# Critique my Soft-start circuit

I've been working on a soft-start circuit to solve an inrush current problem that is causing some sparking when I hotplug my power supply into my board. I played around with different solutions and this is the one I like the best so far. It adds a delay of 250ms between Vin and Vout on the power MOSFET on my load.

My schematic:

My simulation results:

I want to get some opinions from the PROS out there.

Now, since I'm only using M1 to drive M2, there is no reason to use a power MOSFET, am I correct? I could use a small current MOSFET to drive the other one.

Edit:

Updated shcematic thanks to Kaz:

Edit #2:

I'm working on a soft-start circuit to avoid some sparking when I plug it my external power-supply to my board. I'm using 2 P-channel MOSFETS and some RC circuit to control the gate on one of them whicl will help me control the gate on the power mosfet that willl feed my board.

This is my little circuit:

R3 = My Load. I'm using max 8A hence I am representing it with a 3ohm resistor.

R2 and R4 are there to bias my MOSFET's gate with a Vgs goign from 0V to -12V.

The problem I am having is on biasing M1. Currently Vgs is going from -24V to 0V, and I need it to go from -12V to 0V, otherwise I might end up damaging the device. a Vgs of -24V is too high.

The only thing I could think of was shifting the reference point of C1 up by 12V using a voltage divider but I don't know if this is recommendable.

Like this:

My only concern now is the behaviour of the MOSFET, no matter how slow I ramp the Vgs voltage, it is still acting like an ON-OFF switch. Any ideas?

• Is R3 in the schematic? Which node is Vout? Oct 21 '12 at 1:34
• R3 is the Load. I'm drawing a max of 8A, hence the 3ohm resistor. Just to simulate the load. Oct 21 '12 at 1:43
• Also, how about arranging for the capacitor to discharge when you unplug? Otherwise it will not generate another delay if the power supply is plugged in soon after being unplugged. A reverse-biased diode across R1 would let C1 discharge through your R3 load. Add that to your simulation, see what happens.
– Kaz
Oct 21 '12 at 1:57
• Like so? i.imgur.com/Pi4dy.png I am seeing the same behaviour with or without the diode. The Cap is discharging. Oct 21 '12 at 2:12
• No! Use the diode just to short out R1 in the reverse direction. It goes across R1. The cap discharges without the diode, but it discharges through R1, therefore you have an RC time constant there. The idea is to short out R1 in the reverse direction to reduce the time.
– Kaz
Oct 21 '12 at 2:14

There are a large number of off-the-shelf solutions to this sort of problem. The advantage that these schemes have is not only do they manage the hot-plug, they limit the inrush current so that the power supply isn't overloaded.

Your simulation result shows a wicked slew rate after 250ms - that doesn't look at all like soft-start to me. The voltage rise should be a ramp, not a straight edge.

• +1 @GustavoCorona Regarding ramp, see also this TI app note ti.com/lit/an/slyt096/slyt096.pdf Oct 21 '12 at 2:30
• The application of voltage may be sudden, but maybe that is actually all that is needed. The issue to be solved is arcing. The delay is sufficient that the electrical contact has settled, which should reduce the arc.
– Kaz
Oct 21 '12 at 2:37
• I noticed the slew rate, the ramp is fine on the gate for that MOSFET, but maybe the LTSpice model I picked is responding very linearly or just the threshold voltage is not what I am expecting. Take a look at the following graphic: imgur.com/lGwwa <-- green the is the gate voltage. Blue is the Vout. Oct 21 '12 at 2:41
• I like the LM5069. Never used it before. Might be interesting to try it out. Need to find some examples. Oct 21 '12 at 2:57
• With differential sensing (internal vs. external) the gate threshold will be controlled to provide the desired output. No need to muck about with Rs and Cs. Oct 21 '12 at 13:57

You need to install a cap at gate and drain pin of the mosfet. Now you adjust this cap watching an oscilloscope for VDS slope. The slope of VDS is happening at Miller plateau (when VGS is a constant and Cgd is charging).

• Welcome to EESE! I am not sure if you noticed, but this question is 6 years old and already has an accepted answer. It may be better to look for questions which are more recent and which have not already been answered. Nov 21 '18 at 16:58