In Reply to: RE: output impedance posted by hitsware on May 6, 2014 at 10:56:13:
Your argument seems sound to me. It is reasonable to model the output of the amplifier as an ideal (zero-impedance) voltage source in series with a resistor R0. Here R0 is what one would call the "output impedance."
This should be a good approximation as long as the amplifier is not being asked to deliver more current than it is capable of supplying. (If more current than that is asked for, then the amplifier will be overloading and clipping horribly, and so discussions of output impedance then become essentially irrelevant and academic.) So in any measurement, it should always be ensured that the amplifier is not pushed beyond clipping.
There are various ways that R0 in the above model could be measured. One way would be as you said, to connect a load R and adjust R until the measured output voltage is half of the open-circuit output voltage. Another way would be to adjust R until the power dissipated in R is maximised (at fixed output level). The answer in either case will be the same, namely when R is equal to R0. (And the power dissipated in R will equal the power dissipated in R0, if R is set equal to R0.) And so whether judged by voltage or by power, the conclusion would be the same, that the output impedance is R0.
If the M-60 delivers maximum power into 16-17 ohms, then that seems to support the argument that the output impedance is 16-17 ohms.
Chris
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Follow Ups
- RE: output impedance - cpotl 12:25:39 05/07/14 (1)
- RE: output impedance - Duke 15:29:58 05/07/14 (0)