# How can I get this PMOS to turn off?

I have a 2V regulator circuit which is used to drive a microcontroller (EFM32G222F128). On the output of this regulator I have a delay switch type circuit (see image below) which is used in order to allow the reservoir capacitor to charge enough in order to deal with the initial start up current of the $\mu$C.

The circuit as it is works fine, however, should, for any reason, the $\mu$C draw too much current (the regulator is only rated to allow a low current out) the output of the regulator will drop below 2V, thus causing the $\mu$C to lose power.

When this happens the voltage appears to latch at ~1.7V and power needs to be removed for a few seconds before the circuit will power up again properly.

I have put this down to the NX2301P not switching off properly once the voltage has dropped below 2V. This is sort of part of a bigger issue with this delay switch...

Say I remove power to the circuit, the regulator output voltage drops away slowly (a couple of seconds) and once it actually passes the 2V mark it drops more quickly down to about 0.5V where it only trickles away and I never really see it drop to zero without getting really bored waiting. If I try to turn power on during the 'quick discharge' region, the output won't go back up to 2V, it will latch at the 1.7V mentioned previously. I have to wait for the output to level out to around 0.5V before it will be able to go back up to the 2V steady output.

Is there a way I could, preferably using discreet components, improve the delay switch so that if the voltage output of the regulator drops below 2V, the PMOS will turn off and allow the circuit to sort of reset and get up to 2V before turning back on?

• R42 looks a bit too high to me... – Ignacio Vazquez-Abrams Apr 8 '15 at 8:55
• Throughout your circuit you are using resistors with very high values. Leakage currents are not negligible in such circuits, neither are base currents or time constants of RC networks where the resistor is one of your high-value voltage dividers or gate drivers (R42) and the capacitor is a BJT input capacitance or MOSFET gate capacitance (C_GS,T8). While T6 might be able to turn off your PMOS (T8) somewhat fast, T8 will always be slow when being turned on through R42, remaining somewhere between being fully off or fully on, supplying less than the desired 2 V that T24 is supposed to offer. – zebonaut Apr 8 '15 at 8:58

In order to fully shut down the output, you need to bring T8's gate voltage close to its source voltage, and you must make sure this really happens even more with your particular choice of T8 (the NX2301P is rated for low "turn on" gate voltages). T6 needs to be driven into "good" saturation, and looking at the resistors and capacitor at T6's base, this will be hard to achieve.

Note that you must supply current to T6's base (and bring the base-to-emitter voltage well above 0.7 V) in order to turn it on so T8 will turn off, and all this would need to happen while the supply for T6 is around 0.5 V. Try to think of an arrangement where T8 is turned on in an active manner and is turned off using resistors only: Resistor between T8.G and T8.S, active component (transistor) between T8.G and ground.

I don't want to sound discouraging, but I would keep the idea of rearranging R42 as an experiment in thoughts and stop here. Try using a simple and proven concept for building a robust regulator with one pass transistor (instead of a "weak" supply from R24, requiring a soft-start, post-regulator using T8). This different design would, however, be beyond the scope of this particular question/thread.

It looks as if you are forgetting to discharge C29. So the first time you power up, when C29 is fully discharged, it will saturate T6 until Q29 is charged, when T6 will turn off, allowing T8 to turn on.

But if you remove power and re-apply, C29 remains charged, therefore T6 is held off, and the MOSFET remains on.

Provide a discharge path for C29 - perhaps a Schottky diode to your switched 2V rail - then whenever that rail falls below C29's voltage it will discharge C29.

Failing that, you may have to actively drain C29 with an additional transistor as part of the power supply sequencing logic.

(I am not worried by the high resistances per se, as long as you have allowed for worst case transistor leakage currents when calculating their values)