Power supply to LED

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Olszanski

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I have a Christmas ornament which has a dark interior. I decided to buy a white LED (Voltage: 3.6V; Typical MCD: 1100; 20mA (max)) to illuminate the ornament. I attached it to 2 X AA (3 volts), but it is really too dim to light anything. I considered trying to build or buy a circuit booster, but have discoverd that 1) building it is beyond my ability and 2) I can't locate one to buy. I then bought an AA maglite bulb, which worked great but drained the batteries very fast (planned on having it on for ~ 2 hours/day). I'm back to considering the LED, hoping to boost illumination and get better battery life. I considered hooking it up to 3 X AA (4.5 volts) but I am afraid that would be too much. Any suggestions for this simple process of lighting up one 3.6V LED?! Many thanks!
Tony
 
You think about Christmas at the beginning of August?
You just add a resistor in series with the LED to limit the current to less than 20mA. The resistor will have 4.5V-3.6V= 0.9V across it. Calculate the resistance required for 20mA by 0.9V/20mA= 45 ohms. Use a 47 ohm resistor.
The battery will start at 4.5V but drop to about 3.6V fairly quickly, causing the LED to become dim. A 6V battery can have a constant current source circuit instead of a current limiting resistor then the LED won't dim.
 
I've seen some of the three terminal voltage regulators (LM317) wired as constant current regulators. Look at the datasheet for an LM317 and if it can be wired for 20 ma that might be a solution. It really won't be any more efficient than the resistor. Only think I don't know is the min voltage drop across the LM317.

The advantage of a current regulator, if you can find one to work with 3 batteries for a 3.6 volt LED, is that it will hold the 20 ma as the battery weakens (within limits).
 
An LM317 current regulator needs an input voltage at least about 3.25V higher than its output voltage when it is used as a constant current source. So powering a 3.6V white LED the minimum voltage is 6.7V which would be from a 9V battery (its voltage drops to 6V).

A transistor with two voltage-setting diodes at its base to make a current source needs at least 1.3V across it when it is a constant current sink. So the minimum battery voltage for a 3.6V white LED is 4.9V. A 6V battery's voltage quickly drops to 4.8V.
 
What about a simple variable width pulse generator running from 4.5 or 6 volts -PWM? That might actually be quite efficient. Pulse width might be adjusted via variable resistor to keep it simple. Maybe this can be done with a single 555 but if not then a 555 and an LM339 - or other simple arrangements ought to work.
 
Thank you for all your help! I've thought about the resistor before but wasn't sure how to go about it. I think I have enough information here to proceed, so thank you again. By the time the last response from Stevez came in, I knew the conversation was way over my head! I think I understand about 40% of that paragraph! Many thanks.
Tony
 
Tony-don't worry about not understanding. Sometimes we respond, as I did, with a thought that really isn't complete, knowing that someone else may have the knowledge to complete the thought - as Audioguru did.

Don't give up yet - someone may agree with the PWM thought and have a simple circuit for you.
 
Thanks! After a quick visit to my local RadioShack, I found a 4 AAA battery pack, so I picked up a 150 ohm resistor. I hooked everything up and seem to have a nice bright light. Interestingly, I had two LEDs and one of them, hooked up to the same system, is very dim -- must be a bad one?!
 
4 AAA battery cells make 6V. The 150 ohm resistor will have 6.0V-3.6V= 2.4V across it and therefore will have 2.4V/150= 16mA though it.
The actual voltage of LEDs is a little different so each one needs its own current-limiting resistor.
You probably connected two LEDs in parallel and the one with the lowest voltage became bright and hogged most of the current.
 
Oh, no, I'm sorry. I didn't connect the LEDs at the same time. I connected one and it was dim so I was dissappointed. On a whim, I took it off and connected the other and voila! it is bright!
Tony
 
LEDs are all different. Some are very bright because they are good quality, others are very bright because the lens focusses the light into one small spot.

White, blue and very bright green LEDs are destroyed by static electricity and most LEDs are destroyed if the reverse voltage across them exceeds 5V, or if the continuous current through them exceeds 30mA.
 
justDIY.

I built that circuit a while ago by looking in Google for interesting circuits.

It works fantastic. A blue or white LED keeps going on a (dead) AA battery till the voltage drops down to 0.356 Volts.

Amazing how the little toroid or slug can create such an efficient oscillator.

I think a similar principle is used in some of these newer multi LED torches.
 
I find that most of the 3000 - 5000 +, Millicandela LED's have already a very bright output at currents of a few milli amps. ( 1 to 3 ).

Try a white LED on 230 Volts with a 100 kΩ, 1 Watt resistor, you be suprised how bright that is as an indicator. (2.3 mA).

The 20 mA is required for the older type LED's, but especially for batteries you want to keep the current drain as low as possible.
 
RODALCO said:
Try a white LED on 230 Volts with a 100 kΩ, 1 Watt resistor, you be suprised how bright that is as an indicator.
Click to expand...
Two LEDs back-to-back or a single LED with a rectifier diode, plus the resistor.
The peak voltage of the 230V mains sine-wave is 325V. The absolute max reverse voltage for an LED is only 5V.
 
Hero999 said:
It's safe to omit the diode for red LEDs because they aren't destroyed by reverse voltage.
Click to expand...
Where did you get that crazy idea?
Of course they have a max reverse voltage rating. My ordinary red LEDs have 5V as their reverse breakdown voltage like all other LEDs.
 

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Crazy idea?

I know the maximum reverse rating is 5V!

The difference is that red LEDs have a charictaristic avalanche reverse breakdown similar to normal silicone diodes while blue and white ones tend to be destroyed, I don't know about the other colours though. I've powerd a red LED with a 47k series resistor from the 230V mains for hours and it didn't blow.
 
Not if you limit the current to a low enough level.

Suppose it breaks down at 8V:

And I'll calculate the peek power:

I = (230*1.414-8)/47000 = 6.75mA
P = 0.00675*8 = 54mW

Now 54mW isn't going to cook the LED, and that's only the peak power every half cycle.
 
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