I have been working on a small linear power supply, capable of regulated output between 2V to 30V, using the LM338 IC. For the most part it is a simple implementation taken straight from the datasheet.

I want to include a "power" LED on the board indicating when the board is live. The LM338 requires the input voltage to be ~5V above the output voltage. This means that the input power may vary depending on usage scenario from approx. 5V to 35V. I'm not sure what the best way to drive a power LED on the board given this wide ranging input voltage.

Apparently the 7805 5V regulator can tolerate up to 35V input, which would provide a 'known' voltage to drive an LED from. However this seems a little overkill/expensive and would probably have heat issues. I don't think I can use a large resistor either as at lower voltages the LED would not illuminate or be very dim? I did wonder about a constant current LED driver but could not find a suitable part.

I would appreciate some design suggestions on how to include a 'power on' LED in my design bearing in mind the wide input voltage range. Thanks

Clarification Edit: My power supply board has a three way header and 2-pin jumper to change the operating mode from either (A) fixed voltage or (B) adjustable voltage. For the fixed voltages, such as 3.33V, 5V, 12V, a piano DIP switch is used in conjunction with various 0.1% or 1% resistors. For the adjustable voltage operation, the jumper connects the LM338 ADJ pin through a 5K linear potentiometer.


2 Answers 2


Assumption: The requirement is for a "regulated power available" indicator, rather than merely a "mains supply on", since the latter would not be affected by the variable voltage of the power supply as mentioned.

LEDs are essentially current-driven devices, not voltage driven. As long as the supply voltage is at least as much as the rated forward voltage of the LED (plus any headroom for current regulation circuitry), and the current through the LED is regulated to a desired value, the LED will be lit at a constant intensity. Typical indicator LEDs commonly are designed for 20 mA current, but will work excellently at 10 mA.

The easiest way of obtaining a constant illumination from an LED across a wide range of supply voltages, is to use a constant current driver circuit.

This can be made using, for example, an LM317 as a constant current source:


Alternatively, use a constant current 2-terminal device such as the SuperTex CL220 or CL2, simply wired in series with the LED. In other words, it is as simple as using a current limiting resistor with the LED, just with one of these parts instead of the resistor.

  • \$\begingroup\$ When using an LM317 in the above 'current regulator' configuration, would the IC get hot in the same sort of way they do when working as a voltage regulator (i.e., will I need a heatsink if input is 30V)? Also, thanks for the link to the Supertex CL2, I had not seen these -- in TO-92 package form this would be similar cost but smaller board space than a LM317 or 7805 solution which is great. \$\endgroup\$
    – user31299
    Nov 6, 2013 at 14:49
  • \$\begingroup\$ @j-roc The power dissipation by the LM317 would be the difference in voltage times the current (10 mA if you follow my suggestion above). That's 270 mW of heat if you assume a 3 Volt drop on the LED, up to 300 mW if you short out the LED leads. Not really a huge concern, I suspect. The SuperTex parts are small wonders - I've sampled a few of that family of parts, and have never stopped swearing by them ever since. \$\endgroup\$ Nov 6, 2013 at 18:52

Given that the LM338 requires a minimum load current of about 5 or 10mA to work correctly, why not use this for the LED. Adding a PNP transistor as shown below should be OK I reckon: -

enter image description here

If the beta of the the PNP transistor is 400 or greater (BC557C) and R1 is in the order of 1k ohm, the introduction of the PNP will give rise a small offset error in the output voltage. In effect \$\frac{10mA}{400} mA = 25\mu A\$ will modify the effective R1 current from 1.25mA to 1.275mA incurring a static shift in voltage output of 2%. As this application is for a variable/adjustable power supply I don't see this as a problem.

  • \$\begingroup\$ This is an ingenious solution, I like it. It is less expensive than using another linear regulator. However regarding your comment about the 2% Vout shift, this is not ideal as I am using the LM338 in both fixed and variable modes. Please see my clarification edit above. \$\endgroup\$
    – user31299
    Nov 6, 2013 at 15:02
  • \$\begingroup\$ @j-roc what is the highest value of R1 in your fixed applications? \$\endgroup\$
    – Andy aka
    Nov 6, 2013 at 15:10
  • \$\begingroup\$ R1 is always 120R 0.1% for my circuit. I think maybe you mean R2 which is used to adjust the output. In the fixed mode, the highest value of R2 is 750R. \$\endgroup\$
    – user31299
    Nov 6, 2013 at 15:25
  • \$\begingroup\$ @j-roc No I mean R1 - this is the resistor that is "bypassed" by the base current and I said would give a 2% error when R1 is assumed to be 1k ohm. Now that R1 is always 120R, the error will drop to about 0.24% and that is pretty low. Given that Vref is 1.25 +/-4% it certainly should be regarded as miniscule and I would question the need for 0.1% resistors because no matter how good the resistors are there will be a variation that means your output voltage will be up to 4% different from your calculations. \$\endgroup\$
    – Andy aka
    Nov 6, 2013 at 15:37
  • \$\begingroup\$ OK, sounds like that isn't going to be an issue then. Thanks for the clarification. Whilst this is a great solution, after some consideration I have decided to go with the SuperTex CL2 IC suggested by Anindo above (a shame I can't mark both as accepted answers!). \$\endgroup\$
    – user31299
    Nov 6, 2013 at 20:53

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