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Goosing a solenoid??

D

Don A. Gilmore

Jan 1, 1970
0
Hi guys:

I have an application where I have 24 independently-operated push solenoids
(very small tubulars) that are rated at 12 Vdc, continuous operation. To
make them operate properly in my device, I need to deliver a 4x voltage
pulse (48 volts) to the solenoid for a very short period (maybe 10 ms) then
revert to 12 V to hold the solenoid extended while it is in use. This must
occur every time it is fired. I will have both 12 V and 48 V sources
available on the PCB and the circuit is controlled with CMOS.

What is the simplest way to do this? Thanks for all replies.

Don
Kansas City
 
D

Dave

Jan 1, 1970
0
Hi Don,
One solution is to charge a cap to 48V with a resistor that will drop
36 volts when in series with the solenoid. Assuming the duty cycle will
allow the cap to recharge.
+48v
|
resistor
|
CAP-------solenoid
| |
| driver transistor
| |
| |
common---------

Dave
 
D

Dave

Jan 1, 1970
0
OOPS, my drawing with the keyboard leaves much to be desired..

+48v
|
resistor
|
----------solenoid
| |
CAP driver transistor
| |
| |
common---------

Dave
 
J

John Popelish

Jan 1, 1970
0
Don said:
Hi guys:

I have an application where I have 24 independently-operated push solenoids
(very small tubulars) that are rated at 12 Vdc, continuous operation. To
make them operate properly in my device, I need to deliver a 4x voltage
pulse (48 volts) to the solenoid for a very short period (maybe 10 ms) then
revert to 12 V to hold the solenoid extended while it is in use. This must
occur every time it is fired. I will have both 12 V and 48 V sources
available on the PCB and the circuit is controlled with CMOS.

What is the simplest way to do this? Thanks for all replies.

Don
Kansas City
Back in the day of wire matrix impact printers this sort of thing was
very popular as the wire driver (called pick and hold). Sorry that I
don't have time right now to do a more complete search, but here is an
example of how it was done:
http://www.allegromicro.com/datafile/archive/2962.pdf
 
H

Homer.Simpson

Jan 1, 1970
0
Don A. Gilmore said
Hi guys:

I have an application where I have 24 independently-operated push
solenoids (very small tubulars) that are rated at 12 Vdc,
continuous operation. To make them operate properly in my
device, I need to deliver a 4x voltage pulse (48 volts) to the
solenoid for a very short period (maybe 10 ms) then revert to 12
V to hold the solenoid extended while it is in use. This must
occur every time it is fired. I will have both 12 V and 48 V
sources available on the PCB and the circuit is controlled with
CMOS.

What is the simplest way to do this? Thanks for all replies.

Don
Kansas City


This is very similar to how automotive diesel injectors are driven.

For optimal fuel control, one needs to open injectors very quickly
with a fast current rise (often via a boost voltage), hold them
open with a more convenient power source (vehicle battery voltage),
and shut them down quickly via active low side clamping. They are
typically current modulated at four current levels (two for
opening, two more for holding). They are usually low impedance and
will draw high currents if left at either voltage for very long.

You speak of a boost voltage for a period of time and a hold
voltage for the remainder. Is the impedance of these solenoids
sufficient to limit the current to a managable level at 12V?

We use two high side driven MOSFETS. One from the boost rail, the
other from the battery rail. Each is modulated to affect the
desired current levels.

PFETS may the simplest?

48V 12V
---------------------- -------
| | | | | |
| | | | | |
| | R / | |
--- Vz | --| NPN s s
--- | |__| \ ___| PFET _| PFET
| | | | / d d
| | | --| PNP | |
| | | \ | |
|--------------------| V (Diode)
| | | -
| | | |
R | -------------->>-
| d L
| _| O
| s A
| | D
GND------------------------------->>--


Maybe drive the 48V PFET with a one shot triggered off the rising
edge of 12V PFET drive signal (not shown)

There are many custom solutions for this topology. Here's one for a
single voltage rail.

http://www.onsemi.com/pub/Collateral/NCV7510.PDF

Homer Simpson
Spingfield
 
J

Joerg

Jan 1, 1970
0
Hello John, Hello Don,
Back in the day of wire matrix impact printers this sort of thing was
very popular as the wire driver (called pick and hold). Sorry that I
don't have time right now to do a more complete search, but here is an
example of how it was done:
http://www.allegromicro.com/datafile/archive/2962.pdf

A lot of these drivers will be discontinued because the impact printer
appears to have followed the dinosaur. But you may be able to use a chip
meant to drive a stepper motor. To crank maximum horsepower out of a
stepper motor its coils must often also be peaked in a similar fashion.

