1- What heatsink is advised for the power diodes ?
I want to use these one : RURP3020
http://www.fairchildsemi.com/ds/RU/RURP3020.pdf
i think to employ a 0.9°C/w heatsink for 2 rectifier bridges
Is it to much , insufficient or correct ?
2- Do you know the RCore transformer ?
http://www.custommag.com/products/r-core.shtml
http://www.techsavvy.com/industry/file/national/03sfn/lnc01.html
http://www.lenco-elect.com/r_core_transformers.htm
Do you think it worth to use it ?
I want to use these one : RURP3020
http://www.fairchildsemi.com/ds/RU/RURP3020.pdf
i think to employ a 0.9°C/w heatsink for 2 rectifier bridges
Is it to much , insufficient or correct ?
2- Do you know the RCore transformer ?
http://www.custommag.com/products/r-core.shtml
http://www.techsavvy.com/industry/file/national/03sfn/lnc01.html
http://www.lenco-elect.com/r_core_transformers.htm
Do you think it worth to use it ?
Both the rectifiers and the transformer criteria depend a lot on the power they'll have to deliver.
A bit of overkill is never bad on a real class A amp.
You'll find a lot of calculators here and on the net.
Roughly, three times the real power dissipation is plenty to stay in the SOA.
/Hugo
A bit of overkill is never bad on a real class A amp.
You'll find a lot of calculators here and on the net.
Roughly, three times the real power dissipation is plenty to stay in the SOA.
/Hugo
Hi
How did you calculate the gain of the amp (with this one http://www.kk-pcb.com/alephx.html) ??
And if i replace IRF240 by IRFP044N and if i change the value of 0.47R by 0.44R,
must i change value of some resistor of the circuit ?
Thanks.
How did you calculate the gain of the amp (with this one http://www.kk-pcb.com/alephx.html) ??
And if i replace IRF240 by IRFP044N and if i change the value of 0.47R by 0.44R,
must i change value of some resistor of the circuit ?
Thanks.
I don't understand how to simulate the power supply with PSUD.
In a previous repply (p.26), Hifizen had use a constant-current .
Shouldn't we use a resistor who represent the loudspeakers ?
Because result of the simulation doesn't give the same result (fonction of the impedance of the loudspeaker).
I hope you understand what i'm saying !
whazzat
In a previous repply (p.26), Hifizen had use a constant-current .
Shouldn't we use a resistor who represent the loudspeakers ?
Because result of the simulation doesn't give the same result (fonction of the impedance of the loudspeaker).
I hope you understand what i'm saying !
whazzat
Hi,
the PSU is connected to the amp, not to the loudspeaker. Since it is a class A amp the current won´t be affected by the speaker impedance or the power that is sent to the speaker.
Therefore you can simulate the load on the power supply by setting the bias current as a constant current.
William
the PSU is connected to the amp, not to the loudspeaker. Since it is a class A amp the current won´t be affected by the speaker impedance or the power that is sent to the speaker.
Therefore you can simulate the load on the power supply by setting the bias current as a constant current.
William
Hi all,
What do you think about this configuration ? :
-> CRCRC : 22mF / 1.5Ohm / 22mF / 1Ohm / 47mF .
with a 2x25V transformer .
->On the simulation i have 20V at exit.
->i had simulated it with PSU Designer and with a constant current for the load (4.5A).
Does it work ?
(Because i'm not shure of the result)
What do you think about this configuration ? :
-> CRCRC : 22mF / 1.5Ohm / 22mF / 1Ohm / 47mF .
with a 2x25V transformer .
->On the simulation i have 20V at exit.
->i had simulated it with PSU Designer and with a constant current for the load (4.5A).
Does it work ?
(Because i'm not shure of the result)
Hi,
it will work but have a closer look at the power the R's are dissipating:
2 x 1.5 x4.5^2 plus 2 x 1 x 4.5^2 = ca. 100Watts
This is a bit too much for my liking.
It is much better to use a C-L-C configuration with a 1.5 to 2.2mH coil. This gives a lot less power dissipation and more ripple attenuation.
So something like 47mF-2.2mH-47mF would be nice for 4.5A
William
it will work but have a closer look at the power the R's are dissipating:
2 x 1.5 x4.5^2 plus 2 x 1 x 4.5^2 = ca. 100Watts
This is a bit too much for my liking.
It is much better to use a C-L-C configuration with a 1.5 to 2.2mH coil. This gives a lot less power dissipation and more ripple attenuation.
So something like 47mF-2.2mH-47mF would be nice for 4.5A
William
150mOhm. 22 V RMS from unloaded transformer, one set of rectifiers for each rail. 22 VDC per rail out after all filters at 6 A. Things other than the R that affect voltage are transformer secondary resistance and capacitor ESR. In PSUD, replacing the R with 1.1uH gives me an extra volt per rail.
I have multiple primary taps so I can get lower voltage. By having a switch for that and also different bias setting, I'll be able to switch between different power levels and have it optimal for different speaker impedances.
I have multiple primary taps so I can get lower voltage. By having a switch for that and also different bias setting, I'll be able to switch between different power levels and have it optimal for different speaker impedances.
Nixie,
torobars where the only chokes I could fit into my case that had enough inductance combined with low enough resistance and high saturation current. Look at www.intertechnik.de for torobar (the site is also in English)
William
torobars where the only chokes I could fit into my case that had enough inductance combined with low enough resistance and high saturation current. Look at www.intertechnik.de for torobar (the site is also in English)
William