oh, did i mention that you can use it to set output bias without a distortion analyzer?

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- Thread starter unclejed613
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oh, did i mention that you can use it to set output bias without a distortion analyzer?

would anybody be interested in this device in kit form?

yup

If you are worried that the output impedance is varying with load impedance you can measure the output voltage with several loads and plot the result in a graph that would give you the output impedance at any given load impedance.

Regards Hans

Using two real (non-reactive) loads still only gives values for non-realistic cases, IMHO.

BTW, for those who want to use the "two-resistors"-method, this is the formula:

Let V1 be the voltage measured with R1 test load,

and V2 be the voltage measured with R2,

then calculate:

Voltage ratio v = V1/V2,

Resistance ratio r = R1/R2

Rout = R1*(1/v-1)/(1-r/v)

And there is a serious problem here: You must ensure that voltage and resistance ratios differ significantly among each other and be both far from unity, otherwise smallest measurement errors will result in great calculation errors.

Inducing a current and measuring the voltage change is way more insensitive and it is a direct measurement. As I noted, preferably with other than zero voltage and current bias. This is especially relevant for amplifiers with little or no global feedback to control output impedance.

IMHO, output impedance is not a problem (even when it is as high as, say 1/10th of mimimum speaker impedance and is varying with frequency) but its stabilty versus output I/V-bias is (because if it isn't, that's just plain nonlinear distortion).

- Klaus

Otherwise there is nothing that stop you from using other than resistive loads but you would need something better to measure it with that can measure voltage and current independtly, e.g a vector voltmeter.

The accuracy problem that you describe is of course the same when injecting a voltage and measure the current, for amplifiers with low output impedance you are either forced to inject a very low voltage and measure the current or to inject a higher voltage where you risk messing up the feedback loop.

Regards Hans

Obviously this isn't the case. Take eg a simple singe ended emitter follower, zero feedback: It's DC/LF output impedance, assuming low impedance drive, is 1/S + Re.tubetvr said:For a normal amplifier the output impedance should be the same for any [resistive] load

Small signal S (transconductance) now depends on collector current and on temperature, S=Ic/Ut (and that is already heavily idealized). In a push-pull output stage, you then have two terms of this kind, paralled. Unless you bias the output transistor to load invariant, constant output current and constant dissipated power (that is, constant Vce ==> constant power) you will have a load current dependance in the small signal output impedance. But I don't want to sidetrack this thread any further...

- Klaus

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