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Old 22nd October 2012, 05:56 PM   #11
tomchr is offline tomchr  United States
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How about getting one of the chassis mounted Dale/Vishay resistors. The RH or NH series.

~Tom
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Old 22nd October 2012, 09:18 PM   #12
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Those WW resistors are really robust and will take the wattage, but like we've said it's possibly not so good for nearby components. If you get another type of resistor they still need to be mounted such that the heat is sinked away or nearby components should be moved further away. Expencive Caddocks for such PSU duty seems a little luxurious doesn't it? Add to those a proper heatsink and you've got an impressive voltage dropper. Might as well go active...
10mm lead length is a bit short to prevent solder going too soft or even melt. the reason those resistors have steel leads (and not copper) is b/c steel doesn't conduct heat so well. If you have a little longer leads, the solder joint should be fine.
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Old 22nd October 2012, 09:23 PM   #13
disco is offline disco  Netherlands
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For just a test rig why buy a heavier resistor? Are you afraid to pop your caps when it it goes open?

It's unusual to burn 100 volt in the supply of a power amp. As suggested lower AC and perhaps a choke would be better for regulation.
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Old 22nd October 2012, 10:25 PM   #14
alecu7 is offline alecu7  Canada
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Quote:
Originally Posted by DF96 View Post
No, this is not what you explained in post #1. You did not say the derating figure you gave was for case temperature. Derating could be for case or ambient temp, with the former more likely for a metal cased resistor intended for a heatsink and the latter more likely for other resistors. You said "silicon coated" which I assumes means not metal cased.

If you are right, then 7.5W for a resistor rated at 7.8W is asking for trouble. If I am right, then the resistor is fine but the things around it might not be.

150C surface temperature for a resistor OK to 350C is fine. Nothing to worry about.

Yes you're right. It is ambient temperature not case temperature.
My mistake.
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Old 22nd October 2012, 10:38 PM   #15
alecu7 is offline alecu7  Canada
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Originally Posted by SemperFi View Post
Those WW resistors are really robust and will take the wattage, but like we've said it's possibly not so good for nearby components. If you get another type of resistor they still need to be mounted such that the heat is sinked away or nearby components should be moved further away. Expencive Caddocks for such PSU duty seems a little luxurious doesn't it? Add to those a proper heatsink and you've got an impressive voltage dropper. Might as well go active...
10mm lead length is a bit short to prevent solder going too soft or even melt. the reason those resistors have steel leads (and not copper) is b/c steel doesn't conduct heat so well. If you have a little longer leads, the solder joint should be fine.
I think I'll get Caddock resistors. I can mount them on the side panel which is 1/2" thk aluminum. This should allow to dissipate enough heat not to worry about. Talking about money the difference between dale and caddock is about $3. I spent over $500 on this amp. $3 won't make a difference.

Alex
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Old 22nd October 2012, 10:45 PM   #16
tomchr is offline tomchr  United States
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It's still $3 above what's necessary. But it's your money...

~Tom
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Old 22nd October 2012, 10:45 PM   #17
alecu7 is offline alecu7  Canada
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Quote:
Originally Posted by disco View Post
For just a test rig why buy a heavier resistor? Are you afraid to pop your caps when it it goes open?

It's unusual to burn 100 volt in the supply of a power amp. As suggested lower AC and perhaps a choke would be better for regulation.
Well, PSUD2 shows more than 500V in open circuit. And this is what I measured too. My caps are 500V and I know they can take a bit more than that but just to be on the safe side.

I already bought the transformer so I don't know I can change the voltage now. A typical choke (2H/30ohms) will smooth the voltage but will not drop too much voltage. Are there any other options that I didn't think?

Alex
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Old 22nd October 2012, 11:11 PM   #18
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You don't need high spec resistors in the PSU.
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Old 23rd October 2012, 12:32 AM   #19
tomchr is offline tomchr  United States
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Originally Posted by alecu7 View Post
Well, PSUD2 shows more than 500V in open circuit. And this is what I measured too. My caps are 500V and I know they can take a bit more than that but just to be on the safe side.
With 360 V RMS in, you should have nearly 510 V DC with no load on the supply. 510 = 360 * sqrt(2). This will droop a bit once you load the supply. As a back-of-envelope estimate of the supply voltage under load, I use 1.3*VRMS, so you'll end up at about 470 V under load.

This assumes that your mains voltage is exactly at the rated primary voltage for the transformer.

If you wanted 400 V, you would have been better off with a 310~320 V transformer.

Personally, I'd loose the extra volts in a high-voltage regulator as this gives the best ripple rejection and provides a super-low supply impedance. If you're burning the power anyway, you might as well get something useful out of it (other than heat).

The second best option would be to reduce the secondary voltage somehow. If you can find a 40~50 V transformer, you could use it to "subtract" 40~50 V from your Edcor transformer by wiring the 40 V secondary in series with the Edcor secondary - but in opposite phase. Something like an Antek AN-0540 or AN-1240 would probably work just fine. If you want to use RC filtering, you'd probably want to get a 30-ish V transformer, but for LC the 40 V would be just about perfect.

~Tom
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Last edited by tomchr; 23rd October 2012 at 12:35 AM.
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