Originally posted by Randall
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1) "8 ohm, a little over 2.5 v p-p"=0.1W RMS
2) "2.4 ohm, about 2 v p-p"=0.21W RMS
So by going down in impedance we have more than doubled available power, a valuable find, donŽt you think?
3) I would *love* to find how much it puts out into, say, 2 ohm, and even 1.5 ohm.
Since power is so low (so far), I bet you have more usable resistors in your junk drawer.
No need for 5W ones (so far), probably a couple 1W ones in parallel will do for testing.
Worst case one might smoke if overheated ... weŽll worry about that when/if we reach that point.
To make my position clear, I would be more than happy to reach near 1W RMS, go figure. Even surpassing 0.6W or so would be an achievement.
4) and keep posting Vpp values if you wish, but please also measure "RMS Volts" across load with a multimeter.
Preferably a meter which has available 200mVAC or at least 2VAC scales ... even if we donŽt use them, the point being that such meters have a real AC metering system inside, while $10 multimeters, sporting 200VAC and 750VAC scales *only*, have not, as simple as that, just a crude 1N4007 in series with the standard VDC scale ... something we can NOT trust.
5) so in a nutshell: measure similar waveforms but into 2 ohm and if possible 1.5 ohm or so and measure and post "multimeter" AC volts across the load.
Once we find the load, no matter what, which provides maximum power output, then me might try to mod bias and improve that further.
My suspicion, which of course will need confirmation or dismissal, is that the "mystery transformer" or whatever you are using now is very mismatched, IŽm trying to find a better operating point.
Nothing more, nothing less.
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