# Maximum power for a 12V solenoid

I have a 12V 8A 35mm-stroke solenoid from eBay (similar to this listing) which I am powering with a 12V 20A PSU. I am using two of them for a homemade pinball game. At the moment they are hooked up directly to the PSU via a couple of arcade style buttons.

Problem: They have only just enough power to flip a 19mm steel ball up a 90cm playfield.

After doing a little research, I think I can use a similar circuit to this, using a Teensy board and a P30N06LE 30A 60V mosfet (also have some NDP6020P P-channel if that's better suited?)

simulate this circuit – Schematic created using CircuitLab

I have read a few articles which talk about adding a capacitor and resistor into this circuit to provide more initial "power" and then a lower "hold" current. I think the design for the "hold" part adds C1 and R3.

simulate this circuit

My Question

Where and how can I add another capacitor to give it the initial boost of power and what values should I be using for the passive components please?

Update: I have found this article by Paul Rako / Bob Pease which seems like it could help, although mine are 8A solenoids.

• it could be the power supply ..... if you can, temporarily use an automobile battery to test your circuit – jsotola Mar 5 at 22:52
• Can you verify Solenoid DCR is 1.5 Ohms? – Sunnyskyguy EE75 Mar 6 at 0:28
• reduce table angle 3.5 deg? – Sunnyskyguy EE75 Mar 6 at 1:21
• My RC has the same HPF step effect as yours but you need T=RC > flipper time. I chose 75ms.=, It may need to be longer – Sunnyskyguy EE75 Mar 6 at 17:16
• @SunnyskyguyEE75 hi, thanks for your comments. The Solenoid coil is coming up as 1.47 Ohms on my meter. I cannot easily change the table angle without having to do a lot more woodwork. – Alex Holsgrove Mar 6 at 18:52

If you need a faster and more powerful stroke, you will have to go for higher voltage. The capacitor-resistor addition is there to reduce the current after the initial stroke to reduce power dissipation and maybe improve release time and maybe allow it to remain engaged longer.

This is a nice answer about driving solenoids faster: https://electronics.stackexchange.com/a/295529/26987 It shows a schematic of an elaborate driver and explains in text why it is so. But basically the idea is the same: more power comes with higher voltage and therefore current.

• Can the Solenoid run any hotter with a higher voltage ? This may be the limiting factor to voltage and makes it unfeasible. The mass , final velocity² (mv²/2). the final power loss is already at I²R= 8²A*1.5Ω = 96 Watts – Sunnyskyguy EE75 Mar 6 at 17:07
• It will surely dissipate more power, but it possible to use a solenoid at higher power with a <100% duty cycle. – Jurkstas Mar 6 at 22:14
• The old pinball flippers used a dual coil, a heavy gauge starter coil and light guage run coil after 30ms to overcome this since the holding current only needs to be <10%. The heavy gauge cools quicker closer to the metallic former.. – Sunnyskyguy EE75 Mar 6 at 22:21

# Take 2

Generate a 24V supply and pulse 16A (384W) for 30 ms then reduce duty cycle for holding current of 0.8A or a duty cycle of 5%. (1W) Then add a temp sensor to protect the coil. A dual coil or a higher current solenoid would be more powerful with a thicker steel rod. Get a real Pin Ball Solenoid \$20, not a "wanna-be" almost, but not quite pinball solenoid. or make it

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I would reduce angle of pinball table to 3.5 deg or less to reduce energy needed to push 28g ball higher. Assuming PSU has enough current one can raise the voltage.

simulate this circuit – Schematic created using CircuitLab

Unfinished ( anyone want to finish this)

Assumptions: 19mm steel ball 28 g

"This electromagnet: the initial stroke load only about 150g, but the half end stroke is huge(about 2kg)."

Solenoid inductance: Unknown.
Estimate: L=150mH
DCR=1.5 Ω (=12V/8A)
L/R=T=100ms ( no load current risetime)

Newton's 2nd Law: $$\ F=ma \$$ , m=28[g]=0.028[kg]
1st half stroke >= F=150[g], a= 5.4 [m/s²]
2nd half stroke = F=2000[g], a= 71 [m/s²]
Response = 90cm up playfield. elevation angle = unknown,

The standard angle for newer machines is 6.5 or 7 degrees.
Older machines typically are set at 3.5 ° .

Assume Elevation is 3.5 ° . thus 90cm tangent = rise = sind(3.5°)*90cm = 5.5cm
Potential Energy $$\Ep= mgh\$$= Mass[kg] * Gravity[9.81m/s²] * height[meters]
E = 0.028*9.81 * 0.055m = 15mJ

• Thanks you kindly for your answer. The table angle is set at 6.5º and it would be quite a pain to lower this. The flippers are approx 15cm from the bottom edge giving 75cm of travel from the flipper edge to the top of the table. I think that gives E of 44mJ – Alex Holsgrove Mar 6 at 11:04
• Another idea is increase the force by reducing the stroke length by 50% but instead of direct connect to flipper, use a rubber snubber so the solenoid accelerates to 1/2 stroke where it has full force but higher velocity on flipper contact and thus accelerates the ball with full force. just a thought. Bang it with more velocity and force initially using the 50 to 100% stroke length which has more than 10x the force but using a rubber snubber – Sunnyskyguy EE75 Mar 6 at 19:14
• Ah, interesting choice - I will give that a try. I had held the flipper and can certainly feel the difference between the initial stroke VS 50% onwards. I suppose that would mean having it closer to the pivot point of the flipper to give me the same (or at least similar) range of rotation with the flipper? And this would in-turn require more force compared to hitting the flipper arm (under the table) closer to the tip end? – Alex Holsgrove Mar 7 at 17:10
• @AlexHolsgrove Yes 2x longer lever arm reduces torque /2 which is already 10x. I was thinking of hitting the flipper with speed and a vulcanized tire tread in between as a snubber /damper. but maybe that's too hard on it. But the current can be cut 90% when the current rises at end stop due to loss of BEMF.. – Sunnyskyguy EE75 Mar 7 at 18:07
• I just tried to move the solenoid around a little - the piston now makes contact with the flipper mech only after it's 40% or so out of the coil. To get the same 30º ish rotation in the flipper meant moving the point of contact (solenoid piston to flipper mech) a couple of cm closer to the point of rotation. It feels more powerful if I hold the flipper as it fires, but the ball speed isn't noticeably increased. I did adjust the PSU trim pot so that I'm getting 13.9V - again, a very marginal improvement. Interested to try 24V although I don't want to damage the solenoids. – Alex Holsgrove Mar 7 at 18:43