freewheeling diodes to speed up solenoids?

[curlyandshemp]

Interesting theory - first time I've heard of semiconductors doubling as dielectrics.

The free-wheeling diode will delay the drop-out of the solenoid because it provides a path for current generated by the collapsing magnetic field to circulate through the coil in the same direction as when the circuit is energised.

I was thinking in theory what a capacitor was, two conductors separated by a non-conductive material. I seem to recall from my '70s electronics training about something with diodes where the gap between cathode and anode increases with something voltage increases ..blah blah ....brain getting .. foggy....where are my keys?

Ian

 

[Nick B]

I think you might want to speak to solenoid vendor about doing this but if you had a separate DC supply you could drive the solenoid off by applying -24VDC to output. You would need a form C relay and have to factor in additional delay of relay but it would drive the solenoid back HARD.


Nick

 

[oregonsam]

Wow, thanks for all the great responces! Does anyone have an opinion on whether diodes would affect the ON cycle of a solenoid?

 

[Sparkz]

They don't, IF and only if the solenoid is always allowed to deenergize completely before being reengaged.

Meaning: Although a freewheeling diode seriously increases drop off time, at some point during drop off the return spring in the valve will overcome the coil's energy. The valve shuts but there's still some energy left in the coil. If at this point the solenoid is reenergized, the valve will reopen quicker. So, basically all depends on cycle time...

 

[krk]

I came across this circuit more than 20 years ago on a prototype flying shear in a rolling mill.

The engineer who designed the controls was having problems with relay and solenoid actuation speeds and came up with this as a workaround.

IIRC the intended operation was:
On transition of the output from OFF to ON, the 12V coil would be powered with 24V for the duration of the capacitor charging time, decreasing the pull in time.
When the capacitor was charged the coil would be held in at the minimum permissible value through the dropping resistor.

I have absolutely NO idea whether this circuit actually worked as intended (although the machine did...), just offering it up for discussion.

 

[Sparkz]

On transition of the output from OFF to ON, the 12V coil would be powered with 24V for the duration of the capacitor charging time, decreasing the pull in time.
When the capacitor was charged the coil would be held in at the minimum permissible value through the dropping resistor.

I have absolutely NO idea whether this circuit actually worked as intended (although the machine did...), just offering it up for discussion

It should work just fine, as the pull in current level will be reached sooner. Though I have no idea how to size the capacitor...anyone?

A few years ago, I built a compound gun controller with a PWM output, hitting the 12V gun solenoid with 50Vdc during 2ms to open the gun, then using the same 50V supply and the PWM signal at 24% duty cycle just to keep the gun opened. Worked like a dream! Back then my major concern was to open the compound gun as fast as possible, I did not have to worry about drop off. Too bad it never went into production...

 

[Ron Beaufort]

regarding the basic question: how can I speed up the OPENING of a solenoid? ...



I saw the following demonstration done at an Allen-Bradley trade show about 15 years ago ...



picture a large “wheel of fortune” type wheel about 5 feet in diameter ... the thing is mounted vertically (not horizontally like Vanna’s wheel) ... all around the circumference are 6-inch spikes sticking out towards the front (not toward the edge) of the wheel ... the spikes are about 4 or 5 inches apart ... the operator puts an empty Coke can on each spike ... the bottoms of the cans are numbered sequentially 1 through something like 40 ... they start a large Baldor motor which spins the wheel with a drive belt ... faster and faster until the cans are all just a blur ... then the operator asks you which can you want ... you say 26 ... he dials in 26 on his HMI terminal and hits the “go” button ... phhht! ... can number 26 flies off the wheel and lands in a net about 6 feet away ...



how it was done (according to what we were told) ... the wheel has a sensor so that the PLC knows what position the wheel is in at all times ... using immediate inputs and immediate outputs, the PLC picks out the proper position for can number 26 and fires a DC solenoid to blow the proper can off the wheel with an air jet ...



now to our original question: how do you get the solenoid to open FAST ENOUGH to get one can - the right can - and one can only? ...



