thus the reason to put the resistor between the gate and source, to make su=
re it does shut off...?

"Thomas C. Sefranek" <tcs@cmcorp.com> wrote:  Yes, if it is .8 volts or les=
s than 3.3V it will turn on the high side FET
always. (You cant shut it off.)

*
| __O Thomas C. Sefranek WA1RHP@ARRL.NET
|_-\<,_ Amateur Radio Operator: WA1RHP
(*)/ (*) Bicycle mobile on 145.41MHz PL74.4

ARRL Instructor, Technical Specialist, VE Contact.
http://hamradio.cmcorp.com/inventory/Inventory.html
http://www.harvardrepeater.org

-----Original Message-----
From: piclist-bounces@mit.edu [mailto:piclist-bounces@mit.edu] On Behalf Of
alan smith
Sent: Tuesday, September 26, 2006 3:36 PM
To: Microcontroller discussion list - Public.
Subject: RE: [EE] high side FET....remind me on this

And that is due to the fact the output of the CPLD drive is going to be less
than 3.3V?

"Thomas C. Sefranek" wrote: You don't need ANY divider or
interface.
The CPLD should drive the gate just fine.
(Assuming it too is 3.3 volts or more.)

*
| __O Thomas C. Sefranek WA1RHP@ARRL.NET
|_-\<,_ Amateur Radio Operator: WA1RHP
(*)/ (*) Bicycle mobile on 145.41MHz PL74.4

ARRL Instructor, Technical Specialist, VE Contact.
http://hamradio.cmcorp.com/inventory/Inventory.html
http://www.harvardrepeater.org

-----Original Message-----
From: piclist-bounces@mit.edu [mailto:piclist-bounces@mit.edu] On Behalf Of
alan smith
Sent: Tuesday, September 26, 2006 2:29 PM
To: Microcontroller discussion list - Public.
Subject: RE: [EE] high side FET....remind me on this

OK, thanks. I guess then it could be solved by doing a divider on the drive
from the CPLD, or just use something like a Si4788CY to solve it. Lower
RDSon with that package as well.

"Thomas C. Sefranek" wrote: =

-----Original Message-----
From: piclist-bounces@mit.edu [mailto:piclist-bounces@mit.edu] On Behalf Of
alan smith
Sent: Monday, September 25, 2006 3:10 PM
To: Microcontroller discussion list - Public.
Subject: [EE] high side FET....remind me on this

OK, I've got a P channel FET (not my design, just took it over) that is
using 3.3V to switch another 3.3V supply (always on turning on the rest of
the board). Part is a FDC638P. Looking at the data sheet, Vgs is the
voltage differential to turn on the gate, right? So pretty much....you
can't have the same voltage on both the gate and drain to turn it on, just
wont work. Now if I had 5V (or anything more than 4.1V) there, with a
resistor to ground to keep it turned off, that should work.

Huh?

The gate needs only be .8 volts less than the 3.3 volts feeding the source.
To turn it on.
I does not hurt to be brought right to ground (0V.) [-3.3Vgs]
You may want a resistor from the gate to the source to turn it off.
(NOT to ground!)
You got things the other way "round".

*
| __O Thomas C. Sefranek WA1RHP@ARRL.NET
|_-\<,_ Amateur Radio Operator: WA1RHP
(*)/ (*) Bicycle mobile on 145.41MHz PL74.4

ARRL Instructor, Technical Specialist, VE Contact.
http://hamradio.cmcorp.com/inventory/Inventory.html
http://www.harvardrepeater.org


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