Regards, Joerg
 
H

Homer.Simpson

Jan 1, 1970
0
Homer.Simpson said

Hopefully this will come out better. This is my first ASCII circuit.
That's why it looks like crap. ;-)
 
M

Mook Johnson

Jan 1, 1970
0
A solenoid is actually a current driven device.

you might try to make a closed current loop instead of a simple voltage on,
off. While activiting L di/dt will cause high voltage (curent loop
compliance) and will regulate the current to that required to maintain the
device once set.
 
B

Bob Eldred

Jan 1, 1970
0
Don A. Gilmore said:
Hi guys:

I have an application where I have 24 independently-operated push solenoids
(very small tubulars) that are rated at 12 Vdc, continuous operation. To
make them operate properly in my device, I need to deliver a 4x voltage
pulse (48 volts) to the solenoid for a very short period (maybe 10 ms) then
revert to 12 V to hold the solenoid extended while it is in use. This must
occur every time it is fired. I will have both 12 V and 48 V sources
available on the PCB and the circuit is controlled with CMOS.

What is the simplest way to do this? Thanks for all replies.

Don
Kansas City

The cleanest way to do this is with one or more microprocessors and FET
drivers for the solenoids. The solenoids would be run on 48 volts. The
microprocessor(s) would pulse the solenoids on for the required pull in time
then go to pulse width modulation at one fourth duty cycle for the 12 Volt
equivalent drive. There would be 24 input ports and 24 FET driver ports with
the timing for the driver periods controlled by counter registers for each
driver. A couple of PICs could easily do this. The PWM frequency could be
20KHz. Futhermore, being coded, the pulse width, 48V time and other factors
can be easily be adjusted to get the require performance, not easy with
capacitors or other ideas.
Bob
 
J

John Popelish

Jan 1, 1970
0
Don said:
Hi guys:

I have an application where I have 24 independently-operated push solenoids
(very small tubulars) that are rated at 12 Vdc, continuous operation. To
make them operate properly in my device, I need to deliver a 4x voltage
pulse (48 volts) to the solenoid for a very short period (maybe 10 ms) then
revert to 12 V to hold the solenoid extended while it is in use. This must
occur every time it is fired. I will have both 12 V and 48 V sources
available on the PCB and the circuit is controlled with CMOS.

What is the simplest way to do this? Thanks for all replies.

Don
Kansas City
Here are a couple more integrated drivers that almost fit your supplies:
http://www.powerdesigners.com/InfoWeb/design_center/Appnotes_Archive/4032.pdf
http://www.alldatasheet.com/datasheet-pdf/view/STMICROELECTRONICS/L6213.html
 
J

John Woodgate

Jan 1, 1970
0
I read in sci.electronics.design that Dave <[email protected]>
wrote (in said:
OOPS, my drawing with the keyboard leaves much to be desired..

+48v
|
resistor
|
----------solenoid
| |
CAP driver transistor
| |
| |
common---------

Dave

It's still mangled. Use Courier font and don't use tabs.

It's also a circuit that slows down the operation of the relay. It's
better to put the cap in parallel with the resistor. This simple trick
works VERY WELL, and you don't need complex techniques. Use Courier
font:

+48 V----+----R----+---COIL------Switching device----- 0V
| |
| |
`----C----'
The resistor R should be three times the coil resistance, and RC should
probably be around 50 ms for such small solenoids.

This works because initially the cap ISN'T charged, so the 48 V gets
directly to the coil. As the cap charges, the coil voltage drops to 12
V. When the switch opens, the cap discharges through R.
 
J

John Fields

Jan 1, 1970
0
Hi guys:

I have an application where I have 24 independently-operated push solenoids
(very small tubulars) that are rated at 12 Vdc, continuous operation. To
make them operate properly in my device, I need to deliver a 4x voltage
pulse (48 volts) to the solenoid for a very short period (maybe 10 ms) then
revert to 12 V to hold the solenoid extended while it is in use. This must
occur every time it is fired. I will have both 12 V and 48 V sources
available on the PCB and the circuit is controlled with CMOS.

What is the simplest way to do this? Thanks for all replies.
 
P

Pat Ford

Jan 1, 1970
0
Don A. Gilmore said:
Hi guys:

I have an application where I have 24 independently-operated push solenoids
(very small tubulars) that are rated at 12 Vdc, continuous operation. To
make them operate properly in my device, I need to deliver a 4x voltage
pulse (48 volts) to the solenoid for a very short period (maybe 10 ms) then
revert to 12 V to hold the solenoid extended while it is in use. This must
occur every time it is fired. I will have both 12 V and 48 V sources
available on the PCB and the circuit is controlled with CMOS.

What is the simplest way to do this? Thanks for all replies.

Don
Kansas City
there are chips that do this already. look for peak and hold drivers for
automotive fuel injectors.

google can be your friend ( hint
http://www.google.ca/search?hl=en&q=peak+and+hold+injector+drivers&meta=


HTH
Pat
 
E

eromlignod

Jan 1, 1970
0
John said:
I like John Woodgate's trick, but you'll be dissipating three times
more power in the resistors than in the solenoids. Can you stand
that?


