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atypical MOSFET drive question

N

Nick

Jan 1, 1970
0
I'm needing to switch up to a few 100mA to a device, must switch the
source rather than the sink, and only have n-channel mosfets to hand,
specifically 30A STP36NF06's. The supply is 12V and with logic level
gate voltage. Taking into account the voltage drop through the FET in
resistor calculations for the current, this appeasr to work and I'm
wondering what the downside is to using the device to source current
rather than sink it? For example, does the MOSFET need to be derated,
is it likely to fail after a time because it simply shouldn't be used
this way, or is it inefficient but in princple ok? Guidance
appreciated.
 
J

John Larkin

Jan 1, 1970
0
I'm needing to switch up to a few 100mA to a device, must switch the
source rather than the sink, and only have n-channel mosfets to hand,
specifically 30A STP36NF06's. The supply is 12V and with logic level
gate voltage. Taking into account the voltage drop through the FET in
resistor calculations for the current, this appeasr to work and I'm
wondering what the downside is to using the device to source current
rather than sink it? For example, does the MOSFET need to be derated,
is it likely to fail after a time because it simply shouldn't be used
this way, or is it inefficient but in princple ok? Guidance
appreciated.

How do you plan to drive the gate?

John
 
L

leeps

Jan 1, 1970
0
The problem is the gate must be 10 volts above the source(5 volts for
logic level fets) so if your sourcing 12 volts to a device you need 22
volts to the gate, makes things a little more complicated. Other than
that by all means source away.

Joe
 
J

John Fields

Jan 1, 1970
0
I'm needing to switch up to a few 100mA to a device, must switch the
source rather than the sink, and only have n-channel mosfets to hand,
specifically 30A STP36NF06's. The supply is 12V and with logic level
gate voltage. Taking into account the voltage drop through the FET in
resistor calculations for the current, this appeasr to work and I'm
wondering what the downside is to using the device to source current
rather than sink it? For example, does the MOSFET need to be derated,
is it likely to fail after a time because it simply shouldn't be used
this way, or is it inefficient but in princple ok? Guidance
appreciated.

---
Since it's a logic-level device and you're using it for a high-side
switch, the down side is you'll need to drive the gate to about 5V
more positive than the source in order to get it to fully turn on.

That means that if you want to get 12V into your load from a 12V
supply connected to the drain, you'll need to hit the gate with at
least 17V. A high-side driver will do what you want, and here's
one:

http://www.micrel.com/_PDF/mic5011.pdf

Those parts seem to be way overkill for driving a couple of hundred
milliamps into a load, but if that's what you've got they'll
certainly work.
 
N

Nick

Jan 1, 1970
0
Thanks for the quick and informative replies. I thought that there had
to be at least one good reason, and in this case it's the most
fundamental one of being that it actually doesn't work for reasons that
you all explained. Not having time today to test the mosfet in the full
circuit where I've been using them as a sink up until now, I tried the
modified driver side briefly with a gate drive of 12v where it did
work, and also noticed that touching a floating gate that it switched
on, and so assumed that it would switch when driven at 5v. However
trying this now it obviously doesn't work.

Is it correct that P channel devices would also need level shifting on
the gate, and so if having to switch the source one might as well use
an N channel device with suitable gate drive?
 
D

Dorian McIntire

Jan 1, 1970
0
Nick said:
Thanks for the quick and informative replies. I thought that there had
to be at least one good reason, and in this case it's the most
fundamental one of being that it actually doesn't work for reasons that
you all explained. Not having time today to test the mosfet in the full
circuit where I've been using them as a sink up until now, I tried the
modified driver side briefly with a gate drive of 12v where it did
work, and also noticed that touching a floating gate that it switched
on, and so assumed that it would switch when driven at 5v. However
trying this now it obviously doesn't work.

Is it correct that P channel devices would also need level shifting on
the gate, and so if having to switch the source one might as well use
an N channel device with suitable gate drive?

With a P-channel device level shifting is not needed to source current, as
long as the negative rail is considered ground or common, but it will
require a negative drive voltage on the gate instead of a positive drive
voltage.

Level shifting on a MOSFET is only required if the load is attached to the
source of the MOSFET instead of the drain. The gate-to-source voltage is the
only voltage relevant to turning on a MOSFET. With a P-Channel device the
source will be connected to the positive rail and the load will be on the
drain. The difference in potential between ground and VCC is enough to fully
turn on the MOSFET.

This topic is complicated by the fact that the word "source" means two
different things here.

Dorian
 
R

Rich Grise

Jan 1, 1970
0
Thanks for the quick and informative replies. I thought that there had
to be at least one good reason, and in this case it's the most
fundamental one of being that it actually doesn't work for reasons that
you all explained. Not having time today to test the mosfet in the full
circuit where I've been using them as a sink up until now, I tried the
modified driver side briefly with a gate drive of 12v where it did
work, and also noticed that touching a floating gate that it switched
on, and so assumed that it would switch when driven at 5v. However
trying this now it obviously doesn't work.

Is it correct that P channel devices would also need level shifting on
the gate, and so if having to switch the source one might as well use
an N channel device with suitable gate drive?

The P-channel device doesn't need any level shifting. You either
open the gate, and the resistor from it to the source, at +12, turns
off the FET. Ground the gate through a suitable resistor or whatever,
and the FET sources current at almost +12V.

I guess you could call it "level shifting" when your driver turns
off - you might need to use either the N logicFET or an NPN to switch
the gate of the P, and invert the sense of the driving logic.

Good Luck!
Rich
 
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