What do you think of this circuit to activate a solenoid? Does it need anything else? See attached.

Hopefully I calculated this correctly. I didn't factor in the diode (not sure how to), but calculated the capacitor and resistors. When you flip the switch it should initially give it a pull-in voltage of 20V for about 1.5 seconds and then hold the voltage at 8.5V until the switch is turned off.

Do I need a circuit to discharge the cap? Do I need any protection added? Any advice would be great.

Thanks!

 

The resistors, capacitor, and diode are unnecessary. The solenoid coil is rated for 24V, so it will operate at approximately that voltage with no additional resistors needed to drop voltage and/or limit current. The diode would typically be used to protect an active switching device from inductive kickback. In your case, the switch might or might not need that protection.

 

The diode is helpful to reduce EMI and to prolong the life of the switch, it should be rated At least 50V @ 1A (1n4001)

I assume you are trying to reduce the on-state power dissipation in the solenoid? The 4.9 ohm resistor is not doing much, you can eliminate it.
The value of the parallel R/C circuit determines the time constant, consider how fast the solenoid needs to turn on/off?

 

Hi and thank you for your responses.

I think I should keep the diode as there may be other things hooked up to the same power supply later.

I also think the cap and resistors are needed. I don't want to have this solenoid at full power engaged for hours. It will get too hot and waste power. That is the reason for the cap and the higher pull-in voltage with the lower hold-in voltage.

The 4.9 ohm resistor is just a voltage divider in order to get the voltage on the solenoid 20V.

After the initial solenoid activation of 20V, the cap will charge and the circuit will give the solenoid 8.5V. How do I ever get the cap to discharge so I can get the pull-in voltage again for multiple on/offs?

Thanks!

 

Hi and thank you for your responses.

After the initial solenoid activation of 20V, the cap will charge and the circuit will give the solenoid 8.5V. How do I ever get the cap to discharge so I can get the pull-in voltage again for multiple on/offs?

Thanks!

The cap is in parallel with a resistor- open the switch and it discharges by itself into the resistor.

Why limit the pull-in voltage? I see no reason for the 5 ohm resistor?

 

Do you have any Inductance measurements or data sheet for the Solenoid?

My expectation is that you missed the required time-constant by a factor of 50,000 or so...



Yes Grandma, that is a 0.01 FARAD capacitor

Note the power dissipation in the coil and the resistor vs time...

 

[QUOTE="
I think I should keep the diode as there may be other things hooked up to the same power supply later.

I also think the cap and resistors are needed. I don't want to have this solenoid at full power engaged for hours. It will get too hot and waste power. That is the reason for the cap and the higher pull-in voltage with the lower hold-in voltage.

The 4.9 ohm resistor is just a voltage divider in order to get the voltage on the solenoid 20V.

After the initial solenoid activation of 20V, the cap will charge and the circuit will give the solenoid 8.5V. How do I ever get the cap to discharge so I can get the pull-in voltage again for multiple on/offs?

Thanks![/QUOTE]
60Ω In series with a 33Ω load is going to give it a hard time pulling in, if it does at all. And the resistors will get hot instead of the resistors "wasting power"

 

@Sensacell - Thanks! I see. The cap will discharge into the resistor and come off as heat. I agree that the first resistor is not really needed. It was only to voltage divide and mimic the volts in the original application.

@MikeML - I don't have a datasheet or any other info on the solenoid except this : 50616760 solenoid, pinch roller, 24VDC, 33ohm, 727mA, TEAC RL-1614. Are you saying I should switch the cap to .01 F and the resistor to 15 ohm?

@gerty - Is there a better way? I would like to keep the pull-in at 20V and 8.5V hold-in. At least the heat is distributed between two components and not all on the solenoid. Please share if there is a better way using less power during the hold-in times.

Thanks guys!!

* Here is the original schematic where the solenoid was used. I highlighted the jpg. It may lend some ideas about my stand-alone circuit. Let me know what you think. The original power supply was AC out and then they rectified it with that diode bridge. Two 11.5V AC for a total of 23V AC.

 

@gerty : 60 ohms is not in series when it pulls in. this is the reason for the capacitor. before and while the capacitor is charging, the electrons bypass the 60 ohms and go the way of the capacitor path. this is when you get your higher pull-in voltage. after the capacitor is charged then the 60 ohms will be in series and that will be a low hold-in voltage.

second point. more resistance in series does lower power. check ohms law.

[QUOTE="
I think I should keep the diode as there may be other things hooked up to the same power supply later.

I also think the cap and resistors are needed. I don't want to have this solenoid at full power engaged for hours. It will get too hot and waste power. That is the reason for the cap and the higher pull-in voltage with the lower hold-in voltage.

The 4.9 ohm resistor is just a voltage divider in order to get the voltage on the solenoid 20V.

After the initial solenoid activation of 20V, the cap will charge and the circuit will give the solenoid 8.5V. How do I ever get the cap to discharge so I can get the pull-in voltage again for multiple on/offs?

Thanks!

60Ω In series with a 33Ω load is going to give it a hard time pulling in, if it does at all. And the resistors will get hot instead of the resistors "wasting power"[/QUOTE]

 

The resistors, capacitor, and diode are unnecessary. The solenoid coil is rated for 24V, so it will operate at approximately that voltage with no additional resistors needed to drop voltage and/or limit current. The diode would typically be used to protect an active switching device from inductive kickback. In your case, the switch might or might not need that protection.

Just a back emf diode causes slow release on relays (probably not as important on solenoids) - Tyco published an appnote that explains it in detail.

