What are input and ouput impedances?

GroupDIY Audio Forum

Help Support GroupDIY Audio Forum:

This site may earn a commission from merchant affiliate links, including eBay, Amazon, and others.
Do you think that's correct ?
Your definition of "impedance matching" is not conventional. By your definition, power transfer would be at maxium when the load was infinitely smaller than the source impedance and not matched. Load impedance is generally a fixed parameter. If all of the power that can be delivered into a load of a specific impedance is delivered into that load, it is 100% efficient and therefore equally efficient whether the load is 1000 ohms or 4 ohms.

When engineers talk about impedance matching, they're thinking about efficiency. They're worried about the effects of reactance from parasitic inductance and capacitance. If an output coupled with a line that has significant inductance and capacitance caused large peaking at 40kHz, that would result in significant loss of efficiency. Matching input impedance to the the characteristic impedance of the output would be the correct solution to eliminating the effects of parasitic reactances. But for audio frequencies this does not happen because the line is short compared to the wavelength of the signal and therefore matching impedance is generally not necessary. Bridging 100 ohms into 10K or 0.04 ohms into 4 ohms is equally efficient.
 
my understanding is when input and output impedances are matched the power transfer is greatest

I thought perhaps my hint about Ohm's law would lead you to think through the implications, and work out some examples yourself.
To be a little more explicit:
Say you have a voltage source with 0 Ohm output impedance, driving an 8 Ohm load. For 1 volt output, there will be no power dissipated in the voltage source output impedance, since it is 0, and the entire 1V will appear across the load, delivering (1)^2/8 W into the load, i.e. 1/8 W.

Now match the impedances (we'll assume for simplicity that the impedances do not vary with frequency, so we will just discuss the 8 Ohm valie). 8 Ohm output impedance of the source, 8 Ohm load. Now the 1V unloaded voltage of the source is split between the 8 Ohm source resistance and the 8 Ohm load resistance, so 0.5V appears across the load. The power delivered to the load is now (0.5)^2/8, i.e. 1/32 W.
The power dissipated in the source is now also 1/32W. Note that by increasing the source impedance without increasing the open circuit output voltage, not only is less power delivered to the load, but power is wasted at the source, and the total power delivered is only half because the total circuit impedance has doubled.

So you can see that in that simple example matching the output impedance to the load impedance results in 1/4 the amount of power delivered to the load compared to the ideal 0 Ohm output impedance case (given the same open circuit output voltage and same load value). I don't see how transferring 1/4 the power compared to the extreme unmatched case can in any sense be considered the "greatest" power transfer.
 
I thought perhaps my hint about Ohm's law would lead you to think through the implications, and work out some examples yourself.
To be a little more explicit:
Say you have a voltage source with 0 Ohm output impedance, driving an 8 Ohm load. For 1 volt output, there will be no power dissipated in the voltage source output impedance, since it is 0, and the entire 1V will appear across the load, delivering (1)^2/8 W into the load, i.e. 1/8 W.

Now match the impedances (we'll assume for simplicity that the impedances do not vary with frequency, so we will just discuss the 8 Ohm valie). 8 Ohm output impedance of the source, 8 Ohm load. Now the 1V unloaded voltage of the source is split between the 8 Ohm source resistance and the 8 Ohm load resistance, so 0.5V appears across the load. The power delivered to the load is now (0.5)^2/8, i.e. 1/32 W.
The power dissipated in the source is now also 1/32W. Note that by increasing the source impedance without increasing the open circuit output voltage, not only is less power delivered to the load, but power is wasted at the source, and the total power delivered is only half because the total circuit impedance has doubled.

So you can see that in that simple example matching the output impedance to the load impedance results in 1/4 the amount of power delivered to the load compared to the ideal 0 Ohm output impedance case (given the same open circuit output voltage and same load value). I don't see how transferring 1/4 the power compared to the extreme unmatched case can in any sense be considered the "greatest" power transfer.
Thanks for that, I understand your reasoning.

How about you took an 8 ohm source driving into 4ohms, 8 ohms & 16 ohms respectively

4 ohms = 0.0277 Watts (0.33)^2/4
8 ohms = 0.03125 Watts (0.5)^2/8
16 ohms = 0.0277 Watts (0.66)^2/16

Am I getting something wrong with my calculations ?
 
Am I getting something wrong with my calculations ?
No, you are correct for the case where you have a particular output impedance that you cannot change. But why would you not lower the output impedance? You aren't driving a long transmission line.
 
No, you are correct for the case where you have a particular output impedance that you cannot change. But why would you not lower the output impedance? You aren't driving a long transmission line.
Personally I would lower the ouput impedance, but I was just stating what I was taught as a fact. Admittedly by a guy who was originally a Post Office engineer. In the U.K the Post office used to run the whole countries phone system. Although not really relevant (because we are mostly using voltage matching) most pieces of studio gear have an output impedance that we cannot change.
 
