70V questions

[Ross Herbert]

On Wed, 21 Nov 2007 08:20:59 -0800 (PST), coetzee.evert@gmail.com
wrote:

Hi. thanks for the detiailed response JF.

I'm using the following solenoid: www.radionics.ie : RS Part number
search (top of page): 431-7560
(19mm push, 12V)

I'm looking at about 3.5% duty cycle.
It says on the datasheet(
http://docs-europe.electrocomponents.com/webdocs/05c2/0900766b805c2a1f.pdf)
that the wattage increase by the square of the increase in voltage.
12->72 = 6x increase so 36x increase in wattage. (1/36 for the duty
cycle)
7W at 100% duty @ 12V so we're looking at 252W on 72V on a 3.6% duty
cycle.

Unfortunately they don't mention current or resistance in the
documentation but I can go and measure tonight.

According to my RS website page RS 431-7560 is manufactured by
SAIA-BURGESS http://www.saia-burgess.com/2302/2303/2310/2316.asp which
owns the LEDEX brand http://www.ledex.com/ of STA linear solenoids
http://www.ledex.com/linear-solenoids/sta-tublar-solenoids.html

(You will be requires to register)

The STA LEDEX part no. for the RS 431-7560 is 195225-230 for which the
data sheet is Metric STA 20 x 40
http://www.ledex.com/ltr2/access.php?file=pdf/Tubular_20x40_Push.pdf

On this page
http://www.ledex.com/solenoid-applications/linear-solenoids.html you
will find applications for linear solenoids some of which may be what
you are trying to accomplish.

Also, I would suggest that you read as much as you can on the LEDEX
website including
http://www.ledex.com/ltr2/access.php?file=pdf/DriveElectronics_Section_M.pdf

When using your solenoid at the nominal voltage of 12V you can leave
it permanently connected without overheating but if you wish to
increase the voltage then you may be wise to use a LEDEX Hold-in
Circuit Module as described on page 6 of the last document link.

If you need help then I am sure that LEDEX will be able to answer your
queries.

Good luck!

 

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... I was wondering about
this before: 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. Should I add the power of
these together to calculate the duty cycle?
- 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).
- How does the mosfet 'decide' how much current to get through? Will
it just bridge the gap between the source and drain? Or will it put a
multiplier on the voltage or current of the gate? 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?

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?!?
I thought its 24 and that's it. Why would it be 35V sometimes?

Thanks again for your support and HAPPY NEW YEAR!

 

On Jan 7, 4:05 pm, coetzee.ev...@gmail.com 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... I was wondering about
this before: 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. Should I add the power of
these together to calculate the duty cycle?
- 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).
- How does the mosfet 'decide' how much current to get through? Will
it just bridge the gap between the source and drain? Or will it put a
multiplier on the voltage or current of the gate? 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?

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?!?
I thought its 24 and that's it. Why would it be 35V sometimes?

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.

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?

 

[John Fields]

On Tue, 8 Jan 2008 01:22:02 -0800 (PST), coetzee.evert@gmail.com
wrote:

On Jan 7, 4:05 pm, coetzee.ev...@gmail.com 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 before: 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

 

On Jan 8, 6:38 pm, John Fields <jfie...@austininstruments.com> wrote:

On Tue, 8 Jan 2008 01:22:02 -0800 (PST), coetzee.ev...@gmail.com
wrote:

On Jan 7, 4:05 pm, coetzee.ev...@gmail.com 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 before: 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?

 

[John Fields]

On Tue, 8 Jan 2008 11:51:41 -0800 (PST), coetzee.evert@gmail.com
wrote:

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.

---
Not just possible, that's exactly what happened. :-(
---

So I
guess with a pull down resistor on the gate I'd be more safe.

---
Of course.
---

What do you think?

---
Since you're driving the 12V, 100% duty cycle coil with 24V, that
mandates a 25% (or lower) duty cycle so, assuming it's going to be
working 100% of the time, I think a good idea would be to drive it
with something like a 555 wired like this:

+24>----------------------------------------+
|
VCC>--+--------------------------------+ |
| | +------+
[20k]R1 +---------+ | | |K
| |_ | | [COIL] [1N4001]
+-------------7-O|D Vcc|-8---+ | |
| R2 | _| | +------+
+---[82k]---+--6-|TH R|O-4--+ |
| | |__ | | D
+-[1N4148>]-+-2-O|TR OUT|-3---|--G NCH
| | GND | | S
[C] +----+----+ [0.1ľF] |
| |1 | |
GND>--------------+---------+----------+----+

Note that a pot can be substituted for R1 and R2, so you can adjust
the frequency, as required, to accommodate the inductance of the
coil, by choosing the end-to-end resistance of the pot and then
choosing the duty cycle by cranking the pot.

