Resistor Joule Rating

[LMAO]

Anyone knows how to calculate a resistor's "Joule rating"? To be clear, I have a 40 ohm resistor which I use to charge up a 16.8mF cap (inverter cap). Voltage across the cap is about 850VDC. so the current starts at about 22A and takes about 3 seconds for cap to reach 840VDC. I am trying to size the resistor; obviously, it doesn't have to be rated 18kW (22 x 850) . So what should it be, considering the "charge up process" is quite rare and infrequent (maybe once every hour)? How would you calculate the "Joule rating" required for a resistor that gets the job done?

thanks

 

[steve66]

Anyone knows how to calculate a resistor's "Joule rating"? To be clear, I have a 40 ohm resistor which I use to charge up a 16.8mF cap (inverter cap). Voltage across the cap is about 850VDC. so the current starts at about 22A and takes about 3 seconds for cap to reach 840VDC. I am trying to size the resistor; obviously, it doesn't have to be rated 18kW (22 x 850) . So what should it be, considering the "charge up process" is quite rare and infrequent (maybe once every hour)? How would you calculate the "Joule rating" required for a resistor that gets the job done?

thanks

Resistors are rated based on power. I wouldn't try and take any credit for the fact that the peak current only lasts a short time.

The peak power in the resistor will be I^2*R. That would be about a 20KW resistor.

Edit: Obviously, typical resistors are rated much less. And this one would release a huge amount of heat.

By the way: that is a huge amount of energy that would be stored in the cap. Please be very careful.

 

[T.M.Haja Sahib]

Anyone knows how to calculate a resistor's "Joule rating"? To be clear, I have a 40 ohm resistor which I use to charge up a 16.8mF cap (inverter cap). Voltage across the cap is about 850VDC. so the current starts at about 22A and takes about 3 seconds for cap to reach 840VDC. I am trying to size the resistor; obviously, it doesn't have to be rated 18kW (22 x 850) . So what should it be, considering the "charge up process" is quite rare and infrequent (maybe once every hour)? How would you calculate the "Joule rating" required for a resistor that gets the job done?

thanks

I think the resistor is used to control the time taken by a capacitor to fully charged.There is some exponential equation relating voltage,time,resistor and capacitance which may be found in text books on electrical engineering.Using that equation,you can calculate the resistance required.

 

[sgunsel]

Resistors, although appearing to be simple devices, can be complex in many ways. Actual resistor wattage is far from simple - you need to get the manufacturer's specifications and read them very carefully. A "100 watt" resistor may only be capable of dissipating 100 watts under extreme cooling conditions or for a VERY short period of time - it may only be rated for only a few watts in open air on a continuous basis. At high voltages, the resistance can vary based on the applied voltage. And of course temperature also alters resistance, and amps make heat. Wattage can be the real sleeper though, and if at all unsure about power dissipation requirements, it is probably best to oversize massively.

 

[LMAO]

Resistors, although appearing to be simple devices, can be complex in many ways. Actual resistor wattage is far from simple - you need to get the manufacturer's specifications and read them very carefully. A "100 watt" resistor may only be capable of dissipating 100 watts under extreme cooling conditions or for a VERY short period of time - it may only be rated for only a few watts in open air on a continuous basis. At high voltages, the resistance can vary based on the applied voltage. And of course temperature also alters resistance, and amps make heat. Wattage can be the real sleeper though, and if at all unsure about power dissipation requirements, it is probably best to oversize massively.

You are about right; I think the best way is just to give the resistor manufacturer my current curve through resistor and ask them for a 40ohm resistor which can handle this much current every 30 minutes or so in about 40 degree C.

 

[GeorgeB]

Anyone knows how to calculate a resistor's "Joule rating"? To be clear, I have a 40 ohm resistor which I use to charge up a 16.8mF cap (inverter cap). Voltage across the cap is about 850VDC. so the current starts at about 22A and takes about 3 seconds for cap to reach 840VDC. I am trying to size the resistor; obviously, it doesn't have to be rated 18kW (22 x 850) . So what should it be, considering the "charge up process" is quite rare and infrequent (maybe once every hour)? How would you calculate the "Joule rating" required for a resistor that gets the job done?

thanks

The peak will be the issue, not average. Localized heating will hit you if you try to do much time averaging ... like RMS loading an electric motor.

