# Ba9s LED

Discussion in 'The Projects Forum' started by roscirc, Sep 13, 2012.

1. ### roscirc Thread Starter New Member

Apr 30, 2012
9
0
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

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Apr 5, 2008
18,875
3,707
Hello,

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

Bertus

3. ### roscirc Thread Starter New Member

Apr 30, 2012
9
0
They are connected directly to a 110V battery.

Apr 5, 2008
18,875
3,707
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

5. ### roscirc Thread Starter New Member

Apr 30, 2012
9
0
Hi,

So you are assuming that they are connected in series?

Apr 5, 2008
18,875
3,707
Hello,

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

Bertus

7. ### MikeML AAC Fanatic!

Oct 2, 2009
5,445
1,073
Whoa! That lamp is designed to work off the AC line voltage. It has circuitry in the base to take care of current limiting.