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Solenoid pwm driver

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drkidd22

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I'm working on a driver for a solenoid and just want to make sure I got it right, or somewhat. So this is how it works, or supposed to. The signal to the gate driver U55 goes fully on (high) for 10ms before it enters PWM at 24kHz. So for this 10ms Q40 will see a max current of 2A and it's power dissipation will be (2*2*Rds(on)) = 100mW. I used Rds(on) = 0.025. Then 100mW * RθJA = 10. I used 100 for RθJA since my pad will be smaller than 1" and 1oz. With a max ambient of 40C this 10C rise should not be a problem. That's if I didn't miss anything that you guys can catch here.

Now since I'm doing PWM at 24kHz (42us period, 50%DC) I figure I would need 60nc/21us = 3mA current to the gate of Q40. I have not worked with gate drivers before, so I think I might need a limiting resistor on the gate of Q40 to give a current of at least 20mA. So with this the driver will be dissipating (.02*.02*200) = 80mW. 80mW*200 = 16C. Again this 16C rise on a 40C ambient wouldn't be a problem.

If anyone has any suggestions or I missed anything I would appreciate any feedback.
 

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alec_t

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What is the inductance of the solenoid coil? With a 10mH coil, for example, it would take ~4mS for the current to rise to 2A.
 

MaxHeadRoom78

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What is the purpose of the solenoid?, not usual to use PWM unless a proportional valve etc.
Max.
 

spec

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I'm working on a driver for a solenoid and just want to make sure I got it right, or somewhat. So this is how it works, or supposed to. The signal to the gate driver U55 goes fully on (high) for 10ms before it enters PWM at 24kHz. So for this 10ms Q40 will see a max current of 2A and it's power dissipation will be (2*2*Rds(on)) = 100mW. I used Rds(on) = 0.025. Then 100mW * RθJA = 10. I used 100 for RθJA since my pad will be smaller than 1" and 1oz. With a max ambient of 40C this 10C rise should not be a problem. That's if I didn't miss anything that you guys can catch here.

Now since I'm doing PWM at 24kHz (42us period, 50%DC) I figure I would need 60nc/21us = 3mA current to the gate of Q40. I have not worked with gate drivers before, so I think I might need a limiting resistor on the gate of Q40 to give a current of at least 20mA. So with this the driver will be dissipating (.02*.02*200) = 80mW. 80mW*200 = 16C. Again this 16C rise on a 40C ambient wouldn't be a problem.

If anyone has any suggestions or I missed anything I would appreciate any feedback.
You don't normally limit the gate drive current too much. The idea is to charge and discharge the gate capacitance as fast as possible so that the NMOSFET switches as fast as possible. It is best to do a bit of pulse shaping though and a grid stopper is a wise precaution to limit the chance of parasitic oscillations of the NMOSFET. A 22 Ohm resistor mounted directly on the NMOSFET gate terminal should do the job nicely.

spec
 

crutschow

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.......................
Now since I'm doing PWM at 24kHz (42us period, 50%DC) I figure I would need 60nc/21us = 3mA current to the gate of Q40. I have not worked with gate drivers before, so I think I might need a limiting resistor on the gate of Q40 to give a current of at least 20mA. ....................
You want the MOSFET to turn on and off as rapidly as possible to minimize dissipation so you want a lot more than 3mA to drive the gate.
I would shoot for a on/off time of less than 1μs, not 21μs, or a gate current greater than 60nc / 1μs > 60mA .
And as spec noted, you certainly don't want to limit the gate current with a resistor.
 

drkidd22

Member
The only reason I was thinking of limiting the current to the gate was to minimize power dissipation on the fet driver.
 

ronsimpson

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minimize power dissipation on the fet driver.
The gate has a built in capacitor. It takes the same amount of energy to charge up/discharge the gate no matter how fast.
1A for 10nS or 2A for 5nS for 10A for 1nS It is all the same. (assuming no gate resistor)
With a gate resistor; the driver is allowed to have an output of O or VCC and not be in that "analog" place like 1/2 VCC. So the gate driver will have current but near zero volts on the internal parts. current X 0_volts = 0 power. The power loss will be mostly in the resistor.

So the total power from the supply will not change much if you charge up the gate fast or slow.
Using a gate resistor allows the silicon in the gate driver to run cooler. You also have some control over peak current.
 
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