Hi,

I am starting an LED project for my aquarium where I will have 6 channels using MW LDD-1500H 1500mA CC Drivers powering 12no "3W" (3.4-4Vf 750mA) LED's per channel in a series/parallel configuration of 4 strings of 3 LED's. I am looking to put c. 375mA through each string (1500/4) but to avoid thermal runaway have been researching current mirrors.

I have found a technical note by ST which details a suitable balancing circuit 'LED balancing circuit 3' described on page 4 with a diagram and equations on pages 5 and 6. Link to note: https://www.st.com/resource/en/tech...led-balancing-circuits-stmicroelectronics.pdf

I am struggling to grasp exactly how the circuit is working alongside the CC and my maths isn't strong enough to go through all of the equations listed.

In the description it states:

If a LED (for example D1) in one string is open, the current through the string (D1 and D2) is zero. The base to emitter junction of Q1 behaves like a diode and a small amount of current passes through R1. The reference voltage Vr drops, the other base to emitter junction voltages drop, and less current flow through the other strings (see Equation 10).

Specifically I am struggling to understand how much current would be running through the three remaining strings with D1 open and where the remaining current / power would be dissipated given the CC driver will always kick out 1500mA.

Any help or pointers would be much appreciated.

Many thanks,

Steve

I am starting an LED project for my aquarium where I will have 6 channels using MW LDD-1500H 1500mA CC Drivers powering 12no "3W" (3.4-4Vf 750mA) LED's per channel in a series/parallel configuration of 4 strings of 3 LED's. I am looking to put c. 375mA through each string (1500/4) but to avoid thermal runaway have been researching current mirrors.

I have found a technical note by ST which details a suitable balancing circuit 'LED balancing circuit 3' described on page 4 with a diagram and equations on pages 5 and 6. Link to note: https://www.st.com/resource/en/tech...led-balancing-circuits-stmicroelectronics.pdf

I am struggling to grasp exactly how the circuit is working alongside the CC and my maths isn't strong enough to go through all of the equations listed.

In the description it states:

If a LED (for example D1) in one string is open, the current through the string (D1 and D2) is zero. The base to emitter junction of Q1 behaves like a diode and a small amount of current passes through R1. The reference voltage Vr drops, the other base to emitter junction voltages drop, and less current flow through the other strings (see Equation 10).

Specifically I am struggling to understand how much current would be running through the three remaining strings with D1 open and where the remaining current / power would be dissipated given the CC driver will always kick out 1500mA.

Any help or pointers would be much appreciated.

Many thanks,

Steve

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