SOS impedance article Q

[Pringe]

I've just been reading Hugh's excellent SOS article "Understanding Impedance" (http://www.soundonsound.com/techniques/ ... -impedance) and understand it apart from one concept which I am struggling to figure out!

When talking about matched impedance there's the example given of a 0dBm signal from a mixer with 600R output impedance connected to a tape recorder input with 600R input impedance which registers 0dBm also. I can't seem to work out why this is the case! For calculating dB loss with input / output impedances I've been visualising them like a resistor divider and using the formula Av (dB) = 20*log10(Zin / Zin + Zout). Following that through for the voltage matching examples given later in the article gives me the same results as Hugh calculates (0.04dB etc..) but this first example would seem to give a -6dB loss with matched impedances, and with 2 x 600R in parallel (300R equivalent) I calculate a -9.5dB loss. I understand how having two 600R resistors in parallel causes half the total voltage to develop across each of them but can't make that sit with the equation above. I'm obviously doing something wrong but would really appreciate any pointers to show me where!

Thanks in advance.

[James Perrett]

I think you are missing the fact that the 0dBm from the device is measured at the output connector so the effect of the output impedance has already been taken into account in the measurement.

[Pringe]

Ah yes I see, of course that makes sense. Thanks! And would input metering be measured before its input impedance?

[Hugh Robjohns]

...but this first example would seem to give a -6dB loss with matched impedances...

It does. Yes.

The bit you're missing is that you're wearing a 'signal voltage-transfer' thinking-head while looking at a power-transfer interface!

A matched-impedance system is designed to maximise the transfer of power, and that is achieved when the output impedance and the destination load impedance have exactly the same value -- in this case, 600 Ohms.

It's worth remembering that the audio industry acquired this antiquated 600-Ohms interface system from the telephone industry where they wanted to transfer the maximum amount of power from one handset's microphone across miles of telephone wires to another handset's earpiece -- remember this was originally an entirely passive system with passive telephones!

As a result of the matched impedances, half of the source voltage is lost across the output impedance, and half across the destination load impedance -- so the signal voltage at the destination is 6dB lower than that generated by the source. ... but when you're thinking about power transfer, that's not very important!

Of course, the level metering in systems designed to work with matched-impedance interfaces was also designed to take the signal voltage loss into account, as James has already mentioned.

Thankfully, that obsolete system has been eradicated from most audio interfaces, and we now use voltage-transfer interfaces pretty much everywhere. The important characterstic here is that the source impedance is very low, and the destination load impedance very high, so that the maximum possible signal voltage is developed across the destination impedance, with the least amount being lost across the source impedance. Hence we're now maximising the transfer of signal voltage, (not power).

One of the key benefits of this approach is that you can connect a single output to multiple destinations in parallel with a negligible loss of signal voltage. As you've observed, that's very far from the case with a matched-impedance interface system.

HTH.

[Wonks]

As power can be calculated as current (squared) x resistance, if the supply and load impedances have the same resistance, and as they must have the same current flowing through them, then the power at both points will be the same.

[Pringe]

Thanks a lot Hugh for the detailed reply! I was definitely in voltage-transfer mode there, and it makes much more sense now.

Thanks also Wonks