "Instruments" per se, that a wrong generalized statement.

Passive Pickups instrument pickups have an High output impedance.
Piezo Pickups are capacitive and also benefit from an Hi-Z input
So instruments that use those type of pickups will have Hi-Z output

But instruments that use "Active Pickups" will have a Low-Z output (for example a guitar with active pickups)

But there's many instruments that have lower output impedance like most keyboards and drum machines.

in the world of outboard gear like compressors, EQs and preamps - all of which use open circuitry. I recognise that in these pieces of gear, there's going to be a low input impedance, and a high output impedance (measured the inputs and outputs of my TL-C1 valve compressor, and sure enough, ≈6kΩ on input, and ≈1MΩ on output).

Something is not right here , EQs and compressors normally have a low input impedance (10K) and Lower output impedance (20-100 ohm)

 

[Bo Deadly]

I recognise that in these pieces of gear, there's going to be a low input impedance, and a high output impedance (measured the inputs and outputs of my TL-C1 valve compressor, and sure enough, ≈6kΩ on input, and ≈1MΩ on output).

That shouldn't be. Typically output impedance is low and input impedance is high. Modern outputs are typically 20-100 ohms and modern inputs are typically 10K. Microphones are typically higher impedance but still pretty low like 300 ohms. This is because they're either passive or limited by capabilities of phantom power. A high impedance input like for guitar is usually like 1M. Speaker outputs are usually very low impedance like a fraction of an ohm.

Note that you cannot measure output impedance with a meter like you frequently can with inputs. Output impedance refers to how easy it is to source / sink current which for an output means the thing has to be on. Inputs are frequently high impedance like an op amp but with a resistor to set the input impedance and so you can usually measure that resistor with the unit off. But not always. If the input is a transformer, then you can't measure that with a meter either.

The proper way to actually measure impedance of an output or input is to drive it with a signal but through a series resistor of say 100 ohms (or maybe lower if the output can handle it or requires it). Then you measure the voltage difference across the resistor and compute the current in / out at some particular moment (presumably when the voltage difference is high enough to make useful calculations). Impedance is then the just the voltage directly on the device divided by said current. And yes, this works equally well driving outputs as well as inputs.

For example, lets say you put a 2Vpp signal into an output through a 100 ohm resistor, look at either side of the resistor with a scope, freeze the scope display and see that when the inside of the resistor is 1V, the voltage directly on the device output is say 0.180V. That's (1-0.180)/100 = 0.0082A. So 0.180V / 0.0082A = 22 ohms. Meaning the output impedance of the device is 22 ohms.

Note that this value might differ significantly depending on frequency, load or signal level. Meaning the impedance of an output (but less so for an input) can be variable depending on the quality of the output.

 

[Rob Flinn]

Something is not right here , EQs and compressors normally have a low input impedance and Low output impedance...

That is quite a generalisation not really true. If the compressor is an old valve compressor like an LA2a with input transformer then the input impedance is quite low 600. But if you are dealing with a modern unit with a solid state input or a 10k:10k transformer, maybe something like an SSL then the input impedance is quite high & in terms of more than maybe 10k. This is because modern units use voltage matching where as old boxes like Pultecs etc use impedance matching. Impedance matching optimises power transfer which is why you try to have your amp & speaker impedances about the same value but it used to be used for most things back in the 30's, 40's & 50's etc. Voltage matching is used when devices don't need to transfer maximum power for example line level signals & enable you to feed more than one device from one source. Therefore a sweeping statement saying that "EQs and compressors normally have a low input impedance and Low output impedance" is a bit misleading.

 

[scott2000]

For example, lets say you put a 2Vpp signal into an output through a 100 ohm resistor, look at either side of the resistor

Neat! Any chance to see a drawing of what this looks like?

 

[abbey road d enfer]

. Meaning the impedance of an output (but less so for an input) can be variable depending on the quality of the output.

Not forgetting that the impedance would be significantly different whether the unit is turned On or Off...

 

[abbey road d enfer]

What I don't understand though, is what part of a "black box" part of a unit's schematic stops being the input, and starts being the output. I have heard that it's down to what part of the circuit is active, but if that's the case then there will be loads of different input and output stages in something like a compressor, due to all the transistors and opamps in the signal path.

By definition, a stage has an input and an output, so in a multi-stage unit, such as a compressor, there are many inputs and many outputs.
Typically, within the unit, the outputs of individual stages are connected to the inputs of other stages, except for the input and output stages, that receive signal from external sources or deliver to external loads.

