---
With a diode across it, it can't collapse at 200V since the diode is
clamping the voltage across it to about 0.7V.
Since the coil has a resistance of ~ 21 ohms, that means that during
the collapse of the field the maximum current through the coil will
be:
E 0.7V
I = --- = ------ ~ 0.033A = 33mA
R 21R
and the maximum power dissipation will be:
P = IE = 0.033A * 0.7V ~ 0.023W = 23mW
hardly anything to be concerned about since it's less than 1/10th of
a percent of what you normally drive the coil with.
---
---
I wouldn't bother.
---
---
Yes. If the gate is charged positive WRT the source and that charge
is trapped, the MOSFET will stay at least partially turned on for as
long as that charge remains above the MOSFET's threshold voltage.
If the MOSFET Is just sitting there, though, with a charged gate and
nothing connected to it no charge (other than leakage) will flow.
---
---
It doesn't. Once you've driven the gate sufficiently positive, the
only things which will limit the drain-to-source current will be the
impedance of the supply, the impedance of the load, the MOSFET's
drain-to-source resistance (Rds(on)), and the resistance of the load
side wiring.
---
---
Yes. like a switch.
---
---
No. If it's fully turned on all that will appear between the supply
and the load is the MOSFET's Rds(on).
---
---
It may, depending on how quickly you want to turn ON the MOSFET.
That is, since the gate looks like (is) a capacitor which has to be
charged and discharged in order to turn the MOSFET on and off, the
time it takes to do that will be:
T = RC
So the larger R becomes the longer it'll take to charge and
discharge the gate capacitance.
There's also the question of how much current your PIC's I/O can
source and sink, since the smaller the current the higher the port's
resistance will be and the longer it'll take to charge and discharge
the gate capacitance.
---
---
It doesn't say anywhere that it's regulated, only that it's a
switcher.
---
---
If it's unregulated, the output voltage will be load dependent and,
lightly loaded, it'll rise.
---
---
---
Ok, I read up on a few things so some of my questions are answered. I
know now that the gate never draws current becuase it is a FET. It
works with a field, the gate is insulated so no current flows through
it.
---
Hitting the books, huh?
As my friends from Oz say, "Good on you!"
---
I think I find out why things went all wrong the other day. I realised
yesterday that I had the polarities of the 24V reversed. So I had 24V
at the source and 0V at the drain. In addition to mistaken the drain
with the gate pin, I think I had more than enough reason and flaws in
my little system to even break the MOSFET. I can confirm that the
MOSFET is not working now. Or at least: I THINK the MOSFET is not
working. Which brings me to another question: Is there an easy test
for a MOSFET to see if it's functioning properly?
---
An easy test would be:
D<------+
0-15V>----G NCH |+
S [OHMMETER]
| |
GND>--------+<------+
With the gate at 0V the ohmmeter should read infinite ohms or "OL".
As you make the gate-to-source (Vgs) voltage more and more positive
the drain-to-source resistance (Rds) should get closer and closer to
zero ohms.
If you're using a logic level MOSFET the channel should become fully
enhanced with Vgs ~ 3V , otherwise it should become fully enhanced
with Vgs ~ 10V. "Fully enhanced" means the channel's ON resistance
(Rds(on)) will drop to no more than the value specified in the data
sheet for the Vgs specified.
A better test would be:
+24V--+----------------+
| |
[560R] 15V [20R] 30W
| / |
+-------+ +<--------+
| | | |
| | D |
[1N4744A] [10K]<---G NCH [VOLTMETER]
| | S |
| | | |
GND>--+-------+--------+---------+
With the 10k pot cranked to ground,the voltmeter should read 24V.
As the pot is rotated toward 15V, the voltmeter reading should
decrease to almost zero volts and the 20 ohm resistor should get
hotter and hotter.
There's more if you're interested, but I think you've got enough to
keep you busy for a while!