Now let's lab workbench this ... 5 residential irons (about 8 ohms) in series ... rated for maybe 1.5kW each, 7.5kW total, "ONLY" about 250% overload, likely, not going to create damage, inexpensive and available.

 

[LMAO]

The peak will be the issue, not average. Localized heating will hit you if you try to do much time averaging ... like RMS loading an electric motor.

Now let's lab workbench this ... 5 residential irons (about 8 ohms) in series ... rated for maybe 1.5kW each, 7.5kW total, "ONLY" about 250% overload, likely, not going to create damage, inexpensive and available.

residential irons? they would look good in my VFD cabinet no thanks. I am already using 600w resistor and seems to be working fine. I think you are being way too conservative. I have about 22A through the resistor at t=0 and current goes to zero in a couple of seconds...

 

[Speedskater]

Remember that this is a 850VDC system!!!
Many power resistors are not rated at this high a voltage, let alone irons.

 

[LMAO]

Remember that this is a 850VDC system!!!
Many power resistors are not rated at this high a voltage, let alone irons.

you are correct; a 40 ohm, 600w resistor means it can be operated at 155V or lower (P= V^2/R → V = square root(P*R) = 155V). But that's continuous voltage across the resistor. my voltage across the resistor looks like the below image. it drops down to 0 in a couple of seconds and the procedure is repeated quite rarely (maybe once every 30 minutes). So I believe I am safe, don't you agree??

 

[steve66]

you are correct; a 40 ohm, 600w resistor means it can be operated at 155V or lower (P= V^2/R → V = square root(P*R) = 155V). But that's continuous voltage across the resistor. my voltage across the resistor looks like the below image. it drops down to 0 in a couple of seconds and the procedure is repeated quite rarely (maybe once every 30 minutes). So I believe I am safe, don't you agree??


I don't agree. You are subjecting a 600 watt resistor to 20,000 watts. Without knowing more aout the resistor in question, it may work fine forever, or it may die tomorrow. I'm not sure what your definition of "safe" is, but I would say you are lacking any engineering justification that this will work in the long term.

I'm a little concerned about your Capactior too. Is it made to handle such rapid charging?

Steve

 

[gar]

120103-2145 EST

Something does not compute.

The time constant of 40 ohms and 16.8 ufd is 672 microseconds. At 3.4 milliseconds, 5 time constants, the current will drop to 0.006 of its initial value.

For reference i(t) = Iinitial * e^(-t/RC).

Was the initial current measured at at 22 A or just calculated?

Let us assume the initial current was 22 A and it dropped to essentially 0 in 3.4 milliseconds, then without calculating the energy I will guesstimate using 1 time constant that the energy is 22*850*0.00067*0.7 = 8.8 watt-seconds. Peanuts for a 100 W Ohmite resistor once per half hour. A 100 W resistor is about 10 inches long and 850 V would be no problem. Also there is no problem putting 400 V on a 1/2 W resistor about 1/2 inch long.

If the current does take upwards of 1 second to reach near 0 current, then the initial current is much less than 22 A, and something other than the 40 ohm resistor is limiting the current.

A different way to view the energy into the resistor is to calculate the total energy transferred to the capacitor. This is E = C*V^2/2 = 16.8*10^(-6)*0.85^2*10^6/2 = 6 watt-second. Assume an equal amount is dissipated in the resistor as is put into the capacitor, then 6 watt-seconds went into the resistor.

I bet the resistor is cool to the touch to just charge the capacitor. If there is a continuous load current on the supply containing the the capacitor after the capacitor is charged, then the resistor can get quite hot. At 4 A this would be 4*4*40 = 640 W.

I would have no problem putting 6 watt-seconds into a 1 W resistor every half hour.

Check my math and see if it is correct.

.

 

[david luchini]

120103-2145 EST

Something does not compute.

The time constant of 40 ohms and 16.8 ufd is 672 microseconds. At 3.4 milliseconds, 5 time constants, the current will drop to 0.006 of its initial value.

