MOSFET Power dissipation in DC-DC forward converter

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Dear all

Here is the schematic


The formulars am using:


1. Is it right to use Vrms and Irms?
2.The rise and fall time values, extracted from simulations, are around 5 times bigger than datasheets (IRF630). See simulation ss

In the simulations you see Ids on top and Vgate in the bottom.

This gives me Psw around 0.9 W!
That gives me total power dissipation of 3.7W ?! -a lot isnt?

Can this be correct?

Thanks
 

Although a 200V FET won't be first choice for the design, Rdson of IRF630 isn't 0.7 ohm. Also ton and toff aren't correctly extracted from the waveforms. It's easy to make the simulator display the instantaneous power Vds*Id and calculate the switching losses exactly, it's simply ∫Vds*Id dt.
 

That value was found by measuring the voltage during ON period.
Click to expand...
The datasheet specifies 0.3 ohm maximum. I don't know what's wrong, may be it's an incorrect probe adjustment.
Can you highlight how to read them, please?
Click to expand...
The shown waveform is different from the text book model for switching power losses. You should refer to a literature, where the formula is derived. Then figure out, how to apply the model to the shown waveform. Or, as suggested, visualize the actual losses in the simulation and compare with the theoretical value.
 
Hi,
Although very small, there is also a gate charge loss.
Pgate = Qg * Vgs * f

According to ST Microelectronics AN2794, Conduction loss, Pcond = 1.6 * Rdson * (IMOSRms^2)

The Intersil IRF630 datasheet states a typical RDSon of 0.25ohms and max of 0.4ohms, so when you get 0.7ohms, there must be a mistake somewhere.

Hope this helps.
Tahmid.
 
Hi

Thank you for taking the time to answer.

The high VD ON is also present in the simulation results.


That was not what I asked in the first place, but ok.. let me try to brain storm a bit
 

The high VD ON is also present in the simulation results.
Click to expand...
I see about 1 V. But according to your previous simulation waveform Id is 3.8 A, so Rds would be below 0.3 Ohm.
 

I prefer ohms law to calculate the relation of Vds, Ids and Rds. You showed an instantaneous Vds (at end of conduction period) of 1 V, Ids is 3.8 A, so how about Rds?
 

FvM said:
I prefer ohms law to calculate the relation of Vds, Ids and Rds. You showed an instantaneous Vds (at end of conduction period) of 1 V, Ids is 3.8 A, so how about Rds?
Click to expand...
Thats what I actually asked you in #3, if it is wrong or right to use Ids_RMS. As you can see am using Ids_RMS (Measured value, almost the same as simulated). I guess it is wrong then?

Lets say it is 0.3 Ohm. If I want to calculate the power loss during On period, should I use Irms? I want the average power, correct?

Thank you FvM
 

Calculation of losses is a different thing. The formulas in your original post are basically correct. Tahmid has quoted an additional factor of 1.6, by I don't see, where it comes from.

Rds*Ids² is valid for resistors in general. Irms must be determined however, it's always higher than the average value Iav, the factor depends on both, duty cycle and current ripple.

P.S.: Instead of calculating conduction and switching losses separately, you can also average the instantaneous power Ids*Vds over one period and get the total Id related losses.
 
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FvM said:
P.S.: Instead of calculating conduction and switching losses separately, you can also average the instantaneous power Ids*Vds over one period and get the total Id related losses.
Click to expand...
Thanks, I will do that
 

I agree with FvM, Ids*Vds over one period is the total losses including conduction and switching losses. You don't need to calculate them separately. You will need to calculate them only if you need to know which one of them is higher (conduction or switching) to reduce at and achieve high efficiency.

Hope this help. Best wishes
 

FvM said:
Tahmid has quoted an additional factor of 1.6, by I don't see, where it comes from.
Click to expand...

It is stated in ST Microelectronics AN2794 page 17, equation 62. Pcond = 1.6 * Rdson * (IMOSRMS)^2. Here RMS current is taken, not peak.

Hope this helps.
Tahmid.
 
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Hope this helps
Click to expand...
Actually not. I have read the AN before posting my comment. As mentioned below, P = R*Irms² is valid for resistors in general. I guess, the authors of the said application note had an idea why adding a factor of 1.6 in this place. But it's not obvious, and surely not valid in general. If you understood where the factor comes from, please clarify.
 

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