# How to convert 16vDC to 2vDC?



## Janglur (May 7, 2010)

So i'm doing a little home project to make an LED running board for my hallway.  I have a transformer (120vAC to 16vDC 10w) for $1, but all the LEDs I have are 1.9 to 2.4v and <25mA

How would I convert 16vDC to 2vDC?  I understand there are probably several ways, so listing each option and it's advantages and disadvantages is preferred.


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## net-cat (May 8, 2010)

Ohm's Law...
2V / 0.025A = 80Î©

Voltage Divider...
2V = 80Î© / (80Î© + R) * 16V
R = 560Î©


```
16V        560Î©        LED       GND
(+)-------/\/\/\/------->|-------(-)
```


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## Janglur (May 8, 2010)

So if I understand you correctly, if I used an 80 ohm resistor, it would dump the voltage to 2V and allow me to run the LED?

If I wanted to run, say, 10 of these LEDs in a series, I would need an 8 ohm resistor?

And if I wanted 10 in a parralel circuit, i'd split the line into 10 seperate 80 ohm resistors?


Or am I completely wrong?


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## Ricky (May 8, 2010)

Janglur said:


> So if I understand you correctly, if I used an 80 ohm resistor, it would dump the voltage to 2V and allow me to run the LED?
> 
> If I wanted to run, say, 10 of these LEDs in a series, I would need an 8 ohm resistor?
> 
> ...



I'm pretty sure that says 560...

You may be better off just getting a different power supply, though.


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## SnowFox (May 8, 2010)

If you wanted to run 10 in series I think you'd need a higher voltage transformer. If you run them in parallel you could probably do it with a single 56Î© resistor, otherwise use 10 separate 560Î© ones.


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## Janglur (May 8, 2010)

What about 50ohm?  I can't find any 56ohm.  The LEDs operate on 1.9v to 2.4v so it doesn't need to be precise.

Alternatively, couldn't I use a voltage divider?  (Basically split the 16v input to two lines, each with a 100ohm resistor, then recombine?)  I was reading that this should work?

I guess i'm confused by the math in Ohm's Law.

-----
Ohm's Law...
2V / 0.025A = 80Î©
-----
Unless i'm misreading, this says that for a single LED, I need an 80ohm resistor to drop 16v to 2v?

------
Voltage Divider...
2V = 80Î© / (80Î© + R) * 16V
R = 560Î©
-------
And this I just don't understand at all.


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## SnowFox (May 8, 2010)

Janglur said:


> Alternatively, couldn't I use a voltage divider?  (Basically split the 16v input to two lines, each with a 100ohm resistor, then recombine?)  I was reading that this should work?
> 
> I guess i'm confused by the math in Ohm's Law.
> 
> ...



2V / 0.025A = 80Î© is the resistance of the LED.

If there's 2V across the LED that leaves 14V across the resistor, and we know there's 0.025A going through it so R = V/I = 14/0.025 = 560Î©


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## Slyck (May 8, 2010)

Janglur said:


> So i'm doing a little home project to make an LED running board for my hallway.  I have a transformer (120vAC to 16vDC 10w) for $1, but all the LEDs I have are 1.9 to 2.4v and <25mA
> 
> How would I convert 16vDC to 2vDC?  I understand there are probably several ways, so listing each option and it's advantages and disadvantages is preferred.



Try a 1k resistor (brown-black-red). You won't need to drop the voltage any other way for just 16v. That rated voltage of the LED is just what it needs to start conducting. In one of my projects, I have 2 100 ohm resistors on either side of an LED getting 12 volts, but that's with a PWM signal and being used as a Class-A buffer (class a amp with the transistor's c-e junction replaced by the LED) being used to drive a MOSFET. Don't think that counts.

If you need any electronics help from an old hand, just ask!


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## Aden (May 8, 2010)

You don't need to be anal about your resistor values. Just slap on a 1k instead of trying to dig around for exactly 560. If you think it should be brighter, get a 500 and a 100.


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## Janglur (May 8, 2010)

Thanks, Slyck.  I DO need help!

Primarily I want to learn how to do the math myself, and understand ohm's law, so that I can do other projects.

Right now i'm trying to build an LED light strip.

I'm starting simple, as this is a learning project.

So, if I understand you and SnowFox correctly, the LED itself will produce 80ohms of resistance.  So I would include that in my math for the resistor before the LED?

I'm really quite remedial for all this.  Is this correct?


16vDC --> 560ohm Resistor --> LED --> GND?

What would I need for:
16vDC -->  ???  -->  100 LEDs --> GND


Preferably show me the math too, so I can know for future projects.