Reading the thread so far I feel that there is a bit of a disconnect from the reasons why impedance matching, from the days of valve amplification and the true but only part relevant aspects of long signal line matching (telephones).
When valves were the only amplification device available and were very expensive as devices and the effort to operate them (heater and high voltage supplies) to make equipment viable as a 'product for sale' the pressure was on to obtain the best 'amplification' for the number of valves used.
As a valve is (in usual configurations) a high impedance input and a (relatively) low or medium impedance output, the challenge was on to maximise the gain from the different surces and loads. So Transformers were a practical way to achieve this. As valves became cheaper, the possibility of using 'resistance/capacitance coupling to get from one stage to the next became more attractive as an extra valve was cheaper than a transformer, and could offer 'better' performance (flatter frequency response, lower distortion etc).
When transistors hit the scene, the cost of amplification dropped drastically so the need for optimum POWER transfer practically disappeared. Then with the introduction of 'integrated circuits' (multiple transistors in a small package) the cost of amplification dropped again (less circuit board space, fewer extra components (resistors particularly) so all the original constraints were removed.
There is a term 'bridging input' where for practical systems a nominal 10 INPUTS were deemed OK to be attached to a given output. Thus as 600 Ohms had been an impedance reference point, established by telephone experience, a 10K input impedance was then proposed as a 'bridging' input impedance, meaning 10 (or a few more) could be attached to an output that was nominally 600 Ohms so that although the voltage loss (around half a dB per extra input) was acceptable, AND that having 10 would not embarrass the current drive capability of the DRIVING output stage.
Note that the output impedance does not imply the current drive capability of the output.
The output impedance of an 'open loop' op amp is often around 30 Ohms BUT with feedback, the measured output impedance can be a very small fraction of an Ohm, BUT internal current limiting will not permit more than about 25 milliamps of output current. Practically all currently available op amps will current limit to avoid the embarassment of the thing catching fire.
So, in all of this you have to separate the various aspects and work out which parameters can't exist in your specific situation. A power amplifier might have an output impedance of a fraction of an Ohm but it does not have the ability to provide near infinite current into a short circuit ( or very low impedance). If you said it was 0.1 Ohm output impedance. If you attached a 0.1 Ohm load and it was nominally a 20 Watt amplifier into an 8 Ohm load. the required current capability would be say 2 Amps peak which would be far from 20 Watts into the 0.1 Ohm load. Someone else can do maths if they really want.

PS, the output capability of 'line level' (not speakers) outputs needs to factor in driving signal current into the capacitance of the connecting cables (A few thousand Nanofarads) which is then not available to put current into the following input. The balanced output drive chips by THAT Corp 'assume about 1,000 feet of twin screen cable, thus requiring a peak current of IIRC 40 milliamps at 20KHz.
Matt S
 
Ok, if you model your amp as a voltage source in series with a resistor (Or a current source in parallel with one obviously), then the maximum amount of power that you can extract from such a supply occurs when source and load impedances are equal.

However, this is not a condition we WANT for audio, where speaker impedance varies wildly, back emf (and thus damping factor) matter, and we have no electrical transmission lines in play.

What we actually do is design an amp for a certain combination of maximum voltage and maximum current, which lets us calculate a minimum impedance that the thing can drive without hitting current or SOA limits, then we wrap it in negative feedback to lower the already low output impedance of an EF2 or 3 to basically zero, which is what speaker and crossover designers generally assume an amp to be (A voltage source).

The rated speaker impedance is (For solid stage amps) mostly a current and SOA limit, up to those limits the amp is a voltage source.

For glass fet designs, the output transformer leakage inductance means that an unloaded secondary can force the anode of a tube in cutoff WAY up there, so these do have an upper impedance limit due to dielectric breakdown.
 
It is the nature of web forum discussions for multiple posters to share everything they know, or think they know for confirmation, about a given topic.

I wouldn't invest any (more) effort, debating among ourselves, until the OP shows up and acknowledges if anything has answered his question yet.

JR
 
Some basic rules:
If it has a transformer or a voice coil, it has an impedance, which is a calculation of the inductance and resistance (the old 'if a flag pole is this tall and the shadow is this long, what is the hypotenuse?' formula) (the AC part of the equation) which is a measurement in HENRYs and required a voltage and frequency. The combination of all those things is calculated to REACTANCE. But for the most part, it's referred to as IMPEDANCE. This is because the coil of a mic, speaker or transformer has a DC measurement, but with a 1kHz signal going in, the DC part of the coil and magnetic field the AC signal creates has an influence on the "AC Resistance" or inductance. For reference the impedance is generally called "Z", low Z, high Z.