The OP said Milli-Farads, not Micro-Farads

 

[gar]

120103-2255 EST

Sorry about that. I grew up when mfd was microfarad.

So multiply by 1000.

His present resistor won't get very hot from just charging the capacitor.

.

 

[T.M.Haja Sahib]

residential irons? they would look good in my VFD cabinet no thanks. I am already using 600w resistor and seems to be working fine. I think you are being way too conservative. I have about 22A through the resistor at t=0 and current goes to zero in a couple of seconds...

View attachment 6188

You may redraw the above curve with i squared on the ordinate against the time.Multiply the area below the curve with the resistance value.This will give you the joule rating of the resistor.If you want to calculate analytically,use the suggestion in post # 3 and apply integration on the exponential equation for current.

 

[steve66]

If you really want to know if this is reliable, you need a damage curve for the resistor and the capacitor. Everything else is just a guess.

 

[T.M.Haja Sahib]

In that case,the OP has to consult the concerned manufacturers and make it sure before proceeding further.

 

[LMAO]

120103-2145 EST

Something does not compute.

The time constant of 40 ohms and 16.8 ufd is 672 microseconds. At 3.4 milliseconds, 5 time constants, the current will drop to 0.006 of its initial value.

For reference i(t) = Iinitial * e^(-t/RC).

Was the initial current measured at at 22 A or just calculated?

Let us assume the initial current was 22 A and it dropped to essentially 0 in 3.4 milliseconds, then without calculating the energy I will guesstimate using 1 time constant that the energy is 22*850*0.00067*0.7 = 8.8 watt-seconds. Peanuts for a 100 W Ohmite resistor once per half hour. A 100 W resistor is about 10 inches long and 850 V would be no problem. Also there is no problem putting 400 V on a 1/2 W resistor about 1/2 inch long.

If the current does take upwards of 1 second to reach near 0 current, then the initial current is much less than 22 A, and something other than the 40 ohm resistor is limiting the current.

A different way to view the energy into the resistor is to calculate the total energy transferred to the capacitor. This is E = C*V^2/2 = 16.8*10^(-6)*0.85^2*10^6/2 = 6 watt-second. Assume an equal amount is dissipated in the resistor as is put into the capacitor, then 6 watt-seconds went into the resistor.

I bet the resistor is cool to the touch to just charge the capacitor. If there is a continuous load current on the supply containing the the capacitor after the capacitor is charged, then the resistor can get quite hot. At 4 A this would be 4*4*40 = 640 W.

I would have no problem putting 6 watt-seconds into a 1 W resistor every half hour.

Check my math and see if it is correct.

.

You are correct. I checked the resistor temperature using a laser thermometer and it was just a couple of degrees hotter than ambient, even though I was pre-charging it a few times a minute. The resistor, as you said, is about 10 inches.
Your math is also correct except that, as you already know, my cap is 16.8mF not uF.

 

[LMAO]

If you really want to know if this is reliable, you need a damage curve for the resistor and the capacitor. Everything else is just a guess.

Cap is already rated for 1200VDC and even higher, it is an inverter capacitors. There are actually a bunch of them, 16.8 mF is the total capacitance. Resistor is fine too. Resistors' watt rating depends on duty cycle and if you use them rarely you can increase the watt rating.

 

[steve66]

Cap is already rated for 1200VDC and even higher, it is an inverter capacitors. There are actually a bunch of them, 16.8 mF is the total capacitance.

I was more worried about the current rating on the cap. Very fast charging and discharging can be hard on some caps.

Resistors' watt rating depends on duty cycle and if you use them rarely you can increase the watt rating.

Yes, but without the damage curve, you are only guessing how much you can increase the watt rating. Take this resistor for example:

http://www.ohmite.com/cgi-bin/showpage.cgi?product=tap800

Its a 600 Watt (maybe like yours) and it has a 1200W rating for 10 sec. But that's still only about a 1/15th of your surge. The resistor temp. can't tell you much either - you don't have any idea how high the internal temperature is getting. The current pulse isn't long enough for the heat to be transfered to the case.

 

[gar]

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.

.