# Really easy question

Discussion in 'The Projects Forum' started by wazz, Jan 8, 2008.

1. ### wazz Thread Starter New Member

Jan 8, 2008
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You guys be nice now, I am trying to learn this stuff by trial and error. My question is this, is there something else I need to hook up between a 9 volt battery and the two 5 volt lights? I am trying to hook up lights to one of my sons toy trucks, when I attched the lights to the battery it worked fine, but I thought the battery was a little bit to warm, but maybe that is normal, like I said I know hardly nothing about curcuits.
Thanks
Wazz

2. ### SgtWookie Expert

Jul 17, 2007
22,183
1,729
You're probably using PR-3 bulbs (flashlight bulbs). Those drain 300mA; quite a bit of current.

Instead, get some white LED's. These are usually rated for 3.6V @ 15mA.

Connect a pair of them up in series with a 120 Ohm resistor to your battery to get the 15mA current. The battery will last longer if you use a 180 Ohm resistor; that will cut the current to 10mA. The LEDs won't be as bright, but your battery will last about 30 times as long as it does with the incandescent bulbs.

3. ### wazz Thread Starter New Member

Jan 8, 2008
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Thanks, thats great

4. ### SgtWookie Expert

Jul 17, 2007
22,183
1,729
Just got an E-mail flier from Marlin P. Jones & Assoc. They're having a sale on 10mm white 15000 MCD LED's with a 15° viewing angle, 3.8 forward voltage, 30mA max, 20mA recommended, \$0.43 each. These would be perfect for your application.

Since they will be REALLY bright, I recommend running them at even lower current. Super-brite LED's can cause blindness, and kids like to look at lights.

How to calculate the current:
Take your battery voltage, and subtract the voltage of the LED's from that. Since you'll be running two in series, that's 9V - (2 x 3.8V) = 1.4V (ETA: corrected voltages)
Then to calculate the resistor to obtain your desired current (let's say 10mA)
R = E / I
R = 1.4V / 10mA
R = 1.4 / 0.01
R = 140
You'll need to verify that they're not too bright. If you see "spots" after looking at them for a few seconds and then looking away, they're too bright.

5. ### wazz Thread Starter New Member

Jan 8, 2008
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I thought I would be running them in parallel? Not that I really know the difference? I appreaciate the help, maybe you can point me to a tutorial.

6. ### Audioguru New Member

Dec 20, 2007
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If you connect the LEDs in parallel then you are throwing away over half the power of the 9V battery. In series then the battery current is low and you are wasting much less power.

7. ### SgtWookie Expert

Jul 17, 2007
22,183
1,729
What Audioguru said

In addition, if you were running them in parallel, you would need a current limiting resistor for each LED. Let's see what happens:

9V - 3.8V = 5.2V
We'll just go with 10mA for this exercise
R = E / I
R = 5.2V / 10mA
R = 5.2 / 0.01
R = 520 Ohms, for each resistor

Let's compare the power being used in the resistors of each circuit.
P = E x I
P = 5.2 x 0.01 x 2 (for the 2nd circuit)
P = 0.104 Watts

Now for the first circuit:
P = 1.4 x 0.01
P = 0.014 Watts

Get the ratio...
0.104 / 0.014 = 7.43
So, wiring the two in parallel would use up over 7 times as much power just heating up the resistors.

Make sense?

8. ### wazz Thread Starter New Member

Jan 8, 2008
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Yes it does make sense, again I apprecaite the input.
Everybody has to start somewhere, I am at the begining

Wazz