driving a mosfet irf640 correctly

[EmmKay]

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.

 

[Hayato]

I believe that your biggest problem is the Id.

The MOSFET has a 125W maximum power dissipation. You are running it at 100V@11A, which means 1100W (8.8x the maximum). You NEED a heat sink.

 

[misterT]

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.

No, you want to switch the mosfet on and off as quickly as possible. This means driving a good current spike to the gate (you must charge and discharge the gate capacitance to switch the mosfet on and off).
The faster you can switch the mosfet the more efficient your switch is.. and the less heat it will generate. So leave the gate resistor away.

Keep the ground pin of the ir4427 close to the source of the mosfet (to keep the current loop short when driving the gate). And use a good bypass capacitor close to the vcc-pin of the ir4427.

Here is a good paper for more info: https://www.electro-tech-online.com/custompdfs/2010/08/slup169.pdf

 

[crutschow]

As misterT noted, the gate resistor is slowing down the transistor switching time which defeats the purpose of having a high current driver to control the gate. What did you think the gate resistor was for?

 

[audioguru]

The MOSFET has a 125W maximum power dissipation. You are running it at 100V@11A, which means 1100W (8.8x the maximum). You NEED a heat sink.

Absolutely not!
The load uses 1100W, not the Mosfet. The Mosfet is simply a switch that has no current when it is turned off and has a very low voltage when it is turned on.

 

[Hayato]

Absolutely not!
The load uses 1100W, not the Mosfet. The Mosfet is simply a switch that has no current when it is turned off and has a very low voltage when it is turned on.

Indeed, a gross error. Sorry.

 

[MikeMl]

Absolutely not!
The load uses 1100W, not the Mosfet. The Mosfet is simply a switch that has no current when it is turned off and has a very low voltage when it is turned on.

Yes, BUT!!! As the drain of the FET switches, there will be an instant when the FET Drain is at a voltage of half the supply voltage. At that instant, half the supply voltage appears across the FET, and half across the load (they are in series).

The load current at that instant is determined by the resistance of the load, so will be half of what it would normally be, which means that it is dissipating 1/4 of its normal full-on power, or 275W in this example. This means that the FET is dissipating 275W, too.

The whole goal of speeding up the gate drive, is to get the FET drain to slew through the voltage range where the FET is dissipating power as fast as possible. It always takes a finite time for the FET to switch, so hitting the peak dissipation point is unavoidable, but the shorter that time is, or the less often that the FET switches, the less the average dissipation.

 

[audioguru]

The very high value gate resistor guarantees that the Mosfet slews slowly making ramps instead of switching quickly. Then it gets extremely hot.

 

[MikeMl]

The very high value gate resistor guarantees that the Mosfet slews slowly making ramps instead of switching quickly. Then it gets extremely hot.

Here is an illustration: I'm starting with a 10V 10KHz gate drive signal, with fast rise and fall times (dark Blue trace). The series 1K resistor between the signal and the gate slows the rise and fall at the Gate due to the FET input and FET Miller capacitance (lt. Green trace). This causes the Drain voltage to slew slowly through the switching point (Red trace).

The FET power dissipation is the complex expression shown in the lt. Blue trace. LTSpice shows the dissipation in any circuit element. It also will integrate any complex waveform to show the average dissipation. Note that the average dissipation is 46W!. This would require the FET to be on a very large heat sink!!!!

 

[audioguru]

He is using 50kHz, not 10kHz. If you show the simulation at 50kHz then it will be 5 times worse.

 

[MrAl]

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.

 

[ronv]

MrAl is correct. You are pushing it just in power dissapation, but your gate drive calculation is not correct so the switching time is also slower than it needs to be.

 

[RCinFLA]

With 100v miller effect, input is equivalent to about 5000 pF. At 50 kHz with no more then a 2% of total slew period drive you have to move the 3000 pF 12 volts in less then 200 nsecs. This requires a gate drive current pulse of at least about 300 mA.

The IR4427 has an equivalent driver resistance of about 6 ohms. If you want quick rise time you can not afford a series resistor greater then about 22 ohms.

