Hi All,

The LED in the datasheet attached is a Ba9S type (130V) which has internally an array of 7 LEDS.

I am trying to add a resistor to reduce the current into the LEDs.

These are the things that I can not figure out:

1. the array of 7 LEDs in asingle bulb is connected in series or parallel?
2. The 5.9mA refers to the current of the single LED of the whole array (if in parallel)?
3. Can I assume the internal resistor is 22kOhm (130/5.9)?

Thanks,

Ros

 

Hello,

Where does the voltage come from.
If it is directly from the mains, you are violating the ToS.

Bertus

 

They are connected directly to a 110V battery.

 

Hello,

In the calculation of your resistor, you did not take the voltage drop of the leds in account.
White leds will have a voltage drop between 3.5 and 4 Volts each.
So when the voltage drop is 3.5 Volts each, the resistor will be (130 - 7 * 3.5) / 5.9 = 17.88 KOhm (so probably 18 KOhm).

Bertus

 

Hi,

So you are assuming that they are connected in series?

 

Hello,

Yes, that is a practical way to connect them, but if one fails, all will go out.

Bertus

 

Whoa! That lamp is designed to work off the AC line voltage. It has circuitry in the base to take care of current limiting.