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I'm having trouble interpreting the thermal characteristics of the Microchip CL2N3-G Constant Current LED Driver chip.
Here's the datasheet: CL2N3-G Constant Current LED Driver 1

What I want to do is to power a single LED from a 60 VDC power source.
I'm planning on using the TO-92 package.

The problem I'm encountering is when I calculate the temperature rise of the CL2N3-G.
The spec states for the TO-92 package a thermal resistance of 132°C/W.
Here's the equations I'm using:

    60 V - 2.1 V (LED Vf) * 20 mA = 1.152 W  
    1.152 W * 132 (Thermal Resistance) = 158.8°C

This seems like a ridiculously high temperature just to power a single LED. Is this correct or am I missing something?

Thanks for your assistance and insights.

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  • \$\begingroup\$ There are simpler ways of powering a LED from 60V. Is this LED just an indicator for something? Do you really need a 20 mA LED? \$\endgroup\$ – mguima Sep 14 '20 at 6:20
  • \$\begingroup\$ I agree there are many other LED options. But the LED choice was not mine. I'm just working with the hand that was dealt to me. \$\endgroup\$ – DrJackK Sep 15 '20 at 15:15
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Actually, I think you are correct. That chip is designed to drive lots of LEDs in series, which is why such a high voltage is needed. If there's only one LED then the poor chip has to drop 58V, which as you say is over a watt of dissipation which is too much for that package.

If you truly just want to light one LED @20mA and the only rail you have is 60V, then any linear circuit is going to have to dissipate that 1.2W (60V * .02A) somewhere.

The only way around it is to add a switching regulator to generate a more suitable voltage, say 5V, and drive the LED from that. Unlike a linear regulator, with a switcher the power in more or less equals the power out, so you'll only draw about 20mA * 5V/60V = 1.7mA from the 60V rail (0.1W).

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  • \$\begingroup\$ Thank you for your time, analysis and suggestion. It's good to know I was on the right track all along. \$\endgroup\$ – DrJackK Sep 13 '20 at 18:44

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