Burning resistors

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pucho812

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Oct 4, 2004
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Trying to work this out.

Why does a resistor that is sourced for value and wattage from the manufacture smoke when under a minimal load

The resistor is 100ohm 1 watt.
It goes between the psu output of B+(186vdc without load) and a board with 4 x 6922 tubes on it.  There is a 100uf/250v cap to ground in parallel with the psu output and 100ohm/1w resistor.
With no tubes loaded, the resistor gets hot right away and starts to smoke.
I tested the tube board without a bench supply and there is nothing going on to indicate an issue with the b+ feeds to the plates.

The psu is specked to be 185vdc output. I can’t imagine 186VDC and change causing a problem.
 
pucho812 said:
Trying to work this out.

Why does a resistor that is sourced for value and wastage from the manufacture smoke when under a minimal load.

The resistor is 100ohm 1 watt.
It goes between the psu output of B+(186vdc without load) and a board with 4 x 6922 tubes on it.  There is a 100uf/250v cap to ground in parallel with the psu output and 100ohm/1w resistor.
With no tubes loaded, the resistor gets hot right away and starts to smoke.
I tested the tube board without a bench supply and there is nothing going on to indicate an issue with the b+ feeds to the plates.

The psu is specked to be 185vdc output. I can’t imagine 186 and change causing a problem.
likely more than 1W dissipation.

Measure the voltage across the R then calculate W. 

Power equals voltage squared divided by resistance.

JR
 
JohnRoberts said:
likely more than 1W dissipation.

Measure the voltage across the R then calculate W. 

Power equals voltage squared divided by resistance.

JR

Yes. I was more asking how I could be pulling more then 1 watt when resistance is low and no tubes are installed.
 
pucho812 said:
Yes. I was more asking how I could be pulling more then 1 watt when resistance is low and no tubes are installed.
This depends on what of your description is reality. Either this:

pucho812 said:
.... The resistor is 100ohm 1 watt.
It goes between the psu output of B+(186vdc without load) and a board with 4 x 6922 tubes on it.
Then it is in series with the load.

Or this:
pucho812 said:
There is a 100uf/250v cap to ground in parallel with the psu output and 100ohm/1w resistor.
Then it is in parallel with the load and connected between V+ and ground.....
 
analogguru said:
This depends on what of your description is reality. Either this:
Then it is in series with the load.

Or this:Then it is in parallel with the load and connected between V+ and ground.....

Well both are correct. The b+ leaves the psu and the first component is the electro cap to ground  in parallel with the trace and further down the trace is is resistor in series. The other end of the resistor goes into a daughter tube card at where b+ would be coming in.
 
pucho812 said:
Yes. I was more asking how I could be pulling more then 1 watt when resistance is low and no tubes are installed.
Kirkoff suggests that the current is all going somewhere. Look for an unusual low impedance path to ground.

JR
 
This was a good one, since I didn't have schematics. For some it would be a cake walk, for others like myself, a Challenge that I couldn't give up on.  To expedite things, I took a known working daughter tube board and plugged it in, no smoking resistor and the unit works.  This got things back up and running. I did take  the faulty daughter board with tubes causing issues back to the bench and gave it a complete go over. I discovered a shorted film cap and some burned traces. I repaired the traces and replaced the film cap. Now the parts on the daughter board  that would have been the usual suspects were all fine and not shorted out. I  tested the daughter board with a bench supply and nothing out of the ordinary came up. I eventually had a free day to try the faulty daughter board and boom back in business. No more burning resistors and everything is good.  Something to monitor but I think we are good here.  8)

It's always a good feeling when you can work out an issue without having a schematic on hand.
 
One thing that really worried me when I saw your original post was that you didn't seem to get the possible fault scenario and the concept. If you have a resistor and it starts smoking, there is only one reason. Too much energy is being dissipated in the resistor. If the volts are right on one end, then the volts are wrong on the other end. And by logic, probably low. So this points us to a low impedance on the wrong end. In most cases this will be a shorted component. So hopefully you have learned from this exercise, and maybe other readers too!
 
radardoug said:
One thing that really worried me when I saw your original post was that you didn't seem to get the possible fault scenario and the concept. If you have a resistor and it starts smoking, there is only one reason. Too much energy is being dissipated in the resistor. If the volts are right on one end, then the volts are wrong on the other end. And by logic, probably low. So this points us to a low impedance on the wrong end. In most cases this will be a shorted component. So hopefully you have learned from this exercise, and maybe other readers too!

no I understood the concept I just wasn't making the connection because by all accounts there was no load present with the resistor was smoking as tubes were not installed.
 
pucho812 said:
It's always a good feeling when you can work out an issue without having a schematic on hand.
I used to take special pleasure from being able to repair gear without schematics. Generally its the same likely suspects.

Also handy because back in the day hand drawn schematics were not very reliable.

JR

PS; I just got a PM asking about fixing a product I designed 40 years ago...
 
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