I was in the process of building a 2W laser diode driver and I realized after some debugging that I had a 3.3V constant voltage LM1086 chip instead of an adjustable voltage regulator. I intended to build a constant current source that is adjustable up to 1000mA. The following circuit replicates what I put together on the bench using an LM1086 for regulation. Ignore the LM317, I didn't have an TI LM1086 chip to insert into the simulation.

My wrong chip selection made me wonder, can you use a fixed regulator to create a constant current source or do you need an adjustable one?

LM317 Circuit Simulation

As an aside question is this approach in general a good method of driving a 2W laser diode or are there better circuit architectures?


Depending on the regulator you can indeed also use a fixed regulator as a current source. The only disadvantage is that the voltage drop across the shunt resistor (R2 in your schematic) will be higher.

That LM1086 looks suitable to me but it will drop 3.3 V (the regulator's set voltage) across R2. Variable regulators typically will output a smaller voltage when the ADJ pin is grounded (like 1.25 V for the LM317) and will therefore also drop less voltage across R2.

So depending on how much voltage the laser needs, you might have to increase the 10 V supply voltage. For a 10 V supply I guess if the laser's voltage is less than 10 V - 1.3 (dropout voltage) - 3.3 V = 5.4 V this should work. R2 would need to be 3.3V/1A = 3.3 Ohms, not that it will dissipate 3.3 W so use a 5 W resistor!!!. Also the LM1086 will need a proper heatsink.

  • \$\begingroup\$ Laser diode requires anywhere from 1.9-2.3 VDC to operate so it should be easy to supply enough headroom on the input side. What I'm taking away from your comment is that the only difference between the fixed and adjustable version of the regulator is the internal voltage reference? \$\endgroup\$ – Wired365 Oct 18 '17 at 14:29
  • \$\begingroup\$ The LM317 also has a low quiescent current (called adj pin current) of 100uA nearly independent of load current which makes it work better in the adjustable setup than old voltage regulators like the 80xx series. \$\endgroup\$ – τεκ Oct 18 '17 at 14:35
  • \$\begingroup\$ the only difference ... is the internal voltage reference? Sort of, internally in the chip, all use 1.25 V as a reference (it's the silicon bandgap voltage) the fixed regulators have an additional voltage divider to divide the feedback voltage such that the output voltage will be higher. \$\endgroup\$ – Bimpelrekkie Oct 18 '17 at 14:41
  • \$\begingroup\$ @Bimpelrekkie I just took a look at the functional block diagram and I understand a bit better. I guess I misspoke in that I meant the ADJ and GND pin serve the same purpose in both, but as you said one has an internal voltage divider to ensure a higher output than 1.25V. So the actual regulation portion of the chip is exactly the same it just ultimately changes your feedback voltage and the voltage drop across your current limiting resistor. \$\endgroup\$ – Wired365 Oct 18 '17 at 15:09
  • \$\begingroup\$ Yes, that is correct. \$\endgroup\$ – Bimpelrekkie Oct 18 '17 at 15:11


Consider that if the regulator output voltage is V and it's feeding a resistor R, the output current must be V/R. But since this is a linear regulator, the input current is Iout + Iq (quiescent current, from the datasheet), and any excess voltage is dropped across the regulator. So if you put the laser diode between the voltage source and the regulator's Vin, it will see a constant current.

If you were feeling clever you'd also realize that the voltage source and the laser diode are in series, and circuit elements in series can be rearranged without changing the behavior of the circuit, meaning you can put the laser diode to the regulator's GND and now the circuit looks suspiciously similar to the LM317 version.

That's because the LM317 is secretly just a 1.25V linear regulator.

  • \$\begingroup\$ So because it's fixed you have to move the diode to the other side of the regulator or can it still stay where hooked up after the current set resistor? \$\endgroup\$ – Wired365 Oct 18 '17 at 14:20
  • 2
    \$\begingroup\$ The current is no different at the input or output of the regulator so actually it does not matter where the laserdiode is. I would keep it where it is as drawn in your schematic. \$\endgroup\$ – Bimpelrekkie Oct 18 '17 at 14:24

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.