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I have server that won't power up.

I found that if I connect a 50Ω load to the +12 VDC output of the power supply everything works fine.

I only need this additional loading for about three seconds.

I need to momentarily load the +12 VDC rail for about three seconds and then remove the load. I've tried using an N-channel depletion MOSFET with an RC network to the gate of the N-Channel MOSFET but that didn't work.

(The normally closed N-channel depletion MOSFET stayed 'closed.')

I then found that the gate to source voltage needed to be about -2.0 VDC in order for the N-channel depletion MOSFET to “turn off”.

Do you know how I can create a load switch that will be “closed” for about 3 seconds and then “open?"

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  • \$\begingroup\$ Vgs is more like >=+5V not -2V for Nch then use any CMOS gate with a 3s delay filter and diode clamp. In the old days you needed 10% preload to regulate so << 3W for standby supply ? or high power is not bad \$\endgroup\$ Mar 25, 2022 at 1:37
  • \$\begingroup\$ Maybe I'm using the wrong Depletion MOSFET? IXTP3N50D2 \$\endgroup\$ Mar 25, 2022 at 2:03
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    \$\begingroup\$ Is a server power supply needing a 50 ohm load for a few seconds in order to start up a normal condition? Sounds to me like you need to repair or replace the supply. If it's not starting normally there's a reason for that. If parts in it, such as capacitors, are failing and you just add a hack, something else may fail and maybe not in such a benign way. \$\endgroup\$
    – GodJihyo
    Mar 25, 2022 at 13:52

3 Answers 3

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555 to the rescue !!!

One 555 in an 8-pin DIP, a resistor, a capacitor and a diode are are you need to add to a simple 'vanilla' enhancement N-MOSFET to get the behavior you're looking for:

schematic

simulate this circuit – Schematic created using CircuitLab

You don't strictly need the diode for this circuit to operate, but what is does is ensure that the capacitor discharges quickly when the power is turned off, which in turn ensures that you get the same predictable timing when you turn the power back on again.
Without the diode you'd need to wait at least 10-15 seconds for C1 to fully discharge through R1, and note that this would also be the case for a circuit constructed from discrete components instead of the all-on-one 555.

I powered the simulation with a square-wave so you can see the turn-on and turn-off behavior.

Could this be done with discrete components? Sure it could. But why go to the trouble when everything you need is inside the 555.

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    \$\begingroup\$ oh no, not a 555 please , never used one in 40 yrs when a simpler solution existed. (lol) \$\endgroup\$ Mar 25, 2022 at 7:19
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    \$\begingroup\$ @TonyStewartEE75 - "simpler" but uses more parts and then makes you have to worry about the Vgth of your FET and how long it's going to take to switch on & off due to your cap's long slow RC charge curve ... ? No thanks, I'd much rather 'dead-bug' a 555 + 3 parts (2 if you think you can do without the diode) and get a nice sharp switch signal to my FET. \$\endgroup\$
    – brhans
    Mar 26, 2022 at 13:09
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One transistor, a resistor, and a capacitor are all you need.

This is a pretty straightforward circuit. When power is applied, the voltage across C1 is 0 V, the gate is pulled up to 12 V, and Q1 is full enhanced (minimum Rds). As the capacitor charges (downward), the voltage on the gate decreases until the FET turns off. This happens between 1 and 2 time constants.

The part values are not at all critical. The FET is something I pulled from my design library. The max current through RLOAD and Q1 is 0.25 A. Any n-channel, enhancement mode FET rated for at least 25 V and 500 mA will work, but a power device will perform better than a small-signal device, with a shorter transition time from full on to full off. This reduces power dissipation in the FET.

The power dissipation in RLOAD is over 3 W, and is constant for most of the 3 s period. A 5 W resistor is a must, and 10 W is recommended.

enter image description here

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  • \$\begingroup\$ Thank You SO MUCH AnalogKid!!! I tried this today and it worked PERFECTLY. I’ve been working on this for nearly a week and nothing was working. I owe ya! A video link of me testing this is at: youtu.be/J_MldVQ056k \$\endgroup\$ Mar 25, 2022 at 22:44
  • \$\begingroup\$ Excellent! What FET did you use? Also, I forgot to say this in the response - The circuit is not limited to a FET as the switch device. With a different R and C, a bipolar transistor or Darlington will work just as well. At only 250 mA, even a little 2N4401 would be fine. \$\endgroup\$
    – AnalogKid
    Mar 25, 2022 at 22:53
  • \$\begingroup\$ Hello AnalogKid, I’ll get back to you on the FET that I used. I removed it from a 500W power supply. I had to use three 1MΩ resistors in parallel for the R1, 330KΩ resistor. (I didn’t have a 330KΩ resistor) Also, I used a 15W wire-wound 50Ω load resistor. One thing that I really like about this is when it turns of it really turns off, nearly zero current draw. You’re awesome! Thank You!!! \$\endgroup\$ Mar 25, 2022 at 23:38
  • \$\begingroup\$ Glad it worked out. \$\endgroup\$
    – AnalogKid
    Mar 26, 2022 at 3:14
  • \$\begingroup\$ I used a: CSD18536KCS N-CHANNEL MOSFET, 60V, 200A, 2.2mOHM I've added a couple LED's that show the when the load is present and when the load isn't loaded. \$\endgroup\$ Mar 30, 2022 at 22:41
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schematic

simulate this circuit – Schematic created using CircuitLab

Some extra considerations here as button switches are rated < 2A and sometimes much less where as discharging > 1uF caps have low ESR and can reach >xx Amp surges in > 1 us

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