I received this 3.3V or 5V power supply. It's neat because it fits on my breadboard and can provide two separate voltages (chosen using jumpers) on it's two outputs. I originally thought that this was a Step-Down Converter, however, it's been pointed out to be a Linear Regulator.


Here, I have the left-hand side providing 5V, while the right-hand side provides 3.3V. I've used my multimeter to verify it working from a 9V battery.

I am curious what the jumpers in the middle of the board will do. There are two 5V->GND, and two 3.3V->GND. I've made attempts to find information about this specific device, but I haven't found any identifying information, nor have I found any information about it online.

I'm afraid to use this for longer than looking at the voltage on either side because I am unsure the purpose of these jumpers, or if/when they should be bridged for safe/normal operation.

Any help identifying and understanding the purpose of these jumpers in the middle of this device would be greatly appreciated.


The 4x2 header in the middle is there for you to access the voltage rails and Ground with female jumper wire or test clips or whatever. It's for your convenience.

And no, that board does not function as a step up regulator. It's not even a step down switching regulator. It is just two linear regulators.

  • \$\begingroup\$ Are you saying that these are outputs, just like the +,- that are inserted into the breadboard? I can confirm that reading them with the multimeter shows the correct voltage. Glad I didn't put a jumper on them, in that case! \$\endgroup\$ – earthmeLon Dec 5 '16 at 23:51
  • \$\begingroup\$ I will look into the difference between Step-Down converters and Linear Regulators. Thanks for your patience as I learn about these things. \$\endgroup\$ – earthmeLon Dec 5 '16 at 23:52

You have already got your answer but if incase you wish to find about the linear voltage regulator, I guess you have LM1117 for 3.3V that is variable with respect to the register value you use and other one isn't visible to me , most probably it would 7805 or equivalent for 5V.

  • \$\begingroup\$ Thanks for your input. This schematic from another comment seems pretty spot on, but I'm new to all of this. \$\endgroup\$ – earthmeLon Dec 6 '16 at 17:52

The board you show is a simple 2 voltage linear regulator for a breadboard. It will NOT provide boost under any circumstances.

The jumpers left and right allow you to configure the individual regulators. Think of it as enable for +5 and +3.3.

The block of pins (left of center) are output pins so you can put wires and take the supplies to other locations.
When used as shown on a breadboard, the outer (bottom left and right, underneath) pins supply power to your breadboard power rails.

  • \$\begingroup\$ They are not enable pins! \$\endgroup\$ – Passerby Dec 5 '16 at 22:47
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    \$\begingroup\$ They enable either 3.3 or 5 V outputs...don't be picky! The schematic will look much like this: addicore.com/v/vspfiles/downloadables/Product%20Downloadables/… \$\endgroup\$ – Jack Creasey Dec 5 '16 at 22:55
  • \$\begingroup\$ The voltage selection schematic will look something like this: buildcircuit.com/wp-content/uploads/2014/03/schematic.jpg \$\endgroup\$ – Jack Creasey Dec 5 '16 at 22:57
  • \$\begingroup\$ The first is the exact schematic. The second is wrong. Notice that the first routes one or the other or none to the breadboard rail. Putting it in the middle will disconnect the output to the breadboard but does not disable the voltage regulator, or affect any of the headers with those voltages. Not an enable pin!!! \$\endgroup\$ – Passerby Dec 5 '16 at 23:58
  • \$\begingroup\$ OK...not an enable pin...you win. Providing that is the correct schematic. \$\endgroup\$ – Jack Creasey Dec 6 '16 at 0:14

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