I have a circuit where, when first plugged in, I need to delay a few hundred milliseconds to avoid drawing power during contact bounce/not all pins connected.

I haven't done much with designing MOSFET/transistor circuits, but feel this should be a pretty basic circuit to implement.

The below is what I have come up with, but I feel is over-complicated, and takes a while to to discharge between delays. The key specs that I need are: rated to 36 V operation (24 V nominal), 500 mA peak.

Is there a simpler circuit I am overlooking, or any simple improvement I can make to reduce the reset time of the circuit?

Live version of the circuit here


  • 3
    \$\begingroup\$ You must ask a question. What’s downstream of this circuit? If you have an MCU or similar, can’t you solve it via power on reset? \$\endgroup\$
    – winny
    Jan 12, 2022 at 8:00
  • 2
    \$\begingroup\$ What's your question? Please edit it to include specific questions. Please note that asking for product recommendations are off-topic here. \$\endgroup\$
    – NStorm
    Jan 12, 2022 at 8:06
  • \$\begingroup\$ @winny The downstream circuit has a micro-controller, along with other analog components, however by the time the 3.3v rail comes up it will already be too late. The power to this board is provided over the common mode voltage of magnetically isolated differential pairs (think poor mans POE), I have found that when plugging in one side of a pair may make contact first, and current through the unbalanced pair has a high chance of causing a spike on the downstream side of the magnetics. Testing with a bodged in power up delay fixes this problem (confirmed by a couple hundred test insertions) \$\endgroup\$
    – Hugoagogo
    Jan 12, 2022 at 22:27
  • \$\begingroup\$ @NStorm question updated to be a little more clear. \$\endgroup\$
    – Hugoagogo
    Jan 12, 2022 at 22:27
  • \$\begingroup\$ Sounds like an XY problem. How will delaying the 3.3 V rail fix the problem? What about a restart? \$\endgroup\$
    – winny
    Jan 13, 2022 at 7:47

1 Answer 1


The discharge between delays can be solved with an extra diode and resistor to ground. Another diode in positive feedback can help to reduce resistor value.

enter image description here A more simple circuit could be similar to this one, but not so fast when switching on, as there is no positive feedbak.

enter image description here

  • \$\begingroup\$ Hi Bravale, I did have a go with adding a diode in that location but it requires a series resistor otherwise the system will never switch on, it also needs to be pretty high value too, this is the same thing that limits making the 2M resistor smaller. \$\endgroup\$
    – Hugoagogo
    Jan 12, 2022 at 22:21
  • \$\begingroup\$ I see, maybe is better to discharge at the input, but additional resistor is need to have a clear way to ground. That means more components. See edit. \$\endgroup\$
    – Bravale
    Jan 13, 2022 at 7:32
  • \$\begingroup\$ I did simulate that as well and it gives a fast ramp down until the upper mosfet turns off. Thanks for the idea though. \$\endgroup\$
    – Hugoagogo
    Jan 13, 2022 at 11:56

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