Putting Red LED's In Osram Dot it light

The resistor is red-GREEN-black-gold-brown not red-yellow-black-gold- brown as I mentioned earlier. I think the difference is only one ohm or so. The green was light in colour making me think it was yellow, but it does measure 25 ohms.

Dave,

It sounds like the white LEDs are in parallel, drawing (theoretically) about 20 mA per LED.

If you wanted to put your red LEDs in there at the same current draw and wired in parallel, you'd use something like a 43 ohm 1/4 watt resistor. If the LEDs are too bright, use a larger value resistor and gain increased battery life. (56 ohms, approx. 15 mA. 82 ohms, approx. 10 mA. 150 ohms, approx. 5 mA. 820 ohms, approx. 1 mA.)

If you find that one LED is much brighter than the other two with your LEDs wired in parallel, use three individual resistors, one for each LED, to balance the currents. 120 ohms per LED will give you approx.

20 mA per LED. Larger value resistors will lower the current.

Here's how to calculate the resistors and currents:

R = (Batt voltage - LED voltage) / current

Current = (Batt voltage - LED voltage) / R

You want to limit the current through each LED to a maximum of 20 milliamps. High-brightness LEDs put out a decent amount of light at 5 milliamps, or even 1 milliamp. For an outdoor night light, 1 milliamp might be enough if all you want to do is look at a chart at close range, or something like that. Experiment to find the current level that meets your needs. (You might even find that a single LED provides adequate brightness.)

If you have room in that housing to add a miniature switch, you could connect two of the LEDs, biased for fairly high current, switched from the existing power switch. Use your extra switch for a single LED/resistor combo, biased for very low current. That would give you a choice between dim and bright light levels, depending upon your needs.

Have fun!

Tom

I think I have a 47 ohm resistor stashed somewhere, I'll dig that out and try it once I get the LED's tomorrow.

--
From your description, (and neglecting the switch) it appears that
the flashlight is currently wired like this: (View in Courier)


         4.5V      3.1V
        /         /
       +--[25R]--+--[LED>]--+ 
       |         |          |
       |+        +--[LED>]--+
     [BAT]       |          |
       |         +--[LED>]--+
       |                    |
       +--------------------+

I have linked to a photo of the back of the PCB to make it easier to understand the circuit (I'm not very good at tracing out PCB's), hope this helps.

--
Yes, that\'s what I thought.

Including the switch, it\'s wired like this:


.     +--[25R]--+--[LED>]--+
.     |         |          |
.     |         +--[LED>]--+
.  [SWITCH]     |          |
.     |         +--[LED>]--+
.     |                    |
.     +-----[+BATTERY]-----+

Looking at the photo, if you wanted to go with my suggestion it
seems to me the easiest way of getting there would be to rewire the
board, using cuts and jumpers, like this:


.     +--[JUMPER]--+--[LED>]--[R]--+
.     |            |               |
.     |            +--[LED>]--[R]--+
.  [SWITCH]        |               |
.     |            +--[LED>]--[R]--+
.     |                            |
.     +--------[+BATTERY]----------+

If you have the room on the wiring side of the board it\'d be an easy
matter to use surface-mount resistors, but if not it doesn\'t seem
that it\'d be that much more difficult to use 1/8 or 1/4 watters on
the component side of the board.

What do you want to do?

I'm not at all confident on working with PC boards like that, so I might actually make up a whole new unit, point to point construction, leaving the DOT it intact.

I wouldn't mind just installing the red LEDs and a new resistor in place of the old ones, and leave all wiring as-is. I don't mind if the LED's have a shortened life. Not too hard to replace, IMHO. Would I still go with the 47 ohm?

--
               Vbat - Vled
     R = -----------------------
          Iled1 + Iled2 + Iled3

working

I did this with a 56k, blew one of the LED's so I think I'll look at other options than modifying the DOT it. Maybe just getting a cheap flashlight and installing one red LED in that. What resistor value would I need if this flashlight runs on two AA or AAA batteries? I think I measured about 40 mA on one LED before they all went out. Might have been the one that burnt out.

At an earlier point, I believe you had the reasonable response that you'd perhaps just build the thing on perfboard from scratch, following the correct circuit (the one I left above) rather than the flawed one of the original light. You then appear not to have done that, but that makes a great deal of sense, as you might now realize, having tried and fried with the original.

2AA or AAA are 3V.

Typical red LED is 2V nominal. 20mA is typical safe current for generic LEDs. So you have 20mA and one volt = 50 ohms. 51 is close enough, 63 would give more margin of safety, but somewhat less light.

--
Cats, coffee, chocolate...vices to live by

Yes that's right. Much easier to install in an old flashlight lamp base.

Thanks, I still have the 56 ohm resistor I used when I fried the LED. If it tests good I'll use that. I also have the two other red LED's which work, so obviously, I'll be using one of them. Thanks for all your help. Greatly appreciated!

just make a smps boost converter and you won't need the resistors, and the battery will last longer too.

Here's a schematic of a "Joule Thief" circuit ... Should be plenty of room for that pcb you posted.

VCC + -------------| | | toroid | -----|---. ,---- | )|( | | )|( | | -------' '---|--| | | | | | | | | | | LED LED LED | --------->|-->|-->|------ | 1K | | | ___ |/ === |---|___|--|2N3904 GND |>

| | | === GND

I was looking at a circuit similar to that, might give it a try.

I made one using 4 leds, schottky diode, and BUZ11 mosfet. posted it on flickr Here's the link also have the ltspice simulation for it there...