the answer (according to what we were told) ... the secret is to fire the 24 volt DC solenoid with about 40 volts DC - but just for a fraction of a second ... the higher voltage is able to kick the solenoid into action a LOT faster than the rated 24 volts can do it ... as soon as the higher voltage “kick” has done its thing, the voltage is switched down to 24 volts - and then turned off ...



at least that’s what we were told ... and I saw this thing work perfectly - and repeatedly - time after time ... even more impressive was when someone would request something like “every third can” ... the operator would key in the request and hit the “go” button ... phhhhhhhhtttttttt! ... and then when the wheel was stopped, every third can was off the wheel - and in the net ...



I’d call that pretty darn fast ... so maybe something in all of this will be helpful ... if this were my job to do, I’d start experimenting with two DC outputs - one for the 40 volt “kick” - and the other for the regular 24 volt signal ... and if the “time on” is pretty short, maybe all you’d really need would be the higher voltage “kick” ... (survival tip: have some spare solenoid coils handy) ...



anyway ... that’s what I saw ... and that’s what I was told ... hope it helps ...



also ... the method suggested by krk - and ratified by Sparkz - in the preceding posts might be helpful ... it seems to be based on the same “kick-it-open-with-a-higher-than-rated-voltage” idea ...

 

[bernie_carlton]

Here's a 'spike and hold' circuit from The Lee Company

 

[jr9701]

The above posts are correct. I dont see how a diode will trigger a solenoid faster, however The de-energizing time (as already mentioned), will infact be longer then w/ out a diode. I have and still use diodes to surpress short cycling on certain relay circuits that have such issues, but I never used them on solenoids.

 

[Calistodwt]

I seem to recall that the diode across the coil was originally introduced for electronically operated coil cicuits in order to prevent the high back emf generated from the coil magnetic field collapsing from destroying electronic components

 

[Freak]

We fire short cycle solenoids (for gluing cartons) with 240VAC and use a 24VAC coil. Solenoids are on for only a fraction of a second and we don't seem to have any issues with the coil. We did experience some issues getting the actual solenoid to work fast enough mechanically. Now we run high pressure thorugh a small solnoid and it seems to work fine.

Check out companies like Nordson they may have some info.

 

[oregonsam]

Has anyone tried "prefiring" with a low voltage (5VDC or 12VDC) to establish a current in the solenoid, then apply the 24VDC to actually move the solenoid? If the "prefire" doesn't move the solenoid, it should reduce the time to reach the switching voltage.

 

[micahg]

Very interesting, since the current lags the voltage in an inductive circuit, at the instant of connection there is no current and it slowly builds 63% per time constant. So a higher voltage at the begining should reduce the time required to get to the first or second time constant of the lower voltage/current point. (63% of 48 volts is twice as much as 63% of 24 volts)There by shortening the time required for the solenoid to operate without actually overloading the coil.

This could also be done with one output with 2 sources, a diode and resistor and a capacitor. Charge the cap to say 24 vdc then turn the output on and let the cap provide the extra voltage boost as it discharges, then when discharged let the diode furnish the 24 volts to the coil. The boost would be determined by the size of the cap and the inductive reactance of the coil.

24v output>----------------|>|--------/\/\--------|
| diode | coil |
| | |
|------||-------| |
| capacitor |
| |
| |
/ |
\ |
/ resistor |
\ |
| |
-24vdc

The output would have to be able to take -24 volts when not energized. And the refire time would have to be long enought for the cap to charge determined by the RC time constant and the RC time constant would have to be long enough to prevent high current in the resistor during the on time.

 

[micahg]

Sorry about the way my circuit looks, it took all my spaces out when I posted it.

 

[bernie_carlton]

You can use the ladder tool (next to the right end on the second row of tools) to bracket your ASCII drawing. Or you could do the drawing in Paint and 'attach' it or save it is one of the acceptable drawing types (gif jpeg jpg png) then use the 'display pictures' feature in the 'additional options' below the posting box.