# LED Schematics - Help With Checking Math Please

#### CosplayHelp

Dec 6, 2022
4
I am working on a diagram, and I would like to get my math (and feasibility of my project) checked. I originally had it connected to an arduino nano, in order for switches changed to play different sounds, however, I was told I would burn it out. So I am looking into other ways to do so. I just would like to get my math checked on this diagram if possible, and have some help on figuring out how long the battery would last.

I am trying to make sure that my LEDs have the correct resistor values, and would like to find out how much power is coming out back to the negative. I am planning this out first, then adding in the nano later once I figure out how much this is.

I apologize that my schematic is upside down, I am still learning the symbols.
The white diode resistor somehow set to the default when I exported it, but the value was 300 ohms each.

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
13,719
Got a question about driving LEDs?

Use a transistor to increase the current driving capability of your Nano.

find out how much power is coming out back to the negative.
Simply add the currents from all LEDs.

The resistors for the withe LEDs are much to big:

Each resistor drops 6 V (9 V from the battery - 3 V for the LED). 6 V / 4.7 kΩ = 1,2 mA. If you want 20 mA, then R = 6 V / 20 mA = 300 Ω

Your circuit can be improved a lot in terms of component count and power dissipation:
Using the above example, 6 V × 20 mA = 120 mW is dissipated in each resistor (assuming the corrected value of 300 Ω). Connect 2 LEDs in series instead:

The resistor is then R = 3 V / 20 mA = 150 Ω, power dissipated in the resistor is 60 mW. More of the powre is available for the LEDs

For the other colors you can even put 3 or 4 LEDs inseries. Only ever as many that there is some residual voltage for the current limiting resistor. Also take into account that the 9 V battery will quickly fall below 9 V, more 8 V.
A 9 V battery is, assuming you use the typical 9 V blocks) not a good choice. For that many LEDs the battery will quickly be drained. I suggest you use a high capacity powerbank (5 V output as is common for recharging phones), calculate the matching resistors and rebuild the circuit.

#### CosplayHelp

Dec 6, 2022
4
Got a question about driving LEDs?

Use a transistor to increase the current driving capability of your Nano.

Simply add the currents from all LEDs.

The resistors for the withe LEDs are much to big:
View attachment 57219
Each resistor drops 6 V (9 V from the battery - 3 V for the LED). 6 V / 4.7 kΩ = 1,2 mA. If you want 20 mA, then R = 6 V / 20 mA = 300 Ω

Your circuit can be improved a lot in terms of component count and power dissipation:
Using the above example, 6 V × 20 mA = 120 mW is dissipated in each resistor (assuming the corrected value of 300 Ω). Connect 2 LEDs in series instead:
View attachment 57220

The resistor is then R = 3 V / 20 mA = 150 Ω, power dissipated in the resistor is 60 mW. More of the powre is available for the LEDs

For the other colors you can even put 3 or 4 LEDs inseries. Only ever as many that there is some residual voltage for the current limiting resistor. Also take into account that the 9 V battery will quickly fall below 9 V, more 8 V.
A 9 V battery is, assuming you use the typical 9 V blocks) not a good choice. For that many LEDs the battery will quickly be drained. I suggest you use a high capacity powerbank (5 V output as is common for recharging phones), calculate the matching resistors and rebuild the circuit.
Thank you for the clarifications. I will go back through and reformat it and redo the math to a 5v power bank.

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