I finally took the time to start maping out where I'll need to draw my Circuit Scribe ink lines. I also drew where I want to place LEDs and calculated resistor values. I may need to revisit some of this to use less resistors to lower the load on the circuit, but I'm not sure. I'll have to draw out a couple circuits and test.
The LEDs I have need a forward voltage of 3.2 and a max forward current of 20mA. With one LED, I've found that 3 80.6Ω resistors in series gives an appropriate brightness without hitting the max current rating. I'm hoping that the ~130Ω resistance I've placed in a few places will allow a similar brightness for two LEDs in series. Some individual LEDs are placed on the ~130Ω resistance rail as well, and for those I added two additional 80.6Ω resistors in series bringing the total resistance up to ~290Ω. I may have to drop my source voltage to 5V to avoid burning out some LEDs, or add more resistors.
Some calculations follow.
| Measurement | Value |
|---|---|
| Source voltage | 9 |
| LED forward voltage | 3.2 |
| LED max forward current | 20mA |
| Single resistor | 80.6Ω |
For single LEDs I calculated R = ((9 - 3.2) / 0.02) = 290.
I need R / 80.6 = 3.6 resistors. I rounded down to 3. I didn't experience any excessive heat with this configuration.
For two LEDs in series I used four 80.6Ω resistors; two in series, and those two in parallel with two more 80.6Ω resistors also in parallel. The total resistance is R = 1 / (1/80 + 1/80 + 1/(80 + 80)) = 32.
The power, then, is calculated as I = (9 - 3.2 - 3.2) / 32 = 8.1mA.
For driving a single LED off the 32Ω rail, we need to add additional resistors in series to limit the maximum forward current into the LED.
We already know that the resistance required for a single LED is 290Ω. From here, we can subtract 32Ω to determine how many more ohms of resistance we need to add. 290 - 32 = 258. We can then figure out how many 80.6Ω resistors are needed just with division. 258 / 80.6 = 3.2. That's 3 resistors.
Oops! I only drew two additional resistors, which gives us a forward current of about 30mA - about 10mA too many. I'll have to revisit that circuit, or simply use a third resistor. That would give this rail about 274Ω of resistance.
Next up, I should have the corrected circuits drawn out and I'll actually be able to draw them on my canvas!
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