0
\$\begingroup\$

I'm trying to power a large bank of LEDs that require 5V at 5 amps. I plan to use a 12v hobby Lipo attached to a buck convertor to step that down. I also need to power a teensy microcontroller with the same battery. This is the buck convertor I plan to use https://www.amazon.com/gp/product/B01NALDSJ0/ref=ox_sc_act_title_2?smid=AFHAE9RJVUMB&psc=1

My question is will I be able to power the teensy and the LEDs off the same buck convertor or will they interfere with or put too much load on the convertor?

\$\endgroup\$
2
  • \$\begingroup\$ "I also need to power a teensy microcontroller" - Which Teensy? \$\endgroup\$ Commented Feb 18, 2021 at 23:45
  • \$\begingroup\$ @BruceAbbott Teensy 4.0 \$\endgroup\$ Commented Feb 18, 2021 at 23:53

1 Answer 1

1
\$\begingroup\$

It doesn’t sound like your load is very noisy, so you can get away with sharing your power supply, but you need to have enough capacitance on the output side to handle the huge transient current that 5 amps’ worth of LEDs are going to draw when they all turn on.

\$\endgroup\$
2
  • \$\begingroup\$ The LED strip is a bank of 144 and is individually addressable. I plan on changing their colors (but not their brightness) very rapidly. I'm not sure if this will create noise as I believe some LED colors pull less voltage than others. \$\endgroup\$ Commented Feb 18, 2021 at 23:57
  • \$\begingroup\$ Individually addressed LEDs typically have built in PWM drivers, so even if brightness is constant, a given brightness level will be achieved by pulsing the LEDs many times per second. This may produce some noise. The maximum change in current will still be the case where the LED's go from full OFF to full white(all colors) ON, so you should probably plan for that surge unless your design/programming ensures that case does not occur. If you want the option to do highly variable light shows with sharp ON/OFF like strobing effects, etc. It's an option to put the teensy on it's own regulator. \$\endgroup\$
    – K H
    Commented Feb 19, 2021 at 2:27

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.