Dropping voltage from power supply 13.8V -> 9V

Hi guys, I have a 13.8 volt DC power supply used to power a nerf gun. I'd love to get the voltage down to about 9 volts, It's making the gun cycle way to fast and unpredictable. I've read that using a resistor is a bad way to go about it (and I'm not even sure how to do this) but what would the correct way to do it be? Thanks!

Correct way is with a voltage regularor. Using a switching regulator would be the smart way. An example: How much current do you need? 300mA http://www.pololu.com/catalog/product/2102 600mA http://www.pololu.com/catalog/product/2104

Not quite sure of how to measure how much I need. Can it be done with a very basic multimeter?

lets try to back track it first B4 we get to the meter... might save a lot of time. What kind of battery does the original "Gun" use for a power source? If it's anything less that 6 C cells you can do it with a 1A switcher. There are 2 A versions that will give 3 A peak load current and are the same price as well. Then you just don't really care. As long as the device is spec'd for 9V operation and you set the switcher for 9V operation you should have no problems with excessive current drain. As a matter of fact it will draw less current from the 12V source than it supplies to the 9V device because the operative word with a switcher is power and not current .1A @ 9V is .9W and at 12 V the actual power is .075A + the Efficiency loss of approx 80% = .09A X 12V = 1.08W instead of .1A through a linear regulator X 12V = 1.2W. With higher input voltages the efficiency gets a whole lot better. In this case .01A isn't much of a saving but it is less than a linear aand you can buy a module from most anyplace. My choice would be Electrodragon and here is an URL for a $1.70 modulle that should work very well for you. http://www.electrodragon.com/?product=better-than-lm2596-dc-dc-step-down-adjustable-power-supply-module Give it a look...


5-66 rectifiers in series




Runaway pancake, JoeN, those methods just dissipate the extra voltage as heat, and do nothing to conserve battery life. The switching regulator on the other hand will convert 85-90% of the higher voltage into usable current, so more of the battery capacity is able to be taken advantage of.

Everyone in this thread has said "battery" except the OP who said "power supply". If it is a power supply and not a battery then I don't see why using a eight or ten ten dollar part when a 40 cent part will do will actually save you money in the long run. If it is battery powered then that looks like a really nice part. The Electrodragon part looks nice too and cheap. I did a purchase from them and they did deliver some nice items.

Everyone in this thread has said "battery" except the OP who said "power supply".

Not quite "everyone", but right, anyway, JoeN. The man said power supply, so that's why I figured diodes. They could be placed, artfully, aand maybe painted, to look like part of the device, for that matter. I guess. The current draw of the nerf gun is not known; maybe the linear reg is OK, too.

Thanks for your help guys, I have a few diodes here and was testing them as to their effects on voltage drop. They're IN4005. I noticed if I run one on the supply it drops it about .4volt but If I run 2 in a row it drops it like 6volts. It says on the packaging their forward voltage drop is about 1.6v so I'm not sure how this translates to my results and isn't this essentially doing the same thing a resistor would?

The forward voltage across a diode is “constant” (italicised for effect) vs the current through it, not so a resistor.
1N4xxx are only 1A diodes, don’t know if that is enough.
Maybe one of your diodes is bad if it’s going from 0.5V with one and “6V” with two. Diodes, unlike resistors, are polarised devices - you have to connect them the right way round.


It’s possible that a resistor could accomplish this as well, but determining that value would hinge on knowing the current draw of the nerf mechanism (It’s Ammeter Time.)