How can I modify this circuit to sequentially turn on LEDs in parallel, while avoiding voltage drops?
1 = 1 Red LED
2 = 2 Red LEDs
3 = 3 Red LEDs
4 = 4 Red LEDs
5 = 5 Yellow LEDs
6 = 6 Yellow LEDs
7 = 7 Yellow LEDs
8 = 8 Yellow LEDs
9 = 9 Blue LEDs
10 = 10 Blue LEDs
11 = 11 Blue LEDs
12 = 12 Blue LEDs
Since there will be a voltage drop across the diodes, the LEDs at the start of the chain will get progressively dimmer, and increasing the voltage would blow the LEDs in shorter chains.
One modification I know I could already make would be to not use the first 4 diodes on the yellow LEDs, and the first 8 diodes on the red LEDs, as a minimum of 5 yellows and 9 blue will be active for each position. This still leaves a maximum of 3 diodes in each chain, which I still think would cause the lights to be much dimmer.
Another solution I can think of is to have multiple wires coming from each terminal of the switch, each with their own diode, which would ensure all active LEDs had the same brightness (I'm not sure if this would cause the whole chain to be dimmer than shorter chains or not). The downside is this would need 31 wires rather than 13, and that's only if I bridged together all the shortest LED chains of each sequence: individual wires for absolutely every position would mean 79 wires+diodes).
To put this in context, I want to create the health bar from Dead Space 3:
If you can suggest a way of improving the circuit, or totally re-designing it, it'd help me loads!