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I have several batteries in series and the ground point is in the center, such that I can achieve a negative and positive voltage. The OP amp uses both negative and positive voltage. Let's assume I am using two 9v batteries.

If I were to place a SPST on ground, is this sufficient? My confusion/worry is if I have the switch open, then it essentially is creating a 18V source:

enter image description here

With the switch open, the chip will have no source to the battery center tap. This seems safe but the chip would still be connected to both ends of the in-series 18V source (since it required the positive and negative 9v.)

Could the circuit somehow be closed inside of my op amp and pump double the voltage through my chip? Should I be using a DPST like this instead:

enter image description here

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  • \$\begingroup\$ What does the rest of your circuit look like? Perhaps switching the output of the opamp on and off would be enough? Could you post a schematic? \$\endgroup\$ Nov 11, 2014 at 0:01
  • \$\begingroup\$ By the way, your op-amp is running off 18 volts when powered, and the ground is used as a reference point. \$\endgroup\$
    – gbarry
    Dec 10, 2014 at 6:06

2 Answers 2

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Your first drawing would leave your circuit powered by the full battery supply (+/-18 V?).

The second drawing, with the double pole switch will work, but it is more common to put switches in the "hot" wires, rather than on the ground side (or center, in this case) of the circuit.

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You do have to use two switching elements, but not necessarily a DPDT switch. A SPST with several MOSFETs can cut it too:

schematic

simulate this circuit – Schematic created using CircuitLab

This circuit works by controlling two switching elements, two power MOSFETs here, with a single SPST switch. If you used to need switched rated at several amps, this circuit calls for power MOSFETs rated for several amps but a switch rated for a few milliamps.

When the switch is open the two MOSFETs' gates are pulled to their respective sources, preventing both MOSFET from turning on. When the switch is closed the two gate resistors R1 and R2 forms a voltage divider, putting a voltage around 0V at both MOSFETs' gates. M1 will see -9V gate-to-source and M2 see +9V gate-to-source, turning both MOSFETs firmly on.

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