# MOSFET switching with 5V but latching ON with 12V

I am having an issue with a control signal.

I have a CEP50P03 P channel trench MOSFET that is supposed to switch a 12v power supply.

I am having an issue I can not figure out. When I power the board with 5V everything works as it should.

However when I add the GND connection to the 12v side to use the 7805DT voltage regulator the Mosfet will come on and latch on and will not turn off.

I then tried removing the voltage regulator and power the circuit with only 5v and the fet does not switch at all. I have not been able to figure out this weird behavior.

I attached the shematic below. This circuit is used in an automotive rev limiter module where tach signal is a 0-12v square wave used to read rpm that is then converted to a 0-5v square wave the ATMEGA328P reads and if the frequency is above X ammount the Atmega sends a signal to turn the MOSFET off for X time and the mosfet should turn back on.

However like mentioned above when I have 12V coming in the MOSFET latches on and stays on.

• Why are there diodes (D1, D3 and D4) used with the 7805? Moreover, could you please encircle the CEP50P03? – Huisman Apr 28 '19 at 19:55
• Ah... please correct your schematic symbol for the CEP50P03. You drew a PNP transistor :-) – Huisman Apr 28 '19 at 19:58
• Sorry that's what the Eagle library I used had for it. – Hunter Inman Apr 28 '19 at 20:05
• @Huisman I placed the diodes there to see if I was some how having reverse flow through the regulator. No change with them there or not. – Hunter Inman Apr 28 '19 at 20:08
• Sorry but your schematic diagram is really hard to decipher. Doing good schematics would help a lot avoiding errors. However R11 shall go to 12V instead of 5V. Moreover the SIGNAL input conditioning looks wrong too. – carloc Apr 28 '19 at 20:29

1. The CEP50P03 is a PMOS which requires a negative gate source voltage.
2. If the threshold voltage is (typically) -2V, you should not drive it with -2V, but with a bigger negative voltage.

A fast and easy way to get a rough indication what $$\V_{GS}\$$ to use: check the conditions that are given for $$\R_{DS(on)}\$$. In this case: $$\V_{GS} = -4.5V\$$ and $$\V_{GS} = -10V\$$.

First of all, you're using the incorrect schematic symbol for a MOSFET. They are not the same as BJTs, and we just have to guess how you have it connected. MOSFETs do not have an emitter, collector, or base, so I am going to assume you have it connected such that the source is the emitter, the drain the collector, and the gate the base. Please use the correct symbol, especially if you're going to be asking others for help. They are not interchangeable with BJTs.

The MOSFET is not latching, it is simply doing what you've asked it to do. P-channel MOSFETs are switched by a negative voltage between the gate and source. Or, more simply, the gate voltage must be at a lower potential than the source to turn it on.

If you have 12V attached to the source of that MOSFET, then you must drive the gate with that same potential, 12V, to turn it off. This is equal to 0V from gate to source.

Likewise, assuming a turn-on voltage of 10V (the standard for MOSFETs), this is actually -10V gate-to-source, so reusing the 12V example, you would need to drive the gate with 2V, to fully turn on the MOSFET. This is -10V gate-to-source.

The reason your P-channel MOSFET 'latches' (not really latching, just behaving as expected) when you connect 12V, is because no matter what, you have -7V gate-to-source. When there is 5V on the gate, the MOSFET is on because you're keeping -7V on the gate (12V-5V=7V), which is more than enough to turn on any MOSFET. The threshold is usually -4.5V, and lower for logic level MOSFETs. If your ATMega's output goes high, you pull the gate of the MOSFET close to ground potential, driving the gate with -12V.

So your circuit is only capable of driving the MOSFET's gate in two modes: on, and really on.

It works at 5V because now, your gate drive voltage is able to match the source voltage, and you switch it between 0V gate-to-source, and -5V gate-to-source.

You will have to redesign that part of your circuit for it to work correctly. A common trick is to simply use a pull-up resistor between the gate and the source. This will keep the gate pulled up so it will default to the 'off' position. Now, you can simply use an npn transistor to pull the gate to ground to turn the MOSFET on. And who doesn't like making their circuit simpler?

A word of warning: what I described above will only work as long as the maximum gate to source voltage rating is observed. This is typically 20V, so 12V is no problem. But if you had, say, 30V on the source, then you will likely damage or destroy the MOSFET by pulling its gate to ground. This is would cause 30V across the gate and source.

Solution to the problem was too connect R11 to 12v instead of 5v. Too obtain a lower Vgs