Re: MOSFET acting weird, any ideas?

On Jul 22, 4:25 pm, Tim Wescott <t...@xxxxxxxxxxxxxxxx> wrote:
thsuper1 wrote:
On Jul 22, 2:33 pm, Tim Wescott <t...@xxxxxxxxxxxxxxxx> wrote:
thsup...@xxxxxxxxx wrote:
Hi all!
Not a newbie, here, I went to electronics school about a decade ago,
and although I've reeducated myself to this point, I seem to be stuck.
Please help, anybody!
I've got a mosfet that I'm using, driving an irf510 with a tc4420 cmos
driver ic.
I've got the driver being fed a square-wave with a 555.
The Driver is putting out a clean 12volt p-p SW all the way to the
gate leg on the mosfet.
I'm trying to source battery voltage, 12.75v at way more potential
amps than what my circuit is "supposed" to be drawing. around
6-8amps.
The mosfet is only sourcing 2v p-p, and the same square-wave, good and
clean, just not big enough!
I've tried various pull-up resistor arrangements, I've tried using a
transistor instead of the driver ic, and the driver ic works the best.
but nothing over 2v p-p. I'm stumped. I need approximately 12v, I
should be pulling directly form V++,
B
You didn't post a schematic or a link to one, so I'm guessing here:

Is the MOSFET common drain or common source? If you're putting +12V on
the gate and looking for voltage out the source (i.e. if you're using a
common drain 'high side' circuit) then 2V is not unreasonable.

You have to bring the gate 12V _more_ than the source.

--

Tim Wescott
Wescott Design Serviceshttp://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details athttp://www.wescottdesign.com/actfes/actfes.html

So what you're saying, so that I understand, is that if I'm drawing
"high side" or from the drain side? that 2v p-p is normal, and that I
need to draw the gate to a full 12v higher than the source?

I'm not quite making out what you mean - I think your confusion is
confusing me.

If you're connected like the schematic below, and if your gate drive
only goes 12V above ground, then the load not seeing the full 12V is
expected. Only seeing 2V seems to be excessive drop, but if you're
using a really low-resistance load I could maybe see it.

Come to think of it, are you sure your supply voltage is really staying
at 12V? Not just because the battery should, but because you've measured?

12V
+
|
|
||-+
gate drive ||<-
o----------||-+
|
| to load
'---------o

(created by AACircuit v1.28.6 beta 04/19/05www.tech-chat.de)



That means to me that the device (mosfet)takes itself to the potential
of the circuit, and that 2v p-p is all that's available from the
potential of the circuit to the device "potential," so if I draw off
of the "low-side", like in a common emitter amplifier? Please expound
on this, I know all this stuff, it's just like reactivating it, you
know?

The MOSFET needs to see lots-o-volts between the gate and source. In
the circuit above, with the load connected to the source, raising the
load voltage reduces the g-s voltage, and reduces the drive to the MOSFET.

--

Tim Wescott
Wescott Design Serviceshttp://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details athttp://www.wescottdesign.com/actfes/actfes.html

I See. So if:

+ 12v
|______o <- load pos
|
_
^
|______o <-load neg.
gate |
<-drive D
_______G
S
|
|
= ground


What would you expect the output to be if the input waveform is 12v p-
p? The battery is not getting drawn down, yes I have measured, but
thanks for asking. That's the kind of thinking I'm used to. What would
you do to this to make a higher output waveform? take the load off at
a different point? my load is a power step-up transformer.
.

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