0
\$\begingroup\$

I want to power a 5V white LED strip (50cm) from an Arduino (Leonardo, connected through long USB cable to my PC permanently). The idea is to have the LEDs dimmable through the Arduino pwm signal. I have trouble getting enough brightness from the strip.

  • When I connect the strip directly to USB I measure 300mA and get full brightness.
  • When I connect the strip to the Arduino output I measure 30mA and brightness is way too low.
  • When I connect the strip to a ULN2803A Darlington driver (connected to Arduino VCC (4.6V) and output) I measure only 18mA
  • When I connect the strip to the ULN2803A but to an additional 5V USB I measure 95 mA and get a brightness that would be acceptable for my purposes but requires an additional USB connection which I would like to avoid.

Browsing on internet (I'm just a hobbyist) I got the idea to use a boost converter to get a higher voltage on the LED strip. Would that be a good idea or are there simpeler ways to get more brightness?

ULN2803A datasheet

LED strip data: 3528 SMD 30LED power consumption 4W/m (I use 50 cm)

Schematic

The reason for me to think about a boost converter was because the USB is 5V and the Vin of the Arduino is 4.6V so I thought that this difference could be the problem.

\$\endgroup\$
4
  • 1
    \$\begingroup\$ No, a boost converter is not the right approach. Generally you should not power loads through the Arduino, but the main problem with your question right now is that it is not clear precisely what you did in each of these circumstances. You'll need to clearly draw each, and explain how you measured the current. It may prove practically more useful to measure voltage differences across the different parts of the circuit - if you find where the large voltage drop is (apart from the LED itself) you'll know where the limitation is. \$\endgroup\$
    – Chris Stratton
    Commented Mar 6, 2019 at 19:14
  • 2
    \$\begingroup\$ It would also be helpful if you would provide a link to the datasheet for your LED strip and for the Darlington driver. \$\endgroup\$
    – Elliot Alderson
    Commented Mar 6, 2019 at 19:23
  • 1
    \$\begingroup\$ Using a boost converter to apply a voltage >5V to a 5V-rated LED strip will almost certainly destroy the LEDs. Don't do that. \$\endgroup\$
    – Chris Fernandez
    Commented Mar 6, 2019 at 19:30
  • 3
    \$\begingroup\$ use a logic level n-chan fet, like the irlz44n to switch the low-side. \$\endgroup\$
    – dandavis
    Commented Mar 6, 2019 at 19:36

1 Answer 1

Reset to default
2
\$\begingroup\$

A better solution instead of a Darlington is to use an Nch FET with RdsOn*0.15A<=0.15V or RdsON<=1 Ohm with logic level 5V or Vgs(th)<1.5V. 100mOHm Ron FET's have many solutions with Vt<1.5V

Dan's IRLZ44NPBF can certainly do the job.

MOSFET 55V 41A 22mOhm 32nC LogLvl .TO-220-3

Generally, you look for a FET that is rated for 20x the current you need to get the low voltage drop at 4.5V Vgs.

\$\endgroup\$
2
  • \$\begingroup\$ I don't fully understand your answer (RdsOn? Vgs?) but inspired by your original answer I found people using a PN2222A but the IRLZ44NPBF would be better? \$\endgroup\$
    – Emile
    Commented Mar 6, 2019 at 20:11
  • 1
    \$\begingroup\$ Yes,, RdsOn or simply Ron is the FET resistance at some gate voltage 2 to 3x the threshold Vgs(th).. This Ron which is far lower than the BJT that needs 10% base current, and thus can be voltage switched easily with low Voltage drop \$\endgroup\$
    – D.A.S.
    Commented Mar 6, 2019 at 21:07

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.