First circuit

Thursday, Jul 9, 2009. Since I received my big box’o’lectronic components yesterday, I figured I may as well dive right in and start messing around. I figured I’d start simple with some LEDs. But it’s been so long since I’ve done any of this stuff, I’m not sure I remember much. I do remember Ohm’s law though. There are several things you need to know about LEDs, but the two most important are that 1) you can’t just hook them up to a battery, they need a resistor in series with them in order not to fry themselves, and two, they only go in the circuit one way — the two leads are not equivalent.

To figure out what kind of resistor you need, you need to know how much current the LED can take, or what the voltage drop across the LED is. From there, you can figure things out with Ohm’s law:

              V = IR

where V is voltage, I is current (in amperes) and R is resistance, in ohms.

Well, the first problem is the bag of LEDs I picked isn’t labeled very clearly. I picked some cheap red ones for my initial experimentation in case things went wrong and I burned it up. On this bag, it looks like it says, “SAM 3-” Huh? Maybe the S is really a 5, and the AM? milliamps? The “3-” might be the 3 volt voltage drop across the LED?

So, I figured to get 5mA at 3 volts, I = 0.005, V = 3, plugging that in to V=IR and solving for R gives you 3/0.005 = 600 ohms.

I doubled it just for safety to 1.2Kohms. (after all, I’m not remotely sure I’m interpreting that “SAM” as 5mA correctly! Nor sure if I can take that 3 volt drop across the LED and then stuff it through ohm’s law and sort of treat it as the internal resistance of the LED to get a kind of limit on the current.)

The next problem is, which way does the LED go? Googling around I find that the long lead is the positive lead.

Now a power supply. 4 AA batteries in a battery holder in series at 1.5 V each gets me 6V. 6.3V, actually, according to my ohm-meter.

So, I wire up the 1.2Kohm resister in series with the LED — check the polarity — and apply the 6.3 V.
Hey, it works:

My first circuit in a long long time.

I checked the voltage across the resistor, and it’s about 4 V, and across the LED about 2 volts (Hey, oh yeah, Kirchhoff’s law: 2 + 4 = 6 volts of the battery. Starting to come back to me a little bit.)

So, that means, with my doubling the 600 to 1200, that means it makes sense that 1/3rd of the voltage is across the LED and 2/3rds across the resistor. So I probably have a load of 1200 + 600 = 1800ohms,, and current of about 6/1800 = 3.3mA.

Hmm, let me try another LED. One of the expensive white ones this time. The bag these came in seems to be labeled “LoD 5W” Hmm. 5 Watts? That can’t be right, can it? Then I remember I think they are 5mm across, and white — that’s probably what the 5W means. So, no electrical hints there. Well, let’s just hook it up and see what happens.


Hey, it didn’t blow up. And it’s bright. I measured the voltage across the resistor and LED again, and it was about 2.7 V across the LED and about 3.5 across the resistor, so the internal resistance of the white LED is a little higher. Let’s see across the resistor was 3.5, so that means the current was 2.91mA, so the internal resistance of the LED should be 2.7/0.00291 = around 928 ohms — if it makes sense to talke about the resistance of an LED as a constant like that — maybe it varies with current, I don’t know.

Anyway, success so far, however tiny.

~ by scaryreasoner on July 10, 2009.

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