Resistor Joule Rating

[gar]

120104-2116 EST

On peak current. Many years ago I designed a capacitor discharge drawn arc stud welder with about 30,000 ufd capacitance, and peak discharge current was in the 1000 A range. There were 6 separate capacitors so about 160 A per capacitor. In an automotive application there might be one weld every 12 seconds. Given the capacitor is designed for pulse capability in LMAO's application there is no problem at something considerably less than 22 A per capacitor.

.

 

[LMAO]

120104-2034 EST

Revising my previous figures by multiply by 1000 I am predicting about 6 to 8 thousand watt-seconds of energy per charge.

Steve is correct that you need to know the damage point. In general I have considered that you could hit a regular tubular Ohmite resistor with with about 10 times its rated power for about 1 second or more with safety. It took a number of seconds at this 10 times power level to blister the Viterous enamel.

Today I took an Ohmite L25J 10R 10 ohm 25 W and applied 123 V to it. This was 1513 W into a 25 W resistor. It was about 2 seconds to crack the enamel. The ceramic core was not cracked. The resistance was not appreciably changed. This was a continuous application of 60.5 times the resistor power rating for 2 seconds. The watt-seconds were about 3000. So I put about 1/2 the watt-seconds into this tiny 25 W resistor that LMAO is putting into an 800 W resistor.

I think the important question on the 800 W resistor is what long term thermal-mechanical fatigue will occur. But it may not matter if there is a bolt thru the middle of the tubular resistor.

I do question whether an 800 W resistor is in a package as short as 10" unless its diameter is quite large. However, some of the large power resistors are made with built-in metal fins.

On the voltage capability of a resistor that some have questioned. Suppose there were 100 turns to make the resistor, and of necessity the turns require insulation between them. It is not difficult to achieve 1000 V capability between adjacent turns. Thus, if the over the surface distance will support 100,000 V, then that resistor could have that voltage applied to it. The short time voltage rating of a resistor is a function of voltage breakdown, not its power rating. Long ago I made a high voltage scope probe good for over 30,000 V using a resistor about 4" long. This was for spark-plug voltage measurement.

.

Well, this is my current curve and, as you see, the current is not constant and drops down pretty fast. One thing I did not mention is that the resistor is used in a prototype product and I'll be using a different type of resistor for the final product. I am using it because it was the first one I found in our warehouse (we use these as space heaters in our switchgear cabinets). So I am not really worried about it failing.
for the final product, I'll just send my current curves and expected duty cycle to a resistor manufacturer.



Sorry about your cooked resistor. You didn't have to do it!

 

[gar]

120106-1430 EST

Your curve is just the standard exponential decay curve. The energy vs time is the intrgral of

R*(I*e^(-t/R*C))² = R*I²*e^-2*t/R*C

For the total energy it is from 0 to t=infinity.

The integral of e^at is e^at/a

So we have
Energy dissipated in the resistor is -R*I²*R*C*e^-2*t/R*C
evaluated at infinity and minus that at 0 = 0 (-)(-)(I*R)²*C/2 = V²*C/2.

This is the proof that the energy dissipated in the resistor to charge the capacitor is equal to the energy put into the capacitor.

Don't worry about my resistor I have many that may never have a use. I expected it to be damaged. I used that high an overload because that was convenient.

I might suggest an open resistor. Such as a Nichrome coiled wire around a ceramic core as the type of resistor you might want to consider. But obviously you put a mechanical shield around the resistor. I believe HAAS uses a somewhat open resistor for dumping overvoltage on their DC bus. I will try to remember to look the HAAS resistor.

.