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Protection on Step up and Step Down Converter Module Bucks

  1. Aug 10, 2016 #1
    In a circuit that I'm building, I have a power source of 5 volts DC that powers various components. A few of them are LED lights and light bulbs.

    -Background Info-

    In circuit 1, I have a step up buck
    "2A DC-DC boost step-up Conversion module MicroUSB 2-24V 3V to 5v-28V 12v 9v 24v"

    that converts 5 volts to 6 volts to power two light bulbs wired in parellel to each other that are rated at 6.3 Volts
    "Eiko 46 T3-1/4 Miniature Screw Base Halogen Bulb, 6.3V/0.25 Amp"

    and under that same buck of 6 volts, I have another circuit (circuit 2) wired in parrallel to circuit 1, that has 3 LED lights wired in parallel to each other that runs on 6 Volts together (originally they ran on x2 3 Volt 2032 batteries that were wired in series to each other which totalled 6 Volts)

    Circuit 3 on the other hand has a step down buck
    "2x 2A DC DC Step Down Converter Module 4.75-6V to 0.8-4.5V Adjustable Buck Vol"

    that converts 5 volts to 4.5 volts which powers three led lights wired in parallel to each other (they originally ran on x3 1.5 v AAA batteries wired in series to each other which totals 4.5 volts).

    -The main question-

    So my question is, because I'm using bucks to convert my voltage for these led and light bulbs circuit, do I need to protect those devices with rectifying diodes? From what I understand about bucks, they are similar to relays, in that they produce back EMFs, and when a relay switches off the magnetic field collapses and can produce a back emf of 50V or more which can kill devices.

    In these circuits do I need to protect these leds and light bulbs with rectifying diodes? If so, what ratings would you recommend and how should I wire them to the leds/light bulbs?

    Some suggest wiring them in between the bucks and leds,
    whereas others seem to suggest to wire the diodes across the loads/leds, like this


    Here's a picture of my circuit if it helps (on the right side, I apologize for my bad drawings!)

    Last edited by a moderator: May 8, 2017
  2. jcsd
  3. Aug 10, 2016 #2


    User Avatar
    Science Advisor

    The output of a buck converter has a capacitor that isolates the load from the internal inductor. The capacitor is sized so that ripple will be small when supplied through the inductor. The output spike could not be bigger than that ripple.

    You should not need additional protection.
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