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## Janglur (May 8, 2010)

Aden said:


> You don't need to be anal about your resistor values. Just slap on a 1k instead of trying to dig around for exactly 560. If you think it should be brighter, get a 500 and a 100.



So if I understand, putting a 100ohm -->  100ohm  -->  LED is the same result as 200ohm -->  LED?


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## Lobar (May 8, 2010)

Janglur said:


> So if I understand, putting a 100ohm -->  100ohm  -->  LED is the same result as 200ohm -->  LED?



As long as everything's in series, yes.


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## Janglur (May 8, 2010)

Thanks Lobar.  Now, I intend to have around 100 LEDs on this lighting strip.  I imagine this drastically changes things.


Basically, to simplify the entire problem, I need to understand the math for ohm's law and voltage dividing.  How do you get 560 from
2V = 80Î© / (80Î© + R) * 16V
R = 560Î©

I do the math and get 0.0078125?

Once I understand the math, I can figure everything else out from there.


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## SnowFox (May 8, 2010)

Janglur said:


> Thanks Lobar.  Now, I intend to have around 100 LEDs on this lighting strip.  I imagine this drastically changes things.
> 
> 
> Basically, to simplify the entire problem, I need to understand the math for ohm's law and voltage dividing.  How do you get 560 from
> ...


Try: 2V = (80Î© / (80Î© + R)) * 16V

I think if you want to add 100 LEDs you're going to have to do a combination of series and parallel circuits if you're sticking with the power supply you mentioned in the first post. 10watts at 16V leaves you 10/16 = 625mA which isn't enough to put them all in parallel, but 16V isn't enough to put them all in series.


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## Janglur (May 8, 2010)

Okay, Snowfox.

So if I wanted to have a pair of series of 10 LEDs each, that would be:
2V / 0.25A = 8Î©
2V = (8Î© / (8Î© + R)) * 16V
R = 56Î©

So i'd need:

16vDC ---> 56Î© --> 10xLEDs --->GND
|.........................................^
|.........................................|
----------> 56Î© --> 10xLEDs --

?


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## SnowFox (May 8, 2010)

The 56Î© resistor would be if you had the 10 LEDs after it in parallel. The most you could get in any one series loop would be 8 since theres a 2V drop across each one. If you have a group of 10 LEDs in parallel it would draw 250mA, the most the power supply can manage is 625mA, so you need to come up with some configuration that manages to work with this. It's after midnight here and I haven't looked at this sort of thing in years so I can't think clearly at the moment. If you haven't solved it by tomorrow I'll think about it then.

Or you could get a bigger transformer if you don't want the hassle.


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## Janglur (May 8, 2010)

I could get a different transformer, but that'd make this easy.  =D

This isn't really a practical project, it's to teach myself.  The goal is to succeed by learning what I need to know.  (And as an added bonus, have a fun new lightstrip!)  Eventually my projects will get far more complicated.

So, i'm not afraid to fail at this stage.  But I just want to learn how to make the magic electron cooperate.  And the entire resistor ohm voltage problem is confusing me.

So, let's dump it down to 5 LEDs..?


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## Lobar (May 8, 2010)

Janglur said:


> I could get a different transformer, but that'd make this easy.  =D
> 
> This isn't really a practical project, it's to teach myself.  The goal is to succeed by learning what I need to know.  (And as an added bonus, have a fun new lightstrip!)  Eventually my projects will get far more complicated.



Pirate a physics textbook?


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## net-cat (May 8, 2010)

Oh, I missed the part with there being multiple LEDs. If it were just one, a 560Î© resistor in series with the LED would work. But diodes, strictly speaking, aren't ohm's law devices. I'll dig out my microelectronics text book at some point and figure it out...

*grumbles* I used to know this stuff.


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## Janglur (May 8, 2010)

So, I spoke with Slyck, and unless I misunderstand, am retarded, or both (and I know i'm retarded), the following attached diagram should work?


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## Slyck (May 9, 2010)

Here's how to hook up the capacitor and rectifier:

```
[FONT="Courier New"]
             ________           
             |       +|--â”¬--> + to LEDs and resistors
<----------|~       |   |+
<----------|~       | ==== <capacitor
to trans.   |_______-|--â”´--> - to LEDs and resistors
output       rectifier          
[/FONT]
```


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## Janglur (May 9, 2010)

Actually, i've redesigned it, and think I may have it right...?

Opinions?


(Yay, MS paint adventures!)


