Goosing a solenoid??

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

Reply to
Dave
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OOPS, my drawing with the keyboard leaves much to be desired..

+48v | resistor |

----------solenoid | | CAP driver transistor | | | | common---------

Dave

Reply to
Dave

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:

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Reply to
John Popelish

Here are a couple more integrated drivers that almost fit your supplies:

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Reply to
John Popelish

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

Reply to
Don A. Gilmore

Don A. Gilmore said

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.

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Homer Simpson Spingfield

Reply to
Homer.Simpson

Hello John, Hello Don,

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

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Reply to
Joerg

Homer.Simpson said

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

Reply to
Homer.Simpson

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.

Reply to
Mook Johnson

solenoids

then

must

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
Reply to
Bob Eldred

--
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?
Reply to
John Fields

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

Reply to
eromlignod

times

least a

six

total

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

Reply to
eromlignod

I read in sci.electronics.design that Dave wrote (in ) about 'Goosing a solenoid??', on Wed, 4 May 2005:

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.

--
Regards, John Woodgate, OOO - Own Opinions Only.
There are two sides to every question, except
 Click to see the full signature
Reply to
John Woodgate

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

Reply to
Rich Grise

solenoids

then

must

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

google can be your friend ( hint

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HTH Pat

Reply to
Pat Ford

--
Nope, you\'re exactly right.
Reply to
John Fields

NAtional semi samples them. Pat

Reply to
Pat Ford

Hi John, I agree that the cap in parallel with the resistor is a better if the

48V supply can handle the instantaneous current of the turn on of multiple solenoids. It would depend on the criteria for the solenoid actuation and the specs. on the 48V supply.

Dave

Reply to
Dave

--
Don\'t forget the diode clamp across the solenoid!
Reply to
John Fields

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