Questions about Circuitry and Resistors

The voltage across a resistor depends on the current flowing through it, so you'll need to know what current the LED draws.

The voltage drop across the resistor is given by OHM's law: V = IR

V is the voltage drop, I is the current in amps, and R is the resistance in ohms.

This is assuming you plan to wire the LED in series with the resistor.

You need a voltage drop across the resistor of 6 - 2.2 = 3.8 volts. Once you know the current, I, drawn by the LED, the required resistance will be 3.8/I ohms.

 

More info here

 

350 mAh tells you what its energy consumption is, in milli-Amp.hours.

[edit: realised that this doesn't make any sense. Ignore the rest of this post.]

At a voltage of 2.2 volts, we have

E = VIt

so

\(I=\frac{E}{Vt} = \frac{350 \times 10^{-3}}{2.2 \times 60 \times 60} = 4.42 \times 10^{-5}\)

where the current is in Amps. Multiplying by 0.001 on the top line converts milli-Amps to Amps, and the 60 times 60 on the bottom line converts hours to seconds, to make sure the current comes out in Amps.

The resistance you need is therefore

\(R = \frac{V}{I} = \frac{3.8}{4.42 \times 10^{-5}} = 86 k\Omega\)

You need an 86 kilo-Ohm resistor. [edit: WRONG!]

 

Volts = current times resistance (ohms)

Current = volts divided by resistance.

 

86,000 ohms.

In electronic-speak, that's an 86K resistor.

Definitely NOT 86 ohms, which would blow the diode by drawing a current that is way too big.

 

My calculation seems to require a very large resistance. After a quick look on the web, most LEDs seem to draw about 20 mA of current - far more than what I calculated from your figure above. Therefore, the resistance is correspondingly less- typically only a couple of hundred ohms or so.

Are you sure it says 350 mAh?

What other information do you have about your LED?

Is it a green one?

 

Now I'm confused, and I think I've made a mistake. I'm not sure what the 350 mAh thing is now.

Maybe it just means 350 mA is the current, but I don't know why the "h" is there. Hang on a minute and I'll check something...

 

If it is 350 mA, then the resistance you need is:

\(R = \frac{V}{I} = \frac{3.8}{350 \times 10^{-3}} = 10.8 ohm\)

The closest you'll get to that is about 10 ohms, or maybe 10+1 = 11 ohms.

If it's a 1 watt diode, then I can calculate the resistance a different way:

I = P/V = 1/2.2 = 0.45 amps is the current drawn.

So R = V/I = 3.8/0.45 = 8.4 ohms is the resistance you need.

This is close to the same value, but not quite the same.

A 1 Watt LED sounds like too a high power consumption for an LED to me, though, as does a current of 350 mA.

Either this is a VERY VERY bright LED or something is wrong.