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I want to drive a LowSide MOSFET switch. No switching speed needed here.

With a 3.3V µC input on the driver, the MOSFET shall sink current.

I have a npn/pnp stage here. When µC output is HIGH (3.3V), Q2 sinks current, Q1 is activated and the gate of M2 charges up. -> MOSFET ON. When I switch µC output to off, Q2 and Q1 turn both off, and the gate discharges through R6. -> MOSFET OFF. The advantage of this circuit is that it does not consume power, when MOSFET is off.

Is this driver circuit OK? How to correctly dimension R12 and do i need additional resistors somewhere?

enter image description here

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    \$\begingroup\$ Why not use a MOSFET with low enough Vgs to drive directly from the uC? \$\endgroup\$
    – Finbarr
    Commented Aug 31, 2023 at 10:19

1 Answer 1

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Some recommended changes:

schematic

simulate this circuit – Schematic created using CircuitLab

R15 should have a much higher value. Its function is to ensure Q2 remains off in the absence of an explicit input potential, which could occur prior to the microcontroller initialising its output. 100kΩ will do the job, and will draw negligible current from the microcontroller.

Move R15 to the left of R14. With R15 on the right, and with the values you chose, those two resistances behave as a potential divider, severely attenuating the digital source signal.

R12 provides a path for Q2's leakage current, instead of from Q1's base, which could otherwise be sufficient to switch on Q1 when it should be off. It also serves as a current path to evacuate Q2's stored base charge, helping it to switch off more quickly. 3kΩ is fine, but I don't think it needs to be so low.

R4 must be much larger. A value of 10Ω would cause Q2 to try to draw nearly \$\frac{12V}{10\Omega}=1.2A\$ from Q1's base. To switch on Q2, you need only hundreds of microamps of base current.

I don't know what C5 is supposed to do. It only serves to slow down switching, so I removed it.

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  • \$\begingroup\$ Thanks Simon, okay, but R15/R14 divider could ensure switching stability over temperature, because of the Vbe temperature dependancy. Thanks for R4 correnction. Didnt think about that. But R12 also ensures that Q1 can be switched right? Otherwise there is no Vbe drop there. \$\endgroup\$
    – cyrus010
    Commented Sep 1, 2023 at 10:28
  • \$\begingroup\$ @cyrus010 R12 only ensures that Q1 can be switched off, by permitting there to be no base current for Q1, so \$V_{BE}=0\$. When Q1 is on, its \$V_{BE}=0.7V\$, due to base current sunk by R4 and Q2, and R12 plays no role in that state. \$V_{BE}\$ variation with temperature is negligible in this application, and I don't even know what "switching stability" means. \$\endgroup\$ Commented Sep 1, 2023 at 11:43
  • \$\begingroup\$ Okay thank you. So if I remove R12, the following could happen: if the leakage current of Q2 is high enough, it could turn Q1 on, right? When I add R12, the leakage current takes the path over R12 and not over Q1 base and R4. With "switching stability" i mean the following: The µC or whatever output has a defined maximum LOW output voltage. Lets say it is 0,5V with higher temperature. This already can cause some current through Q2, but here it is maybe negligible. \$\endgroup\$
    – cyrus010
    Commented Sep 7, 2023 at 9:54
  • \$\begingroup\$ @cyrus010 Correct about R12. For 0.5V low output, this is a problem with the signal source, not this circuit. If you're really worried about it, replace Q2 with a MOSFET with \$V_{GS(TH)} \approx 1.5V\$, or use emitter degeneration with Q1, to raise the switching threshold and reduce gain. \$\endgroup\$ Commented Sep 7, 2023 at 10:19

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