Couple questions on vf of a light.

iced_theater

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Oct 12, 2005
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Green River, Wyoming
I had bought a 1 watt spotlight from Wal-Mart, and just barely took it completely apart to see how hard it would be to modify it for a different LED. The light runs off of 4 AA batteries, 2 X 2, which if I understand right would be 3 volts going to the LED if no resistor were in place.

The LED that's in it is a RVOH which recommends a minimum vf of 3.03 and a max of 3.27 volts. That would run about perfect in direct drive with a pair of L91 AA's. But there is a resistor still on this light which is Gold, Gold, Violet, Yellow which comes to 4.7 ohms according to a site.

So how would you figure out the vf of the light with that resistor in place? And would I notice a big difference if I just take the resistor out and put a piece of wire in it's place?
 

eebowler

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iced_theater said:
And would I notice a big difference if I just take the resistor out and put a piece of wire in it's place?
I'm sure it wouldn't be long before you see a :poof: If you do decide to try it, leave the resistor there and 'short' it with the wire for very short intervals to see the difference.

iced_theater said:
The light runs off of 4 AA batteries, 2 X 2, which if I understand right would be 3 volts going to the LED if no resistor were in place.
Not necessairly. The cells might be laying down 2 X 2 but most likely are connected in series. A series connection would give about 6V output to the LED if no resistor was there. The 4.7 ohm resistor limits the current to about 1.3A (current = V/R ) and the internal resistance of the cells reduces the current even more.

Do you have a multimeter? That would make it so much easier.
 

iced_theater

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Oct 12, 2005
Messages
819
Location
Green River, Wyoming
Here's a couple of pics. Too bad, I replaced it with wire before you said anything eebowler :p oh well. The reason I thought they are in parallel is because there are 4 wires, 2 positive 2 negative with the positives going to the switch and the negatives going down the board and being connected by the resistor. So is this 6v from what you can see?

P8160104600x449.jpg

P8160105600x449.jpg
 
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