 

[shot]

That shouldn't be. Typically output impedance is low and input impedance is high. Modern outputs are typically 20-100 ohms and modern inputs are typically 10K. Microphones are typically higher impedance but still pretty low like 300 ohms. This is because they're either passive or limited by capabilities of phantom power. A high impedance input like for guitar is usually like 1M. Speaker outputs are usually very low impedance like a fraction of an ohm.

Note that you cannot measure output impedance with a meter like you frequently can with inputs. Output impedance refers to how easy it is to source / sink current which for an output means the thing has to be on. Inputs are frequently high impedance like an op amp but with a resistor to set the input impedance and so you can usually measure that resistor with the unit off. But not always. If the input is a transformer, then you can't measure that with a meter either.

The proper way to actually measure impedance of an output or input is to drive it with a signal but through a series resistor of say 100 ohms (or maybe lower if the output can handle it or requires it). Then you measure the voltage difference across the resistor and compute the current in / out at some particular moment (presumably when the voltage difference is high enough to make useful calculations). Impedance is then the just the voltage directly on the device divided by said current. And yes, this works equally well driving outputs as well as inputs.

For example, lets say you put a 2Vpp signal into an output through a 100 ohm resistor, look at either side of the resistor with a scope, freeze the scope display and see that when the inside of the resistor is 1V, the voltage directly on the device output is say 0.180V. That's (1-0.180)/100 = 0.0082A. So 0.180V / 0.0082A = 22 ohms. Meaning the output impedance of the device is 22 ohms.

Note that this value might differ significantly depending on frequency, load or signal level. Meaning the impedance of an output (but less so for an input) can be variable depending on the quality of the output.

Thank you for this!
Seems like by far the most elegant approach how to measure impedance.
But it's the language problem for me, since I'm not native english speaker. I didn't quite understand the procedure.

So for an output, we put 100ohm resistor on the output. Then we measure the voltage AFTER that resistor, right? And we compare that with the voltage before the resistor and do the math, right?

On balanced outputs I suppose this should be done on both hot and cold. If yes, do we have to add or divide those results for transformer balanced output impedance?



Luka

 

[Whoops]

Therefore a sweeping statement saying that "EQs and compressors normally have a low input impedance and Low output impedance" is a bit misleading.

I was referring “Low” when compared to the OP 1Mega figures.
It’s much lower than that, but you’re correct it was a generalization just to help the Op understand that compared to a DI input it’s much lower, and also much lower than the Ops stated reading so something was wrong with the measurement.
You’re correct 10k are common for line inputs and 20 to 100 ohms for line outs.

I’ll edit my post to it’s clearer

 

[moamps]

The simplest way to measure the output impedance and the closest to real operating conditions is as follows:
The device for measuring the output impedance Zout is switched on and set to give the nominal output voltage Vout without any load. Then the output is loaded with some resistor Rload and the voltage Vload on it is measured.
The output impedance is then simply calculated using the following formula:
Zout=Rload*(Vout-Vload)/Vload
This calculation is valid if the output impedance does not have a significant imaginary component (reactance). This measurement can also be made with an ordinary DMM meter, provided that the signal frequency should be up to 400Hz (almost all DMM meters accurately measure the AC signal up to that frequency).
Rload should be chosen so that Vload is approximately Vout/2 (a trimmer pot can be used here).
If this is achieved exactly, then Zout=Rload.

 

[moamps]

..... Impedance matching optimises power transfer which is why you try to have your amp & speaker impedances about the same value but it used ....

That is not correct. Output amplifiers are generally designed to have a very low output impedance and to actually be an ideal voltage source.

 

[ccaudle]

Impedance matching optimises power transfer which is why you try to have your amp & speaker impedances about the same value

If you work out power losses in a simple example using Ohm's law you can see that is not correct in a general sense. Impedance matching has use in radio frequency connections, but is not applicable to audio systems at less than telephone system lengths. Amps driving modern speaker systems generally have less than 0.1 Ohm output impedance up to at least low kHz frequencies, they are not close to matched.

 

[Bo Deadly]

Neat! Any chance to see a drawing of what this looks like?

This is a simple diagram of how to measure AC impedance of an audio device:



This should work equally well for inputs, outputs, balanced, unbalanced, transformer or otherwise.

There is one exception to this (that I can think of at this particular moment) which is that an output that drives pin 3 should probably not be connected to ground as shown in the drawing.

