How do I determine the power reqirements of a resistor with pulsing. .g. if a 2W resistor can handle 2A continuous, can it handle 20A at 10% duty cycle, or at 1% duty cycle. What about the ontime, is there a difference if using 250us on time/25ms offtime versus 250ms ontime/25s offtime. Will it vary from resistor to resistor.

 

Yes. All of this is covered in the data sheet for the resistor...

 

if a 2W resistor can handle 2A continuous, can it handle 20A at 10% duty cycle

Possibly.
There are several parameters which will affect the handling ability: resistor shape, size, thermal capacity, air circulation, ambient temperature .....
For example, if the thermal capacity is small then in 250ms the resistor may overheat with 20A going through it (unlikely), whereas in 250us that would be extremely unlikely.
'Consult the datasheet' is the usual advice.

 

How do I determine the power reqirements of a resistor with pulsing. .g. if a 2W resistor can handle 2A continuous, can it handle 20A at 10% duty cycle, or at 1% duty cycle. What about the ontime, is there a difference if using 250us on time/25ms offtime versus 250ms ontime/25s offtime. Will it vary from resistor to resistor.

In a resistor the power varies as the square of the current so, for example, at 10% duty-cycle it could handle √20 = 4.47A max. if it is rated for 2A continuous.

The maximum on time for this to work depends on the thermal time-constant of the resistor, as noted. Likely this time is less than a second.

 

Possibly.
There are several parameters which will affect the handling ability: resistor shape, size, thermal capacity, air circulation, ambient temperature .....
For example, if the thermal capacity is small then in 250ms the resistor may overheat with 20A going through it (unlikely), whereas in 250us that would be extremely unlikely.
'Consult the datasheet' is the usual advice.

Excellent clarification Alec,

I could just see the complaint, ... "They said I could run at 20 amps if I was at 10% duty cycle, 2.4 hours per day."

 

Hi,

The average power in a resistor is:
P=d*I^2*R

where
P is the power in watts,
d is the fractional duty cycle (0.25=25 percent 'on'),
I is the current in amperes,
R is the resistance in ohms.

If your resistor can handle 2 watts with 2 amps when the duty cycle d is 1.00, then we have:
2=1*2^2*R

so R works out to 1/2 ohm even though that may not be the real resistance of the device.

Using this information and again using:
P=d*I^2*R

we get:
2=d*I^2*1/2

and solving for d we get:
d=4/I^2

At 20 amps this comes out to:
d=4/20^2=4/400=1/100=0.010

which is a 1 percent duty cycle.

This is all based on the average power though, and that assumes that the frequency is higher than the thermal time constant of the resistor, which means we have a second constraint to think about: the resistors thermal time constant and it's relation to the minimum frequency.

If, with a given power input, the resistor heats up to 100 degrees C in 5 seconds and then the temperature stays relatively constant, that means the thermal time constant is about 1 second. That would put the min raw frequency at about 1 Hertz.

To test this you could apply some power (in the same environment as the intended application) and wait for the temperature to get within about 0.993 of it's final temperature, then divide the time it took to get there by 5, and that would be the time constant. The min frequency should then be f=1/(tau*Pr), where tau=t/5 and t was the time it took to reach 0.993 of it's final temperature with the given power input, and Pr is the power ratio FullActualPower/FullTestPower (both at 100 percent duty cycle).
In practice you can probably get away with a little lower frequency than this, but you can not go too low. A cycle time of 1 full day would be too long for example for a regular resistor.

 

Components can have a max current spec separate to max power dissipation.

That is because the internal wires etc have PTC of resistance.

So a part that is safe for 1A continuous may destroy itself at 100A for 1% duty cycle.

 

The relationship between maximum current and duty cycle is not linear because power is proportional to the square of the current.

At 50% duty cycle, the equivalent current is 1.4 times the current at 100% duty cycle.
At 10% duty cycle, the equivalent current is only 3 times the max current.
At 1% duty cycle, the equivalent current is 10 times the max current.

 

The relationship between maximum current and duty cycle is not linear because power is proportional to the square of the current.
...

No, no, no. Mr Al has it right.

Both the voltage applied to the resistor and current through the resistor are being interrupted by the pulse width modulator (switch). During the time the switch is closed, the power in the resistor is E*I = P. During the time the switch is open, I=0 and V=0, so P=0. During the time the switch is closed, the power in the resistor is EI or I^2R

You have to integrate (average) the power across the Off period and On period.

Look at the following example. During the time the switch is closed, the current through the 1Ω resistor is 10A, I(R1) is shown as the blue trace on the upper plot pane. Peak power is 100W = V(a) * I(R1) shown as the red trace on the lower plot pane. Power could also be calculated as I(R1)^2*R1.

The average power in resistor R1 is determined by integrating the power vs time plot, which LTSpice does for you. The result is shown in the inset box. As expected, since the switch is closed 20% of the time, the Power is 20% of 100W, or 20W.