I have been working on a project for ACDC Constant LED driver with 1.2A, 36V specifications. While going through the internet, I found one of the circuit and couldn't understand the working the of the same. Few things I understood but I am not sure of those. Herein, I am attaching the PDF file I found.LED Driver

Another thing I was not able to understand is the how can change the circuit to work as per my specifications.

Can anyone help me understant the working of the same?


The LM358 is operating as a differential amplifier, with the feedback path going through D12, U2, U1, TX1, and the rectifiers and output filter. It modulates the primary such that the voltages at its two inputs are equal. The voltage at the left of R12 is lower than that at the right. The wiper of pot VR1 is adjustable between that lower voltage and the top of D13, which is 0.6 V higher. So U3A is comparing the voltage at the low side of the load to a voltage that decreases across R12 in direct proportion to the load current, plus an adjustable percentage of 0.6 V. Because R12 is 1 ohm, the math is simple; 0.6 V equates to 0.6 A, the max output current available if the pot is all the way to the left.

BTW, the pot pin numbers probably are incorrect. I don't know if this is a true standard, but it is ((very)) common practice for trimpots to be numbered like this:

1 - Counter-clockwise terminal

2 - Wiper

3 - Clockwise terminal

For this circuit you want the clockwise terminal connected to D13, so that turning the pot clockwise increases the output current.

  • \$\begingroup\$ As pointed by you that the maximum current that I can get through this is 0.6A. But supposing that I want to get to 1.2A, what components might I need to work / change to get to that? Should the transformer also be redigned as per the specificatiosn or just the passive component and diodes will be enough? \$\endgroup\$ – DodZi Mar 15 at 12:18
  • \$\begingroup\$ If I change the value of R12 to 0.5Ohms, will I be able to get 1.2A CC? \$\endgroup\$ – DodZi Mar 15 at 12:22
  • \$\begingroup\$ Maybe go for a smaller increase first, just to test the concept. Put a 10 ohm in parallel with R12 and see if the available current goes up 10% -ish. NOTE - Remote control power circuit design is dangerous. Update your will, and please do not burn down your house. \$\endgroup\$ – AnalogKid Mar 15 at 18:33
  • \$\begingroup\$ If not this, then what generally is the way to increase the current in such circuits? I have understood the primary side completely and found that nothing on that side can increase or decrease the current. Feedback from the secondary side is only responsible for this change. So, if for example someone in there circuit wants to increase the current, then whats the general approach? \$\endgroup\$ – DodZi Mar 15 at 19:16

It's pretty obvious how it works if you understand off-line switching power supplies, and if you don't then getting it working won't be so easy or safe.

Current reference voltage source and measurement resistance are D13 and R12 respectively, so maximum control current is Vf(D13)/R12.


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.