I've been trying to figure out a design someone made for controlling a precision high current LED, and it's got me in a bit of a pickle. For the life of me, I can't figure out this exact op amp circuit config:

enter image description here

It looks at something between a constant current sink and a unity-gain buffer. I'm wondering if the design is using a voltage reference schem in the LM358 datasheet: enter image description here

The shunt used looks like it's configured incorrectly as well.

I need to adjust the circuit to match a new LED.

I found these links helpful for getting a partial understanding:

Opamp topology without a stable voltage reference

How does this constant current sink actually work?

Can someone help me understand what's going on here? R9 and the intended application lead me to believe this is set up as a constant current sink but I'm royally confused. Thanks for your help :)


2 Answers 2


It's designed to deliver a constant current through the LED when the transistor is on. When the transistor is off, no current flows and the LED is off.

In order to adjust the LED current, adjust R9 accordingly. Adjusting R17/R18 will work as well.


I figured out what I needed to know about the design problems:

  1. Excellent video that covers everything about current sink op amps: https://m.youtube.com/watch?v=ceQDIfWKSK4

  2. The original designer made a weird choice of using a 2.5 Vref shunt instead of a standard 1.24V one. This means to get a few hundred milliamps of current, a 20 ohm sense resistor needs to be used, instead of a more sensible 10ohm one. It also means the shunt has a strange configuration. (Current output is determined by the equation I = Vref/Rsns, where Vref is the voltage at the input of the op amp and Rsns is the 20ohm resistor.)

  3. The op amp is missing a low-pass filter and current limiting resistor used to prevent oscillations in this kind of a circuit.

  4. This circuit is incorrect most of all because its designed for a couple hundred mAh current output to the LED, while also sourcing all of that current from the op-amp directly. A better configuration is to use the op-amp to drive the Gate of the FET, and then a P channel FET for fast on and off switching of the LED. (LSO_LED)

  • 1
    \$\begingroup\$ Remember that answers float up and down based on votes and user sorting preferences so your "Thanks" at the start doesn't make much sense as your answer appears directly below your question in my page view. Use the @username syntax to ping a particular user. If this post is the correct answer to your own question then you can accept it to mark it solved. Welcome to EE.SE. \$\endgroup\$
    – Transistor
    Commented Nov 12, 2019 at 21:41
  • \$\begingroup\$ Okay, I understand. I'm new here so I'm still figuring out how things work. \$\endgroup\$ Commented Nov 13, 2019 at 23:27

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