I like it too. Thanks, John W. & Dave. I also like the idea of the
solenoid driver chip, but it seems that they run three or four bucks
apiece.

The solenoids are 2 watts. So does this mean that I'll need at least a
6-watt resistor (3 * 2W) for each solenoid circuit? No more than six
of the solenoids ever fire simultaneously. So I guess the max. total
heat loss would be

(6 * 2W) + (6 * 6W) = 48 watts

Am I all wet on this?

Don
Kansas City
 
J

John Fields

Jan 1, 1970
0
that?


I like it too. Thanks, John W. & Dave. I also like the idea of the
solenoid driver chip, but it seems that they run three or four bucks
apiece.

The solenoids are 2 watts. So does this mean that I'll need at least a
6-watt resistor (3 * 2W) for each solenoid circuit? No more than six
of the solenoids ever fire simultaneously. So I guess the max. total
heat loss would be

(6 * 2W) + (6 * 6W) = 48 watts

Am I all wet on this?
 
E

eromlignod

Jan 1, 1970
0
John said:
OK, great. Now, I don't have the solenoids in hand yet, but I'm
thinking the coil resistance should be Rc = V^2 / P, or

Rc = (12V)^2 / 2W = 72 ohms

So the resistor should be about 3 x 72 = 216 ohms. For an RC of 50 ms,
this would make my capacitor about

C = t / R = .050s / 216 = 231 uF

If I have room for such large components, this should do nicely.
Thanks everyone!

Don
Kansas City
 
P

Pat Ford

Jan 1, 1970
0
eromlignod said:
that?


I like it too. Thanks, John W. & Dave. I also like the idea of the
solenoid driver chip, but it seems that they run three or four bucks
apiece.

The solenoids are 2 watts. So does this mean that I'll need at least a
6-watt resistor (3 * 2W) for each solenoid circuit? No more than six
of the solenoids ever fire simultaneously. So I guess the max. total
heat loss would be

(6 * 2W) + (6 * 6W) = 48 watts

Am I all wet on this?

Don
Kansas City

NAtional semi samples them.
Pat
 
R

Rich Grise

Jan 1, 1970
0
OK, great. Now, I don't have the solenoids in hand yet, but I'm
thinking the coil resistance should be Rc = V^2 / P, or

Rc = (12V)^2 / 2W = 72 ohms

So the resistor should be about 3 x 72 = 216 ohms. For an RC of 50 ms,
this would make my capacitor about

C = t / R = .050s / 216 = 231 uF

If I have room for such large components, this should do nicely.
Thanks everyone!

I would recommend that when you _do_ get the solenoids, you actually
measure their resistance. The "2 watt" spec sounds a little "loose"
for me to want to base a series resistor on.

Besides, I want to know. :)

Good Luck!
Rich
 
F

Frank Bemelman

Jan 1, 1970
0
Don A. Gilmore said:
Hi guys:

I have an application where I have 24 independently-operated push solenoids
(very small tubulars) that are rated at 12 Vdc, continuous operation. To
make them operate properly in my device, I need to deliver a 4x voltage
pulse (48 volts) to the solenoid for a very short period (maybe 10 ms) then
revert to 12 V to hold the solenoid extended while it is in use. This must
occur every time it is fired. I will have both 12 V and 48 V sources
available on the PCB and the circuit is controlled with CMOS.

What is the simplest way to do this? Thanks for all replies.

Is is possible to connect it to 48V only, and switch from initial 100% duty
cycle to something much lower duty cycle PWM, to drop to a current that is
equal to 12V/Rsolenoid?
 
J

John Woodgate

Jan 1, 1970
0
I read in sci.electronics.design that eromlignod <[email protected]>
wrote (in said:
that?


I like it too. Thanks, John W. & Dave. I also like the idea of the
solenoid driver chip, but it seems that they run three or four bucks
apiece.

The solenoids are 2 watts. So does this mean that I'll need at least a
6-watt resistor (3 * 2W) for each solenoid circuit?
No more than six
of the solenoids ever fire simultaneously. So I guess the max. total
heat loss would be

(6 * 2W) + (6 * 6W) = 48 watts

Am I all wet on this?
Not wet, but that would be true only if you really need to keep full
rated voltage across them for more than a few seconds. Once the solenoid
armature has come into position, you may well be able to hold it there
with far less than 12 V. The spec may tell you. It depends on whether
what the solenoid moves applies a restoring force. If there isn't much
of a 'push-back' force, you might need only 3 or 4 V to hold position.
That would cut the power down a lot.

2 W and 12 V is 144/2 = 72 ohms. 3 V/72 ohms is 42 mA. 48 V times 42 mA
is 2 W total for solenoid PLUS resistor.
 
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