I hate losses, so I tend to neglect them
I have read this at least once per day during the past week or so. All I can figure out is that the bias current is about 1.3A in the original F5. But I can't find any specific information on the best value for the bias current in the F5T V2 or V3. All I found is this: "Just run the amplifier into a reasonably low impedance until it gets good and hot – as hot as you plan to let it get - ever. Then adjust the bias to a point below where the idle current starts to really take off. You should find that this point is around 0.4 volts across the 1 ohm resistors.". But before I can do that I need to know how to build the thing first.
Can someone tell me how many output pairs and how much bias current per pair is best, and why?
For 100W of ClassA into 8ohms you will need a bias current of ~2.6A
max Pout = {2.6*2}² * 8/2 = 108W
You also need a maximum output voltage of 40Vpk
To get to those values you will need a PSU that does not sag below ~44Vdc excluding ripple.
I would expect a PSU that when fully biased delivers ~±46Vdc to ±48Vdc
Assuming the higher figure the output devices will have a quiescent dissipation of 2*48*2.6 = 250W.
8output devices (4pr) each dissipating 31.2W should be sufficient, if you have a big enough heatsink.
I built a 100W Krell Klone and used 6pairs because that fitted onto 3 big sinks (wrapped around three sides of a cube) better than using 4pr, or 5pr.
max Pout = {2.6*2}² * 8/2 = 108W
You also need a maximum output voltage of 40Vpk
To get to those values you will need a PSU that does not sag below ~44Vdc excluding ripple.
I would expect a PSU that when fully biased delivers ~±46Vdc to ±48Vdc
Assuming the higher figure the output devices will have a quiescent dissipation of 2*48*2.6 = 250W.
8output devices (4pr) each dissipating 31.2W should be sufficient, if you have a big enough heatsink.
I built a 100W Krell Klone and used 6pairs because that fitted onto 3 big sinks (wrapped around three sides of a cube) better than using 4pr, or 5pr.
That's the peak power, right?
Effective power is P-eff = I-eff² x R, where I-eff is effecive current and R is load resistance.
Peak current is I-peak = 1.414 x I-eff. The bias current must be equal to the peak current (or higher) for class A.
For 100W effective output power at R=8Ohm, I-eff=3.54A is needed. This corresponds to I-peak=5A. So I need 5A bias for 100W@8Ohm effective power.
Is this correct?
No.
That was the power.
If I want to refer to something else, as in instantaneous peak power, I will spell it out.
You are asking for 100W into 8ohms.
Power = IV = I²R = V²/R This applies to constant DC currents and voltages.
If you want to use AC current and AC voltage the formulae become
Power = IrmsVrms = Irms²R = Vrms²/R
for a sinewave become
Power = IacVac = Iac²R = Vac²/R
for peak levels of a sinewave become
Power = IpkVpk/2 = Ipk²R/2 = Vpk²/R/2
That was the power.
If I want to refer to something else, as in instantaneous peak power, I will spell it out.
You are asking for 100W into 8ohms.
Power = IV = I²R = V²/R This applies to constant DC currents and voltages.
If you want to use AC current and AC voltage the formulae become
Power = IrmsVrms = Irms²R = Vrms²/R
for a sinewave become
Power = IacVac = Iac²R = Vac²/R
for peak levels of a sinewave become
Power = IpkVpk/2 = Ipk²R/2 = Vpk²/R/2
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How did you get these values? The FQA12P20 data sheet says max. power dissipation is 150W @*25°C, with a derate of 1.2W/°C. A dissipation limit of 40W would correspond to a temperature of about 117°C, and with a 20W limit temperature could go up to about 133°C. Is this realistic?
(I have no clue how the FET temperature relates to the heatsink temperature)
You have no clue of how an amplifier works.
Read up on temperature de-rated SOAR (Safe Operating Area Region) and how reactive loads affect that. ESP is just one web site that discusses this in detail.
Once you understand, then you can do some calculation. Read David Eather and then
Download Bensen's spreadsheet, or Jan Didden's SOA, or my modified versions of Bensen.
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