 

Well, if you really want to save the power (the solenoid should handle being energized at pull-in current for some period mentioned in the datasheet); you'll want to:

• get rid of the 5 ohm resistor, pull-in current will be brief
• determine the worst case hold-in voltage experimentally to choose an appropriate series RC

Why are you operating the solenoid from 23V? How did you determine the hold-in voltage? Is your capacitor 4.7uF or 4.7mF? I don't think either value would give you the 1.5 seconds.

And yes, you need to worry about discharging the cap so the solenoid will energize if you try to cycle the switch quickly or there could be other momentary power interruptions.

 

Just a back emf diode causes slow release on relays (probably not as important on solenoids) - Tyco published an appnote that explains it in detail.

I understand that. It's a trade off to protect the switching device which is why I stated it might not be needed for a normal switch if it can survive any arcing.

 

Well, if you really want to save the power (the solenoid should handle being energized at pull-in current for some period mentioned in the datasheet); you'll want to:

• get rid of the 5 ohm resistor, pull-in current will be brief
• determine the worst case hold-in voltage experimentally to choose an appropriate series RC

Why are you operating the solenoid from 23V? How did you determine the hold-in voltage? Is your capacitor 4.7uF or 4.7mF? I don't think either value would give you the 1.5 seconds.

And yes, you need to worry about discharging the cap so the solenoid will energize if you try to cycle the switch quickly or there could be other momentary power interruptions.

Hi DL. Ok will get rid of 5 ohm resistor since no one likes it.
I actually need to use this AC power supply. See my jpeg from previous post. Not sure what my dc would be after I rectify it.
The hold in voltage someone told me.
I'll recalculate my RC. It's probably wrong.
This solenoid won't need to do rapid on/offs. It is more activate and hold for long times. I would still need cap drain? U have a circuit?
Also should I consider switching with a midget? Because of arcing with the manual switch?

A lot of questions I know. Sorry. Thanks!!

 

@MikeML - I don't have a datasheet or any other info on the solenoid except this : 50616760 solenoid, pinch roller, 24VDC, 33ohm, 727mA, TEAC RL-1614. Are you saying I should switch the cap to .01 F and the resistor to 15 ohm?.

I have never met a solenoid like that didn't have an inductance of at least 0.1H. Refer back to the simulation in post six. The current in the s0lenoid starts out at 0.68A, and within 1/4s is down to 0.48A. Note that to create even such a short delay before the current drops, I had to make C1 10,000uF. The delay circuit will discharge in about the same time as it charges.

Note that t0 make the final power dissipation (7.5W, red trace) in the solenoid about 1/2 of the initial power dissipation(15W), I had to make R1=15Ω.
Note the power dissipation in R1 is about 4W, so the total heat reduction is not that much. If you can prove to yourself that the solenoid will hold in with an even lower current, then R1 could be even bigger, but likely not all the way to 56Ω.

 

Your circuit will work. Drop the 5 ohm resistor and select a capacitor that will give you minimum pull-in current for sufficient time. 56 ohms and 4700uF will give you an RC time constant or two of high current; that's a quarter to a half second. If that isn't enough, make the cap larger.

I checked a couple datasheets for pull-in and hold-in currents and the datasheets were devoid of useful specs. You should determine pull-in time and hold-in current requirements experimentally. It seems odd that hold-in voltage would be a third of pull-in.

If you need the capacitor to be discharged at power off because it doesn't discharge fast enough on it's own, I recall seeing a Bob Pease designed circuit to do it...

 

Your circuit will work. Drop the 5 ohm resistor and select a capacitor that will give you minimum pull-in current for sufficient time. 56 ohms and 4700uF will give you an RC time constant or two of high current; that's a quarter to a half second. If that isn't enough, make the cap larger.

I checked a couple datasheets for pull-in and hold-in currents and the datasheets were devoid of useful specs. You should determine pull-in time and hold-in current requirements experimentally. It seems odd that hold-in voltage would be a third of pull-in.

If you need the capacitor to be discharged at power off because it doesn't discharge fast enough on it's own, I recall seeing a Bob Pease designed circuit to do it...

lol...my phone auto corrected MOSFET to midget. sorry about that.

thanks mikeml and dl324. i will experiment with values around those ranges and report back soon.

i have a 24V DC power supply i can use temporarily, but ultimately i will need to rectify this AC one. i don't have a choice and have to use it. can i check back in and get help with the diode bridge rectifier for that? do you guys know anything about such?

thanks!!

 

@gerty : 60 ohms is not in series when it pulls in. this is the reason for the capacitor. before and while the capacitor is charging, the electrons bypass the 60 ohms and go the way of the capacitor path. this is when you get your higher pull-in voltage. after the capacitor is charged then the 60 ohms will be in series and that will be a low hold-in voltage.

second point. more resistance in series does lower power. check ohms law.

This is DC correct? You have the power marked + -. A capacitor does not pass dc ,nor will it give you "higher pull in voltage".

i have a 24V DC power supply i can use temporarily, but ultimately i will need to rectify this AC one

Now you're talking about an AC supply? Which do you have??
This will make a huge difference on what'll work..

 

@gerty : this guy is using DC (battery) and says the current flows through the caps until the potential diff matches the battery.

Are you saying this circuit is wrong too? :

http://electronicdesign.com/analog/what-s-all-solenoid-driver-stuff-anyhow

 

@gerty

how about this : " Capacitors are used in DC circuits to provide "bursts of energy." "

http://dev.physicslab.org/Document.aspx?doctype=3&filename=DCcircuits_KirchhoffCapacitors.xml

 

@gerty

how about this : " Capacitors are used in DC circuits to provide "bursts of energy." "

http://dev.physicslab.org/Document.aspx?doctype=3&filename=DCcircuits_KirchhoffCapacitors.xml

The problem, as mike pointed out is to get 1.5 seconds you need a very big capacitor.
There are other ways to do it, but not quite as simple.