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Power Switching with mosfet

J

Jamie

Jan 1, 1970
0
Jon said:
I only have an n-channel enhancement mosfet(2N7002) and I'm trying to use it
to switch power to an IC.

I can only bring the gate voltage about 1.7V above the V_DS. I'm not
switching the entire voltage to the IC.

Some question I have, What is the difference between a high side mosfet
driver and a low side? A low side brings the drain within a few mV above
ground but the high side brings the source within only several hundred mV
below the drain. Shouldn't it be symmetric?

i.e., whats the difference between

V
|
Mosfet
|
Load
|
G

V
|
Load
|
Mosfet
|
G

I'm calling the first one a high side and the second a low side, I think
thats the correct terminology. Both mosfets are the same as is the load.

The low side works fine but the high side doesn't. It seems that in either
cause it should be the same. V_DS is the same in both cases? Obviously this
isn't true but I can't seem to understand why?

The way I'm looking at it is basically as two resistors in series so the
voltage drop across each should be exactly the same.

So whats wrong?

Thanks,
Jon
You most cases, you need a P-channel for high side.
and N-channel for low side.
so that would mean, the high side S (source) would be at the
+ rail end low side (S) at the - rail etc...

Just think upside down :)
On the high side you would pull to common to turn it on and
low side, apply + to turn it on.
 
P

Paul Hovnanian P.E.

Jan 1, 1970
0
Jon said:
Its a Pic24.

How can it end up above the rail? You mean input to the IC? Since the
mosfet might not pull the down completely or pull up completely?

If thats the case then how does one go about solving that problem? In my
application my inputs the the IC will be at the rail since I'm using a
pullup and very little current will flow. It will only be a few mV I imagine
but maybe that will make a difference? R_DS is about 10 ohms which may or
may not be significant? What kinda wiggle room do I have to work here?

Oh, I think I can just use the mosfet to power up the pullups too though...
so when the mosfet is triggered it will power all pullups used for
inputs/outputs and so should reduce there voltages be the same amount... But
assume I just didn't think about that ;)

That'll work for the high side switching case. Now all you have to do is
solve the drive level problem. If you decide to go with low side
switching, all of the external inputs will have to follow the PIC
ground, or you'll have to be certain that it (the PIC) will tolerate
inputs held below the ground rail.
 
P

Paul Hovnanian P.E.

Jan 1, 1970
0
Tam/WB2TT said:
Paul,
You hit the nail on the head. From personal (unintentional) experience, at
least for digital ICs, opening up either the ground or the Vdd lead can lead
to unexpected and bizarre operation.

Some devices have internal I/O voltage protection in the form of diodes
to the supply rails. Cut the power or the ground to the chip but keep
the externals tied either high or low and some interesting sneak
circuits result.
 
J

Jon Slaughter

Jan 1, 1970
0
Paul Hovnanian P.E. said:
That'll work for the high side switching case. Now all you have to do is
solve the drive level problem. If you decide to go with low side
switching, all of the external inputs will have to follow the PIC
ground, or you'll have to be certain that it (the PIC) will tolerate
inputs held below the ground rail.


What I did was put the ground of all the inputs to the drain of the
mosfet(low side) so it does follow it(i.e., my ground is really the the
drain).

I get confused when you say "below the ground rail" as nothing is below the
ground rail. I see what you mean though(it sounds almost like I would be
using a negative supply). I think as long as I treat the drain as ground
then everything will work?
 
P

Paul Hovnanian P.E.

Jan 1, 1970
0
Jon said:
What I did was put the ground of all the inputs to the drain of the
mosfet(low side) so it does follow it(i.e., my ground is really the the
drain).

I get confused when you say "below the ground rail" as nothing is below the
ground rail. I see what you mean though(it sounds almost like I would be
using a negative supply). I think as long as I treat the drain as ground
then everything will work?

Yes. If 'the ground rail' is the ground for the entire system, not just
the one IC. Your original post about switching power to 'an IC' was
somewhat misleading in this respect.
 
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