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schematic

simulate this circuit – Schematic created using CircuitLab

Dear Group Members,

I have a question which might have been asked before, but I couldn't find the suitable answer for my application, due to which I need more help in my case.

my uC with 3,3V logic is required to turn ON the N-channel Mosfet (as a safety measure, that means Mosfet will operate only in error conditions and not with a certain frequency) which has different reference potential point as compared to uC's Ground (GND). In my Opinion, I cant overload my uC with source and sink currents of Mosfet (might be around 700mA - 1.5A depending on Mosfet), and I think I would need the same reference potential as that of Mosfet to turn it on. The reference potential of Mosfet is shown with small triangle symbol with OpAmp which has same voltage references as that of Mosfet.

In the shown circuit, I have a plan to pull up the GPIO to turn OFF the Mosfet (nearly -5V at Gate of Mosfet, Rail to Rail OpAmp) and pull down the GPIO to turn ON the Mosfet (nearly 0V at Gate of Mosfet).

My question is if this solution is elegant and if it will work as expected or maybe I am missing any detail. In other case, any other more elegant solution will be highly appreciated. Priority is to have a small size and less costly solution.

Note: the microcontroller 3,3V logic is generated with an LDO supplied with reference voltage point of Mosfet (0V) and -5V. That means microcontroller ground (GND) is same as -5V point of Mosfet.

EDIT_1: Resistor R2 moved from Mosfet Source to Gate Side as Pull Down

EDIT_2: 3.3V LDO added

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    \$\begingroup\$ It would be a lot clearer if you drew the circuit with unambiguous reference conenctions. For instance, if the MCU 3.3 volts is in fact ground then show it as such. \$\endgroup\$
    – Andy aka
    Jan 17, 2018 at 10:03
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    \$\begingroup\$ Is the +3.3 volts from the LDO regulator actually meant to be -1.7 volts i.e. 3.3 volts above -5 volts? \$\endgroup\$
    – Andy aka
    Jan 17, 2018 at 11:57
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    \$\begingroup\$ As others have mentioned, it seems there is still something wrong. If the MCU is fed with -5V and +3.3V as shown, it means it sees 8.3V. I don't think your MCU would stand that without blowing up. \$\endgroup\$
    – dim
    Jan 17, 2018 at 12:01
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    \$\begingroup\$ Also, is the opamp really powered between ground and -5V ? because if the MCU feeds the opamp input with +3.3V, it will be outside its voltage supply range, which is wrong with most opamps (not TL081, though - but you probably shouldn't use TL081 here, since the common-mode input range is so bad and it can't even be powered with only 5V). \$\endgroup\$
    – dim
    Jan 17, 2018 at 12:05
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    \$\begingroup\$ Are you aware of the existence of (optical isolated) gate drivers? \$\endgroup\$
    – Jeroen3
    Jan 17, 2018 at 13:26

2 Answers 2

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

enter image description here

You will get better performance from the 74VHC1GT04 than an op-amp. You can also get rid of R1 also. It looks a lot simpler now and you get a decent drive speed to the MOSFET gate.

My question is if this solution is elegant and if it will work as expected or maybe I am missing any detail. In other case, any other more elegant solution will be highly appreciated. Priority is to have a small size and less costly solution.

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

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  • \$\begingroup\$ Hi @Andy, in current configuration another resistor from R6 should go to 0V i think, so that at equal inputs (when GPIO is connected to ground) the difference of inputs to operational amplifier is zero, and so the output to turn on Mosfet. Your suggested solution is very simple. i find it better. I would like to know would it also work with lower negative voltages like -8V. or there the Base- Emitter diode of PNP and NPN might not allow it ? \$\endgroup\$
    – HerrderElektronik
    Jan 22, 2018 at 16:18
  • \$\begingroup\$ In your configuration a resistor of equal value to R6 should go the the LDO output. If the -5 volts shifts down to -8 volts, providing the LDO regulator generates 3V3 higher than the -8 volts (i.e. -4.7 volts) it should work fine. You have to begin to think about the power dissipated in the LDO regulator in case it overheats because it will drop 3 more volts across it when lowering -5V to -8V. If current is a few mA it shouldn't be a problem. \$\endgroup\$
    – Andy aka
    Jan 22, 2018 at 16:25
  • \$\begingroup\$ Thank you @Andy. MC74VHC1GT04 doesn't work unfortunately for supply voltages higher than 7V. That means i couldn't supply it with -8V (Gnd) and 0V (Vcc) to convert logic levels -8V and -8V+3.3V to -8V and 0V outputs. Infact I can see from other manufacturers too, there are hardly buffers in voltage range of 8V \$\endgroup\$
    – HerrderElektronik
    Jan 23, 2018 at 11:55
  • \$\begingroup\$ @HerrderElektronik you got me there! I suspect that there will be something out there that could work at 8 volts. \$\endgroup\$
    – Andy aka
    Jan 23, 2018 at 12:10
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    \$\begingroup\$ There are MOSFET driver chips made by International Rectifier (now Infineon). The IR2110 springs to mind. Try investigating what they offer. I have a feeling that there will be a device capable of taking in a 3.3 volt logic signal from the GPIO. \$\endgroup\$
    – Andy aka
    Jan 23, 2018 at 13:26
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Assuming that your values of resistor are all correct, in order to toggle the Gate properly, remove R2 from where it is in your circuit, because with R2 where it is at the moment, it will not toggle correctly. Now place it as a pulldown to -5V on the Gate. If you put anything between the Source of the FET and GND (or in your case, -5), then it simply will not toggle.

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

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