What type of switches are you using for?
The two pins bicolor LED (white/green) need to change polarity in order to change colors.
Without using a DPDT switch (Double Pole Double Throw ) you can’t use both colors of this LED.
Switches
You are doing the arithmetic based on assumed forward voltages. You can't simply choose them.
You must design for the worst case, i.e. that the diodes will exhibit the maximum quoted voltages.
In your first example, 2 x white plus 1 x 3mm green, the voltages you have to assume are the maximum figures; 3.3 + 3.3 + 3.5.
In the event that the particular examples of l.e.ds coveniently exhibited the forward voltages you would like, the difference between their sum and the battery voltage is so small that they wouldn't stay bright for long, unless of course you use a very large capacity battery so that its terminal votage doesn't drop too rapidly.
If you could use a 12V battery you would not run into the problems to which I refer.
I am fairly new to building electronics, I had assumed(and made an arse out of myself) ...
I can up to a 12volt batt with little modification to the design, and redo the math for the resistors at max power.
Is this the correct components list? (except the number of resistors)
Yes. I could post links to everything if that would help?
I see now there’snt any ''two pins bicolors LEDs'', you are using only ''3 pins LEDs''.
Try using a SPDT switch (Single Pole Double Throw) and LEDs connected in parallel. (each LED may use a series resistor = 470 ohms – 680 ohms )
:-D Some examples to connect 3 LEDs:
Let us suppose we have 9V.
Two white LEDs in series would need about 6.6V @20mA so the resistor value could be (9V-6.6V) / 20mA = 120R (nice value!)
The same applies for the two green LEDs and their series resistor could be 120R.
For one green LED, (9-3.3)/20 = 285R (that is 270R or 330R)
For one red LED, (9-2)/20 = 350R (that is 330R or 390R)
Obviously, the last two configurations suffer of relatively high power loss since only one LED is driven by 9V.
I'm shooting for only having 3 LEDs lite up, or when th switch is thrown it switches to a different three.I think you know that if you have 12V you can light 3 white (or green LEDs) connected in series. In this case, the resistor value would equal to:
(12 - 3*3.3) / 20mA = 105R (that is 100R or 120R) you can even use higher value to save power.
Also on 12V one can light 5 red LEDs (in series) with the same current:
(12 - 2*5) / 20mA = 100R , or any higher value to save power.
Note: if you use less number of LEDs in a branch, the power of the removed LED will have to be dissipated (lost) by the limiting resistor (in this case, its value will be higher than the ones calculated above)
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