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How do I calculate what resistor I would need? Answered

Ok, so I'm working on a project. I need to know how I can reduce 18volts down to 12volts. If someone would explain how to calculate this myself I can figure out what resistor (or series of resistors) I need.


I need to take the voltage down from 12v to 2v with a resistor.

Which do I use?

Hey, Tim,

one way is to connect a 1,000 ohm and a 5,000 ohm resistor in series across your battery, And the LED in parallel to the 1,000 ohm resistor. This is not the most efficient circuit.

I've drawn the circuit, and will try to share the pic tomorrow

A more efficient method would depend on the specific LED specs (current). If you can find that info, post it.

A voltage divider usually isn't a good choice for a current source. To get any useful current from that divider would require very low resistance values (like 2 ohm & 10 ohm), which makes the divider very inefficient. In the drawing above, the most current the 5000 ohm resistor will pass will be around 2 milliAmps (0.002 amps).

Dividers are good as voltage sources (like a reference), but that's for high impedance stuff that draws little current.

A series resistor with the load can work, but that requires a constant load, and series devices always draw the same amount of current. A voltage regulator, as others have suggested, is probably best.

It sounds like Tim is an electronics newbie, and I'm guessing that the LED he wants to drive is a little one. like an indicator on a panel. An LED draws little current, so I made a guess on resistance. A slightly larger LED might require 500 and 100 ohm in place of 5000 and 1000. The divider circuit is easy for someone new to electronics.

If anyone wants to explain voltage regulators, zener diodes, etc to Tim, feel free.

here's a basic volt reg. Anyone care to explain how to choose components for it? If I once knew, I've forgotten. Actually, I'd like to know this stuff, too. GMOON?

volt reg.jpg

T_Dan: The pic is there, if you click on it. Just the thumbnail is clipped.

Those work well. They make other output versions, including a 3.3V. They even make small ones the size of TO-92 transistors, like a standard 2N2222, for lower current requirements. An LM317 would be slightly more complicated, 'cause it's adjustable over a wide range.

I think the first capacitor would be much larger if it followed a transformer/bridge type power supply. Not so with a battery. The second cap (0.1uF) is probably for decoupling, shunting higher frequencies. Pretty standard.

The 1K resistor is there to make the LED work correctly with 5V.

TimC115: I hate to ask you for more details. It sounds like a proprietary thing you'd like to protect, and I'm guessing you're not here as rep for a major corporation. So maybe it's inappropriate to ask for more info...

Does adding / adjusting the size of a current limiting resistor change your LED effect? Does differing voltage really change the color? I'd have thought an LED hue is fixed, and only the brightness would change...unless you're using RGB LEDs...

phoey. pic got clipped. supposed to have 9v batt on left, 1k resistor and LED on R.


I have made and sold a simulated firefly for 20 years that comprises a 12 volt CPU fan on 2 spun 32 gauge wires, that pulls around a green LED that is 2.2 volts. To get the proper yellow/green hue, I push up the voltage, closer to 3, but right at 3 makes it pure yellow. I need to have it a little less. It's a novelty item, and cannot be comprised of a lot of heavier components.

If it's only an LED, then a pretty standard approach would be a single series resistor (current limiter). There are online calculators for that kinda thing.

It's helpful to remember that a series resistor together with a load form a voltage divider. Current limiter or voltage dropper (different ways of saying the same, or similar things), using a fixed resistor can work in some circumstances.

With a varying load or any complexity (like multiple LEDs with logic), something like an LM317 would be much preferable.

I am not an electrician, but an artist. I designed the simulated firefly for Disneyland's Pirates of the Caribbean. I have tinkered with a standard green LED, putting 3 volts through it for the desired hue, which becomes more yellow. 3 volts seems too yellow, though, which is why I'm thinking 2 volts might be perfect. My original used 12 volt grain of rice incandescent. I have used 2 C batteries for what I'm looking for, and oddly, some come out more yellow, and others just right. It's just a touch below 3 volts, maybe, to get the mixture of yellow/green.

If you know how much current the LED draws, you can calculate resistance for a series resistor. Is this info on the package? or can you use a multimeter to check current?

Google "voltage divider circuit" especially look at images. This is what I'm describing. Also learn the terms series and parallel.

Regulator and/or Zener diode

I don't know much about electricity, but I need to convert 12 volts to 2 volts, for an LED special effect I'm making. Which resistor would I use?

how are your math skills? it's not quite as simple as you might hope, but we can explain if u will stick with us.

