# How can I solve this MOSFET question

## Control of an N-channel MOSFET with a microcontroller

simulate this circuit - scheme created with CircuitLab

Dear group members,

I have a question that may have been asked before, but I couldn't find a suitable answer for my application, so I need more help in my case.

My uC with 3.3v logic is required to turn on the N-channel mosfet (for safety reasons this means that mosfet only works under fault conditions and not at a certain frequency) which is a different compared to the ground of the uC Has reference potential point (GND). In my opinion I can't overload my uC with source and sink currents from Mosfet (can be between 700mA and 1.5A depending on the Mosfet) and I think I would need the same reference potential as that of Mosfet to turn it on . Mosfet's reference potential is shown with a small triangle symbol with OpAmp, which has the same voltage references as Mosfet's.

In the circuit shown, I plan to pull up the GPIO to turn off the Mosfet (almost -5 V on the Gate of Mosfet, Rail to Rail OpAmp) and pull down the GPIO to turn the Mosfet on (almost 0 V on the Gate of Mosfet) .

My question is if this solution is elegant and if it works as expected or if I am missing details. In other cases, any other more elegant solution will be highly appreciated. Priority is a small and cheaper solution.

Note : The 3.3 V logic of the microcontroller is generated with an LDO that is supplied with the reference voltage point Mosfet (0 V) and -5 V. This means that the ground of the microcontroller (GND) corresponds to the -5 V point of Mosfet.

EDIT_1 : Resistor R2 was called Pulldown moved from mosfet source to gate side

### Andy aka

It would be much clearer if you were to draw the circuit with unique reference connections. If the MCU is actually 3.3 volts grounded, display it as such.

### Andy aka

Should the +3.3 volts from the LDO regulator actually be -1.7 volts, i.e. 3.3 volts above -5 volts?

### dim

As others have mentioned, there still seems to be something wrong. If the MCU is powered at -5V and +3.3V as shown, it means it is seeing 8.3V. I don't think your MCU can take this without blowing up.

### dim

Is the operational amplifier really operated between ground and -5 V? Because if the MCU feeds the opamp input with +3.3V, it is outside of its power supply range, which is wrong with most opamps (although not TL081 - but you probably shouldn't use TL081 here as the common mode input is range so bad and it can't even run on just 5V).

### Jeroen3

Do you know the existence of (optically isolated) gate drivers?

### Andy aka

I would redraw the circuit to make it clearer then I would see that you could just use a TTL to 5 volt logic level converter like this: -

The 74VHC1GT04 offers better performance than an op amp. You can also get rid of R1 too. It looks a lot easier now and you get a decent drive speed to the MOSFET gate.

My question is if this solution is elegant and if it works as expected or if I am missing details. In other cases, any other more elegant solution will be highly appreciated. Priority is a small and cheaper solution.

Your current op amp configuration won't work without adding another resistor from which R6 is -1.7 volts to provide a bias point halfway up the logic voltage range.

### Master of electronics

Hello @Andy, in the current configuration another resistor should go from R6 to 0 V, so that with the same inputs (if GPIO is connected to ground) the difference between the inputs to the operational amplifier is zero and the output Mosfet switches on. Your proposed solution is very simple. I find it better. I would like to know if it works with lower negative voltages like -8V too. or the base-emitter diode of PNP and NPN may not allow this?

### Andy aka

In your configuration, a resistor with the same value as R6 should go to the LDO output. When the -5 volts go down to -8 volts, the LDO regulator, provided it produces 3V3 higher than the -8 volts (i.e. -4.7 volts), should work fine. You need to start thinking about the power dissipation in the LDO regulator in case it overheats as it drops another 3 volts as it drops from -5V to -8V. If the current is a few mA this shouldn't be a problem.

### Master of electronics

Thank you @Andy. MC74VHC1GT04 unfortunately does not work for supply voltages above 7V. That said, I couldn't supply -8V (Gnd) and 0V (Vcc) to convert the -8V and -8V + 3.3V logic levels to -8V and 0V outputs. In fact, I can also see from other manufacturers that there are hardly any buffers in the voltage range of 8V

### Andy aka

@HerrderElektronik you have me there! I suspect there will be something out there that could work at 8 volts.

### Andy aka

There are MOSFET driver chips from International Rectifier (now Infineon). The IR2110 comes to mind. Try to research what they offer. I have a feeling that there will be a device that can pick up a 3.3 volt logic signal from the GPIO.

### MCG

Assuming your resistor values ​​are all correct. To properly toggle the gate, move R2 away from where it is in your circuit, as R2 won't toggle properly where it is right now. Now put it on the gate as a pulldown to -5V. If you put something between the source of the FET and GND (or -5 in your case) it just won't toggle it.

Try that and see if it works. Assuming your op amp is supplying the correct gate voltage, it should be fine.