Hi,
I am driving an irf640 mosfet with ir4427. Vg is 12V . for the mosfet Qg is 83nC. from Microchips documentation Ig = Qg/T(trans)
the driving pulse is 50Khz. so T(Trans) = 1/50000. From the above equation I calculated Ig to be .00415 Amps.
The gate resistor value would be 2891.5 Ohm.
Am I correct in my above calculations.
VDD is 100V and Id = 11Amps........
The problem is that the mosfets get very very hot just after 1 min and I have to switch off!. Currently I do not have the luxury of a scope as it is out on loan!!!!..
Any help is appreciated
cheers .. Emmkay.
Hello there,
The problem is the switching power in the mosfet is far too high. The load power doesnt have as much to do with it, it's the mosfet power during the switching intervals. Lets take a quick look at that and derive some formulas that will make choosing a reasonable value for the gate drive current quick and easy.
There is a certain maximum power consumption level that a mosfet can consume given it is allowed to switch from totally on to totally off. This occurs in a situation similar to your original one where the switching period is set to about half of the total period. The equation for this is quite simple:
PTmax=(Epk*Ipk)/6
where
Epk is the max voltage (100 in your case), and
Ipk is the max current (11 in your case), and
the 6 comes about by estimating the power in the mosfet for the above mentioned conditions.
From this max power calculation we can derive the total switching power estimate by simply multiplying PTmax times the duty cycle when we consider the duty cycle to be the time the transistor is switching on divided by the entire cycle period. This equation becomes:
PT=2*F*PTmax*Ton
where
PT is the total switching power that the transistor has to dissipate, and
F is the switching frequency (50kHz), and
PTmax is calculated from above, and
Ton is the time it takes to switch the mosfet on or off (which we will calculate next).
This allows us to estimate the total power the mosfet will have to dissipate just because of the switching on and off itself and does not include the conduction or leakage losses. As we will see, it is good to know this switching power loss information.
To estimate the power dissipated we need to know Ton too. This is calculated from the gate charge and the gate drive current as:
Ton=Cg/Ig
where
Cg is the gate charge, and
Ig is the gate current.
Now we have a way to estimate switching power dissipation from the gate charge, gate current, peak voltage Epk, and peak current Ipk, and the switching frequency F.
For example, with a gate charge Cg=83nC and a gate current Ig=0.00415 we get turn on and turn off times of:
Ton=Cg/Ig=83e-9/0.00415=0.00002 or 20us. That's quite a long time for a mosfet BTW.
Next, the max power figure is:
PTmax=(Epk*Ipk)/6=(100*11)/6=183.3 watts.
To calculate the actual switching power now we use:
PT=2*F*PTmax*Ton=2*50000*183.3*20e-6=366.6 watts.
This estimate is a bit fictitious as the max can only be as great as 183.3 watts so in this case the power would really be 183.3 watts, but even with 183 watts the transistor will get very very hot and eventually burn up. In any case, we can take the first estimate to be about 367 watts numerically and then go from there.
Lucky for us, the equations are all linear in Ig and PT, so when we multiply Ig by a factor X we can divide the power PT by X too and maintain the same exact relationships.
For example, to lower the switch power PT from 367 watts to 36.7 watts (a division by 10) we would multiply the gate current Ig by 10. Doing this we get IgNew=Ig*10=0.00415*10=0.0415 amps. That provides us with a faster switching time and about 10 times less power dissipation for switching, but even 37 watts isnt that good, so lets do it one more time:
IgNew2=IgNew*10=0.0415*10=0.415 amps
and this takes us down to about 3 or 4 watts for switching, which is far less than when we started.
We'll double check the Ton time:
Ton=Cg/IgNew2=83e-9/0.415=200e-9
so we're down to about 200ns and that's about where we want to be, or even less, 100ns if we can. To get down to around 100ns we would need about 0.8 amps, but typical mosfet driver chips go up to 1 amp so that's even better if we want to get really good with the switching power dissipation.
So to sum up, a workable design should be obtained with about 400ma gate drive, with 800ma even better, and if you feel like buying a mosfet driver chip made just for this purpose you can get even a little better yet.
In closing, there is one more little formula derived by combining the above formulas, and this formula allows you to 'dial in' your desired switching power dissipation:
Ig=2*F*Cg*Epk*Ipk/(6*PT)
where
Ig is the gate current that will be required to obtain the designs target PT,
F is the frequency,
Cg is the gate charge,
Epk is the max voltage,
Ipk is the max current,
PT is the target switching power dissipation.
Note that it is a good idea to check Ton after using this last formula too just to make sure you have a reasonable turn on time period.
For example, if we want to try to achieve 4 watts switching power dissipation we would substitute all the values and calculate Ig. Using the quantities from before, we would get:
Ig=2*F*Cg*Epk*Ipk/(6*PT)=2*50000*83e-9*100*11/(6*4)=0.380 amps gate current.
Next we would check Ton to make sure it is reasonable for the mosfet.
BTW you will also have to calculate the conduction losses (Rds=0.18 ohms for the IRF640) and that will come out to around 22 watts times the time the transistor is turned on divided by the total time period, and add that to the switching losses. A heat sink will still be required however. If you want to get a little more fancy, you can calculate the gate power dissipation too and add that also.