Parallel MOSFETs burning despite relatively low rating and kickback diodes

Thread Starter

design_12121

Joined Nov 29, 2017
16
Hi guys.

I'm currently working on a project that switches a 140A semi-inductive load using a 13.2v lithium battery.
I was planning on using a 5v arduino to switch the mosfets with pwm.

My problem is connecting the gate to the positive terminal of the battery(13.2v) switches the circuit, disconnecting it turns it off. But connecting it to a 5v supply blows 1 mosfet and leaves the circuit on. Why?!

5v is above the gate threshold, it's not above Vgs. My current limiting resistor is still giving each mosfet more than 100nA. It doesn't even last until I disconnect the load so it can't be the diodes' fault.


I have 8 IRFB7437 in parallel on a large heatsink, which is overkill.
Vds: 40v
Id: 190 package limited
Vgs : +-20v
Vgs(th): 3.9v
Igss : 100nA

I also have 2 SR506 schottky diodes for extra kickback protection.

The pulldown resistor is 1kohm and the current limiting resistor is 120ohm.

Thanks
20171129_141347.jpg
 

OBW0549

Joined Mar 2, 2015
3,566
My problem is connecting the gate to the positive terminal of the battery(13.2v) switches the circuit, disconnecting it turns it off. But connecting it to a 5v supply blows 1 mosfet and leaves the circuit on. Why?!
The reason is that with only 5V gate drive, the MOSFET needs substantial drain-to-source voltage (around 10 volts) to conduct the high current you're demanding (see figs. 3, 4 and 5 of the data sheet). The result is extremely high power dissipation in the MOSFET, leading to its quick demise.

This is why the MOSFET was OK when you drove the gate with 13.2 volts, which was more than adequate to turn the MOSFET completely ON at 140 amps.

You need to drive the gate with a higher voltage.
 

Thread Starter

design_12121

Joined Nov 29, 2017
16
The reason is that with only 5V gate drive, the MOSFET needs substantial drain-to-source voltage (around 10 volts) to conduct the high current you're demanding (see figs. 3, 4 and 5 of the data sheet). The result is extremely high power dissipation in the MOSFET, leading to its quick demise.

You need to drive the gate with a higher voltage.
I have 8 mosfets in parallel. Wouldn't that translate to 17.5A (140/8) each?
 

OBW0549

Joined Mar 2, 2015
3,566
MOSFET characteristics vary widely, even among units of the same type. They cannot be counted on to share current equally. One is bound to carry a lot more than the rest; leading to its quick failure.
 

crutschow

Joined Mar 14, 2008
34,280
MOSFETs have a positive Rds temperature coefficient (i.e. they increase in ON resistance as they get warmer), so they will tend to balance their load when placed in parallel
Thus running those MOSFETs in parallel should be okay.

But if that diode across the MOSFET is your "kickback diode" it's in the wrong place.
It won't do any good connected from the MOSFET drain to source, and the inductive load will quickly zap the transistor (the inductive load will generate a large positive spike at the MOSFET source when its switched off).
You need to have the diode go from the MOSFET source (anode) to the plus supply voltage (cathode).
 

Thread Starter

design_12121

Joined Nov 29, 2017
16
They blew instantly, which made me think the temperature coefficient does not make a difference.
I notice the diode issue now. Will correct it ASAP.

I currently have an NPN transistor switching it with the 13.2v supply from the battery and it is working okay. Going to see if it works with PWM tomorrow.
 

cmartinez

Joined Jan 17, 2007
8,218
MOSFETs have a positive Rds temperature coefficient (i.e. they increase in ON resistance as they get warmer), so they will tend to balance their load when placed in parallel
Thus running those MOSFETs in parallel should be okay.

But if that diode across the MOSFET is your "kickback diode" it's in the wrong place.
It won't do any good connected from the MOSFET drain to source, and the inductive load will quickly zap the transistor (the inductive load will generate a large positive spike at the MOSFET source when its switched off).
You need to have the diode go from the MOSFET source (anode) to the plus supply voltage (cathode).
Question, how about a 15V TVS diode in parallel with the load? Wouldn't that work as well and also prevent from excessive voltages of any polarity?
 

Sensacell

Joined Jun 19, 2012
3,432
Question, how about a 15V TVS diode in parallel with the load? Wouldn't that work as well and also prevent from excessive voltages of any polarity?
No, you typically place a catch diode rated for the peak load current across the load.
If the load is inductive, there could be significant energy stored, enough to fry a TVS diode.

A standard diode theoretically clamps the spike to one diode drop above the positive supply rail.
 

crutschow

Joined Mar 14, 2008
34,280
how about a 15V TVS diode in parallel with the load? Wouldn't that work as well and also prevent from excessive voltages of any polarity?
Unless you need a faster decay of the inductive current than a diode can provide, the TVS would be a more expensive option.
And I don't see how it will prevent excessive voltages in general as it would be connected directly across the inductive load. :confused:
 

Thread Starter

design_12121

Joined Nov 29, 2017
16
Update:

I only have 4 MOSFETs left but they work with PWM and relocated diodes.

Using a ULN2803 to ground the pulldown resistor. Arduino to control the ULN2803.

The only downside is if the Arduino shuts off, the MOSFETs will be on. I'll likely use an additional PNP transistor to fix that.

Thank you guys.
 
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