Hi all,
I have a test rig for my kt88 set amplifier.
The configuration is:
Edcor XPWR105 (360V) 82uF 250R/13W 820uF 250R/13W 820uF 10k/5W 8uF
After the second 250R I get ~400Vdc for kt88 and after 10k I get ~300Vdc for 6N1P driver. Voltages are very similar with PSUII program.
My question is: is it normal that both 250R resistor reached ~150°C in about 5 minutes? After 5 more minutes temperature is the same. Resistors are Dale wire-wound silicon coated (# CW010250R0JS73). From their spec sheet it says derating power at 150°C is 60% (13Wx0.6=7.8W) I'm dissipating 7.5W so I should be ok from the wattage point of view. But what about temperature? It seams excessive to me. Apparently these resistors can go to max 350°C.
Should I replace them with caddock 25-30W?
Thanks,
Alex
I have a test rig for my kt88 set amplifier.
The configuration is:
Edcor XPWR105 (360V) 82uF 250R/13W 820uF 250R/13W 820uF 10k/5W 8uF
After the second 250R I get ~400Vdc for kt88 and after 10k I get ~300Vdc for 6N1P driver. Voltages are very similar with PSUII program.
My question is: is it normal that both 250R resistor reached ~150°C in about 5 minutes? After 5 more minutes temperature is the same. Resistors are Dale wire-wound silicon coated (# CW010250R0JS73). From their spec sheet it says derating power at 150°C is 60% (13Wx0.6=7.8W) I'm dissipating 7.5W so I should be ok from the wattage point of view. But what about temperature? It seams excessive to me. Apparently these resistors can go to max 350°C.
Should I replace them with caddock 25-30W?
Thanks,
Alex
7 + watts into a component just 45mm by 9.5mm will no doubt make that component very hot. Probably not an issue for such a resistor, but make sure it's kept a distance from other sensitive components such as electrolytics. Also they must be mounted a bit off the circuitboard, not right onto it, if you use pcb that is.
I think u can get heatsinks for those resistors, or make your own.
Kinda waste dropping so much power in a resistor. How bout using a choke?
I think u can get heatsinks for those resistors, or make your own.
Kinda waste dropping so much power in a resistor. How bout using a choke?
Resistor is about 10mm higher than the pcb. and it's ~10mm to the 82uF cap. From what I've been reading solder melts @182°C which seams too close to the 150°C. As a test load I have 2.8K/100W(2x 5.6K in parallel) in place of kt88 and 33k/25w caddock for 6n1p.
Other results are 45°C on 10k/5w and 41°C on 33k/25w. 10k resistor is also dale and similar to 250R. 33k has a small heatsink.
Other results are 45°C on 10k/5w and 41°C on 33k/25w. 10k resistor is also dale and similar to 250R. 33k has a small heatsink.
The problem is the amount of heat you are dissipating relative to the size of the component. Its thermal resistance to ambient will be quite high so you get a large temperature rise ( a quick calculation shows it is about 17 degrees per watt). One possibility is to use two 500 ohm parts in parallel but physically separated from each other so they share the dissipation. This should nearly halve the temperature rise.
Cheers
Ian
Cheers
Ian
The derating spec is probably for ambient temperature, not surface temperature. If you put a straight line between 13W at around 20C and 0W at 350C you will get around 60% at 150C. The resistor should be happy at 7.5W, but adjacent components might not be. Thermal cycling could affect solder joints eventually. I would ensure that the resistor has long legs to lose a bit of heat. In the olden days very hot resistors sometimes had screw terminals for connections so solder melting (or oxidising) was not an issue.
Since I need to buy new resistor (you suggest to get 2x 500) should I not get caddock and mount then on the case? The sides are 1/2" aluminum.
Alex
derating spec is for the case temperature not ambient. Ambient is 25°C. And this is exactly what I did/explained in post #1. 60% from 13W is 7.8W.
I'm more concerned with the high temperature that I measured.. I think I'm gonna go with caddocks for my piece of mind.
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.alecu7 said:
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.
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.
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
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
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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