Designed a 20 LED Flashlight. Opinions?

[gmork]

I designed this flashlight after playing with my first project and learned a few things about output, viewing angle, etc. Still learning though, so heres my idea to get some feedback. :?

Power Supply

9.6V 900 mAh Ni-Cd battery pack.

Individual LED Info

Style= 5mm 16000 mcd super pure white :twisted:
Voltage Drop= 3.4 volts
Current= 25 mA
Viewing Angle= 15 degrees

Project Info

10 banks of 2 LED's, with one 115ohm 1 watt resistor per bank.
The 2 LED's are wired in series and the banks are wired in parallel.
All mounted into a 6 volt flashlight housing.

-----------------------------------

R = V/I

R = (9.6 - 2*3.4) / 0.025 =
= 2.8 / .025
= 115 ohm resistor

-----------------------------------

I = V / R

I = 2.8 / 115 ohms

I = .024 or 24mA

----------------------------------

Total Current = 10 * 24 = 240 mA

----------------------------------

Do my calculations seem accurate? I know the battery pack voltage varies, as I have seen around 11V from it. Should I do the calculations using the higher voltage for the resistors?

R = (11 - 6.8) / .025
= 4.2 / .025
= 168 ohms

If I leave in the 115ohm resistors each LED will be drawing approx 37mA and using 4.2V. Still within the safe zone??? (probably not, eh) :shock:

Any thoughts would be appreciated.

I have drawn up a schematic, can post it if you want.

[/u]

 

[Russlk]

The 11 volt reading is right off the charger. It won't stay there long, so I would not worry about it. The power dissapated in the resistors is: .24X.24X115 = 6.6 watts, the power used by the LEDs is: .24 X3.4X2 = 1.6 watts. You are wasting a lot of power! A current limited switcher is a better design. There are ICs for driving LEDs, do a google search.

 

[FRED6298]

If the operating curent for the LED = 25 ma, The current is the same
for the two LED's in series. Then R = 9V/25 ma...> 360 ohm.

Ideally, the battery should last for 3.6 Hrs, given the capacity = 900 ma/hr
with a drain current =0.25 amps.

Is this acceptable for your application ? If not, there are more efficient
high brightness LED's out there !

 

[Russlk]

Fred, Gmork's calculations were correct, how did you figure?

 

[gmork]

The 11 volt reading is right off the charger. It won't stay there long, so I would not worry about it. The power dissapated in the resistors is: .24X.24X115 = 6.6 watts, the power used by the LEDs is: .24 X3.4X2 = 1.6 watts. You are wasting a lot of power! A current limited switcher is a better design. There are ICs for driving LEDs, do a google search.

You have me interested in building a better design, now to search and learn about Led drivers hehe

 

[gmork]

Hmmm had another idea, if I run 3 leds in series, at 3.2V from the 9.6V battery supply, I wouldn't need a resistor then right? And I could run several parallel banks with that. :?

R = 9.6 - 9.6 / .04= 0

Drawing 20 mA per led I could run 3 banks of 3 leds with a total draw of 180 mAh. More efficient.

 

[Russlk]

The problem is that the LEDs are very low resistance, so the current would be high enuf to burn them out. You have to have a current limit circuit.

 

[John Sorensen]

A switcher is a good idea, but as a minimum you should consider using a current source. That way you don't have to make gross and compromising assumptions about the size of the current limiting resistor, based on a battery voltage that will in actuality be every value except the one you assumed. You'll still need resistors in the parallel legs, but they can be much smaller, because they will be used only to balance the legs (to ensure current sharing) rather than for limiting current.

There's a lot of ways to skin that cat, but a LM317 in a TO220 package should be suitable, and it takes only one current programming resistor.

j.