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I am trying to drive a vibratory pump from a 220 V boost converter (the common 150 W DC/AC 12 V to 220 V boost converter found on ebay) and a MOSFET switched by an Arduino/optoisolator at 50 Hz. The pump can be considered a solenoid and a video description is available here.

I arbitrarily chose a 2N60 MOSFET because it was the first one I found that was cheap and had a 600 V breakdown voltage. I assume there is a better and cheaper MOSFET I could have chosen (feel free to suggest a better choice).

If I can trust some of the blogs I have read, the pump has a measured inductance of 854 mH and a series resistance of 165 Ω. The power draw is rated as 53 watts. I am told that a simple flyback diode will reduce pump pressure and that a diode plus a 10 W 220 Ω resistor will allow full pressure although the resistor will get a little hot. I am thinking that a zener will improve the power dissipation but I am not sure how to choose good values for the resistor and zener.

pump flyback circuit

I am trying to keep the circuit cheap, physically small, and not outrageously hot for a closed box.

Is there a better way for me to do this? I assume that a 600 V breakdown voltage still needs some flyback protection.

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  • \$\begingroup\$ You are aware your converter has an AC output, which needs to be rectified and filtered properly, and which will give you about 300VDC. \$\endgroup\$
    – Janka
    Commented Mar 23, 2018 at 18:52
  • \$\begingroup\$ Yes it is rectified. The output is 20Khz and I am assuming that a 40khz dc rectified power rail can get away without a filter. It's an automobile so the input voltage may be as high as 15v. I don't yet know what the exact output voltage will be. I will adjust the results to allow for a voltage higher than 220. \$\endgroup\$
    – Ed McGuirk
    Commented Mar 23, 2018 at 19:02
  • \$\begingroup\$ What's the diodes you use for the rectifier? \$\endgroup\$
    – Janka
    Commented Mar 23, 2018 at 19:14
  • \$\begingroup\$ I don't now if it's good choice but I have purchased a GBU808. Could I use a bridge of FR207 diodes? \$\endgroup\$
    – Ed McGuirk
    Commented Mar 23, 2018 at 19:18
  • \$\begingroup\$ The GBU808 is a 50/60Hz rectifier. The datasheet says Single phase, 60Hz, resistive or inductive load. You cannot use it for kHz applications, it would simply burn out because the diodes need too much commutation time. The FR207 is good up to 100kHz (datasheet says test conditions are 1MHz, but it's a good idea to stay wide away from that.) \$\endgroup\$
    – Janka
    Commented Mar 23, 2018 at 19:27

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220V through 165ohms is about 1.3A. The rise time for 0.865H and 165 ohms is about 7msec, so at 50Hz on/off the current will pretty-much reach the steady state. So every time you turn it on, you're putting 1/2 i^2 L = 0.7J into it, and when you turn it off you need to pull 0.7J out: That's about 35W that your flyback will be dissipating.

The point about reduced pump pressure with just a diode may be because the pump wants to dump the energy fast (into a zener or resistor) so it can return to its previous position in time for the next stroke. Just a flyback diode lets the current flow mostly unimpeded, so the solenoid stays pulled for longer.

If the pump really does want a nice, smooth sinusoidal drive, you could try to find a 120uF high-voltage AC capacitor and see if you can create a 50Hz parallel resonant circuit. Two warnings: Check that value, I am working on my phone and it might be wrong. And second: Make sure you get a high voltage AC capacitor, i.e. for a motor start.

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    \$\begingroup\$ I came up with similar flyback wattage but I thought I must be wrong. That implies that more than half of the power dissipation of the 53W pump comes from flyback power. However the pump is a resonant solenoid at 50hz so the actual working inductance may not be the same as the low voltage measurements in a dry pump. \$\endgroup\$
    – Ed McGuirk
    Commented Mar 24, 2018 at 0:28
  • \$\begingroup\$ I suspect that "53W" comes from driving it from sinusoidal AC, i.e. mains power, which doesn't work the same way. There instead of dissipating the reactive power, it's just put back into the mains. \$\endgroup\$
    – Bob Jacobsen
    Commented Mar 24, 2018 at 0:30

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