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Old 06-29-2019, 07:15 PM   #1
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Default Power indicator light.

Here is a mod I did several years ago and had to replace it today. Basically, it is a 12 volt LED that is hooked up to the storage light circuit. When its on it means you have 12 volts running through the trailer. Its always on when you are plugged in with shore power (i.e. AC). When you not plugged in, its only ON when the main switch is on and you are running off the battery. Its a nice visual reminder.
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Old 06-29-2019, 07:33 PM   #2
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Very clever! Any idea how much current it draws? .02Amps?
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Old 06-29-2019, 08:24 PM   #3
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No power consumption info was availabe but its probably pretty low. Its a marine dash indicator light that I bought off ebay.
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Old 06-29-2019, 08:31 PM   #4
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I'll try to find out as even a low current adds up to a lot over 24 hours a day. Hope to report back with good news.
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Old 06-30-2019, 12:19 PM   #5
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Around 20 milli-amps (0.02 amps) is what I'm finding. A drain, but not much of one.
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Old 06-30-2019, 12:54 PM   #6
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Thanks! I agree that ~.02A would be almost negligible since it is only 0.5 Amp-hours per day. I posted a question on Amazon for a similar 3/4" LED and the only response has been that the current is 0.1 Amps which would be 2.4 Amp-hours per day which would seem like too much to waste on just a light. Since a deep cycle battery is ~40 Amp-hours and you can only discharge to half without damaging the battery, that would mean 20 Amp-hours are available. To me anything close to 10% seems like too much. Since you measured .02A on your lights, is there any identification on the light so I could try to get the same ones? Is there a brand or pn? Thanks!
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Old 06-30-2019, 03:27 PM   #7
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Unfortunately there is very little information on these. I got them from Ebay. My estimate was based on other LEDs on amazon that looked similar to the one I purchased.
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Old 06-30-2019, 04:59 PM   #8
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You probably already know this, but you can actually measure the draw with an ordinary multimeter:


1. Make sure everything in the RV is off.



2. Disconnect the battery, using the battery disconnect switch.



3. Put the red (positive) multimeter plug into the 10 amp socket on the front of the multimeter.


4. Unhook the negative terminal of the battery but not the red positive one. Using an alligator clip or tape, clip the multimeter's red lead to the negative cable (the one that would've been connected to the battery).

5. Clip the black lead to the screw post of the negative terminal on the battery. This is hooking up the multimeter "in series" – i.e., the current has to flow from the negative battery cable through the multimeter to the negative terminal.

6. Turn the multimeter on to the 10 amp setting. It should read "zero."

7. Reconnect the battery, using the battery disconnect switch. You should get a fairly low reading – around 0.2 -- the background draw of the carbon monoxide monitor, the smoke detector, and the propane detector.

8. Turn on the device you are trying to measure. So, for example, if the background number is 0.10 amps, and the meter jumps to 0.30, you know that whatever you are testing draws 0.20 amps.

9. Disconnect the battery. Reconnect negative cable to negative screw post. Reconnect the battery. Restore the red plug to the V Omega mA socket on the multimeter.
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Old 07-01-2019, 01:52 PM   #9
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I poked around on Amazon and found some customer photos of units which had been disassembled. There are three LED chips wired in series with a resistor marked 221 which would be 220 Ohms in series with the LEDs. Each of the three internal LEDs would have a forward voltage drop of probably 1.8 to 2.0 VDC. For 1.8V each times 3 would be 5.4V. Subtract this from the battery voltage and divide by 220 to get the current in Amps. Using a fully charged battery of 12.6V would mean 7.2V divided by 220 equals .033A. If the forward LED voltage is closer to 2.0V, then the current would be 6.6V divided by 220 which is .03A. The lights should be noticeably brighter and draw more current when the Tow Vehicle is charging them to 13.5VDC. Notice 13.5V - 5.4V = 8.1V divided by 220 Ohms would be .036A. I hope this helps the geeks like me out there! Some people on Amazon have asked about placing two lights in series and this could be done but the brightness and current would be greatly reduced (far less than half. Consider 13.5V - 10.8V equals 2.7V divided by 220 equals only .012A through the lights while driving (one third of the current). They would likely be much dimmer and draw less current when operating from the battery and not being charged.
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