I built a minimal LED driver circuit using a LM317LZ and a 12 Ohm resistor between OUT and ADJ (identical to e.g. Understanding this LM317 LED Driver circuit). According to the datasheet this should give 100 mA output current but in fact it it is around 60 mA. Input voltage is about 40 V, the LED is a COB LEB which has aprox. 32 V forward voltage at 100 mA. The voltage drop across the resistor is 0.55 V. What am I doing wrong?

  • \$\begingroup\$ What's the real voltage going to the LED when powered with your regulator? \$\endgroup\$ – JRE Dec 1 '20 at 15:12
  • \$\begingroup\$ @JRE about 29 V \$\endgroup\$ – Christian K. Dec 1 '20 at 15:13

LM317LZ has internal thermal overload protection. TO-92 can not dissipate 600 mW. (40-32-1.25)*0.1=0.675. Your chip overheated.

  • \$\begingroup\$ Good point. The datasheet just mentions that power dissipation is "internally limited". Thank you. \$\endgroup\$ – Christian K. Dec 1 '20 at 15:30
  • \$\begingroup\$ T0-92 can dissipate 150 mW in normal condition. Get in TO-220. If not enough, heat sink can be set \$\endgroup\$ – user263983 Dec 1 '20 at 15:40
  • \$\begingroup\$ @user263983 Get a switcher if you need to drop that much voltage! \$\endgroup\$ – winny Dec 1 '20 at 15:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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