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Peak and RMS current - when to use which for thermal?

**shadewind**:

I have a current sense resistor on the low side of a boost converter switching MOSFET for current mode control and the average current through this resistor (and thus also the MOSFET) would be D * I(Lavg) where D is the duty cycle and I(Lavg) is the average inductor current. Also, the current through the flyback diode would be (1 - D) * I(Lavg).

From what I understand here, it is the average (or rather RMS) current that matters. But when does the frequency become so low that it is the peak current that matters? Some components have current limits for other reasons than thermal (inductor saturation et.c.) but thermal is what my question is about.

**Zero999**:

The RMS is not the average current but the current which results in the same average power dissipation as the same DC current so 1A RMS will produce the same power into a 1R resistor as 1A DC: 1W.

The point at which the peak current becomes important depends on the thermal time constant of the resistor. I've not done much research into the subject but I'd think the peak current becomes important below the lower cut-off frequency which is equal to 1/(2pi*t) where t is the thermal time constant.

**Alex**:

In other words to the ones Hero used, it depends on how quickly the resistor can get rid of the heat to prevent itself from overheating to destruction.

If the rate of heat dissipation to ambient is lower than your rate of power dissipation on the resistor, then obviously you cannot sustain it without damage. As you have guessed, you need to start looking at pulse characteristics. These figures closely describe what would happen if the resistor was thermally insulated from ambient, which is the case due to the thermal mass of the resistor's internal materials. If you are familiar with thermal impedance networks (there is an episode on here), then this behaviour is equivalent to adding capacitors to the nodes hence increasing the thermal time constant.

Back to the boost converter, a pulse of juice through your sense resistor has finite energy. That will be equal to the plots of I through the resistor multiplied with V across (hence instantaneous power versus time) and then integrate this plot to find the dissipated energy per pulse, or more importantly the maximum energy over a given period of time. It doesnt matter if your pulse has an amplitude of 1 ExaWatt on a 1/4W resistor if it has no time duration; energy is zero. Having added some margin, check with the datasheet. Very modern DSOs will do this work for you, but you can use a workaround (excel, approximations with pen and paper etc).

**shadewind**:

--- Quote from: Hero999 on September 10, 2011, 10:46:58 am ---The RMS is not the average current but the current which results in the same average power dissipation as the same DC current so 1A RMS will produce the same power into a 1R resistor as 1A DC: 1W.

The point at which the peak current becomes important depends on the thermal time constant of the resistor. I've not done much research into the subject but I'd think the peak current becomes important below the lower cut-off frequency which is equal to 1/(2pi*t) where t is the thermal time constant.

--- End quote ---

Yes, I know the difference between aveage and RMS, I just messed it up this time.

**shadewind**:

--- Quote from: Alex on September 10, 2011, 12:17:54 pm ---In other words to the ones Hero used, it depends on how quickly the resistor can get rid of the heat to prevent itself from overheating to destruction.

If the rate of heat dissipation to ambient is lower than your rate of power dissipation on the resistor, then obviously you cannot sustain it without damage. As you have guessed, you need to start looking at pulse characteristics. These figures closely describe what would happen if the resistor was thermally insulated from ambient, which is the case due to the thermal mass of the resistor's internal materials. If you are familiar with thermal impedance networks (there is an episode on here), then this behaviour is equivalent to adding capacitors to the nodes hence increasing the thermal time constant.

Back to the boost converter, a pulse of juice through your sense resistor has finite energy. That will be equal to the plots of I through the resistor multiplied with V across (hence instantaneous power versus time) and then integrate this plot to find the dissipated energy per pulse, or more importantly the maximum energy over a given period of time. It doesnt matter if your pulse has an amplitude of 1 ExaWatt on a 1/4W resistor if it has no time duration; energy is zero. Having added some margin, check with the datasheet. Very modern DSOs will do this work for you, but you can use a workaround (excel, approximations with pen and paper etc).

--- End quote ---

Yes, now, when thinking about, it makes sense. I assume the resistor has some temperature limit that it can handle. If the power dissipation is larger than the heat dissipation, this means that the temperature will rise as a function of time and if this critical temperature is reached, then it's a problem. If the RMS power dissipation is more than the heat dissipation, then it won't work no matter how short the pulses are.

Is my thinking correct?

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