I know I am 5 years late, but I wanted to clear something up that was told to you wrong. You can divide Voltage up by using resisters in a series. I would explain but chances that you read this is slim to none. Just look up KVL (Kirchhoff's voltage laws). I hate I did not get to this five years earlier and it's sad that people pretend they know what they are talking about which leads to being misinformed.

That is not possible to use a resistor to reduce 18v to 12v, it will still be 18v, even if you use a high resistance resistor... You will need to use and 12v regulator.

Apparently I have completely misunderstood the point of a resistor. I thought they were designed to RESIST a portion of the current. Why, with my LEDs am I able to turn 30v down to 28.7v? It hasn't fried my LEDs so I know it's working.

no, you understand resistor correctly...it reduces current....not voltage...you misunderstand "current" and "voltage" With your LED's you are lowering the current to an acceptable level but the voltage doesn't really change

I forgot to detail voltage and current:

Voltage is the potential energy of an electric charge. Consequently, the degree of voltage is expressed as high or low.
Current expresses the amount of electricity (electric charge: Q) flowing per unit time.

Resistance indicates how difficult it is for current to flow. Even when the voltage is the same, the lower the resistance, the larger the amount of current that flows.
Power P (W) = Voltage V (V) x Current I (A)

I feel like a dummy. Haven't paid much attention to this stuff since getting out of high school. lol. Thanks. I guess my best suggestion so far has been a 7812 IC. I guess I'll do that. Anyone know how much they run at Radio Shack? Funny, lately most of my hobby projects have been involving Home Depot/Lowes and Radio Shack. lol.

Well, before purchasing any device, try to find out just how much current it will both source or sink.
And don't feel like a dummy....many of the quite young ones here have corrected me when I forgot something I had learned over the years :-) We help each other and so the intelligence is in the whole, not in any one cell.

well now i have a new problem I wasn't thinking of...it never occured to me that a car battery pushes more amps than 9v batteries. :( So it seems the only way for me to get my project to work, I need to either figure out if it is possible to increase the amperage (which I dont believe can be done)...or just lug around a car battery. Apparently this glow plug draws 24amps. The next problem is finding a car battery. I dont wanna pay $70+ for one. :( I think this project is doomed. The whole of the electrical component is basically a motorcraft ZD 11 glow plug, wiring from pos and neg to the corresponding poles on the powersource. Includes some form of on/off along the way. Thats about it. So my problem now is amps. blah.

if you take a certain input, and increase the amperage, you will decrease the voltage, and vice versa. The amps have to be there for what you are doing, as it is like getting a gallon of gas from a dixie cup :-)

So then how do I increase the amps? If I understand you, then I could use two 9v batteries and MAYBE toss up enough amps and bring it down to 12 volts? Am I understanding that right?

No, for one thing, it is very difficult with DC. AC can be ramped up and down pretty easily....a transformer suffices. For instance, a step down transformer will lower the voltage and raise the amperage. The reverse happens in a step up. But working with DC is a whole other ball game. One can step down voltage with a regulator, or step down the current with resistors, but ramping up the current is not so easy.....and I am afraid I am at fault for misleading you there.....I was speaking generally, and had forgotten you were using DC currents.

Ah, thats ok. I think the easiest solution here is find a used car battery and rig it up someway to be portable. Unless there is some other way I could achieve 24amps. ok, did some looking around before posting this, I think I may have garnered an understanding. I need 12v and 24amps. How about a combination of series and parallels? I have to figure out how many amps a 6v battery produces, but lets just say, for the sake of argument that it's 12amps. Would I achieve the 12v/24a by doing this?... Connecting one + and one - on two batteries creating one set in series. Doing that same thing with another two batteries. Now I'd have two sets in series. Now, if I connected both + terminals to the positive on my plug and same for the - poles. Wouldn't that give me the power and amps needed? Am I understanding this yet?

Hmm, a deep cycle marine battery(ies) might suite you better, although it will be a bit pricey. Parallel connections will multiply the amperage while the voltage will increase when in series

Exactly, so if I hooked two batteries in series I would then achieve the 12 volts but it would only be, as our example, 12 amps also. So if I hooked in parallel to that another series of two 6 volts that would bring the total to 12 volts and 24 amps right? if I understand it correctly that sounds like that would work just fine. I would just have to figure out what the actual amperage of a 6v lantern battery is to see if it is practical. I think I'm finally getting the hang up this. I appreciate the help. Now my next question is...if I achieve the 24amps, since the plug draws 24 amps, how long would this power source last me? 1 continuous hour? I do appreciate the help, once I get this completely in operation I'll post pix and vids of it and show you what you helped me to create. :-P

First a few details:

Lantern batteries will not have a high amperage rating; a few examples: a Universal emergency light battery at 6 v 4.5 A, with others rated at only 1.5 A Then we have at this link, where you will find the typical Lantern battery Ah ratings and one last thing, many lantern batteries are not rechargeable, in case that makes a difference.

