0
\$\begingroup\$

For LCD, I have to make back-light voltage provision.

As per below image, LEDK1 to LEDK4 pins are internally grounded in LCD module itself.

But I have to add series resistor to limit current.

LED Pins

LED specification in LCD is as follows:

Blacklight

Is it good option to have common resistor to all LED's? What value should I choose if VF=3.2V (typ), IF-25 mA (max) for each LED.

led

What value of If should I consider to put it in below equation: Rs= Vs-Vf/(If)? Is it for single LED (25mA) or for 4 LED's (100mA)?

\$\endgroup\$
1
\$\begingroup\$

Nothing changes. Calculate the resistor as normal (Vin- Vf) / If. The leds were chosen to match each other, and even if they are in parallel with a single resistor it is okay if you pay attention to the specs.

As it says, the remark is the the leds are in parallel, and they still give a 25 mA max current, so that's the maximum you want, not 4 x 25 mA.

\$\endgroup\$
  • \$\begingroup\$ Thank you sir..I have attached one more image..pls go through and suggest. \$\endgroup\$ – Electroholic Mar 2 '16 at 10:36
  • \$\begingroup\$ @electroholic the update shows additional notes that indicate I'm wrong, but spher shows that the schematic may already have a resistor for this. Check for R5, a 3.9 ohm resistor. \$\endgroup\$ – Passerby Mar 2 '16 at 10:54
  • \$\begingroup\$ Sorry My bad. I attached wrong link. Updated now. Please check. \$\endgroup\$ – Electroholic Mar 2 '16 at 11:07
  • \$\begingroup\$ @Passerby looks like the "Power Consumption" rating of 256mW confirms that it's 20mA per each parallel LED, 80mA total. With that said, I've just spent 20-30mins digging through Google results & datasheets, but I still can't find anything that states whether of not there's already an internal resistor on these LEDs. \$\endgroup\$ – Robherc KV5ROB Mar 2 '16 at 13:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.