Author Topic: load switching with mosfet power loss calculation  (Read 3797 times)

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waspinator

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load switching with mosfet power loss calculation
« on: June 05, 2016, 02:55:29 am »
I'm trying to turn on a 24V rail powering a 2.4A device using an N-channel MOSFET (IRLB8743). I'm having an issue understanding the basics of how it's working

The MOSFET has an RDS of 3.2mOhms, at a VGS of 5V. The load is 10Ohms.

I tried hooking it up as shown.

The current going from the 24V power source to ground is about 0.22A, or 10X less than I expect it to be, and the MOSFET gets to about 100 degrees Celsius in about a minute while attached to a 9.6 °C/W heat sink.

On Semi's application note for MOSFETs in Load Switch Applications says that PLOSS = ILOAD2 * RDS(on).

I'm not exactly sure how to get ILOAD, but if just using Ohm's law it should be I=V/R or 24/10 = 2.4A. I'm actually only getting 10X less than that though. But even if ILOAD = 2.4A, than PLOSS would be 5.76 * 3.2mOhms = 18.4mW. That seems like a very small amount of power, so why is the MOSFET getting so hot?

What am I not understanding here?

blueskull

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Re: load switching with mosfet power loss calculation
« Reply #1 on: June 05, 2016, 03:09:18 am »
Of course, it is a source follower. You need the MOSFET to be turned on completely. What you need is a P MOSFET, 3 resistors, and an NPN transistor to form a high side switch, or use the same N MOSFET with a charge pump gate driver.

waspinator

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Re: load switching with mosfet power loss calculation
« Reply #2 on: June 05, 2016, 03:38:07 am »
Of course, it is a source follower. You need the MOSFET to be turned on completely. What you need is a P MOSFET, 3 resistors, and an NPN transistor to form a high side switch, or use the same N MOSFET with a charge pump gate driver.

can you point me to a place where I can learn more about driving MOSFETs? The places I've seen explaining how to use MOSFETS seem to suggest that applying VGS is enough to close the "switch". The datasheet for example says that it's RDS value is 3.2mOhm at a VGS of 5V, and that it's  VGS(th) is 1.8V.

Would my best bet be to get a gate driver? Any suggestions on which one would work well with a FQP30N06L or IRLB8743?
« Last Edit: June 05, 2016, 03:43:47 am by waspinator »

blueskull

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Re: load switching with mosfet power loss calculation
« Reply #3 on: June 05, 2016, 03:44:24 am »
can you point me to a place where I can learn more about driving MOSFETs? The places I've seen explaining how to use MOSFETS seem to suggest that applying VGS is enough to close the "switch". The datasheet for example says that it's RDS value is 3.2mOhm at a VGS of 5V, and that it's  VGS(th) is 1.8V.

Would my best bet be to get a gate driver? Any suggestions on which one would work well with a FQP30N06L or IRLB8743?

Your problem is not on the gate driver, but the circuit topology. Google MOSFET high side switch.

waspinator

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Re: load switching with mosfet power loss calculation
« Reply #4 on: June 05, 2016, 04:17:51 am »
oh I see. So all I need to do is put the load above the transistor and it should work. I didn't realize that the effective VGS becomes so much higher in the high side configuration. I don't see why you would ever use a MOSFET in the high side configuration, but is that the only time you would use a MOSFET driver? (other than when the VGS is naturally very high)

this video cleared it up for me I think.

Thanks for pointing me in the right direction.

blueskull

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Re: load switching with mosfet power loss calculation
« Reply #5 on: June 05, 2016, 10:15:46 am »
oh I see. So all I need to do is put the load above the transistor and it should work. I didn't realize that the effective VGS becomes so much higher in the high side configuration. I don't see why you would ever use a MOSFET in the high side configuration, but is that the only time you would use a MOSFET driver? (other than when the VGS is naturally very high)

Depends on your load. If its a load without ground reference, then you can use low side switch. If you need ground reference, then you need a high side switch.

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Re: load switching with mosfet power loss calculation
« Reply #6 on: June 05, 2016, 11:19:36 am »
Love Cypress PSOC, ATTiny, Bit Slice, OpAmps, Oscilloscopes, and Analog Gurus like Pease, Miller, Widlar, Dobkin, obsessed with being an engineer

waspinator

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Re: load switching with mosfet power loss calculation
« Reply #7 on: June 05, 2016, 04:19:39 pm »
Hmm, so if I wanted to control power going to two devices that have their grounds connected to each other, would low side switching be a concern?

For example switching a Raspberry Pi at 5V and a Arduino at 3.3V, and then connecting them together through a USB cable.

When both MOSFETS are off would the Arduino's ground be actually higher than 3.3V and have current flowing backwards?
« Last Edit: June 05, 2016, 04:22:13 pm by waspinator »

blueskull

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Re: load switching with mosfet power loss calculation
« Reply #8 on: June 05, 2016, 04:33:20 pm »
If the 2 devices don't switch on/off simultaneously, the ground current loop can and will fry your boards. Do NEVER do this.

Ian.M

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Re: load switching with mosfet power loss calculation
« Reply #9 on: June 05, 2016, 04:35:23 pm »
That obviously wont work and may actually damage something if there are any other ground referenced connections.   If you need simple high side switching of moderate currents at voltages less than 20V, use P channel MOSFETs with a gate driver or level shifter to drive them.

waspinator

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Re: load switching with mosfet power loss calculation
« Reply #10 on: June 05, 2016, 04:41:40 pm »
so would this kind of configuration work?

Edit: I think I had the P-MOSFETS connected backwards

« Last Edit: June 05, 2016, 04:48:38 pm by waspinator »

Ian.M

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Re: load switching with mosfet power loss calculation
« Reply #11 on: June 05, 2016, 04:48:27 pm »
No.  You've got your MOSFETs backwards and their body diodes will always conduct.

For a P-MOSFET high side switch the source goes the supply and the drain to the load.   Also, to turn them off, the gate must be taken right up to the supply rail so Vgs=0.

blueskull

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Re: load switching with mosfet power loss calculation
« Reply #12 on: June 05, 2016, 05:09:06 pm »
R1 and R2 is chosen so that when R2 is pulled down, the voltage between G and S of the P-MOSFET will fully turn the device on, while will not fry the gate.
R3 is chosen so that the NPN BJT can pull R2 close to ground potential, while not sucking too much current from control port, usually from 1K to 10K.
For a 24V system, I recommend R1 to be 10K, and R2 to be 15K, that gives ~10V gate voltage (being pedantic, -10V) between gate and source.

waspinator

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Re: load switching with mosfet power loss calculation
« Reply #13 on: June 05, 2016, 11:41:04 pm »
Could I control a 52V rail using 5V logic?

Something like this wouldn't work for high voltage (>25V) rails because it would exceed the maximum VGS the FQP27P06?

But would something like this work? The Vgs would be -21V, which would be below the +-25V limit

« Last Edit: June 06, 2016, 12:53:02 am by waspinator »

blueskull

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Re: load switching with mosfet power loss calculation
« Reply #14 on: June 06, 2016, 02:29:14 am »
The second works, you are getting the point.

Smf