Re: Amp output Z

On Wed, 13 Apr 2005 01:41:33 GMT, "Harry Dellamano" <harryd@xxxxxxxxxxxxx> wrote:

>
>"Jim Thompson" <thegreatone@xxxxxxxxxxx> wrote in message
>news:tlvn511djoqu9enbu0skimii42ulthedps@xxxxxxxxxx
>> On Tue, 12 Apr 2005 16:56:34 +0100, John Woodgate
>> Yep, I should have reduced it to what the meter actually shows.
>> However I'm pretty sure you can't get the answer you want.
>>
>> ...Jim Thompson
>Hey Jim, your killing me!
> Are you saying that we cannot measure Zo because there is a coupling cap
>there? Given that the unit is not stable with an open load but I can measure
>the voltage on the either side of the coupling cap and vary the load.
>Instead of varying a resistive load, why don't we vary a capacitive load?
>That should cause a voltage divider with the coupling cap and yield readings
>to prove the Zo is about 70 ohms, not 12 ohms.
> Regards,
> Harry
>

You don't have enough measurements to fully determine the output inpedance.

It looks to me as though the locus of possible output impedances is a circle with center
at about 10+j47 and with a radius of about 24.

If there is no relative phase shift between E2A and E2B, the the solution is an output
inpedance consisting of a 28.27 millihenry inductor in series with a 13.1078 ohm resistor.
The reactance of that inductor just cancels the reactance of the 5.6 uF capacitor, and the
13.1078 ohm resistor provides the drop you see when you apply 300 and 600 ohm loads (the
open circuit voltage would be 15.1335 volts).

If there is phase shift between E2A and E2B, then the solution lies somewhere else on the
circle diagram.

.

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