# Setting Voltage supply for a solenoid

• Erik_clifton102

#### Erik_clifton102

I'm currently making a solenoid valve for a home project, I've been testing the solenoid with a DC Lab power supply, I have set the desired 12V but after connecting it drops and I get a high current with a small voltage.

I understand that it will only draw the voltage required to push the current through the circuit but i want a higher voltage to create a quicker plunger movement.
I have also ran into problems of over heating as there is a spike in current over the desired value.

How would I increase voltage and decrease current without the need of large and expensive resistance to make sure I get correct voltage and current I desire.

Some details
Desired voltage:12V
Desired Current:12A
Solenoid resistance:0.5Ohms
Total turns:225
wire Diameter:1mm

## Answers and Replies

Can your DC supply put out the required current for the stated voltage and coil resistance. Use Ohms law to calculate the current the coil will take. Current equals voltage divided by resistance. Now work out the current the coil will require from the supply.

Erik_clifton102
Can your DC supply put out the required current for the stated voltage and coil resistance. Use Ohms law to calculate the current the coil will take. Current equals voltage divided by resistance. Now work out the current the coil will require from the supply.
Using Ohms law, it works out to be 24A, my power supply can go up to 40A. I don't require 24A, I only need 12A. But am I able to maintain the voltage of 12V whilst only using half of the current its capable of suppling.

... I have set the desired 12V but after connecting it drops and I get a high current with a small voltage.
Maybe your solenoid is rated for AC, where current must be limited by frequency and solenoid inductance, not winding resistance. That would explain the symptoms.

For a DC solenoid, you will need a fly-back diode to protect the switch and the supply.

We need a link to the datasheet.

Erik_clifton102 and DaveE
Maybe your solenoid is rated for AC, where current must be limited by frequency and solenoid inductance, not winding resistance. That would explain the symptoms.

For a DC solenoid, you will need a fly-back diode to protect the switch and the supply.

We need a link to the datasheet.
Its a home made solenoid

I'm not sure you have a grasp on what is happening here. Apply 12 volts to a coil as yours is described and you WILL draw 24 amps providing your power supply is capable of delivering it. You can insert a series resistance but in the end you will have half the voltage across the coil with the target current.

Erik_clifton102
Desired voltage:12V
Desired Current:12A
Solenoid resistance:0.5Ohms
Using Ohms law, it works out to be 24A, my power supply can go up to 40A. I don't require 24A, I only need 12A. But am I able to maintain the voltage of 12V whilst only using half of the current its capable of suppling.
I'm not understanding this. The first quote shows an incompatibility between the specs (V, I, R), and the second quote seems to imply that the 0.5 Ohms value is really what you want. What do you get when you measure the resistance of your solenoid (after subtracting out the baseline resistance you get when you touch the DVM leads together to get your measurement DCR)?

And does your power supply show you the current being drawn? Does it have a variable current limit knob maybe?

Erik_clifton102
Double the turns (to 450) - that will get you 12V/12A (if your other numbers are correct). I suspect that it's still going to get pretty hot (144W dissipation in the windings).

Erik_clifton102
The problem with DC solenoid valves is that it takes more current to start the pull-in than it does to maintain that on state.

You need a high initial voltage to quickly get the current flowing through the coil. Then, once the solenoid has been pulled in, the voltage can be allowed to fall. To do that, you need two different voltages.

I have built voltage boosters into the DC supply to solenoids. While the solenoid is off, the booster slowly builds up to three times the supply voltage in the output capacitor. When the solenoid is connected, the high current flows, then falls to normal because the multiplier can only produce low current compared to the normal supply.

Averagesupernova and berkeman
Here is a schematic for a boosted supply designed to provide a fast start for a solenoid. The power output of this circuit must be switched to one solenoid only.

V1 and E1 simulate a centre-tapped transformer with 9 V peak either side of the grounded centre-tap.

Full-wave rectifier, D1 and D2, provide 6.3 Vrms DC to the output, to hold the solenoid once it is in.

D3 to D5, with C1 and C2 form a half-wave voltage multiplier, while there is no load connected, that pumps up the output voltage on C3 to about 40 volts. C3 then provides the energy needed to initially pull in the solenoid. It takes a few seconds to build up that boost voltage.
The value of C3 must be selected to reliably pull in the solenoid.

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DaveE, Erik_clifton102 and berkeman
Hi Eric, Fixed your problem yet. I suspect you need a better understanding of the issue. If your coil resistance is 0.5 Ohm as you quote, when you connect your supply set to 12 volts the current will be 24 amps. This will no doubt cause overheating and a very low magnetic force ( depends how many turns of wire do you have) because or supply volts drop.
I suspect your power supply set at 12 volts cannot maintain the voltage(reason unknown). I think you might have to rethink your solenoid winding. Have you tried using a car 12v battery to operate your solenoid.That should have enough capacity to either operate as you want it to or burn it out.

