electronics question

Mirlyn

Mirlyn

Well-Known Member
I want to step down a 12v DC line (in a car) to a 4v line to power a set of LEDs. I have a limited experience in electronics, but don't know what I need for this. Anyone got any good sites to read about the basics and how to design a circuit to do this?

Thanks! :headbang:
 
I was going to say easier and probably cheaper to buy than to build. A simple resistor would probably do the trick. We used to put in line resistors to all the gauges on old VWs when we upgraded the generators to twelve volts (they came with six volt systems through 65 or so) so you could have decent tunes and good driving lights. I'll have to look through the old library, it's been years since I fooled with anything like that.
 
What you need is a simple voltage divider. It's really easy to construct.

here is a handy calculator and schematic to hlep you understand what is happening. Essentially, you need just two resistors in series, and then you draw power across one of them to power the LED's.

The information you need is the total resistance of your group of LED's. Since LED's are voltage driven, I assume you'll be wiring them in parallel so that they all get 4V? Or is it a group that's already wired and says it needs 4V across the whole gang?

In either case, you will need to know the resistance of the group to determine the correct resistor values so that under load you get a 4V output. If the group has a high resistance (like, 100 ohms or more) then one resistor simply needs to be twice the value of the other. Draw power across the larger of the two, and you'll get a 4V output. If the LED group has a lower combined resistance, then you will need to adjust values accordingly. That url I provided should help you do that.
 
outside looking in said:
The information you need is the total resistance of your group of LED's. Since LED's are voltage driven, I assume you'll be wiring them in parallel so that they all get 4V? Or is it a group that's already wired and says it needs 4V across the whole gang?.
Click to expand...


It'll be powering at least two of these @ ~4v ea.

Another idea I had was some kind of battery-backup for them. Basically the batteries get charged whenever the car is running, and the lights run the batteries down instead of draining the main car battery. Thats not necessary though (just another zany idea), as I don't think the LEDs will drain power that fast in the event I forget to switch them off.

I'm somewhat numb right now, so I'll read into that link a bit more in-depth tomorrow. Thanks! :)
 
In my van, I have a deep-cycle battery that runs a 700watt inverter.
I have it wired to a switch that I can flip to switch from my main
cranking battery to it, so it can be charged while on the road.
It's worked pretty good so far.:)
 
I believe there is an even easier way, now that I know the LED's current rating as well as voltage rating (actually, now that I recall that LED's are SS devices which have an inherent voltage drop, not an inherent resistance per say, I don't think the voltage divider would be the best method to use). The goal is for each LED to draw 20mA or current, and the voltage rating is saying that it will have nominal voltage drop across the LED of 3.7V.

So, run the two LED's in parallel. The voltage drop is still 3.7V, but now you want to draw a combined current of 40mA (.04A). All you need to do is add a resistor in series with the two paralled LED's so that the proper current is running through the circuit.

The voltage drop across the two LED's is 3.7V. That leaves (12 - 3.7)V across the series resistor, or 8.3V. The current flowing through the resistor is the same as the combined current running through the LED's, which is .04A. To find the resistor value that creates this current flow, use the following equation:

R = V/I

So, for V = 8.3V and I = .04A, we have

R = (8.3V)/(.04A) = 207.5ohm.

A 200ohm resistor would work just fine. The power draw will be quite low, so you don't need to worry about the resistor's power rating.

If you want more LED's, wire them in parallel and calculate the appropriate value for the resistor. 3 LED's would require .06A, so 8.3/.06 = 138ohm.

For a quick reference, if you're using standard 10% resistors, use the following values:

2 LED's = 200ohm
3 LED's = 150ohm
4 LED's = 100ohm
5 LED's = 82ohm
6 LED's = 68ohm

The wiring for two would look like this:
Code: 
+ ________/\/\/\/\/\______________LED____________  _
        series resistor       |         |
                              |___LED___|
For three:
Code: 
+ ________/\/\/\/\/\______________LED____________  _
        series resistor       |         |
                              |___LED___|
                              |         |
                              |___LED___|

And so on.
 
Awesome! Its starting to make sense now. :D

Looking at 200Ohm resistors...I'm finding them with different wattage ratings. 1/5/10/20W. Is this for the projected load on the resistor for the circuit behind it?


Thanks again for your help! :headbang:
 
Yes, those wattage ratings are for the amount of power the resistor will have to dissipate as heat due to the current flowing through it.

Power (P) = (I^2)*R

So a 200 ohm resistor with .04A of current would need to dissipate (.04^2)*200 = .32Watts of heat. A half-watt or larger resistor should be fine, which are pretty common, which is why I said not to worry too much about it. Resistors in that impedance range usually are 1-watt or greater I believe.

For the other configurations:

2 LED -- .32 Watt
3 LED -- .54 Watt
4 LED -- .64 Watt
5 LED -- .82 Watt
6 LED -- .98 Watt

Any resistor above those power ratings (i.e., a 1 watt resistor in any case) would work fine.
 
Doh! Misread that part...sorry. :)

Thanks for your help! I'll give it a shot tomorrow night. :headbang:
 
Ok, got the setup almost ready to install...but I've noticed the 1W resistors are getting extremely hot (using two sets of 200 Ohms with two of the above LEDS on each set, as the calculation shows). It'll burn you if the circuit is left on for only a few minutes. Should I replace them with 5W ones, or do you think they'll be alright?
 
Not much of anything, really. A friend asked how to do some custom accent lighting here and there in a car using low voltage stuff. I figured LEDs would be the way to go. I always wanted to know how the math works, so I played around with a few different circuits here and there. :shrug:
 
You must log in or register to reply here.
Back
Top