# DC Voltage Supply

new here! And I know just enough to be dangerous!

I'm building a toy for my grandchildren that consists of LED lights, switches and knobs, small motors and sound circuits, etc. I used 3V as a spec when ordering components, so that the unit could run on two 1.5V batteries.

I want to add a power supply to minimize battery use. I have lots of power supplies laying around, 5V to 24V "wall warts." I could use one of those, and step it down somehow, or buy a 3V supply. But here's the rub:

As various switches and components are turned on and off (while using battery power supply), I notice LED lights brighten and dim. Something to do with voltage (or amps?) varying as the total load across the batteries changes, I presume.

What type of power supply should I use so that the voltage stays constant, regardless of how many LEDs or motors or sound boards are in use at any given time? One that would always supply 3V, whether the boys cranked on everything at once, or just a few components at a time...

Or is there a 5V to 3V step down circuit commercially available that would properly regulate the voltage (as I have a very nice 5V power supply that I could use, one that came with a now-discarded computer accessory).

Keep in mind (if possible) that I'm not looking to spend too much, as this "toy" is already considerably over budget! $10-$30 would be great.

Thanks! Mark G

• What batteries are you using and what is maximum current taken? Does it have to work on batteries? If so then how much room do you have and how long do you expect it to work for? Jun 15 '16 at 17:37
• A cheap and dirty solution might be to add 3 diodes just in series to the whole circuit. Of course they should be rated for the current consumed. Jun 15 '16 at 17:42
• You need a regulated power supply. This keeps the voltage constant regardless of current drawn (within rated current.)
– user114104
Jun 15 '16 at 17:49
• If the batteries are not holding 1.5 volts, then they are the problem. If you are using AA, try C. If you are using C, try D. A power supply will not have that problem if it can supply enough current (which the current batteries can not).
– Mark
Jun 15 '16 at 21:47

You can get a ready-made buck regulator with an output of 3 V from places like Ebay for a few dollars. You will need to get one with sufficient output current capacity for your circuit, and make sure you select a "wall wart" with a DC output and also of sufficient current capacity. If you are stepping down from, say, 9 V to 3 V then you need the wall wart to be able to provide 1/3 -and-a-bit of the current needed at 3 V.

Buck regulators are more efficient than linear regulators, so you probably won't need to add a bulky heatsink to it, but do make sure it is not possible for little fingers to touch it, and give it some space to dissipate heat.

You can get a socket which matches the plug on the wall wart. Centre-positive connections are the most common - the socket itself is not polarised.

A simple single-pole double-throw switch will enable the user to switch between using the mains adapter and the batteries (technically "cells" rather than "batteries"). If the description of the switch says "make before break" then you do not want it. The chances are that you won't find a SPDT slide switch for any less money than a DPDT slide switch (less than a dollar); the latter is also perfectly usable.

You may not need a fuse in there, but for something that children will be playing with it might be a good idea. You can get an in-line fuse-holder for a 20 mm glass fuse if you don't want to make up a circuit board for it. The rating of the fuse will have to be determined by you. A slow-blow fuse (e.g. T2A) should be fine (someone will correct me if I'm wrong).

simulate this circuit – Schematic created using CircuitLab

• Appreciate all the replies! Awesome!! To be clear, I was not looking for a solution for the batteries. Andrew has a complete grasp of what I'm after. Thanks! In fact, the circuit you so graciously provided was already in place when I asked the question (switching between battery and power supply). Jun 23 '16 at 15:04
• Oops, hit "return." Jun 23 '16 at 15:04
• And again... I understand the concept of "break before make." Good thought. I'll have to figure out how to test for that in the switch I already have in place. Fuse is a great idea, I didn't think of that. I'll have to test for amp draw, too, I suppose. I think I have a meter that can do that. I appreciate Smith's answer, but I don't want to start with a chip (if I understood correctly). I don't want to build anything, just order it off of Amazon! Jun 23 '16 at 15:15
• My power supply spec's at 3A at 5V. If I remember my electrical math correctly: "Buck Amp'ing" down to 3V should give me 3A plus the "margin" Smith recommends, yes? I'm powering dozens of LEDs, plus some small sound circuits and a couple of very small motors. I can't imagine all that together is more than 3 amps! Thanks again to all that responded... Jun 23 '16 at 15:15
• Guys, what do you think? This looks to be just what I need. I can use my 5V-3A supply, and it looks to have built-in volt AND amp meter, plus a "fuse" of sorts. I could even put it behind some plexiglass so the boys could view how switching on and off the various components affect the current draw: amazon.com/DROK-Regulator-Stabilizer-Converter-Transformer/dp/… Jun 23 '16 at 15:31

I suggest a BA30, which is a linear DC-DC regulator. You can see the selection on digikey.com if you search for BA30, and you can find a chip you like and read its data sheet. Here is a quick design checklist to use it:

1. What is your max current draw? Make sure it is within the range available by the BA30 with a healthy margin as well (50% margin is great, try not to go under 25% margin). Note that different packaging offers different amperage ratings. A thru-hole TO-220 package will provide more current than a small surface mount package.

2. What is your input voltage? Make sure the BA30 can accept it.

3. Quick thermal check - take your max current draw and multiply it by your power supply voltage minus 3V. This equals the number of Watts the BA30 will be dissipating itself. Watts = Current Draw * (Vsupply - 3V). Since you may be running at 5V, then you are probably in good shape. But test to make sure it doesn't get so hot that it is dangerous, because linear regulators will run warm.

4. Put input and output capacitors by it within the BA30's datasheet.

That's the basic way to get 3V. For fixed output linear regulators, 3V is offered but there isn't a lot of selection. There is more selection out there for 3.3V and 5V. The Digikey search tool is your friend here.

Bonus if you can get the BA30 manufacturer to sample you some for free. You can always ask. Find out the specific chip you want, go to the manufacturer's web page and see if you can get a free sample.