Also, if you can't find a pot with the end-to-end resistance you
need, you can always do this:


V>---+
|
[R1]
|
[POT]<--
|
[R2]
|
GND--+

Since we don't know the inductance of the coil it's impossible to
determine the pulse repetition frequency but, assuming it's 100
millihenrys:

Version 4
SHEET 1 880 716
WIRE 160 -32 -752 -32
WIRE 256 -32 160 -32
WIRE 160 16 160 -32
WIRE 256 16 256 -32
WIRE -272 176 -624 176
WIRE 48 176 -48 176
WIRE 160 208 160 96
WIRE 256 208 256 80
WIRE 256 208 160 208
WIRE 160 224 160 208
WIRE -624 240 -624 176
WIRE -576 240 -624 240
WIRE -464 240 -496 240
WIRE -272 240 -464 240
WIRE 16 240 -48 240
WIRE -464 304 -464 240
WIRE -432 304 -464 304
WIRE -320 304 -352 304
WIRE -272 304 -320 304
WIRE 112 304 -48 304
WIRE -16 368 -48 368
WIRE -464 400 -464 304
WIRE -416 400 -464 400
WIRE -320 400 -320 304
WIRE -320 400 -352 400
WIRE -320 464 -320 400
WIRE 16 464 16 240
WIRE 16 464 -320 464
WIRE -624 496 -624 240
WIRE -16 496 -16 368
WIRE -16 496 -624 496
WIRE -752 512 -752 -32
WIRE -624 512 -624 496
WIRE -320 528 -320 464
WIRE -752 640 -752 592
WIRE -624 640 -624 592
WIRE -624 640 -752 640
WIRE -320 640 -320 592
WIRE -320 640 -624 640
WIRE 48 640 48 176
WIRE 48 640 -320 640
WIRE 160 640 160 320
WIRE 160 640 48 640
WIRE -752 688 -752 640
FLAG -752 688 0
SYMBOL nmos 112 224 R0
SYMATTR InstName M1
SYMATTR Value IRF7468
SYMBOL ind 144 0 R0
SYMATTR InstName L1
SYMATTR Value .1
SYMATTR SpiceLine Rser=20.7
SYMBOL voltage -752 496 R0
WINDOW 0 42 56 Left 0
WINDOW 3 42 84 Left 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V1
SYMATTR Value 24
SYMBOL diode 272 80 R180
WINDOW 0 -43 35 Left 0
WINDOW 3 -101 -5 Left 0
SYMATTR InstName D1
SYMATTR Value MURS120
SYMBOL Misc\\NE555 -160 272 M0
SYMATTR InstName U1
SYMBOL voltage -624 496 M0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
WINDOW 0 -73 53 Left 0
WINDOW 3 -65 76 Left 0
SYMATTR InstName V3
SYMATTR Value 10
SYMBOL res -448 288 M90
WINDOW 0 0 56 VBottom 0
WINDOW 3 32 58 VTop 0
SYMATTR InstName R1
SYMATTR Value 8e4
SYMBOL res -592 224 M90
WINDOW 0 0 56 VBottom 0
WINDOW 3 32 56 VTop 0
SYMATTR InstName R2
SYMATTR Value 2e4
SYMBOL cap -304 528 M0
WINDOW 0 -33 32 Left 0
WINDOW 3 -39 58 Left 0
SYMATTR InstName C1
SYMATTR Value 1e-6
SYMBOL diode -416 384 M90
WINDOW 0 0 32 VBottom 0
WINDOW 3 32 32 VTop 0
SYMATTR InstName D2
SYMATTR Value 1N4148
TEXT -610 666 Left 0 !.tran .2 uic


--
JF

 

[John Fields]

On Tue, 08 Jan 2008 17:25:49 -0600, John Fields
<jfields@austininstruments.com> wrote:

---
Since you're driving the 12V, 100% duty cycle coil with 24V, that
mandates a 25% (or lower) duty cycle so, assuming it's going to be
working 100% of the time, I think a good idea would be to drive it
with something like a 555 wired like this:

+24>----------------------------------------+
|
VCC>--+--------------------------------+ |
| | +------+
[20k]R1 +---------+ | | |K
| |_ | | [COIL] [1N4001]
+-------------7-O|D Vcc|-8---+ | |
| R2 | _| | +------+
+---[82k]---+--6-|TH R|O-4--+ |
| | |__ | | D
+-[1N4148>]-+-2-O|TR OUT|-3---|--G NCH
| | GND | | S
[C] +----+----+ [0.1ľF] |
| |1 | |
GND>--------------+---------+----------+----+

Note that a pot can be substituted for R1 and R2, so you can adjust
the frequency, as required, to accommodate the inductance of the
coil, by choosing the end-to-end resistance of the pot and then
choosing the duty cycle by cranking the pot.

---
Or, change the frequency by selecting an appropriate 'C', duh...


--
JF

 

Hi John

I thought I had all I needed but now I have one last embarrassing
question to ask...

Should the 5V ground be connected to the 70V ground? I.e should the
5V gnd be connected to the mosfet source? What is the reason for it to
be attached not attached?