Electronic – LED Poster configuration

No, it doesn't work that way. The two LED strings are in parallel. That means the both see the 3.3V independently. Let's say the LEDs drop 2.0V, which appears is what you are assuming from your numbers. That means when SRC is 5V, there will be 3V on R1. 3V / 150Ω = 20mA as you said. However, when SRC is 3.3V, the same logic still applies. The LED will still take about 2V, so there will be 3.3V - 2V = 1.3V accross R1. 1.3V / 150Ω = 8.7 mA, which will be the current thru D1 when SRC is 3.3V.

If this extra 9 mA is of no consequence to your 3.3V supply, then you can simply leave R1 and D1 on the board. If the 9 mA matters, then you have to do something. Removing either R1 or D1 would do it.

Another thing to consider is that you don't need 20 mA thru D1 when debugging. You probably got that from the datasheet, which shows 20 mA as the maximum. That's a common value for small LEDs. However, it will still be plenty bright enough to see on your bench at 5mA. Since R1 is dropping 3V, it can be 600Ω to allow 5 mA to flow. Then at 3.3V you only get 2.2 mA thru R1 and D1. That's a lot less than 9 mA as you originally had it. If 2.2 mA is tolerable, then you need to do nothing more.

No, a 1.5 V battery is not going to light a string of LEDs. It can barely light some kinds of single visible LEDs. Also, you have the LEDs hooked up backwards in your diagram, so what you propose definitely won't work for several reasons.

Look in the datasheet for whatever LEDs you have, and find what forward voltage they will typically have when run at near full current. The voltage you will need will be about that times the number of LEDs you have in series. For example, let's say these are common green LEDs and drop 2.1 V each. 24 of them in series will require over 50 V, and 42 of them around 88 V.

However, you don't want to drive LEDs by controlling their voltage. You want to control the current instead, making sure that the system can supply a little more than the expected forward voltage. This is because LEDs inherently work on current, and because the current can change a lot with small changes in the voltage.

One way to achieve what you want is to use a boost converter to drive the string of LEDs. Instead of regulating the output voltage of the boost converter, you instead configure it to regulate its output current. A small sense resistor between the LED string and ground will give you a voltage signal proportional to the current thru the LED string. This becomes the feedback signal into the boost converter, which it then tries to regulate to a fixed value.

Take a look at the schematic of my KnurdLight where I did exactly that. In this case the circuit is lighting a string of 4 white LEDs, connected between P1 and P2 in the schematic. R6 is the current sense resistor. The voltage generated accross R6 is compared to a internal 600 mV reference built into IC1. 600 mV / 30 Ω = 20 mA, which is what these LEDs are rated for.

L1 is the boost converter inductor, and Q2 is the switch, which is driven by the microcontroller IC1. The extra parts R1, R2, Q1, R3, and R4 are there to prevent the boost converter from creating too high a voltage in case of the LED string getting disconnected or the wire broken. In that case, the current would be 0 and the feedback would make the boost converter keep increasing its output voltage until something would get damaged.