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I've built a comparator circuit that I want to use to drive a PMOS transistor. I only need negative output, but I'm on a single 12V/24V battery supply. I know I can create a virtual ground and split the supply, but since I only need negative output or 0 output is there a way to accomplish this without splitting the supply?

What would happen if I were to ground the V+ terminal and apply the battery positive to the V- terminal? In simulation this seems to work, but I haven't seen any information about it so I'm not sure what to do. I'm using an OPA454 TI operational amplifier (I know it's overkill for a PMOS, but the comparator I ordered (LM211QD) isn't in yet.

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  • \$\begingroup\$ +Vsupply goes to +24V and -Vsupply goes to 0V of the battery \$\endgroup\$
    – Andy aka
    Commented Sep 17, 2013 at 19:55
  • \$\begingroup\$ Just to clarify: you have a battery supply that provides 12V and 24V and you want to produce a "negative output". Negative with respect to what? With respect to a virtual ground that is 12V above the lowest voltage in the circuit, or negative with respect to the negative terminals of the batteries (the lowest voltage in the circuit)? \$\endgroup\$
    – Joe Hass
    Commented Sep 17, 2013 at 20:51
  • \$\begingroup\$ Sorry for the lack of clarification. I only need the op amp to output negative voltage or zero voltage, depending on the inputs. I have made two circuits, 1 for a 12V supply, and the other for a 24V supply. I want the voltage to be negative with respect to ground, whether that's virtual or actual. I have created a circuit with a virtual ground, but since my design only needs one side of the output I'm basically wasting 6v/12v of range. Andy I understand that the V+ is normally connected to battery positive, but I want the circuit to work in the negative voltage range and that won't work. \$\endgroup\$
    – user29193
    Commented Sep 18, 2013 at 12:54
  • \$\begingroup\$ You're not making much sense. Show the circuit of the "PMOS transistor" thing so we can see what it really needs to be fed. What is the PMOS transistor supposed to switch or regulate? Based on what input? \$\endgroup\$ Commented Sep 19, 2013 at 19:28
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    \$\begingroup\$ PMOS transistors do not require negative voltage. (Enhancement-mode) PMOS-s require a gate voltage which is more negative than the drain terminal in order to turn on. Can you add a schematic to the question? \$\endgroup\$
    – Kaz
    Commented Sep 19, 2013 at 23:27

2 Answers 2

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If it is only necessary to drive the MOSFET for a brief moment, you can use a concept like this:

schematic

simulate this circuit – Schematic created using CircuitLab

While the output of the comparator is high, the capacitor gets charged. Then when the output of the comparator switches LOW, the capacitor is still charged and the right plate will be pushed below 0V. The capacitor cannot discharge through the diode so it will stay charged for a while with only leakage current flowing.

Usability depends on exact specification of your components, but you can play with capacity, an extra pull up resistor and the cathode voltage on the diode.

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I'm not aware of any op-amp or comparator that can drive its output beyond its supply rails. So you will have to designate as "ground" a voltage higher than the negative supply. Swapping the supply connections of the op-amp (if that's what you are suggesting) is likely to destroy it.

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  • \$\begingroup\$ I'm not trying tooutput beyond my supply rails, the supply rail voltage is overkill for the region that the PMOS needs to operate in, but the output is positive w/ respect to GND for the OP amp and I need the op amp to drive a PMOS so it needs to output negative w/ respect to GND. \$\endgroup\$
    – user29193
    Commented Sep 23, 2013 at 19:52
  • \$\begingroup\$ Sorry, I was going with the information provided. Perhaps you would explain how you are using this PMOS device. PMOS transistors are typically connected source to the high side of the supply; you won't need to drive the gate negative with respect to ground. \$\endgroup\$
    – user28910
    Commented Sep 23, 2013 at 21:03

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