MOSFET heat and math

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gstavrev

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Hi,
I am interfacing a few motors and a microcontroller. I am using a MOSFET driver MIC 4427YN and a PIC18. Data sheet is here.

I burned one somehow last week so I am being more careful. I am noticing that driver heats up significantly after about 15 sec. I think this must be normal?

My battery source is 12V and the motors are drawing 350mA at first, and it smoothes down to 120mA in a second.

I am reading the data sheet for power Dissipation on page 7 and it has some formulas, such as P = I² * R and P = f * C * V² for capacitive loads. I am clueless on the more complicated one. I have two capacitors to smooth the power drops so the microcontroller doesn't restart.

Can you help me figure out the math and see how hot the controller will get and if it will burn?

Thanks,
-George
 
You also need the thermal resistance for your MOSFET, expressed in °C/W, for junction to case (Θjc) and for Θca (case to ambient) and the Θ for the heat sink you'll probably be needing.

So then you have a series thermal circuit with "thermal resistances" and °C being analogous to voltage. You want to keep the junction temp below some value, probably <150°C.
Ohms law without the ohms.
Spreading the heat generated in the tiny volume of the MOSFET junction over a large surface area exposed to ambient temp is the goal.
Service lifetime for circuits and a lot of other things halves for each 10°C rise above ambient.

Can't open your attachment; my problem. Can you excerpt parts of the sheet?
 
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The driver IC is tiny so it can't dissipate much heat. The datasheet shows it heating when it drives a capacitive load. The heating is worse when the load capacitance and/or frequency are high.

A Mosfet has a specified gate-source capacitance and a Q capacitance that includes the added capacitance due to drain to gate feedback (Miller effect).
 
Are there any capacitors across the motor? Those would significantly increase the driver dissipation.
 
Will this make it unusable for PWM? I am adapting some stuff from a robotics book. Let me upload the schematic I have.

BTW, Θjc = 42C/W. I'm having trouble finding case to ambient.

..
I don't have a capacitor across the motor, just 2 to prevent on the voltage source.
 
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Slewing 1000 pF of capacitance from linearly from 5 V to zero in 100 nS takes a square pulse of current equal to [1000 pF *5/(10^-7)] = 5 mA each time the waveform changes direction.
Hi freq. + hi pF = hi current = hi power.
 
I have attached the schematic. It is copied straight out of David Cook's Intermediate Robot Building.

I think my best shot is to separate the motors so each motor has its own driver and then add a heat sink?
 

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Wait a minute. You don't have a Mosfet. You are trying to drive your motor with a Mosfet driver IC that is made to drive Mosfets, not motors.

When it drives a Mosfet then it heats only during switching.
When it drives a motor then it heats the entire time it is turned on.

An L298 is a motor driver IC. it has a heatsink tab that is bolted to a heatsink.
 

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that makes sense. I would have to get the individual mosfets, n and p channel? What is the purpose then for the driver?
 
The Mosfet driver charges and discharges the high gate capacitance of a Mosfet so it switches very quickly. It has an output of up to 1.5A for a moment.

A P-channel Mosfet pulls a load high.
An N-channel Mosfet pulls a load low. They invert the input signal.
 
gstavrev said:
BTW, Θjc = 42C/W. I'm having trouble finding case to ambient.
Click to expand...

So if you had this chip mounted on a perfect heat sink [Θca = 0°C/W], and your Tj max was 150°C and your ambient was 25°C, you could dissipate (125-25)*1/42 = ~ 3 watts.

If the rms values of the current through *the voltage across this chip is > 3W, give it up unless you want to go to liquid cooling.

Θca can be found by finding a Θca for similar shaped objects, then scale up/down the surface area. 2x the surface area of the object with the known Θca = 1/2 the Θca for your package.
 
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Willbe said:
Slewing 1000 pF of capacitance from linearly from 5 V to zero in 100 nS takes a square pulse of current equal to [1000 pF *5/(10^-7)] = 5 mA each time the waveform changes direction.
Hi freq. + hi pF = hi current = hi power.
Click to expand...
I believe that should be 50mA.
 
Sounds like Cooke needs to be strung up by his short ones. I hate it when amateurs make money off BSing beginners!

Fact of the matter is you can use very small FETs if the Rds is very small as well. I get FDS6333s for $0.20 each (in reels) and at and Rds of 0.1 I can easily use two to drive a motor at 1A.

FET drivers on the other hand, while some can deliver up to 10A, can only do so for for a short time to charge the FET gate.

Dan
 
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