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I'm working on a project where I want to switch a load (utility LED light, +- 20W) using a motion sensor. The PIR can only operate at a maximum of 12 volts, while the load can be operated at a voltage between 12-36 volts. I want to power this circuit using a lead acid battery, with the light on the 'unregulated' output and the PIR regulated (with a buck converter) to something like 5V. The PIR also outputs a 3,3 volt signal, so I made a circuit to boost this to the battery voltage. I was wondering if a circuit like this would work or if the '0V' line would be different between the battery negative and the buck converter negative output.

The signal boost circuit, the 3,3V supply simulates the PIR output Signal boost circuit The whole circuit, with gate driver and pulldown being the circuit in Whole circuit simplified

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  • \$\begingroup\$ Yes. Same ground. Consider ditching the BJT inverter-pull-up-scheme if 3.3 V is enough to drive your FET. \$\endgroup\$ – winny Sep 19 '17 at 13:46
  • \$\begingroup\$ It strongly depends on the exact converter. For most of them this is true, but no one can tell you without knowing its schematic. Some DC/DC converters which can output constant current have current sense resistor in the negative path, but it would not be a problem if you power the PIR sensor only. \$\endgroup\$ – Todor Simeonov Sep 19 '17 at 13:47
  • \$\begingroup\$ Aha, so a buck converter without current control will probably be fine. Thank you! Unfortunately the mosfet needs about 10V, it's an IRF540 mosfet. \$\endgroup\$ – Rick van Schijndel Sep 19 '17 at 13:57
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According to the comments the ground is probably the same if some constraints are satisfied:

  • Non-isolated buck converter

  • Buck converter without current regulation, current regulated converters have a shunt at the ground line changing the voltage

  • Buck-boost converters have a negative (afaik) 'ground' voltage and thus are not usable

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