If it's a newer device that has no transformers, it may still be referred to as IMPEDANCE, but it is generally just resistance.

AND FYI: NEVER leave any device hooked up to a signal path if the power is off. It won't hurt anything, but the circuitry inside the device that is off changes from a nice stable impedance to a complex collection of resistors, diodes, capacitance and inductance that will wreak havoc with the tone. I have had people call me up and say "Yesterday it sounded really good, but today the mids sounds mushy". My first question is "What is hooked to the console that is not powered on?" The answer is 99% of the time is "OH". This includes a digital interface, cassette recorder, 2 track recorder, CD recorder, DSD recorder, etc.

-Mics can have an output impedance of 25-300 ohms output
-Guitars typically 5K ohms to 20K ohms, but with the output pots they can vary fro the pickup impedance to 250K ohms
-Speakers typically have 2-32 ohms
-audio equipment with transformers 150-600 ohms input and output
-audio equipment without transformers usually 10K ohms to 50K ohms input and 5-200 ohms output

When matching things with no transformers, low Z output into high Z input.

When matching things with transformers, you have to follow the manufacturer's suggestions, as a tube device feeding into a solid state device may require a 600 ohm resistor to "terminating" the output. Loading it correctly will greatly effect the way it sounds. Same goes with some inputs.

Matching impedances generally gives you the best transfer of signal from one device into another.

Some other rules, devices that lose signal easily from any kind of load (like a mic or pickup) usually follow the "bridging load" rule, where the source is say 150 ohms so the device that it plugs into needs to be 10X that or 1500 ohms. This rule will usually only drop the signal output level of the device by a dB. This is why a mic pad is designed with so many resistors, so it loads the mic with 1500 ohms and loads the input transformer with 150 ohms. This is important for the mic and preamp to retain it's specifications. Though, when two coils (mic + transformer) are connected they also have interaction between them from reflections, etc. that can effect the sound, and when you introduce a pad, the solid resistance loads for each of them tend to "calm" the interaction between them down and changes the sound.

Another rule, what ever you do to one side of a transformer will reflect onto the other side at the ratio of the windings, so if your transformer is a 1500 to 10K transformer, the 10K side needs to be loaded with 10K so the other side is actually 150. It has a 6.6 X ratio, so if the 10K is loaded with 15K, the other side will look like 2.3K.

I'm sure someone will correct somethings, but I'm trying to simplify 50 years of learning into one page.
 
I wouldn't invest any (more) effort, debating among ourselves, until the OP shows up and acknowledges if anything has answered his question yet.
Why do you care? Let everyone make a speech. I like to read comments like that. I'm much more likely to learn something new from comments like that.

If you want to be a comment nazi, go to stackexchange.
 
Power transfer usually refers to a situation where a source of impedance Zs is connected to a load of impedance Zl. Then the max available power to the load is when Zl=Zs.
However, the efficiency is low, at 50%, with as much energy wasted in the source than is available for the load.
This does not apply to power amps and speakers, because the very low impedance of the amp exists only as long as the amp operates in its linear operating region. If you load a power amp with a 0.1 ohm load, the amp either goes into protection, or the PSU collapses or it goes in flames (unless you run it with very low signal).
Maximum power transfer could be seen as maximum efficiency, which happens when the load tends to infinity, since the losses in the source tend to zero.
In the end it's a matter of semantics.
 
I wouldn't invest any (more) effort, debating among ourselves, until the OP shows up and acknowledges if anything has answered his question yet.

JR

This discussion became really interesting and useful so it doesn't matter that the OP disappeared and doesn't seem too interested.
This thread can be really interesting for anyone that wants to learn more about impedance and how to measure it, and will be a good read for anyone in the future.

So it would be great if the debate and opinion sharing continues.
 
Last edited:
Wikipedia has a good chart, which makes it clear that the Zout=Zload optimum condition is actually when the ratio of power in the load over the maximum power is one (e.g., voltage and current in the load is maximum).

File:Maximum Power Transfer Graph.svg - Wikimedia Commons

There is a blurb of text which clears this up as well (which is really just repeating what Abbey mentioned):

It is important to note that under conditions of maximum power transfer as much power is dissipated in the source as in the load. This is not a desirable condition if, for example, the source is the electricity supply system and the load is your electric heater. This would mean that the electricity supply company would be wasting half the power it generates. In this case, the generators, power lines, etc. are designed to give the lowest source resistance possible, giving high efficiency. The maximum power transfer condition is used in (usually high-frequency) communications systems where the source resistance can not be made low, the power levels are relatively low and it is paramount to get as much signal power as possible to the receiving end of the system (the load).
 
It would be really interesting to see how to find the optimal load impedance for a power amplifier design that has a negative output impedance.:)
 

Latest posts

Back
Top