On Jan 8, 6:38 pm, John Fields wrote:
On Tue, 8 Jan 2008 01:22:02 -0800 (PST),
wrote:
On Jan 7, 4:05 pm, wrote:
Hi JF
I would have been happy to report that my 5V-24V switch worked with
the solenoid.
And it did...Until later that night that I discovered that my solenoid
must have gotten warm and melted in the middle and now the metal thing
is stuck in there. (that's what I deduce at least. After pulling out
the metal shaft, I can see brown discoloration in the inside of the
solenoid).
So I'm pretty sure I fried it with some sort of exhaustion. So now I
have the following questions:
- Is is possible that the diode can break it...
---
No.
---
I was wondering about
this befo If the magnetic field colapses and force current in the
opposite direction, does that mean the duty cycle extends further than
the amount of time that it is ON? So it is on for 0.5 seconds at 24V.
Then it colapses for 0.1 second at -200V.
---
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.
---
Should I add the power of
these together to calculate the duty cycle?
---
I wouldn't bother.
---
- Is is AT ALL possible that the MOSFET will allow some current
through if I have nothing connected to it (not necessarily negative
but just nothing).
---
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.
---
- How does the mosfet 'decide' how much current to get through?
---
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.
---
Will
it just bridge the gap between the source and drain?
---
Yes. like a switch.
---
Or will it put a
multiplier on the voltage or current of the gate?
---
No. If it's fully turned on all that will appear between the supply
and the load is the MOSFET's Rds(on).
---
And then the
question that goes with this is: does is matter what size resister I
put in between my 5V pic and the gate?
---
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.
---
One comment: I thought in the begginning that the gate is the middle
pin (like the base of a transistor). So I had my gate and drain the
wrong way around for about 10 iterations (and of course I was
scratching my head as to why the stupid thing didn't work). After that
it worked when I swopped it but maybe that introduced the timebomb
slow death for my solenoid.
-The final question I have is about my powersupply. Now it says it is
a 24V, 800mA, regulated power supply. (http://www.maplin.co.uk/
module.aspx?ModuleNo=48484&doy=7m1 --- the 24V one) This is a
switched mode PSU. Now, the bizarre thing is that despite all this 24V
talk theres a sticker on the power supply that say 24V and then also
"Typical voltage 21-35V" ---WTF?!?
---
It doesn't say anywhere that it's regulated, only that it's a
switcher.
---
I thought its 24 and that's it. Why would it be 35V sometimes?
---
If it's unregulated, the output voltage will be load dependent and,
lightly loaded, it'll rise.
---
Thanks again for your support and HAPPY NEW YEAR!
---
:-)
---
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! 
--
JF
Thanks again for the info. This gives me another possible scenario for
why I toasted the solenoid.
I was 'driving' the gate with 5V+ a resistor and a switch. I didn't
have a pulldown resistor. I gather that it may be possible that the
field still existed on the gate even after I opened the switch. So I
guess with a pull down resistor on the gate I'd be more safe. What do
you think?