I've bought from Aliexpress 10 pieces of infrared SMD led with no datasheet but the folowing specs written by the seller on the bag:
Infrared 60 degrees 940nm 3W
1,3-1,8V
650-750 mA
I used my multimeter on the diode mode to measure voltage drop and I got 0.965-0.968 V
My power source will be 10 AA batteries of 1.5V (or 4 cell phone batteries of 3.7V) passed through a buck converter 13-24V to 12V 3A.
To estimate the value for the resistor I use the formula R = ([Source Voltage]-[Led Voltage Drop]x[Number of leds in series])/LedCurrent
My questions are:
1.Which values to choose for best led brightness without burning them and without wasting a lot of power on the resistor, as I got an interval for current and for voltage I measured 0.968V but the seller gave me the interval 1.3-1.8V
For Led Voltage Drop the correct value is something in the seller interval (1.3, 1.6 or 1.8V) or the measured value (0.968V)?
For Led Current should I choose the average of 700mA ?
2. Based on the reply for the above question I can establish how many leds I can put in series to draw on the 12V power source. For the extreme values I got between 6 and 9 led, but I can't predict if I get the best brightness or a weak one. I attached some of the options below.
Thank you for the answer. Using [Led Voltage Drop] = 1.5V and [Led Current] - 0.7A I got the following results:
Chain of 8 leds in series R = 0 ohm
Chain of 7 leds in series R = 2.142 ohm
Chain of 6 leds in series R = 4.285 ohm
Chain of 5 leds in series R = 6.42 ohm
Is it right that if I chain 8 leds in series I do not need any resistor, as calculated at point 1?
Is it correct that if I use the smallest resistor value, like in the 7 led in series chain I will waste the least power on the resistor (P=UxI) ?
Should I have a better ratio Brightness/[Dissipated Power] for a chain of 7 leds in series over two parallel chains of 5 leds in series (total 10 leds)?
Try to find the real forward voltage first.
With 2x AA battery power one LED with 2.7ohm 1W resistor in series and measure the supply voltage and voltage across the LED.
Then tells you that NONE of the LEDs may be from the same batch of LEDs from the same manufacturer. Therefore you need to measure the voltage drop for each LED and sort them so identical voltage drops are put into series strings.
I did the measurement. Voltage drop between 1.35-1.33V, the higher the source voltage - the higher the led voltage drop. The AA batteries are discharging fast. Source voltage gave me somethinng strange: while the LED was on Source Voltage = 2.79V (2 AA batteries). If disconnected from the LED Source Voltage = 3.1V.
Led intensity measurement gave 0.87A, and I think it's a mistake, because it does not match the computed value of 0.54A, and the LED didn't burn (rated for 0.75A max).
Voltage at resistors' end (4 pieces of 10 ohm in parallel - measured 2.7 ohm) 1.31 V. Resistors's voltage + LED voltage don't add up to source voltage: 1.31+1.33 = 2.64V, different from measured 2.79V.
I realize that I didn't take into account the source internal resistence.
I read somewhere that during use the AA battery voltage is decreasing from 1.6V to 0.9V. To have always the same voltage I use the LM2596S step down power suppy, to make sure that when the batteries are full I don't give too much voltage to the leds, and to drain the batteries to the last drop without any decrease in the led intensity. And also because the part was only 1.68$.
New measurement:
Source voltage = 4.93V from cell phone battery 3.7V through boost converter to 5V
Led voltage drop = 1.41V
Resistor = 5.5 ohm (from 2 resistors of 10 ohm in parallel)
Computed intensity = 0.64 A
Something in the system started to smell like burnt plastic, but the led is still working and got very hot. The boost converter is rated for 1A but unfortunately for now my resistors are rated only for 0.25W (all shops closed this time of year).
Should I make the computing for the series chains using [Led Voltage Drop] = 1.35V as in the previous test nothing melted?
I dont know how you calculated that. Intensity is not measured in amps.
So the residual voltage across the resistors is 3.52 volts, and the current 3.52/5.5 -= 0.64A
and the power in the resistors is .64*3.52 = 2.2 watts
so .25W resistors will get hot.
and to the LED 1.41 * 0.64 = 0.9W
Sorry, but in my physics lession in high school the current intensity was measured in Ampere units. I don't know how is translated in English - Amperage?
The LED is rated 3W, so it's a scam if at 0.9W it starts to melt. Or maybe the resistors/the boost converter produced that smell, I couldn't find the source.
Later edit: you were right - the resistor started to melt
Thanks, I didn't know this is available. I think the 500mA model will do (LDD 500h).
With this device I can chain in series as many LEDs I like up to 37 pieces (56/1.5V), with unknown voltage drop, and it will simply work, with nothing burning?
My bad.The bottom is indeed metallic and I can glue them on an aluminium band, but no more traces and no more PCB, a lot of wires from one leg to another.
