1
\$\begingroup\$

enter image description here

I want to power four LEDs using a 12 V, 7 Ah SLA battery.

I connected four LEDs in series and connected them to the battery, and the light glows.

Can I use this for a long time or does it need some additional components? Will the big capacity battery harm the LEDs?

\$\endgroup\$
9
  • 1
    \$\begingroup\$ Can we have a link to the LEDs, or a photo of the LEDs wiring to the SLA battery? \$\endgroup\$
    – tlfong01
    Jul 24 at 14:27
  • \$\begingroup\$ it is taken from 220v ac bulb Philips 10w f6500 \$\endgroup\$
    – Sasi Kumar
    Jul 24 at 14:36
  • 2
    \$\begingroup\$ You need at least a series resistor to prohibit thermal runaway \$\endgroup\$
    – Jens
    Jul 24 at 14:45
  • \$\begingroup\$ it is taken from 220v ac bulb Philips 10w f6500. i have nt permanently connected just checked with 12v sla battery by using only 4 leds to match arround 12v .tried 5 leds in series,glowing but brightness is low.6 leds are not turning on.updated image \$\endgroup\$
    – Sasi Kumar
    Jul 24 at 14:45
  • \$\begingroup\$ how much value can I use \$\endgroup\$
    – Sasi Kumar
    Jul 24 at 14:51

1 Answer 1

2
\$\begingroup\$

Philips make their own LEDs. LEDs all have different forward voltages so they must have a current limiting circuit. Didn't you notice that the datasheet you posted and all other LED datasheets show a wide range of forward voltages from 2.6V to 3.2V? Then four 2.6V need no more than 10.4V (12V will explode them) and four 3.2V need 12.8V (they barely light with 12V).

\$\endgroup\$
1
  • \$\begingroup\$ Such LEDs typically have a fairly "soft knee" so forward voltage is highly dependent on current as well as temperature and color. One reference from Rohm shows an LED that passes 1 mA at 1.8V and a still "safe" current of 20 mA at 2.2V. rohm.com/electronics-basics/leds/led-circuit-configuration The 2.6 to 3.2V variation is also +/- 10%. However, I/V has a positive temperature coefficient, so thermal runaway can happen. \$\endgroup\$
    – PStechPaul
    Jul 24 at 22:59

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.