I am building a mini pinball machine, which is about 10x22cm. To drive the flippers I plan to use this 5v mini push/pull solenoid: https://www.sparkfun.com/products/11015 , along with momentary SPST buttons and a 5V 2A wall power adapter.

Those are the only electronic components in the machine. I am not using an Arduino or any other controller. In this scenario, is it advisable to use a diode (or any other components) to protect the power supply?

If diodes or other components are advisable, I will be grateful if you could explain at a very basic level why they are needed and where they should be inserted into the circuit. The schematic below is pretty much the current extent of my electronics knowledge. I am interested in expanding my knowledge but find many of the technical terms used on this site overwhelming.

Here is the circuit that I have so far:

Simple pinball flipper circuit

  • \$\begingroup\$ "Necessary" is open to interpretation. If you do nothing more than what you show, it's likely that your switches will encounter some arcing each time they open up. Whether or not you consider that a problem is a different matter. You can use a diode to mitigate that problem, but just a diode by itself means it takes more time for the solenoids to dissipate their energy. You can use zeners and diodes to improve that. Or design an RC snubber. Or ... well, ... details matter. And perhaps the arcing is fine with you. No idea. I'm less worried about the 5 V supply, though. \$\endgroup\$
    – jonk
    Feb 24, 2019 at 19:23
  • \$\begingroup\$ Well, this mini pinball machine is just a playable prototype for testing out how all of my laser-cut and 3D-printed parts fit together, geometry of the gameplay, etc. before I scale it up to full size and install real pinball components into it. I'm probably not that worried about the arcing for the prototype, except that I wouldn't mind learning how to add the diode to this circuit just for the sake of completeness. Where would I add it? In my drawing above, would I connect it on either side of the flipper button? \$\endgroup\$ Feb 24, 2019 at 23:03
  • \$\begingroup\$ As you can probably already see from those writing here, a lot depends on the solenoid details and switch design. Can you provide some details? \$\endgroup\$
    – jonk
    Feb 24, 2019 at 23:28
  • \$\begingroup\$ Well, there is a spec sheet for the solenoid at cdn.sparkfun.com/datasheets/Robotics/… . As for the switches, the only documentation I can find is that they're momentary SPST and rated for 250VAC @ 1A. Oh, and the power supply is 5VDC at 2A. \$\endgroup\$ Feb 24, 2019 at 23:45
  • \$\begingroup\$ Sheesh. That datasheet takes some care to read. I don't see a specification for the release time. Just a duty cycle period. I can see that it is \$4.5\:\Omega\$. If I knew the release time I could estimate the inductance (just multiply the release time by the coil resistance and it is often close.) Here, I'd get over 10 Henries if I assumed 3 seconds release. But that's just part of their duty cycle spec and I'm sure it releases faster than that. But it may be a Henry or so. (Which suggests that Mr. EE75 probably picked out a good value for Spicing it. \$\endgroup\$
    – jonk
    Feb 25, 2019 at 7:35

1 Answer 1


YES. Arc temperatures exceed 5000'K when the switch opens for a few microseconds, and slowly erodes contact silver content and rises in resistance. Better contacts have a palladium-silver alloy which is more robust at higher temps. Cheap contacts have neither silver nor palladium.

This is why DC relay contacts with inductive loads are de-rated to ~25% of rated resistive loads.

Plan B

Upon reflection, I think a resistor snubber will work best. The diode method reduces EMI too much and slows down the flipper action to T=L/Rs

So measure solenoid series resistance (DCR) with a DMM and choose a resistor across the inductor about 50x this value. That will only increase current a bit but sped up solenoid response and reduce arc voltage to V=IR for solenoid current and snubber R value.

It depends on your DC solenoid.
If coil draws 0.1A at 5V, it is 5/0.1=50 Ohms ,
then use a 50x 50=2500 Ohms/=50% tolerance 1/2 W

plan A

The diode current or power rating must be similar to the coil so it conducts when the switch opens bypassing or shunting the arc voltage to 1 diode drop to the opposite rail. It is normally from the switch in reverse polarity to the opposite supply rail ( + or Return)

Shown below as High and Low side switch. FYI only.

Otherwise, without a diode, high EMI impulse noise is created where the contacts arc. You can hear this between channels on any AM/SW radio. ( as long as the flipper isn't too loud) hah.

Solution 1N400x x= number dont care about reverse voltage rating number) directly across switch in reverse polarity.


simulate this circuit – Schematic created using CircuitLab

The Power supply current does not spike since the diode acts as a bypass switch to the same DC current when the flipper is activated, so the current simply declines smoothly. But without a diode the supply must absorb any transient arc voltage drop ( which it usually does) and this also creates more radiated EMI. So use any 1A diode.

Plan B on left

enter image description here

  • 2
    \$\begingroup\$ That location for the diode doesn’t work unless it’s a Zener. It’s opposite the current direction through the solenoid. A capacitor or snubber there would be much more advisable. \$\endgroup\$ Feb 24, 2019 at 19:33
  • \$\begingroup\$ Brain fart. thanks Yes a Zener or Diode will work. \$\endgroup\$ Feb 24, 2019 at 19:40
  • \$\begingroup\$ A large cap also needs large R for good dampening and this slows the flipper response time drastically, and no longer snaps back to position and slower acceleration. What RC values would you suggest? I think they are only useful for AC switches tinyurl.com/y3os44v3 @EdgarBrown \$\endgroup\$ Feb 24, 2019 at 20:05
  • \$\begingroup\$ I see 1uF and 500 Ohms might be perfect too, . for 1H, 10 Ohms \$\endgroup\$ Feb 24, 2019 at 20:13
  • 1
    \$\begingroup\$ All that is needed is to reduce and slow down the voltage buildup to avoid arcing. You don’t need too much capacitance to achieve that, the exact value would depend on the inductance but I would start at around 1nF 100ohm. That would be much faster than any mechanical response (and probably slow enough to stay out of the AM band). \$\endgroup\$ Feb 24, 2019 at 20:19

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