There are 6v batteries like this Fischer-Price 6v 9 A, 6 Ah example but for that weight and maybe even price you could probably go with a golf cart battery with a 25-75 amp rating

if I achieve the 24amps, since the plug draws 24 amps, how long would this power source last me? 1 continuous hour?

This depends entirely on the battery's A/h rating. A battery rated for 1 Ah will put out 1 A of power for one hour, or 1/2 an A for 2 hours. Discharging more quickly than rated for is not recommended...the internal resistance would not stand it, and your battery could quickly overheat and leak.

The easiest way to quickly determine the total battery amp hours required is to first determine total watt-hours required by all loads, and then divide by the nominal DC system voltage. This resulting number will indicate the amount of amp hours needed to operate all loads for a given period.

You can then get the Ah rated battery(ies) that will comfortably handle this load.


I think i've come to the conclusion that this project is not a feasible as I had hoped. I cant use a powersource that is too heavy to move quickly over uneven ground because I want to make this device universal for all the exploits of my friends and myself. I guess the best bet is find a cheap car batter somewhere. The upside is, I'm learning about powersources. :-P And according to that chart you linked to, it says 6v lantern batteries are 0.3A (300mA). Oh well, time to find a car battery someone is getting rid of.

Yea, I'll look into that. I have to weigh voltage/amperage vs price. I know one car battery will be enough. Are marine batteries 12v? If not, will the combined weight be less than a car battery?

Yeah, you probably wouldn't want to spend the $79-$249 + for a new one.And "RV batteries can be really Deep cycle too, and yes they both are normally 12v. Just something extra to look out for in the "used" dept. since brand new is pretty pricey.

Yea, especially since the whole project is just for fun...not that big a deal. lol. Thanks for the help though. I've learned alot and I appreciate it.

if you take a certain input, and increase the amperage, you will decrease the voltage, and vice versa.

For converting AC with a transformer (different winding ratios), yeah.

As a general rule of thumb, no. Unless you mean for a certain load.

Yeah, sorry.....I was thinking transformer when I wrote that....

this old man regrets his error


Well, it's true enough in other limited circumstances (like if the current source is adjustably regulated.) But that's not the usual case.

We just wouldn't want someone to increase the voltage on a typical setup and expect the current draw to decrease....the "vice versa" situation.

of course, thank you.

Measure the resistance of the circuit that needs 12V (use a multimeter). Add a resistor in series that is half that resistance.

Ok, maybe I should explain further. I'm using a motorcraft glow plug (diesel engine equivalent of a spark plug) to make what I am calling an igniting rod. The plug, brand new, is supposed to have a resistance of 0.5ohms. The plug itself says 11v on the side so now I believe that I need to drop it to 11 volts not 12. I was going off the fact that it's powered by a 12v vehicle battery then again it is supposed to be connected to some circuitry in the engine so there may be some form of limiting elsewhere in the engine. So, with a source of 18v (two 9v) and a resistance of 0.5ohms and a draw of 11v. What is the resistor or series of resistors I would need? I apprediate the help.

At 11V and 0.5 ohm, you glowplug uses 5.5A. Two 9V batteries of the usual type won't be able to supply 5.5A, or not for very long, anyway. You can probably connect them directly to your glow plug without worrying about damaging the plug, but it may not get hot enough to do anything useful. If you have more powerful batteries, you can do like kiteman said and use a 0.25ohm resistor. This is the sort of circuit with nice dependable resistances that a dropping resistor WILL work with. You'll need at least an 8W resistor, and it's going to get pretty hot too. (8W dissipated by the resistor, 16W by the glow plug...)

According to what I have found online, this particular glow plug draws 24amps. Right now I have four 9v batteries and two 6v lantern batteries. Any ideas? I really dont wanna have to lug around a car battery. Impractical.

well, let me tell you this a 9volt cannot deliver 24 amps a couple of 9 volts can definitely not deliver 24 amps (internal resistance) how long does it take to get it hot (in seconds)? Maybe the most practicle solution is a capacitor and a small charging circuit working off of a couple 9 volts?

just use a 7812 chip to do it for you. (You can get free samples)

you can use a voltage divider with 2 resistors, but I'm not sure if you even need to lower the voltage for the glo plug, it should just get hotter faster.

um, it will get hotter faster with more current, not necessarily more voltage. but yes, I seem to have forgotten that a voltage divider could be rigged up, if there is somewhere for the "divided off voltage to go". :-)