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I've modelled an undervoltage lockout circuit that should trip when the voltage is around 7 V. Attached is an image of the circuit.

At 12 V, the positive end of the comparator is less than the negative end, so there is no voltage at the gate, meaning Vgs is negative. At 7 V, the positive end is greater, so there's 7 V at the gate and Vgs is around zero. However, after performing a DC sweep, I noticed that there is still voltage at the drain below 7 V.

enter image description here

(Green is voltage at the drain, while blue is voltage at the gate.)

I've tried switching out the transistors, but to no effect, so I'd appreciate any help on this matter.

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    \$\begingroup\$ There is no load in your schematic. Where are you measuring the voltage? What happens if you add a load? \$\endgroup\$
    – Ian Bland
    Commented Feb 6, 2023 at 1:57
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    \$\begingroup\$ It also might work better to add a resistor from gate to source, and pull the gate low with a diode. \$\endgroup\$
    – PStechPaul
    Commented Feb 6, 2023 at 2:29
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    \$\begingroup\$ @IanBland It seems to behave as expected after adding a load. Thanks for this, I'm quite new to circuit simulation so your comment helped me a lot \$\endgroup\$
    – dechair3
    Commented Feb 6, 2023 at 2:54
  • \$\begingroup\$ @dechair3 it's an easy thing to overlook if you're new to simulation. Glad I could help! \$\endgroup\$
    – Ian Bland
    Commented Feb 6, 2023 at 16:23

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Simulating your circuit with a .step card with a variable load on the output shows that you need to have a load resistor. Even 1 Mohm is enough.

General rule: SPICE simulations should never have floating nodes as you have in your schematic.

Red traces: R7 = 1e12 ohms
Green traces: R7 = 100k.

enter image description here

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