First Watt-ish Chip Amp, Variable Impedance - diyAudio
 First Watt-ish Chip Amp, Variable Impedance
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 12th February 2005, 02:47 PM #1 diyAudio Member   Join Date: Jan 2004 Location: KC Metro First Watt-ish Chip Amp, Variable Impedance Nelson Pass's F1 amplifier got me to begin exploring current amps which lead me to variable impedance which lead me to realize I obviously don't get some basics. I was looking at http://sound.westhost.com/project56.htm Figure one is intuitive for both voltage and current amps. Figure two and the voltages into 8 and 4 ohm loads has me stumped. The 18.8 with no load is straight forward enough, but I can't get to the 12.8 and 9.7 volt calculations. My guess was that for quick and dirty calculations gain is going to be set by r1 and (r2+r3) in both open and under load calculations. If so (and it appears not to be), I would have expected Load and r3 to give me the reported voltages under load as a simple voltage divider with 18.8 as the Vin. I suspect that I am grossly over simplifying. Can anyone point me in the right direction?
 12th February 2005, 03:36 PM #2 diyAudio Member   Join Date: Dec 2003 Location: Near Seattle One thing to keep in mind is that he measured those values and didn't calculate them because he doesn't know the output impedance. You're doing those calculations assuming negligible output impedance compared to the load. That's a usual "rule of thumb" for voltage sources (most opamps). But in this case you have 3.75 ohms output impedance. So you can't ignore it because, it's basically like having 3.75 ohms in series with the output and before R1,R2,R3,Rl. Try adding a 3.75 ohm resistor between the output and R1 and then recalculating and see what you come up with. -- Danny
diyAudio Member

Join Date: Jan 2004
Location: KC Metro
Quote:
 Originally posted by azira But in this case you have 3.75 ohms output impedance. So you can't ignore it because, it's basically like having 3.75 ohms in series with the output and before R1,R2,R3,Rl.
The 3.75 ohm impedance was calculated using the voltages with and without load. It is effectively the unknown that I am trying to figure out how to calculate, with the voltage under load as the intermediate calculation that I can't figure out.

My goal is to sketch out a current amplifier with an 80ohm output impedence (following the F1's lead here if not all the way to class a and zero feedback). To do that I need to know how to calculate

1) The no load voltage (no problem)
2) Vout under load (don't understand... )
3) Output impedence (well document in Project 56 based upon the availability of 1 and 2)

Being able to calculate Vout under load where output impedence is *not* known, is the missing link.

diyAudio Member

Join Date: Dec 2002
Location: Calgary, Alberta
Quote:
 Originally posted by tgorham3 Being able to calculate Vout under load where output impedence is *not* known, is the missing link.
You missed the point of azira's reply: RE obtained those results empirically. Unless you win the lottery by someone here having modelled a power op amp and it is the same one you intend to use, your efforts are best spent on making a test jig and seeing what you yourself can come up with.
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