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I'm trying to turn on a laser (3.3v - 4.5v, 100mw) using a 9v battery and LF33CV as voltage regulator(TO220 package, 3.3v output). After 20 seconds the voltage regulator is getting so much hot. I was wondering if it's natural and how can it be fixed?

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update1: Here is the test circuit in LF33CV datasheet, is it possible because of not using, capacitors?

enter image description here

update2: Laser draws 0.1 ampere when connected to a DC power.

update3: Original circuit (below picture) was changed with an equivalent circuit. enter image description here

update4: Thermal data: enter image description here

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    \$\begingroup\$ where did you find that "schematic"? \$\endgroup\$ – jsotola Oct 10 '19 at 9:17
  • \$\begingroup\$ sengpielaudio.com/calculator-ohm.htm \$\endgroup\$ – Bruce Abbott Oct 10 '19 at 9:21
  • \$\begingroup\$ @jsotola I draw it in altium designer, I didn't found the lf33 part so used tip42 which has similar foot print. Base is input of voltage regulator and emitter is output of voltage regulator. \$\endgroup\$ – Mehran Oct 10 '19 at 9:23
  • \$\begingroup\$ It's a bit of a train wreck going on here. \$\endgroup\$ – Andy aka Oct 10 '19 at 9:29
  • \$\begingroup\$ use 5V input or so \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Oct 10 '19 at 12:57
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Let's do the maths!

9 V - 3.3 V = 5.7 V dropped over the LDO

0.1 W = I * 3.3 V -> I = 0.03 A through the laser module

P = 5.7 V * 0.03 A -> P = 0.171 W dissipated in the LDO (approximatey)

UPDATED P = 5.7 V * 0.1 A -> P = 0.57 W dissipated in the LDO (approximatey)

55 C/W junction to ambient -> 55 * 0.171 = 10 C above ambient for a TO-220

UPDATED 55 C/W junction to ambient -> 55 * 5.7 = 31 C above ambient for a TO-220

I would guess you're drawing power for more that just the laser module from the LDO or the laser module is consuming more than 100 mW. I realise i am assuming that input power equals output power for the laser but i think it'd need to be a factor of 2 or more out before it'd explain why it gets too hot to touch.

UPDATED The LDO should reach about 55C which i would consider quite hot to the touch as it's ~18 C above body temperature.

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  • \$\begingroup\$ When laser is connected to a dc power it shows 0.1 ampere. \$\endgroup\$ – Mehran Oct 10 '19 at 9:30
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    \$\begingroup\$ @Mehran that makes more sense then, so, it becomes 5.7 V * 0.1 A = 0.57 W. then 55 * 0.57 = 31.5 C above ambient. If we call ambient to be 25 C then the case temp of the the LDO will be ~55 C which will be quite hot to touch. \$\endgroup\$ – hooskworks Oct 10 '19 at 9:35
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    \$\begingroup\$ At 55 C it's not too hot as far as the LDO is concerned but if you do want to cool it off then a small heat sink will help. Something like this one should be enough to (approximately) half the 35 C temperature rise so i'd expect 35-40 C for the LDO with a heat sink but no forced airflow. \$\endgroup\$ – hooskworks Oct 10 '19 at 11:22
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    \$\begingroup\$ The LF33CV data sheet does state that a 2.2uF output cap is "required" for stability. Obviously a heat sink will help to dissipate the heat. Also check if the laser diode's data sheet calls for a current limiter, (possibly a resistor as a minimum). \$\endgroup\$ – Nedd Oct 10 '19 at 11:39
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    \$\begingroup\$ @Mehran It isn't overheating. It sounds like the metric you're using is 'if my finger can touch it or not' which is not a valid way to determine if something is overheating. This specific voltage regulator has its own power limiting circuitry, so you cannot induce it to overheat. The worst that will happen is it will reduce the output current until the power dissipation is something it can handle. But regardless, however hot it is getting from the load you've described is no where near 'overheating' for this part. \$\endgroup\$ – metacollin Oct 20 '19 at 2:30

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