Re: Amp output Z

The Phantom wrote:

> On Thu, 14 Apr 2005 00:06:16 +0100, John Woodgate <jmw@xxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
>
> >I read in sci.electronics.design that The Phantom <phantom@xxxxxxx>
> >wrote (in <0q4r51pg9l8v7sin5c9jenvd4v6opa9k90@xxxxxxx>) about 'Amp
> >output Z', on Wed, 13 Apr 2005:
> >
> >>In the real amp it must be something other a pure resistance of 10 ohms
> >>to account for the voltages he measures.
> >
> >I see you have assumed that they are accurate. But do you not think that
> >your inference from them is, on reflection, improbable? How could such
> >an inductance arise?
>
> As I said in another post, this problem is not well described.
>
> In his original post, Harry said:
>
> "The output voltage of the amp is
> 10.0Vrms at 400 Hz with no loads connected."
>
> Then in another post, he said:
>
> "Hi John,
> I am looking for Zo not Ro. I cannot measure the open circuit output
> voltage because the amp is unstable with no load. This is a customers design
> that I have to redesign. He believes that the output Z is about 10 ohms
> which is easy to measure by Ro=dVo/dRL. I believe it to be 70 ohms but
> cannot prove it. The spec requirement is less than 25 ohms."
>
> So, which is it? Is the output voltage of the amp with no loads connected equal to 10.0
> Vrms, or is the amp unstable with no loads connected? My best guess is that in the OP he
> was talking about the simulated amp, and in the "Hi, John" post he was talking about the
> *real* amp. It would help avoid confusion if he gave more detail and tell us whether he's
> talking about the simulation or the real amp.

Amen to that. I think the 10 ohms output resistance is a guesstimate too.


> If that 10 ohm resistor is in place, then you can't have the voltages he gives us for E2A
> and E2B without some inductance in the circuit.
>
> But if you allow the 10 ohms to become .54874 ohms, and the open circuit voltage to be
> 14.927 volts, then you can get the 14.81 volts with a 600 ohm load, and 14.5 volts with a
> 300 ohm load. And in this case, it's actually 14.81 V @ 6.747 degrees, and 14.5 V @
> 13.301 degrees. But, he's not seeing the phase angle. Too bad; if he were, that would
> help figure out what's going on.

Multimeters don't see phase angle - lol !


Graham


.

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