RMS current through a switched inductor

Hi,

I am trying to calculate the RMS current through a MOSFET for the sake of calculating its power dissipation. The MOSFET is acting as a low side driver with an inductive load. Now most of the app notes and articles out there model the inductor current as ramp for calculations. This is ok for high frequencies of switching (in KHz). But I think for low frequencies (say 100Hz) the exponential nature of inductor current must be considered. Otherwise, we would underestimate the current, especially for higher duty cycles. Therefore, I tried to calculate the RMS current using standard RMS current formula using integration.



I evaluated the integral in this formula. But when i plug in the actual values, the term under square root goes negative. I don't know whether I am evaluating the integral incorrectly or I am approaching this whole thing in the wrong way. Any help would be highly appreciated. If there is any other way of calculating the RMS current, that too would be helpful.

If the FET cannot carry a current of V/R, then I would think that the design is marginal. How close to R/L is your pulse? If its 4R/L then the current will be 95% of the DC value.
Frank

I am worried about whether I am providing enough heatsink for the FET or not. Unless I know the exact power, I cannot be sure of that. The inductor is 4.5mH with 4 ohms of resistance. The frequency is 300Hz with a duty cycle varying from 50% to 90%.

T= L/R = 4.5 X 10^-3/4 = 1.1 mS . F= 300 HZ, 1/F = 3.3mS. At 90% mark, pulse width = 3mS = 2.7 X T. So putting this into your formula. Gives e^- (TR/L) = e^ - (1/2.7) = e^ -.37 = 1/e^.37 = 1/1.44 = .69. So 1-.69 = .31, so the square bracketed bit becomes V/R X .31 .
Well that is the answer but I feel its wrong. We all know that the current is 63% of its final value when the time period = T, so if we say the exponential term is e^ (L/TR), then the numbers become:- e^-2.7 = 1/ e^2.7 = 1/14.8 = .067, 1- .067 = .932, so the square bracketed term becomes V/R X .932, which looks right.

What do you say?
Frank

chuckey said:

T= L/R = 4.5 X 10^-3/4 = 1.1 mS . F= 300 HZ, 1/F = 3.3mS. At 90% mark, pulse width = 3mS = 2.7 X T. So putting this into your formula. Gives e^- (TR/L) = e^ - (1/2.7) = e^ -.37 = 1/e^.37 = 1/1.44 = .69. So 1-.69 = .31, so the square bracketed bit becomes V/R X .31 .
Well that is the answer but I feel its wrong. We all know that the current is 63% of its final value when the time period = T, so if we say the exponential term is e^ (L/TR), then the numbers become:- e^-2.7 = 1/ e^2.7 = 1/14.8 = .067, 1- .067 = .932, so the square bracketed term becomes V/R X .932, which looks right.

What do you say?
Frank

Click to expand...

I think you inverted when you shouldn't have in the first analysis. Your second one is correct. This is in line with your observation in #2 that for such long pulses, the difference between the RMS current and the "always on" current is going to be too small to bother with. I would size the heatsink to take the full "always on" current of V/R.