0

Given:

  • I have some home alarm peripherals (motion detectors, glass break sensors) which operate at 12V and draw a nominal current of 20mA when in alarm state.
  • I have a Particle Photon (3.3V logic level) which will accept up to 5.6V on its VIN pin.
  • I have a 12V @ 2A switching power supply.

Can someone describe to me an efficient circuit which would allow me to power all of the peripherals @ 12V and the Arduino (Particle Photon) at 3.3ishV?

I have on hand an LM7805 linear regulator, which I believe will work, but will be inefficient. I also have on hand a DC-DC step down converter (4.5V-28V IN, 0.8V-20V OUT @ 2A) but I am a novice and don't know how to use it for the scenario described here.

  • 2
    "I am a novice and don't know how to use it for the scenario described here." - Put 12V on the IN, connect the OUT to a volt meter, twiddle the knob till it reads 5V. Disconnect the volt meter. Connect it to the VIN of your photon. – Majenko Jul 12 '16 at 20:11
  • You could use the arduino's built in regulator, no? – tuskiomi Jul 12 '16 at 20:34
  • 1
    How inefficient will the 7805 be, in the big picture? Depends on the current draw. How much current will be drawn? – Dampmaskin Jul 12 '16 at 21:23
  • 1
    @Dampmaskin For the 3.3V Arduino (Photon) current draw will not exceed 500mA. For the 4x 12V peripherals, if all were in an "alarm" state simultaneously, current draw will not exceed 110mA. – loopforever Jul 12 '16 at 21:51
  • 1
    Ok, dropping 8.7V at 500mA wastes 4.35W so OK you're right, that is inefficient. A buck converter is probably the best approach then. – Dampmaskin Jul 13 '16 at 10:04
1

@Majenko remarked:

Put 12V on the IN, connect the OUT to a volt meter, twiddle the knob till it reads 5V. Disconnect the volt meter. Connect it to the VIN of your photon.

@loopforever added:

For the 3.3V Arduino (Photon) current draw will not exceed 500mA. For the 4x 12V peripherals, if all were in an "alarm" state simultaneously, current draw will not exceed 110mA.

@Dampmaskin added:

Dropping 8.7V at 500mA wastes 4.35W so OK you're right, that is inefficient. A buck converter is probably the best approach then

@SDsolar added:

If you use a screw-terminal barrier block you can have both 12V and your lower voltages available at the same time

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.