Hmm, It seems that there is a little too much current going through those yellow LEDs. This is because the yellow LEDs are rates for only 2V as you said, instead of the 3V the White LEDs are. To compensate for this, you need to add some resistance to each line of yellow LEDs. To figure out the resistance, you need to use Ohm's law

**V = I R** Where

*V* is the

*voltage supplied minus the voltage used*,

* I* is

*current* needed for the LEDs, and

*R* is

*resistance*.

In this case the voltage supplied is 9V, and we use 6V (3*2V yellow LEDs)

9V - 6V =

**3V**
LEDs are typically rated for 20 mA or

**0.02 A**
Rearranging Ohm's law gives us

**R = V / I**
substituting out values, we get

R = 3V / 0.02A

**R = 150Ω**
This means we need to add a 150 Ohm resistor to each line of yellow LEDs. Now, because you will have 2 resistors in parallel, we reduce this to 1 resistor. To find the equivalent resistance of 2 resistors, we can use the product over sum method:

View attachment 334053
(150 * 150) / (150 + 150) =

**75Ω**
This means you can simple add one 75Ω resistor before your yellow LEDs to reduce the current to a tolerable level for the yellow LEDs.
Hope this helps!