More generally, this method measures the AC impedance of a particular conductor. So there are instances, like the above mentioned, where you need to consider that they might be different. So technically you might need to measure both pins 2 and 3 separately if the input / output is balanced.

Similarly, for an output you might want to choose a source resistor (Rs) that is representative of what the output would see as a load in practice. If you use Rs of 100 with a simple op amp output and drive it to a high level, it will not be able to keep up and the impedance will measure higher than it would in real life. In this case you might use 10K because that is what inputs are typically. Imagine your driver as being like the load an output would see under normal operating conditions. Or you might deliberately load it with lower values to see how it handles the load (such as because you want to use an op amp as a headphone amp).

The method described by moamps would work equally well albeit for an output only of course.

 

[Rob Flinn]

That is not correct. Output amplifiers are generally designed to have a very low output impedance and to actually be an ideal voltage source.

Yes, I agree but my understanding is one gets maximum power transfer when the speaker impedance (ok it's nominal & fluctuates with frequency, but in an ideal world) matches the output impedance of the amplifier or aerial matches transmitter impedance.

 

[Rob Flinn]

If you work out power losses in a simple example using Ohm's law you can see that is not correct in a general sense. Impedance matching has use in radio frequency connections, but is not applicable to audio systems at less than telephone system lengths. Amps driving modern speaker systems generally have less than 0.1 Ohm output impedance up to at least low kHz frequencies, they are not close to matched.

In the real world yes, but my understanding is when input and output impedances are matched the power transfer is greatest.

 

[Rob Flinn]

I was referring “Low” when compared to the OP 1Mega figures.
It’s much lower than that, but you’re correct it was a generalization just to help the Op understand that compared to a DI input it’s much lower, and also much lower than the Ops stated reading so something was wrong with the measurement.
You’re correct 10k are common for line inputs and 20 to 100 ohms for line outs.

I’ll edit my post to it’s clearer

A 10k input impedance is considered (in my book at least) a high impedance not a low impedance. This is because normally you use these sorts of 10k inputs in voltage matching setups which are loZ into hiZ. Ok, apprerciate crystal or piezo mics/pickups & guitar outs are hiZ out but they are not normal line level (convertors, mixers, outboard) sources & as you already said like 1M or greater loads.

 

[Bo Deadly]

In the real world yes, but my understanding is when input and output impedances are matched the power transfer is greatest.

That's only true for high frequences like RF and above where you might get reflections on the line if the source and load do not match. Those reflections equate to losses. For anything below 100kHz the will (probably) be no reflections and therefore all of the source power will be transferred to the load. In this case, the source impedance should be as low as possible although this is done primarily for fidelity and not efficiency.

 

[Rob Flinn]

That's only true for high frequences like RF and above where you might get reflections on the line if the source and load do not match. Those reflections equate to losses. For anything below 100kHz the will (probably) be no reflections and therefore all of the source power will be transferred to the load. In this case, the source impedance should be as low as possible although this is done primarily for fidelity and not efficiency.

Yes, I understand that voltage matching is better for audio purposes. Could you explain to me the scenario if theoretically one used a 1kohm i/p speaker on a near to zero o/p impedance amp compared to say some 4ohm speakers ? In your opinion is the power transfer exactly the same ?

 

[Bo Deadly]

Yes, I understand that voltage matching is better for audio purposes. Could you explain to me the scenario if theoretically one used a 1kohm i/p speaker on a near to zero o/p impedance amp compared to say some 4ohm speakers ? In your opinion is the power transfer exactly the same ?

I don't understand the question. With a zero impedance non-HF AC voltage source producing a certain level, the power delivered into a 1000 ohm speaker vs a 4 ohm speaker will not at all be the same because the load is completely different. Please rephrase.

 

[Rob Flinn]

I don't understand the question. With a zero impedance non-HF AC voltage source producing a certain level, the power delivered into a 1000 ohm speaker vs a 4 ohm speaker will not at all be the same because the load is completely different. Please rephrase.

That's my point. It's my understanding that when the input and output impedances are more or less matched the power transfer is closest to maximum. i.e when the zero impedance amp feeds the 8 ohm speaker you get a much better power transfer than if you feed the zero impedance amp into the 1Kohm speaker. Therefore it seems to me that the statement I made originally (Impedance matching optimises power transfer) is not only applicable at high, above audio frequencies. Do you think that's correct ?