output impedance bridge

[unclejed613]

i'm thinking of building a device to test output impedance, and it will be a mostly "stand alone" unit but with a scope and spectrum analyzer tap output. once i get prototype completed and tested, would anybody be interested in this device in kit form? it will contain an oscillator (single frequency and swept) and amplifier, test load, metering circuitry,and calibration circuits. the device gives a direct reading 1mV/milliohm indication of dynamic output impedance at any frequency from 1-100khz or with a spectrum analyzer can give a trace of output impedance vs frequency.

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

 

[Burnedfingers]

would anybody be interested in this device in kit form?

yup

 

[KSTR]

Do you plan to overlay both common mode temporary voltage and current? Background is the availibility of four-quadrant biased measurements to obtain small signal Zout at various voltage/current combinations (which can be quite revealing).

- Klaus

 

[tubetvr]

Why not use a much simpler and more reliable way of measuring output impedance? Measure output voltage with 2 known load impedances, e.g 4 and 8 ohm and calculate the output impedance from the voltage difference.

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

 

[KSTR]

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

 

[tubetvr]

For a normal amplifier the output impedance should be the same for any load, it can always be characterised by the voltage drop at a given load compared to the open loop voltage, it is only if you have a reactive element in the amplifier where this is not true.

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

 

[KSTR]

For a normal amplifier the output impedance should be the same for any [resistive] load

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.

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

 

[unclejed613]

what i have found is that the direct measurement method shows up impedance variations caused by gain changes in the crossover region (which is why the method can be used to set optimum bias). you can also use the method to find the physical output resistance (without breaking the feedback loop). it measures directly how an amplifier reacts to a back EMF from the load (i.e. damping factor).