sm

Ag951 - I didn't see your post. We must have posted at the same time! It seems that a watch battery may be the route to go...

sat968

You will want a resistor in series with the LED to limit the current flowing through the LED. If you assume 2.8V across the LED and the battery at 12V, that leaves 9.2V across the resistor. Since you want 20mA (.02A) to flow through the LED (and thus the resistor), the value of R is V/I = 9.2/.02 or 460 ohms.

The power dissipated in the resistor is P = I^2R or .02 * .02 * 460 or .184W. So you will want to find a 1/4 Watt resistor (or 1/2 watt to be conservative) with a minimum of 460 ohms (470 ohms is a standard resistor value).

Ag951

Isn't what sat968 said really complicated?
Just use the watch battery. Radio shack has 3.3V batteries that are designed to provide ~20mA.

led

It isn't, just put a 470Ohm 1/4 Watt resistor before the led. The battery will need to be replaced every so often vs. never having to do anything ever again.

sm

Thanks everyone for the help. I actually understood what Sat968 was talking about! Now, do I solder the resistor in series with the + or - side of the LED?

Ag951

One side of the led should have the lip smoothed down. That is the end you want towards the negative terminal.
For a series circuit, it doesn't matter if it's +, led, R, - or +, R, led, -. The current will be consistent through the whole series.

I was thinking about that design and if I remember my EECE 211 class correctly (and I might not, it's been a few years), I don't think it'll work. If the circuit is:
+____PR___LED___-
Won't the entire 12V drop over the power resistor? Diodes don't have significant resistance, so the diode won't have much of a voltage drop across it, will it?
Now I'm going to have to dust off my DMM and measure some diodes tonight.


I'd rather have to replace a cheap battery every year than wire it directly into the car. You don't have to run any wires, or cut anything.
And the circuit is
+___LED___-
What could be simpler?

rcldesign

First of all, a transformer can't do jack with DC... nothing at all. Please re-read any physics materials you may have read regarding transformers.

Second, you don't need a voltage divider for this at all. You could hook an LED up to 10,000V if you wanted to... you just have to make sure the current isn't too high (in this case, you'd need a 500K ohm resistor that happens to be good to 200 watts) (Ag951 is correct, the 12V drop will be over the power resistor)

Many LEDs that are said to be "12V", etc. are meant to be used without a resistor in line - they typically have some sort of resistance built in.

For this circuit, here is the super simple math. You need to solve for R

V = .02*R + .7 ... don't forget the .7V drop across an LED!! In this equation, V is the voltage you're hooking it up to... 12ish in a car, 1.5-3 with watch battery, etc.

The 'voltage' specification mentioned for the LED refers to the LED's 'transistion' voltage... it will need at LEAST that much voltage applied to it for it to turn on....

And the circuit simply is:

+ -----/\/\/\/\-----|>|------ -
R LED

Keep in mind this won't blink. To make it blink is a bit more complicated. You can use a 555 timer, or make a ring oscilator with 3 NOT gates, two resistors, and a capacitor.

The final consideration is the resistor's power dissipation. So that you don't cook the resistor, choose its wattage as was described earlier by someone else. (P = IV where P is the wattage you need, I is .02 and V is the battery voltage from above minus .7V. Always round up.

Now, if you hooked a diode (LED) directly up to the battery, it will fry. LEDs have no resistance. Period. They introduce a voltage drop across their P-N junction, but there is no current loss. So, with an LED hooked up directly to a battery, we see by Ohm's law:

I = V / R... I = 12 / 0... division by zero = infinity, and therefore you have an infinate amount of current.

Of course, if V is close to or less than the transistion voltage, (2.x V) then you won't cook the LED because there won't be enough potential across the PN junction to make the electrons jump the junction, and no current will flow.

Anyway, enough of a lesson... here's a schematic for a blinking LED. The LED will blink every 1.64 * R * C seconds. You can determine your own R and C to suit your tastes.

Ag951

Brings back memories, now where did I put that copy of SPICE...

keith

You may think about wiring it up the way I did on my last car - get the factory LED door lock *****, put your blinking LEDs in there, then run battery power and ground the negative to 15 power on the ignition switch.

That way, the LEDs blink whenever the key is off. Perfect.

sm

First off, thank you all for the lessons. You guys pack a lot of knowledge. And those diagrams? Wow.

If the LED I buy from Radio Shack is of the blinking type, would I still need all that gadgetry?

I like your idea. But what is "15 power"? Is that a ground or a power/live wire? The name sounds confusing.

Thanks folks

keith

15 power is ignition switched power (on when key is turned)

30 power is battery-direct power (always hot)

sm

Forgive my ignorance. If I connect one lead to the battery, and the other (ground) lead to the 15 power line, then wouldn't I get power on both leads of the LED when the key is ON? Isn't that bad?

Ag951

Timer gadgetry: no, it has the timer circuit built in. It'll be smaller and require much less work on your part. The downside is it might be a few bucks more expensive than do it yourself, and you probably can't select the period and pulse width of your square wave. (How quickly it blinks and how long it stays on per blink).
Voltage control gadgetry: yes, if you use your car battery.
If you use the watch battery, all you need is the battery and the blinking LED.

I'm not sure what the 15 and 30 line are, but if they're at different potentials then there's not a problem connecting a circuit across them.
Voltage is a relative term. "Ground" in one circuit could have a higher potential than the 12V lead in another. Voltage (or potential difference) is the difference between the potentials of two leads. Current will flow from the node with higher charge density to the lower one. Voltage is somewhat analogous to pressure in a pipe or hose. If you open a 50psi hose, it'll spray out, since it's 35psi above external pressure (greater potential). But if you open it in a tank of 80psi water, water will flow into the hose.
If you take a 12V battery and 5V battery and short their negative terminals, you can get 7V out of them across their positive terminals, causing current to flow into the 5V's positive terminal.

keith

15 and 30 are both 12v. That's the point. WHen the key is on, 12v is on both sides of the LED, so there is no flow of current. Turn the key off, and 15 acts as a ground and the LED blinks.

That way "arming" and "disarming" are automatic, and occur when you want them to!

(This is not originally my idea, Steve Petty clued me in on it - but it's THE way to do it! GREAT solution)

sm

Keith,

I think I'll be going with your idea (along with a starter kill and DME switch, especially after hearing about Tom's stolen car). But if you use 15, will the accessories (radio, sunroof, etc.) now be constantly on, or still require the key to be in the "ON" position?

Thanks.