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## Janglur (May 9, 2010)

erm, try2


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## Slyck (May 9, 2010)

Janglur said:


> erm, try2



Looks okay! I presume the rectifier and capacitor aren't included here but will be in the final cabob? (as per my last post)


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## Janglur (May 9, 2010)

Rectifier??

I don't know where the capacitor goes.  I'm getting some help from an ASE Master Certified automotive tech right now on this.  =3

http://img222.yfrog.com/img222/7920/1273380449janglurdiagra.jpg

This is where we are as of now


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## Janglur (May 9, 2010)

OK, how about this?

http://www.furaffinity.net/view/3830343


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## Janglur (May 9, 2010)

Okay, finally got my roomy (an industrial electrician) to help me.

He tells me that using a parralel circuit would eliminate the need for capacitors and aside from using more wire, would be overall easier, and allow me to use a single resistor as opposed to a metric dickton.

Here's what we got as the final draft:


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## Janglur (May 9, 2010)

And now, i'm an idiot!

Me and my roomy both failed to notice the transformer is AC->AC, not AC->DC.  X.x


Back to the drawing board!


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## CerbrusNL (May 9, 2010)

you could use a diode bridge to "convert" the AC to DC,
info: http://en.wikipedia.org/wiki/Diode_bridge


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## SnowFox (May 9, 2010)

The AC-DC thing isn't a big deal, just use a diode bridge with a capacitor across it. I agree that putting the LEDs in parallel would be better if possible, but at 25mA each any more than 25 would overload the transformer.


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## Janglur (May 9, 2010)

ALRIGHT let's try THIS then!

Is this optimal?

This time I calculated for the manufacturer's maximum forward voltage on the LEDs, added a heavy duty capacitor, and am not stressing the power source as much.

Good?  Advice?  I'm still learnin' here!


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## Janglur (May 9, 2010)

Note:  I fixed the capacitor mistake.  It's between -/+ in parralel as per correct now.


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## SnowFox (May 9, 2010)

Janglur said:


> This time I calculated for the manufacturer's maximum forward voltage on the LEDs, added a heavy duty capacitor, and am not stressing the power source as much.



Are you working on different ratings for the LEDs now? I can't really read the resistor value on your diagram.

I'm trying to test this stuff out myself, but it's not really working like I expected when you put them in series. I'm wondering if it's because LEDs don't really have a fixed "resistance" because the current goes up exponentially with voltage.


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## Janglur (May 9, 2010)

Just zoom in on the picture?  Pretty much any browser should be able to.

90ohm 1/4watt 5% carbon resistors

The LEDs are 20mA 1.9v (typical) 2.4v (max) forward voltage



How are you 'testing', btw?


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## SnowFox (May 9, 2010)

I think what you've got should be fine. You could probably fit more LEDs in each loop if you wanted, and you could probably have a single resistor in series with the whole circuit if you've got one that can take the current.

I was testing with my own collection of LEDs. I tried putting 8 in series, even though the voltage across each one was about right, the current seemed way too high.


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## Slyck (May 10, 2010)

..Yea. Diode bridge = rectifier. Just a difference in terminology like capacitor / condensor, tube / valve, etc.


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## thegeekguy (May 15, 2010)

If the blinking wouldn't annoy you, you could just put a ceramic capacitor accross the input and ignore the rectifier. The led's are diodes, so this circuit would only turn on when the voltage is positive, which is only about half the time. This causes the LED's to blink at 30 hertz (I assume you are using line voltage) This isn't good for reading, but would be acceptable for a hall light. With a big enough capacitor, you could diminish the effect. Be sure, however, to avoid electrolytic capacitors in an AC circuit, as they are polarized. Just a thought, you could probably learn more by using a diode bridge.


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## Slyck (May 16, 2010)

thegeekguy said:


> If the blinking wouldn't annoy you, you could just put a ceramic capacitor accross the input and ignore the rectifier. The led's are diodes, so this circuit would only turn on when the voltage is positive, which is only about half the time. This causes the LED's to blink at 30 hertz (I assume you are using line voltage) This isn't good for reading, but would be acceptable for a hall light. With a big enough capacitor, you could diminish the effect. Be sure, however, to avoid electrolytic capacitors in an AC circuit, as they are polarized. Just a thought, you could probably learn more by using a diode bridge.



Having a cap across an AC potential will act as a load. Oops! Unless you mean like a 100n (0.1u) cap for transient suppression. Doing this will make you a better person but isn't really needed.:grin:

Don't go above 0.1u for this, though, to keep the amount of power going across the cap to a minimum. Remember: Capacitors can sop up DC to be used later but with AC they act as a ballast with the reactance depending on the frequency of the AC and the value of the cap.


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