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.
Just a little question.
I know it may be little on the weak side, but will the mini dissipante 3U 300mm be enough for the F5 (with stock Bias). The sinks are rated at 0.4c°/w and it seems like Papa used sinks like that for the reference design (3Ux300mm).
I know it may be little on the weak side, but will the mini dissipante 3U 300mm be enough for the F5 (with stock Bias). The sinks are rated at 0.4c°/w and it seems like Papa used sinks like that for the reference design (3Ux300mm).
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.
Hello guys
I did order my F5 parts and I will get two pairs of IRFP9240PBF and IRFP240PBF what kind of matching you guys recommend?
Thanks
I did order my F5 parts and I will get two pairs of IRFP9240PBF and IRFP240PBF what kind of matching you guys recommend?
Thanks
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.
You can't get 29.4V (4.2V + 25.2V) across the feedback resistors with an F5 operating on 25V supplies. The sch you linked to shows 23V supplies.
Are you building an F5t with enhanced power output?
Are you building an F5t with enhanced power output?
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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
you still can't get 25.2V from 23V supply rails.
Even when you increase the supply rails to 25V you still can't get 25.2V across the feedback.
25W into 8ohms is equivalent to 14.14Vac and 1.77Aac, not 25.2V
Even when you increase the supply rails to 25V you still can't get 25.2V across the feedback.
25W into 8ohms is equivalent to 14.14Vac and 1.77Aac, not 25.2V
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
You cannot just add regulation voltage %, it's not there (transformer is supplying high current).
So,
14vac * 1.42 - 1.4v (full wave rectification voltage drop) = 18.48 DC rail voltage.
So,
14vac * 1.42 - 1.4v (full wave rectification voltage drop) = 18.48 DC rail voltage.
Power is Amperes times Voltage.
P=IV
for ac this becomes P=Iac * Vac
For a sinewave P= Ipk * Vpk / 2
if you want to use peak values of a sinewave you must use the correct format.