Goosing a solenoid??

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.

"Don A. Gilmore" schreef in bericht news:Czdee.37826$ snipped-for-privacy@tornado.rdc-kc.rr.com...

solenoids

then

must

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?

I had exactly that problem when automating a tape deck system (and it used exactly those same voltages).

Charge a large capacitor to 48 volts through a resistor selected to give sufficiently fast recharge but not so low as to dissipate large amounts of power when the solenoid is on.

Feed the top of the solenoid from the +ve of the capacitor and from the

Don't forget Mr. Diode:

D1 ------! ! !

D1 keeps the voltage from spiking up to a huge value when the switch turns off. Connecting it as shown causes the 36 or so volts on C1 to be the limit on the spike on the coil.

Car fuel injectors are also pulsed to get them open quickly.

I couldn't get the link to work, John. But you have piqued my curiosity.

Don Kansas City

The schematic has a gross error.

...Jim Thompson

-- | James E.Thompson, P.E. | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | Phoenix, Arizona Voice:(480)460-2350 | | | E-mail Address at Website Fax:(480)460-2142 | Brass Rat | |

Can't you just add one more diode (D2 connected to +12V)? Then R can be made as large as possible, limited only by desired recovery time, and the resistor dissipation drops way down.....

D2 +----|

Wouldn't that kind of depend on how good your +12 supply is at sinking reverse current?

Thanks, Rich

Aaaarghhh! Found it and fixed it. Thanks, Speff. :-)

As it's shown it turns off faster, but the recycle time is slower... so pick your druthers ;-)

...Jim Thompson

Faithfully repeated in the next version.

Best regards, Spehro Pefhany

can

time,

sinking

Why? Provided C charges slowly, which it will, at release time the coil's supply node will be at 12v minus D2's drop. The flyback voltage seen by the driver then will be 12V-Vf(D2), plus D1's forward drop, or 12v.

Jim Thompson's arrangement turns the coil off faster. Recycle time isn't much different -- clamping by D2 prevents reverse-charging of C, so that node stays near 12v during flyback for both circuits.

Regards, James Arthur

Interesting pseudonym. The original "Dagmar" was a couple of years ahead of me in high school. The Egnor family barbershop... always good for an illegal poker game ;-)

...Jim Thompson

-- | James E.Thompson, P.E. | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | Phoenix, Arizona Voice:(480)460-2350 | | | E-mail Address at Website Fax:(480)460-2142 | Brass Rat | |