I have connected 64 led strips and 8 shift registers (connected to 64 sensors) to 5V and ground of arduino. Voltage received at led strip is 3.1V and at registers is 3.2V. I plan to add an led display as well a small buzzer to it. Can I connect it to the same 5V? What should I make sure in this in terms of voltage supply so that there is no less supply as well loss of power? What if I add a 9V battery instead?

Note: I am just a beginner in this topic.

  • How big are the LED strips? You do know you can't draw more than 500-800mA from the 5V pin, don't you?
    – Majenko
    Commented Apr 11, 2022 at 15:24

1 Answer 1


I guess you are currently powering all this through a USB port on your PC. You are lucky if you haven't fried anything on the Arduino. When the 5V pin goes down to 3.1V this means that either you have fried the power electronics or you are currently totally stressing the power supply out. Power supplies, which are overloaded often drop the voltage significantly, or just fail outright.

All the LEDs, shift registers and sensors draw current. With this much LED strips you can easily get into the territory of many amps of current needed for driving them (depending on the number of LEDs and the current draw of each LED). A PC USB port will only provide 500mA by default. An Arduino can handle a current output from the 5V pin of about 1A, including the current, that the Arduino itself needs. So lets say 800mA. When you provide power through the barrel jack this also depends on the voltage of your power supply. The linear voltage regulator dissipates the voltage difference to 5V as heat, so supplying a higher voltage means more heat, means it can handle less current.

What to do now:

  • You need to power your LED strips, shift registers and sensors directly from your power supply. The current for them should not go through the Arduino. If you use a power supply with a higher voltage than 5V, you need to buy a voltage regulator for 5V that is capable of providing the total needed current (+ at least 20% headroom to be safe and reduce heat). Go for a switching "buck" converter type, not for a linear converter. They are way more efficient and provide less heat. There are ready-to-use modules available on the typical sides (for example the one named like the biggest river). Often they have a small potentiometer on them, which you can spin with a small screwdriver to set the wanted output voltage (here 5V).
  • Make sure, that the used power source is actually capable of providing the total needed current. For this you first need to calculate how much current the project will draw. For example: Standard addressable LED strips with WS2812 will draw 60mA per LED when on full brightness white. Multiply with the number of LEDs to get the total current for them. GO through all components of your project and write down the current draw (look at the datasheet for such info). Add it all up. Then get a power source (and if needed voltage regulator) that can handle at least 20% more than that.

What if I add a 9V battery instead?

As written above you will need an external voltage regulator for your project. Also: If you mean the standard 9V block batteries, then just forget it. They are not meant for high power applications and can only provide very low current. You should look in the direction of LiIon or LiPo batteries. When you use them please also buy a fitting charger module (since these are rather particular about how they want to be charged).

But before choosing a battery you first need to know how much current your project draws. Then you need to set a time range how long the project should run on the battery without charging. Then you can choose a battery based on that.

  • Then as per your suggestions, total current supply or Icc of each component used in this project adds to 4436mA. Should I multiply it with 4 to be able to run the circuit for 4 hours? Commented Apr 17, 2022 at 13:33
  • Yes, then you have the needed battery capacity in mAh. Also adda bit headroom on top
    – chrisl
    Commented Apr 17, 2022 at 14:13
  • By headroom on top, do you mean some extra mAh of battery for better life? 5V needs to be supplied to the circuit which buck converter will do the job. Is output current of battery a factor too or just voltage and mAh are required to decide a battery for this project. Commented Apr 18, 2022 at 2:43
  • Yes, I mean some extra mAh. You need a buck converter which can take the batteries voltage, can convert it to 5V and can provide the needed current. Thats enough info to find a fitting one. You can even use multiple of them if you don't find one which this current rating. Make sure its current rating is higher than needed (you don't want to drive it at its limit the whole time).
    – chrisl
    Commented Apr 18, 2022 at 8:06
  • About battery current: you need to transmit power, meaning voltage multiplied by current. With lets say 5A at 5V you have 25W to transmit. When the battery is higher voltage than it need to provide less current. For example: if the battery is 10V, then for 25W you only need 25W/10V=2.5A. The battery needs to be able to provide at least that amount of current without dipping the voltage too much.
    – chrisl
    Commented Apr 18, 2022 at 8:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.