 

[EmmKay]

Thank you all, The responses have all been very very helpful in understanding the requirements for mosfet driving. While the amount of information on the web about mosfet driving is more than plenty, I must say that MrAl's response worked best in explaining things and from now is a permanent fixture on my workbench. I did know that fast switching is the key to successful mosfet usage, but a particular application note threw me off completely, by replacing the gate resistor with 18R 1/4W resistor I saw a dramatic improvement (reduction) in the temperatures. The mosfets were fan cooled with no heat sink, I could barely touch the mosfet before. I will add a heat sink to the mosfets just in case eventhough after todays running I am confident that the fan is all that would be required!.


So thank you all again and especially MrAl for that very helpful detail.

Emmkay.

 

[MrAl]

Thank you all, The responses have all been very very helpful in understanding the requirements for mosfet driving. While the amount of information on the web about mosfet driving is more than plenty, I must say that MrAl's response worked best in explaining things and from now is a permanent fixture on my workbench. I did know that fast switching is the key to successful mosfet usage, but a particular application note threw me off completely, by replacing the gate resistor with 18R 1/4W resistor I saw a dramatic improvement (reduction) in the temperatures. The mosfets were fan cooled with no heat sink, I could barely touch the mosfet before. I will add a heat sink to the mosfets just in case eventhough after todays running I am confident that the fan is all that would be required!.


So thank you all again and especially MrAl for that very helpful detail.

Emmkay.

Hello again,



You're welcome, and i see you are already using a mosfet driver chip so you're good to go

Oh yes, a fan does wonders for getting rid of heat. Usually the choice is to use a heat sink first and a fan second, but even a small fan blowing on a resistor can lower it's temperature by quite a bit without any kind of heat sink. We just have to be careful that the fan does not fail without our circuit detecting it or else the design overheats badly.

 

[EmmKay]

Hi All,
Rather than starting a new thread, Perhaps I might be able to get an answer here..!! If not I will start a new thread....

On a circuit I have two voltages.... Hv (between 60 and 110VDC at 12A) and Logic (5VDC and 2A). Is there a specific method in combining the two grounds i.e. the power ground and the logic ground or can I just connect them together with a 0R resistor at one point . At the moment I have just connected the two ground planes with a 0R resistor at a point in the circuit.

 

[MikeMl]

What is the value of the resistor? Methinks you have broached all of the issues with "single point ground", "ground loop", "common-mode noise", "ground-currents". There is enough stuff out there on these topics for several dozen Doctoral Dissertations....

 

[EmmKay]

Never knew it would be such a mine field. The resistor is 0R (infact a piece of wire!!!!). Some people have suggested a 1uF 250V film capacitor, while others have suggested just connecting the two grounds together with a piece of wire. The piece of wire works no problem, but for my own info, I wanted to know if a better or THE RIGHT way of doing this exists..

 

[MrAl]

Hello,


Normally what you do is analyze the current flows, and make sure that the analog ground is not connected such that a large current flow in the other parts of the circuit can cause a voltage drop that will bother the analog part of the circuit. That's the whole ball game in a nut shell. If the analog part gets biased by the high current part or by the logic switching then that is going to show up as a disturbance in the analog measurements or analog signals in general.
One way of doing it is to connect the logic grounds all together separately from from all the analog grounds, then connecting them together at some point where the current flows separately to the two parts of the circuit. It's something like a star connection, but in this case there would be two stars eventually connected together with only one lead, not two or three or some wide area pc board trace.
For a quick example, below shows the analog and digital going to a common point but that point should be as close as possible to the power source, possibly also with adequate bypassing at that point too.
Notice the wire they share is very short if any length at all, and that is the only place where the digital signals can have any effect on the analog, and since voltage drop is related to distance (length) of the wire, the shorter that short section is the better. Any capacitor bypassing will also help keep the analog circuit at a constant voltage too, but that may not be possible.

ANALOG o---------+--------o DIGITAL
                 |
                 o
                 PS

 

[vincehhh]

To turn of your fet you kneed to charge up your gate capacitance.

time to turn on, not considering the FETs specified turn on speed is (gate resistor) x (gate capacitance).

Initial gate current should be in amps not mA but we are only talking about nanoseconds

So we need to minimise the resistance and inductance in the path from the gate to the gate driving circuit as this slows down the rate the gate capacitance charges up => slows down the rate the FET turns on.


If the DS resistance if quoted as 0.5 ohm, if it is not turned on hard enough will have many times that value and this is the time the FET is dissipating the most heat (I²R)

Hope this helps from my limited experience.