Calculate simple mosfet power

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kakiayam

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Hi all,

Would like some advise on some power calculation of a simple mosfet. Currently the mosfet BSC100N06LS3 is generating a lot of heat and I would like to calculate the power it is generating.

My simple calculation does not make sense, or I am just confused.

Power = (Voltage drop x Voltage Drop) / Rdson
= (10.27 x 10.27) / 0.010 Ohm
= 10547.29 W

I used the Rdson max to calculate the power. The result does not make sense. Can anyone help to clear my mind on this? Appreciate it.




Thanks.

KaKiaYam
 


If your load is so large to cause the 10 V voltage drop across the Rdson of 0.01 Ohm, then your calculation is correct.

In well designed circuits the MOSFET acts as a switch, and the load current may be accordng to your MOSFET specification e.g. 10A DC.
If the MOSFET is well open and really has RDson of 0.01 Ohm, then the dissipation heat to remove from the MOSFET will be P = RI*I= 10 x 10 x 0.01 = 1 W.
If the MOSFET is hotter, then either the current is higher, or it is not open to RDSon of 0.01 Ohm.
Check the gate voltage with a scope and use a lighter load ( a lamp or a 1... 10 W resistor) to see if your MOSFET is well open and its load current correct.

- - - Updated - - -


If your load is so large to cause the 10 V voltage drop across the Rdson of 0.01 Ohm, then your calculation is correct.

In well designed circuits the MOSFET acts as a switch, and the load current may be accordng to your MOSFET specification e.g. 10A DC.
If the MOSFET is well open and really has RDson of 0.01 Ohm, then the dissipation heat to remove from the MOSFET will be P = RI*I= 10 x 10 x 0.01 = 1 W.
If the MOSFET is hotter, then either the current is higher, or it is not open to RDSon of 0.01 Ohm.
Check the gate voltage with a scope and use a lighter load ( a lamp or a 1... 10 W resistor) to see if your MOSFET is well open and its load current correct.

In your schematic you indicate the output voltage +4.73 V and gate voltage +6.8 V both against ground. This means the gate-source voltage is only ~2 V which may not open the MOSFET to get RDson of 0.01 Ohm. Use a potentiometer in place of the 6.8 V Zener, and adjust the gate-source voltage to a specified value (logic MOSFETs need 4 V or more).
 
Okay, I guess my understanding of mosfet is incorrect. I assume the Rdson to be the resistance between D and S at ALL circumstances when the gate is driven. So the parameter Rdson max in the datasheet only shows the max resistance when the mosfet is fully switch on (saturated). Am I correct in this? But if that's the case, based on the circuitry above, the mosfet should have been fully turn on, considering a Vgs of ~+2V07. I am more of the BJT guy, but in terms of mosfet, all I know is that it turns on and off when the Vgs parameter is satisfied.

In any case, i am still scratching my head thinking of how to reduce heat at the mosfet while at the same time provide a decent voltage drop from the incoming of 15V. Any suggestions will be very much welcome. Appreciate it guys.

KaKiaYam

- - - Updated - - -

would find ways to estimate the current drawn by the load, thanks.
 

Hi,

...fully switch on (saturated). Am I correct in this?
Click to expand...
Yes, you are.

...the mosfet should have been fully turn on, considering a Vgs of ~+2V07..
Click to expand...
In datasheet you see V_GS(th) to be 1.2/1.7/2.2V (min/typ/max) with a drain current of 23uA.
This is the point where the fet is getting slightely conductive. Here the fet is not fully saturated.
So the 2.07V are well within specification - not saturated.

In any case, i am still scratching my head thinking of how to reduce heat at the mosfet while at the same time provide a decent voltage drop from the incoming of 15V. Any suggestions will be very much welcome. Appreciate it guys.
Click to expand...
With a linear regulation it is not possible to reduce power loss. You can work with Zener diodes, resistors, BJTs, fets... it is always the same: P_tot = V_drop x Current.

One way to improve power loss is to use a switching regulator.
Look for "step down" or "buck" converters.

Good luck
Klaus
 
You are not fully turning on the MOSFET, thus, it is operating in LINEAR mode, like what one would have for a power amplifier output. Thus, your resistance is higher and you are dissipating significant heat from the V*I across&through the MOSFET.
Reading the datasheet for the part, you need to have Vgs of between 6V to 10V to get the RDS down to the 10milli-ohm level. At just over 2V, you are barely turning it on and the RDS on isn't even on the graph shown in the datasheet.
Adjust your resistor and zener diode, if that is the topology you want to use, to get the VGS at least to 6V. Also, make sure that you've got plenty of heatink or PCB space allocated for heatsink purposes - that part wants quite alot of heatink space if you are using it to anywhere approaching its amp capacity.
 

The total power loss of a mosfet in working is the result of conduct loss add switching loss.except for the explanation above, if you wanna know the exact power loss which make your mosfet heat,you need a oscilloscope to show the switching current(Igs) and switching voltage(Vgs) to know their product relate to switching time.
hope this helpful.
 

Hi,


Because this is a non switching application all you need to know is current and voltage. All you need is a DVM.

Klaus
 

KlausST said:
Hi,



Because this is a non switching application all you need to know is current and voltage. All you need is a DVM.

Klaus
Click to expand...
would you please explain it in detail?
 

FvM said:
Can you explain where you see switching involved in the circuit?
Click to expand...
Got it，in this application the mosfet would not switch frequently.
Thank you very much.
 

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