All leds will be in parallel, they have forward voltage of 1.2V and forward current is 100mA. The flashlight has a power pack with 3 AAA 1.5V batteries, but I converted it to use 2-123 lithium batteries at 3.0V each. Will this voltage be ok? I will be using it with IR photography, and want as much IR light as i can get out of it. (invisible light)
I obviously do not know much about electrical currant, so I apologize for such a basic question. There are no resistors or other components in this circuit.
Steve
Is this an existing circuit or a proposed circuit because if you power up the circuit you described you'll destroy all the leds immediately.
This is an existing circuit. A small flashlight with 28 white LEDs on a circular board, all in parallel, using 4.5v. I was just going to desolder and replace the LEDs with the IR LEDs. Since I am electronically stupid, I knew enough to ask this question. Can you guide me to a correct voltage range? I don't know if I can manage a resistor on the board, it would have to be tiny in size.
Thanks...
Need a link for the flashlight vendor website and a closeup photo of the circuit bd.
How did you get the forward voltzge and current ?
Ningbo Nangang Electric Co. Ltd
28 Led Aluminum Flashlight
Model Number:NG-9104
battery specs (CR123A)
3V/ 1500mAh
IR LED SPECS (1 LED)
High intensity
Perfect for security systems and electronic gates
Forward voltage (V): 1.28V, rated at 100mA
ITotal = 28 x 0.100 A = 2.8 A
Runtime = Battery mAh / Load current = 1500 mAh/ 2800 mA = 0.535 hr = 0.53571 x 60 min/hr= 32 min.
I'm curios about the resistance of the two spring contacts in the flashlight. Do you have a multimeter (DMM) that you could use to check to see if the resistance is more that one ohm ?
The leds in the flashlight are not IR leds, they are white leds, which probably draw 20 mA each, which gives a total current for 28 leds of 560 mA. The CR123 battery with a mAh rating of 1500 could power that for almost 3 hours.
White Leds Specs
Figure 7 shows the current-voltage curves from a group of randomly selected white LEDs. Applying a voltage of 3.3V to these LEDs (upper dotted line) produces forward currents in the range 2mA to 5mA,
http://www.maximintegrated.com/en/app-notes/index.mvp/id/3070
As stated, the forward current of the white leds when powered by a 3V battery is about 2 mA
28* 2 mA=56 mA (0.056 A)
Runtime of 28-LED flashlight with CR123 battery rated at 1500 mAh = 1500 mA /56 mA= 26.7 hours
Based on the information provided, it appears that the current limiting is accomplished by using a battery with a voltage that prevents the white leds from drawing more than 2 - 3 mA, hence making the standard current limiting resistor unnecessary.
The same configuration with red leds might yield different results altogether.
The situation is drastically different for the IR led (5mm)
http://roithner-laser.com/datasheets/led_div/IR333C.pdf
SEE Fig. -4 (Forward Voltage vs Forward Current
at 2.8V the 5mm IR led draws 2.6 A (because the forward voltage is 1.2 V instead of 3.2 V)
At this point you should stop and stare at that curve and ask yourself ;
"Why does the curve STOP at 2.8 V ?" (that's funny. It's ALMOST as if it CEASED TO EXIST AT THAT VOLTAGE)
That's exactly what that means. If vaporizes the junction at that voltage. If you connect an IR led directly to that battery it will be toast before you can say "puff".
Just for the record,
if you replace all 28 white leds with IR leds, and power them with 2.8 V they would draw 28 * 2.8 A =78.4 A
Needless to say you wouldn't be able to find a battery that would fit in your flashlight that could supply that but there are many
RC Lipo batteries that could. Obviously 2.8 V (let alone 3V) is not the right voltage for an IR flashlight because you need 1.2 V, NOT
3 V. This is not a problem electrically because a 1S (one cell) Lipo battery has a voltage of 3.7 V
Since you need 1.2V then to get the 3.7 V down to 1.2 V you need a current limiting resistor.
Since the total current for a 28 IR LED flashlight = 0.100 A (100 mA) * 28 =2.8 A, then the dropping resistor would be
RDropping = (Vin-VLoad)/ILoad =3.7V-1.2V)/2.8 A = 2.5V/2.8 A = 0.89 ohms
PR= IR*VR=2.5 V * 2.8 A = 7 Watt
5 ohms /6 = 0.83 ohms ==> SIX 5 OHM /2 W Resistors in PARALLEL = 0.83 ohms /12 Watts ( close enough)
GOOD NEWS
Your flashlight will work with the following components without burning up the leds.
Battery = 1S 25C Lipo (3.7V fully charged)
Resistor = 0.89 ohm/7 Watt (SIX 5 ohm /2 Watt resistors in PARALLEL = 0.83 ohms 12 Watts (close enough)
Total Load current = 0.100 A * 28 = 2.8 A
BAD NEWS
If you followed your original plan to desolder the white leds and replace them with IR leds,
it would try to draw 78.6 A and fry all the leds AND cook the battery in less than 1 ms.
You won't find a round battery that can source 3 A at 3V that would fit in your flashlight.
Here's a battery you could use: ($25)
The Parts List
28 5mm Hi-Output IR leds @ $2.5/each = $70
1 1S 65C Lipo battery = $25
QTY: 6 5 ohm / 2W resistors $3
$98 => ~ $100
Thanks. I guess I will scrap this, and just buy an IR flashlight for a lot less.