To drive 6 LEDS in parallel

[vijayaummadi]

Hi,

In my project I need to drive 6 LED's in parallel with the 3.3V voltage supply. Can I drive all 6 LEDs in parallel with a single MOSFET? Each LED requires only 5mA and The gate of MOSFET will be driven by Micro-controller. How far this approach is correct? Please let me know ASAP.

Thanks in Advance!

Vijaya

 

[KlausST]

Hi,

6 x 5mA current should be no problem for a MOSFET. But in detail it depends on the MOSFET type.

Another question is the voltage of the LEDS and the voltage variation from LED to LED.
Usually you need a current limiting resistor for driving LEDs.
If the LEDs internal serial resistance is high enough you may connect all LEDs in parallel with one resistor.
But for equal brightness you may use a resistor for each LED.


We need more technical data from LED and FET.

Klaus

 

[vijayaummadi]

Yes Thanks for the reply

I am using Amber LED LNJ447W84RA1 and Mosfet BSN20-7. Attached datasheets for your reference. I want to use 6 LEDs in parallel configuration with Mosfet control by Micro and If I connect one current limiting resistor of 200 ohms; will the circuit is fine? Please guide me

 

[D.A.(Tony)Stewart]

2nd edition
The Vf of these LEDs is only 1.95 nom@5mA.
ESR= ΔV/ΔI= 20Ω below 5mA and 10Ω above , not 100Ω


Using the Vt or threshold voltage at 1mA V=1.86, I have estimated the ESR from above. To drop from 3.3V to 1.95, current limiting resistors may be selected based on ESR of 20Ω for 5mA.

If you put 6 LED in parallel , there is a low probability of thermal runaway. But above 5mA the knee of the curve varies for different batches and hence there is a wide variation in Vf min-max @ 10mA.

The safer design isolates diodes with a small R to normalize the variation in ESR. then one R may be used to change the current in all of them (7R's), or use 6 fixed R's to limit 5mA in each.

To drop 30mA from 3.3 to 1.95 requires 45Ω.

You may test 6 in parallel with one 45Ω resistor will see even brightness but I cannot guarantee the margin to thermal runaway* without more data from Panasonic.

* Thermal runaway is a mechanism when power semiconductors with NTC voltage drop are put in parallel. The device with the lowest ESR draws more current. If the power dissipation is enough to reduce ESR or Vf then current becomes hogged by device. Generally it is not a problem in low power devices and perhaps not here either, but without more calculations with supplier tolerances,

In 5mm 30 deg 590nmD LEDs, I get parts with over 15,000 mcd @ 20mA in bulk orders, whereas this design is only 6x 30 mcd @ 30mA.

Variations of this suggestion include adding 10Ω in series to each LED with 43Ω to feed this parallel array. The 10Ω raises the voltage from 1.95 to 2.00 and reduces the variation in each string. The 43Ω @30mA=1.3 accounts for the drop from 3.3 to 2.0

 

[Audioguru]

LEDs must all have their forward voltages the same (matched) if they are in parallel. How will you know if they are all from the same batch?
Buy many LEDs (thousands) and measure them all. Pick ones with the same forward voltage.

I have two white LEDs. One has a forward voltage of 3.0V and the other has a forward voltage of 3.29V. If I connect them in parallel and use one series current-limiting resistor then the low voltage LED is extremely bright and might burn out soon and the higher voltage one is dim. If I had one LED with a low voltage in parallel with 5 LEDs with a higher forward voltage then the lower voltage LED will probably instantly burn out because it will use most of the current.

 

[KlausST]

Hi,

the 270 ohms that SunnySkyguy correctely calculated are for individual current resistor.
For 6 LEDS you need 6 resistors.

Then i´d expect even brightness.

*********
If they really have internal 100 ohms resistance, then there is a voltage drop of 500mV at 5mA.
This should give good even brightness.

then you can use a 260 ohms / 6 = 43 Ohms single current limiting resistor.

*****
re calculation:
with 100 Ohms, 1.95V and 5mA the internal true LED voltage should be about 1.95V - 100 ohms x 5mA = 1.45V.
If this voltage varies 0.1V then the resulting current (proportional to brightness) should be (1.95V -0.1V - 1.45V) / 100 Ohms = 4mA.
This is 80% of the wanted 5mA. and 80% of the desired brightness. It will be hard to recognize.

I´d say: try it and see if a single resistor works. If you are not satisfied, then use individual resistors.

Klaus

 

[D.A.(Tony)Stewart]

TY Klaus, I edited my former suggestions.