I'm planning to block dc on the F5 input too.. I guess the fact the F5 is a dc amp makes the current limiting more important... so I think I will go rogue and avoid the current limiting circuit.
Thanks.
I'm really fascinated by this amp, how simple it is.
if you take the current limiting circuit ... it really very simple.
I decided to go with 2w non-inductive wirewound for R3-4
and R9-12 with two 10w ordinary wirewound so making 20w.
If you measure your operating temperature and adjust bias to a maximum of 65C output transistors and 55C heatsinks, it should work.
To make sure that the heatsinks have a good chance to work at their best, make sure the amp has 30cm airspace beneath it and at least 100cm above, this room is necessary to let the heatsinks make the convection chimney effect and move air.
To make sure that the heatsinks have a good chance to work at their best, make sure the amp has 30cm airspace beneath it and at least 100cm above, this room is necessary to let the heatsinks make the convection chimney effect and move air.
Just to correct myself. for R9-12 pair ~8W is what is needed at peak if you listen very close to clipping ... so 6W for the pair should be very safe.
Thanks
R11||R12 is in series with R4. That gives a total of 60r from output to ground.
When the F5 is driven to 25W into 8ohms the dissipation in R11||R12 is {14.14Vac/60}^2 * 50r * 50% * 0.5 = 0.69W
The 50% is the duty cycle of the upper and lower halves, the 0.5 is due to two resistors in parallel.
The peak instantaneous dissipation is 4 times that. i.e ~2.8Wpeakinst
When the F5 is driven to 25W into 8ohms the dissipation in R11||R12 is {14.14Vac/60}^2 * 50r * 50% * 0.5 = 0.69W
The 50% is the duty cycle of the upper and lower halves, the 0.5 is due to two resistors in parallel.
The peak instantaneous dissipation is 4 times that. i.e ~2.8Wpeakinst
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if you are lstening very close to clipping then the average power delivered to your speaker load is roughly 10dB to 20dB below the amplifier's maximum output.
The dissipation in the feedback resistors is also 10dB to 20db below the instantaneous peak dissipation. and then divide by 4 again to get the average. = 7mW
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I used this approach to get the 8W per pair :
Voltage (V) = Current (I) * Resistance (R)
Power (P) = Voltage (V) * Current (I)
and I calculated the voltage drops to give the power per resistor on the feedback network.
so : 0V - 10 ohms - 4.2v - 50 ohms - 25.2v ( 18v * 1.4 )
so for the pair of 50ohms : is 21v that gives 8.82w
for the 10ohms : 4.2 v that gives 1.764.
Hello Andrew.
My post wasn't very clear... 25.2v in the input voltage on ( R9 and R10 ) so it drops to 4.2v and goes into R3.
Thanks
14vac * 1.4 + 20% (+/- transformer regulation ) = 23.7v very close to my crude 18vac * 1.4
so 0v <-- 10ohms <-- 3.95v <-- 50 ohms <-- 23.7v
it gives 19.75v into 50 ohms = 7.8W
and give 3.95v into 10 ohms = 1.56W
