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I am trying to drive a pinball flipper using the circuit below, but after switching the flipper on a few times the MOSFET failed (short to ground) and the voltage level shifter appears to have been damaged (3 to 3.3V on the output pin rather than the expected 5v).

Any help in determining the cause of failure and suggestions for safeguards to prevent flippers being stuck on in the event of a failure would be appreciated.

Driver Circuit

My values are:

  • VDD1 = 50 Volts
  • L1 = flipper solenoid (~2.2 Ohm during initial kick, then 78 Ohm hold resistance)
  • D1 = 1N4004-T
  • Q1 is a IRL540NPBF 100V 36A N-Channel MOSFET
  • U1.1 is a buffer from a 74AHCT125 level shifter, where VCC is 5V.
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  • \$\begingroup\$ The basic circuit is correct. The gate threshold for this part is 1V, so 5V is plenty. What is the value of R1? Is it possible that R1 is too low and pulling the gate down? \$\endgroup\$
    – slightlynybbled
    Commented Mar 15, 2016 at 2:31
  • \$\begingroup\$ I'm not sure if that's indeed the root cause of your problem, but a fast Schottky diode should be used instead of 1N4004. \$\endgroup\$
    – Nick Alexeev
    Commented Mar 15, 2016 at 2:34
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    \$\begingroup\$ Possibly this? electronics.stackexchange.com/questions/221433/… Also, Nick is spot on about the 1N4004. \$\endgroup\$
    – user16324
    Commented Mar 15, 2016 at 12:56
  • \$\begingroup\$ slightlynybbled The R1 resistance is 10K. NickAlexeev I actually have a second diode on the flipper itself that is part of the original hardware. Don't know the part number but D1 is just a secondary diode but I will look into getting some Schottky diodes. BrianDrummond I don't think decoupling is the issue as I am using the original pinball power supply form the machine the flippers came from. \$\endgroup\$
    – ks0ze
    Commented Mar 16, 2016 at 1:55
  • \$\begingroup\$ @slightlynybbled Based on this answer (electronics.stackexchange.com/questions/62415/…) the gate threshold is only the minimum voltage for the device to start conducting. Based on the figure 3 of the datasheet (irf.com/product-info/datasheets/data/irl540npbf.pdf) I think I would be fine up to ~20A with a 3.3V drive if I'm reading that correctly. I think you may be onto something though because at power on the FPGA pin would be high Z until programmed, which could cause unexpected behavior on the buffer? \$\endgroup\$
    – ks0ze
    Commented Mar 16, 2016 at 16:07

2 Answers 2

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You need to slow down the turn off of the mosfet or control the rate of voltage change over s-d at turn off to give your protection diode time to turn on. Using the 74AHCT125 which is a push pull driver lowers the gate voltage very fast this turns off the transistor fast leading to a large fast voltage spike from the solenoid or even from stray lead inductance.

Remove r1 and add a series resistor between the output of the 74AHCT125 and the mosfet gate start around 4.7k this will slow down the turn off and on of the mosfet.

Normally mosfets are set up to switch very quickly to reduce the amount of time the device spends in the switching region in order to reduce power dissipation. This is important for some applications where it is switching many thousands of times a second. For your application it does not matter as you will be switching a few times a second at most.

Another method of controlling the voltage spike is to wire a snubber across the s-d of your mosfet. This consists of something like a 100ohm resistor in series with a 330nf 100V capacitor, again you are not switching this on and off rapidly so there will not be a lot of power dissipation.

If the coil is a dual winding flipper solenoid with a separate pull in and hold winding and a limit switch to turn off the pull in winding you need to ensure there are diodes across both windings. If there is not a diode across the pull in winding you will get arcing across the limit switch causing all kinds of nasty spikes on your drain.

Those interested in the suitability of 1N400X diodes as clamps should look at http://www.cliftonlaboratories.com/diode_turn-on_time.htm. I have been using them in this application for 40 years.

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  • \$\begingroup\$ Will the resistor on the gate be enough if I drive the mosfet for 40ms for the initial swing then drop down to pulsing at 20us 50% duty cycle for holding? \$\endgroup\$
    – ks0ze
    Commented Jul 9, 2016 at 2:43
  • \$\begingroup\$ In that case you would need a schottky diode across the coil to limit power loss in the diode. Is your flipper coil dual winding? your very different on resistances suggest this. \$\endgroup\$
    – RoyC
    Commented Jul 9, 2016 at 8:38
  • \$\begingroup\$ You say you are using the original pinball power supply. I remember, from a brief period in the gaming machine industry, that big solenoids were driven from full wave rectified AC unsmoothed in any way, saves the cost of capacitors and lets you use triacs as switches. Is that the case here?. \$\endgroup\$
    – RoyC
    Commented Jul 9, 2016 at 9:07
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    \$\begingroup\$ The supply is filtered with a 0.1uF and 100uF cap. Yes, it is a dual winding coil and I probably don't need to pulse the coils because the End of Stroke switch should enable the second coil and prevent burnout but I would like to use the same circuit to drive other playfield coils such as an electromagnet that will require a PWM with 2% duty cycle so I figured I'd try it on the flippers as well. \$\endgroup\$
    – ks0ze
    Commented Jul 9, 2016 at 17:03
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Two possibilities come to mind:

  1. The FET was damaged by the inductive kickback of the solenoid when switched off.

  2. The FET was fried due to excessive dissipation when on.

#1 will happen if D1 was not properly connected or itself damaged. 1Nxxxx is a poor choice of diode for this purpose. It is very slow and is meant for 50 or 60 Hz rectification. Something faster would be better, and its peak current spec needs to match the highest possible solenoid current.

#2 would be if the FET isn't being turned on all the way. It looks like you're trying to drive it with 5 V on the gate. You didn't link to the datasheet, so I didn't look it up. However, a FET that can do 100 V and 36 A is very unlikely to be specified for only 5 V gate drive. It probably needs 10-15 V to get the nice and low Rdson mentioned in the datasheet. Surely the datsheet tells you at what gate voltage the Rdson promise is made.

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  • \$\begingroup\$ If not both 1 and 2 helps, add an RC snubber across the MOSFET to limit the dV/dt and peak current. Heck, you can have an RC across the coil to if it rings against D-S capacitance. \$\endgroup\$
    – winny
    Commented Jul 5, 2016 at 20:59
  • \$\begingroup\$ #1 would be more likely. Pulsed current handling of the 1n4004 is 25-30A depending on who makes it too close for comfort for my liking. \$\endgroup\$
    – RoyC
    Commented Jul 8, 2016 at 23:27
  • \$\begingroup\$ #2 is just not correct this device with just 5V on the gate is pretty close to its min rdson at the specified current and well within its safe operating region. The datasheet is linked in one of the comments. These modern fets are remarkable bits of kit. \$\endgroup\$
    – RoyC
    Commented Jul 8, 2016 at 23:29

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