Delayed switch (closed for three seconds then open)

Question

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?"

Answer 1

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:

^{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.

Answer 2

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.

Answer 3

^{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