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Author Topic: Resistor value  (Read 1630 times)

meechingman

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Resistor value
« on: March 11, 2009, 03:35:18 PM »

Hi

Here's one for the electronics experts.

Having installed and wired a set of lights on my Smit Nederland, I'd like to power them off a simple distribution board, powered by the boat's 6.4v battery.
All the lights are 3v GOW. To keep things simple, what value resistor could I put in series with each lamp to let them run properly?
Andy
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stallspeed

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Re: Resistor value
« Reply #1 on: March 11, 2009, 04:27:13 PM »

If you have an even number of identical element bulbs the solution is to wire two in series for 6 volts.
For an odd one left over just use an extra bulb or use the R=V/I rule and get the I from a meter with two connected to the six volts.
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meechingman

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Re: Resistor value
« Reply #2 on: March 11, 2009, 05:01:14 PM »

I wish I'd remembered that before spending hours getting the 3 mast lights wired up in parallel and neatly taken down the mast so I only needed one pair of leads to power them!

For the moment, I'm testing it all out with a couple of AA's, and this would offer a simple solution, but I've just discovered that the two aft deck floodlights are 6v bulbs.

Any further thoughts?

Andy
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stallspeed

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Re: Resistor value
« Reply #3 on: March 11, 2009, 06:02:45 PM »

......a low ohm potentiometer?
.......lm317t + 2 resistors wired for 3 volt output?
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portside II

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Re: Resistor value
« Reply #4 on: March 11, 2009, 09:00:46 PM »

Wait for it ?
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portside II

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Re: Resistor value
« Reply #5 on: March 11, 2009, 09:01:19 PM »

Wont be long now ?
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portside II

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Re: Resistor value
« Reply #6 on: March 11, 2009, 09:02:45 PM »

Aw come on Dave , your missing an oppotunity here . And i would like to know too  :D :D .
daz
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OMK

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Re: Resistor value
« Reply #7 on: March 11, 2009, 09:31:17 PM »

Assuming that each GOW pulls 40mA (0.04A).
First you have to deduct the bulb voltage from the battery voltage. (6.4 volts minus 3.0 volts = 3.4 volts).
Now divide the remaining voltage by the current draw (3.4 volts divided by 0.04 amps = 85 ohms). The nearest common value would be 82 ohms (grey, red, black and gold bands).
One other thing to take into account, and that's the actual wattage rating of the resistor. The worse scenario being that the wattage value would be too low, would overheat, ending up open-circuit.  So to calculate watts you simply multiply the volts by the amps. In this case, your 3-volt GOW multiplied by 0.04 amps = 0.12 watts (one-eighth watt). In reality you would need plenty of headroom, so best to multiply the wattage value by at least a factor of two (0.12 x 2 = 0.24 watts, or one-quarter watt). Two GOWs in parallel would draw 80mA, which in turns means the resistor would be best happy with a rating of at least 0.5 watts (one-half watt).
The reason you must be generous with the wattage rating is due to the heat factor (read wasted battery energy). For instance, if the resistor wattage value is underrated, it means the resistor itself will start to struggle, won't be able to cope and start to get hot. Which ultimately means a lot of the energy in your battery is being converted into wasted heat, rather than light.

You say that your floodlights are 6v bulbs. No problem. You no doubt have it figured by now that it's okay to stick the 6v lamps straight across the battery. (i.e: no dropper resistor needed). 2 x 3v bulbs in series is another way to do it (as you now know)  ;)

So the answer to your question is: Each series resistor would be 82-ohms, 1/4 watt.
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malcolmfrary

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Re: Resistor value
« Reply #8 on: March 11, 2009, 10:11:10 PM »

A Smit Nederlad (if its the one I remember), is a fairly big boat.  Carry on using 3 volts (or a couple of rechargeables, slightly under-running the bulbs will let them last much longer) for those, and run the 6 volt ones off the 6 volt supply, maybe with a 1N4001 diode in series to lose about 0.5 volt.  If you want to run the lot off 6.4 volts, feed the 3 volt ones through the highest power rating 3.9 volt zener diode you can get your hands on wired in facing the "wrong" way.  Its a simple, down and dirty way of losing a few volts without too much calculating, and has the benefit that if a bulb goes, it does not increase the voltage across the rest, taking them along.
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meechingman

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Re: Resistor value
« Reply #9 on: March 12, 2009, 09:17:01 AM »

Thanks for the info guys.

My solution was to run pairs of lights, P&S navs, deck lights etc in series, and the floods on 6v. That left the 3 mast lights and a searchlight. They didn't want to work together at all, so the searchlight was given its 3v via a spare 7805 and a small pot. I can vary the brilliance on that one which is nice. The mast lights are running off 3v from 2xAA cells.

I might leave things like this as, should things go pear shaped at night on the lake, and the main fuse blows, at least I'd have 3 mast lights still on to let me see where the tug is!

Thanks again, especially for the formula for calculating resistances and the advice about the zener.

Andy G
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