Sylvania DOT-It household power mod.

[JakeyCakez]

So, First of all this is my first post.

so if any of you have bought, or looked at sylvania's Dot-It you'll notice that it runs on batteries, I think it's 3 AAA's. Well for my purposes I Needed it to run on 120VAC (household power)

regulating 120VAC down to 4.5VDC isn't practical or efficient, so I found an old phone charger that was the closest but over 4.5VDC


I found an old Motorola cellphone charger to use as a power supply, it's output is 6VDC, I ran the LED's directly off of it by accident once, it must have been at least as bright as my Inova X5 on fresh cells.

And here's the useful part..
In order to keep from frying the LED's I had to add resistance so they would draw the same amount of power they did on batteries.

using a meter i found how much it drew on batteries..

45mA @ 3V = 135mW
using the formula R= I/E I get 66 Ohms

to avoid reducing the life of my LED's I need to keep the power input the same so by I = P/V gives
135mW @ 6V = 22mA

So I now know how much current I need to draw at 6V all that's left is to find the value of resistance I have to add to the circuit.

6V @ 22mA through a circuit by R= E/I gives me 273 Ohms.

Since my circuit already has 66 Ohms I find the difference to know how much to add.

66-273 = 207

So I need to add 207 Ohms, Since you'll never find the exact since resistors and it'll be a real PITA to put them together til you get it just keep the values with in 5%, also a little too much is better than not enough. a 220 Ohm resistor would be perfect for this application.

by 6E/286Rx6E = .125P
your LED light will now consume 125mW @ 6V. Now your LED's should have the same life expectancy they did when they ran on batteries, and now you can leave it on all the time if you wish.


This can be used anytime you want to run something on a higher voltage (to a certain point). I wouldn't try to use it on anything that has more than a 5 volt difference. ex running a light made for 1.5V on 12V.

I hope someone can understand this; and it helps them :3

 

[horizonseeker]

how did you hook up the connections? solder directly onto the light or somehow create a plug for the charger?

thanks.

 

[Christoph]

:welcome: Thanks I will have to try this out.

 

[Blindasabat]

Based on previous work I've done, those numbers don't sound right. Did you read the 45mA and 3V at the LED or at battery input?

I'm getting different numbers than you if I use LED-Pro (forget where I got the program, use Google, it is available free several places and is simple to use) and assume 45mA at 4.5v from the batteries:
45mA @ 4.5V requires a ~28 ohm resistor to run an LED at a presumed 3.25V. Round to common 30 ohm.

I think is safer to just supply it with 4.5V to the battery terminals at a current it can handle to make your DC input act like 3AAA's. In this case, all you need to do is get the 6V down to 4.5V via the resistor at a reasonable amperage range close to what you measure from fresh batteries:
Assuming 45mA and 6V input and 4.5V desired into LED-Pro and I get 33.3 ohms.

Try 200 ohms across the 6V power with your DMM and you will probably get something around 15mA at 3V.

 

[JakeyCakez]

Based on previous work I've done, those numbers don't sound right. Did you read the 45mA and 3V at the LED or at battery input?

I'm getting different numbers than you if I use LED-Pro (forget where I got the program, use Google, it is available free several places and is simple to use) and assume 45mA at 4.5v from the batteries:
45mA @ 4.5V requires a ~28 ohm resistor to run an LED at a presumed 3.25V. Round to common 30 ohm.

I think is safer to just supply it with 4.5V to the battery terminals at a current it can handle to make your DC input act like 3AAA's. In this case, all you need to do is get the 6V down to 4.5V via the resistor at a reasonable amperage range close to what you measure from fresh batteries:
Assuming 45mA and 6V input and 4.5V desired into LED-Pro and I get 33.3 ohms.

Try 200 ohms across the 6V power with your DMM and you will probably get something around 15mA at 3V.

I would have much rather used a 4.5V power supply, but I could find one. so I made do with what I had. if you have a 4.5V powersupply, you can ignore everything I did and plug it directly into the light and you're good to go.

For my calculations I used the formula's stated, I didn't actually get the meter readings from the meter that I used, but they should have been close.

And I didn't calculate the drop across my resistor either, I figured all that mattered was to keep the power delivered the same, and everything would work great:duh2:

 

[JakeyCakez]

how did you hook up the connections? solder directly onto the light or somehow create a plug for the charger?

thanks.

I soldered them directly to the light the only resistor i could find that was the right value was like a 4 watt wire wound. That light will never accept AAA batteries again.

 

[Mr Happy]

I hope someone can understand this; and it helps them :3

I think your calculations are a bit wonky here.

Here's how I think they should go:

You say the Dot-It draws 45 mA at 3 V. Firstly are you sure about this? Three fresh AAA cells would supply about 4.5 V, and this would not drop very much at 45 mA. Let me assume 4.5 V for the sake of argument, and stick with 45 mA.

Next, your power supply is delivering 6 V, which is 1.5 V too much compared to the batteries. So you need to lose 1.5 V in the series resistor.

What this means is that 45 mA through resistance R equals 1.5 V. To calculate R requires Ohm's law, where:

R = V / I = 1.5 / 0.045 = 33 ohms

Notice that power is not involved in any of these calculations yet.

The last step is where power comes in. We need to know how much power will be produced in that series resistor in order to make it big enough. To find that we use:

W = V x I = 1.5 x 0.045 = 0.07 watts

This basically means that any power rating will do, from 1/8 watt upwards, though a 1/4 watt would probably be best.

Do you follow the logic above? Unfortunately what you have done in your first post is to make lots of calculations, but using the wrong equations in the wrong way, which means your answers have come out all wrong.