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Isindil
Joined: 30 Dec 2006
Posts: 13
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 Posted: Mon May 12, 2008 11:38 am Post subject: |
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I've made a diagram for the complete sensor bar now. Will the circuit work with a 27 ohm resistor for the power indicator LED (the 2.2v one) and a 4.7 ohm resistor for each of the IR LED arrays (the 1.3v ones)?
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Frivolous Sam
Joined: 18 May 2008
Posts: 16
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| Posted: Sun May 18, 2008 10:39 pm Post subject: |
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| Isindil wrote: | | Will the circuit work with a 27 ohm resistor for the power indicator LED (the 2.2v one) and a 4.7 ohm resistor for each of the IR LED arrays (the 1.3v ones)? |
Let's work it through. Ohm's Law tells us V=IR and therefore R=V/I. You know the current required for each one and you know the voltages they drop so you can work out the resistance needed.
For the indicator LED:
The LED is dropping 2.2V, leaving 0.8V for the resistor. The LED needs 30mA (typical).
0.8/0.03=26.7 ohms. The nearest standard resistor value is 27 ohms (5% tolerance).
For each set of IR LEDs:
Each IR LED is dropping 1.3V, leaving 0.4V across the resistor. You say they're getting 100mA each.
0.4/0.1=4 ohms, which is close to the standard value of 4.7ohms (5% tolerance).
So the answer to your question is yes. You could change your circuit so that only the IR LEDs are in parallel and then you would only need one 4.7ohm resistor. You will be drawing 230mA. Typical capacity rating for alkaline batteries is 10,000mAh but at 690mW (3V*230mA) you will get a lot less so I would expect you to have 3-4 hrs usage.
I think you would be far better off using a resistor (or resistors if you don't change your circuit) in the range 15-20 ohms instead of the 4.7 ohm resistor. This would run the IR LEDs at 20-30mA so you would get less IR power (and therefore less distance) but far greater battery life (up to 100 hrs instead of 3-4). It all depends whether you find that you need the max range. If you're in a typical room I think you'll find that 20-30mA will be plenty. |
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