I have searched many places but have failed to get a clear and convincing explanation about why we match input and output impedance in audio circuits. For example: lets consider a simple audio circuit where microphone is connected to amplifier and then to a speaker. So here, what are input and output impedances and why is it important or why do we care about matching the input and output impedances? Please enlighten me. Appreciate it!

  • \$\begingroup\$ There are really several questions in here: What is source/load impedance? Why would we match them? Why would we match them in audio circuits? It's difficult to write an answer that covers all of that. \$\endgroup\$ – Phil Frost Jun 21 '13 at 3:28

We typically don't match impedances in audio circuits. For example, the output impedance of an audio preamp is typically relatively low whilst the input impedance of an audio power amp is typically relatively high.

The output impedance of an audio power amplifier is typically very low whilst the nominal impedance of a speaker is of the order of 8 ohms.

While any EE is probably familiar with the maximum power transfer theorem, which shows that maximum power transfer occurs when the load equals (or is conjugate to) the output impedance, it does not follow that the the output impedance ought to match the load.

In the first case, the output impedance is fixed and the load is the variable. In the second case, the load is fixed and, in that case, maximum power transfer occurs when the output impedance is zero, i.e., it is not desirable to match the impedances.

  • \$\begingroup\$ What two cases are you talking about in the last paragraph? \$\endgroup\$ – Phil Frost Jun 21 '13 at 3:06
  • \$\begingroup\$ Agreed, 100%. I've been working in the pro-audio field as an EE for 15+ years and nobody matches impedance of analog audio signals. \$\endgroup\$ – user3624 Jun 21 '13 at 4:09
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    \$\begingroup\$ @PhilFrost, the first case is the question of what load impedance, for a given source impedance, yields maximum power transfer. The second case is the question of what source impedance, for a given load impedance, yields maximum power transfer. \$\endgroup\$ – Alfred Centauri Jun 21 '13 at 10:52

Matching source and load impedances is rarely done nowadays; and never between amplifier and loudspeaker, where you want the lowest practical source impedance (including cable) to improve damping factor (a measure of the ability of the amp to control the speaker's position.

It used to be important, when gain was expensive, and better matching could eliminate an amplifier stage, when amplifier stages came in glass bottles.

It is still important when driving audio signals over long distances - hundreds of metres and upwards, when you have to start treating the audio cable as a transmission line instead of a simple connection.

It is also still important with very low level signals, where "every electron is sacred" - usually in microphone amplifiers. This is worth a little more explanation.

Microphones tend to fall into extremes of impedance : either extremely low source impedance, or extremely high, and each brings its own problems and solutions.

But first you have to understand how an amplifier can be modelled in terms of either input noise voltage and input noise current, or input noise resistance and noise figure.

Noise resistance is simply noise voltage divided by noise current; and noise figure is the excess noise introduced by an amplifier over the noise generated by a perfect resistor equal to the noise resistance.

Now the noise figure is an ideal : if the source impedance is equal to the noise resistance, the amplifier will add noise equal to its noise figure.

However if the source impedance is much higher, (meaning the source current is lower for the same signal power) the noise current will dominate. Or if the source impedance is much lower, then the source voltage is much lower for the same signal power, and will be swamped by the amplifier's input noise voltage.

In either of these cases, if you can transform the source impedance to equal the amplifier's noise impedance, you will improve the overall noise performance to the NF of the amplifier itself.

And this is one reason why microphone transformers have not disappeared altogether.

One scenario where this applies is vacuum tube amplifiers. The concept isn't at all obsolete; exactly the same applies to FETs. Now vacuum tube amplifiers tend to have very high input impedances, and that translates into fairly high noise voltages (go away and come back when you've worked out the Johnson noise for a 1 megohm resistor!) but VERY low noise currents.

Combine them and you can find triodes with state of the art noise figures (even today!) but at very high noise resistances. So to achieve low noise amplification, you may need to transform your source impedance up by a factor of 100 or so. The simplest way is with a 10:1 voltage ratio transformer.

Similar constraints apply inside a capacitor microphone, where the source impedance is extremely high. Indeed some capacitor microphones have never abandoned the tube; not from laziness, not from some hi-fi fad, but just because it's the right tool for the job.

Now it is possible to be a professional audio engineer for decades and never have to worry about this. But the basic physics still applies, and there are still a few situations where it matters. Not as many as in RF design though!

  • \$\begingroup\$ Vacuum tube amplifiers also have a higher source impedance, so getting all the power out of them possible, and into say, a loudspeaker, merits some sort of matching. \$\endgroup\$ – Phil Frost Jun 21 '13 at 12:07
  • \$\begingroup\$ @PhilFrost Matching amp/speaker impedance causes the variations in impedance over freq of the speaker to me mirrored into the overall frequency response of the system. Put a different way, you might get ideal power transfer but the overall audio quality will be terrible. Of course, if you're using tubes then audio quality isn't your primary concern. :) \$\endgroup\$ – user3624 Jun 21 '13 at 16:22

You will get maximum power transfer if the load impedance matches the source impedance, but it is not critical. As you can see in the diagram, even with a 2-1 mismatch, you will get 8/9 of the power. Reducing the source impedance will increase it further. Rematching the load will increase it further still. Variations in the load impedance will have least effect on power o/p if the load is matched.

BUT, output stages are like power supplies. They are meant to be loaded to the load they are designed to handle, NOT for maximum power transfer. If you whack a resistance of about .5Ω across your electricity supply you will get the maximum possible power out of it for a SHORT TIME! enter image description here


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