Erik_clifton102
I would have to agree with you, i still need to do some more research on the subject to understand it better

Here is a schematic for a boosted supply designed to provide a fast start for a solenoid. The power output of this circuit must be switched to one solenoid only.

V1 and E1 simulate a centre-tapped transformer with 9 V peak either side of the grounded centre-tap.

Full-wave rectifier, D1 and D2, provide 6.3 Vrms DC to the output, to hold the solenoid once it is in.

D3 to D5, with C1 and C2 form a half-wave voltage multiplier, while there is no load connected, that pumps up the output voltage on C3 to about 40 volts. C3 then provides the energy needed to initially pull in the solenoid. It takes a few seconds to build up that boost voltage.
The value of C3 must be selected to reliably pull in the solenoid.

View attachment 302920
Thank you for your comment, I have found it to be very useful.

Baluncore
I'm not understanding this. The first quote shows an incompatibility between the specs (V, I, R), and the second quote seems to imply that the 0.5 Ohms value is really what you want. What do you get when you measure the resistance of your solenoid (after subtracting out the baseline resistance you get when you touch the DVM leads together to get your measurement DCR)?

And does your power supply show you the current being drawn? Does it have a variable current limit knob maybe?
I've measured the resistance across the solenoid and it come back with 0.5ohms. my power supply has both variable voltage and current. It switches between constant voltage and constant current, which from my understanding means it is suppling the maximum output for one of them depending on the value of the other as they are directly proportional.

I suspect your power supply set at 12 volts cannot maintain the voltage(reason unknown).
The current limiting on the supply is the reason for the voltage falling.

Many adjustable voltage laboratory power supplies have an adjustable current limiter associated with the current meter. Check that the limiter is set sufficiently high to allow the current required.

berkeman
The current limiting on the supply is the reason for the voltage falling.

Many adjustable voltage laboratory power supplies have an adjustable current limiter associated with the current meter. Check that the limiter is set sufficiently high to allow the current required.
Ill try explain myself to the better as I believe I haven't explained it well enough for you guys. (note: I am definitely not a professional in subject and haven't taken any courses, I only have knowledge from my own research).

My power supply is a Powertech MP-3091 0-15V 0-40A, "switching mode power supply". For my testing i have a very basic circuit which is connected directly to the power supply, the only two components in the circuit is the solenoid and a switch which is rated at 12V and 20A.

Ill try and explain what happens when I use the power supply to the best of my ability, before I connect the circuit to the supply, I'm able to turn the current up which allows me to then set the voltage to my desired 12V,(note: the current on the display remains at zero as there is no draw yet). the display shows C.V. which from my understanding means constant voltage, once the circuit is connect and I turn the switch on, the voltage drops and current obviously is drawn, the display then shows C.C. which again I believe means constant current.
As expected, the solenoid does heat up.

what could I do to make sure that my voltage is constant. would I need to create a controlling circuit to regulate voltage or complete change my supply method.

I would like to just say big thank you to all the help and comments I've received.

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what could I do to make sure that my voltage is constant. would I need to create a controlling circuit to regulate voltage or complete change my supply method.
That supply regulates it's output voltage to what you set on the front panel. If the current reaches the set limit, the regulated voltage is reduced to prevent excessive current.
To get high current from that supply, you must wind the current setting up to the maximum. You should also be using the rear binding posts, not the banana-plug sockets on the front panel.

I would also check the 0.5 ohm solenoid resistance. Some of that 0.5 ohm may be in the test connections. Wire tables contain the resistance per length, so you can compute the resistance expected from the length of wire used.
If you record the voltage and current while the solenoid is connected, dividing the voltage by the current will give you the resistance of the load.

For a fixed current, the strongest magnetic field comes from the greatest number of turns of wire around the solenoid core. If you wind the wire close, and tight, you will get the maximum possible number of turns possible, and so the strongest field. Also, use the minimum diameter core.

Erik_clifton102
For a fixed current, the strongest magnetic field comes from the greatest number of turns of wire around the solenoid core. If you wind the wire close, and tight, you will get the maximum possible number of turns possible, and so the strongest field. Also, use the minimum diameter core.
If I was able to have a current that constantly fluctuates, would you know the resultant of that?

If I was able to have a current that constantly fluctuates, would you know the resultant of that?
In or close to current limit. Or a bad power supply.

once the circuit is connect and I turn the switch on, the voltage drops and current obviously is drawn, the display then shows C.C. which again I believe means constant current.
What current is displayed on the power supply in this condition? Is it at the max rated current of 40A? If so, you have not correctly measured the resistance of your coil (see our comments about DVM/connection wire resistances, etc. in earlier replies).

If it shows less than 40A in this current-limit situation, be sure you have turned the Current Limit knob all the way to full, and if you have then your Power Supply likely has an issue.

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I have a weird feeling about this thread.
We have no picture or other evidence of the possible construction.
Maybe the wire used to wind the solenoid was not insulated.

Tom.G, DaveE and berkeman
I'd like to see a pic of the power supply. Make and model.

Boy I feel silly. Lol. Couldn't find that.