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I'm using the LT3477 IC for operating up to 6 LEDs in series. The LED that I'm using has Vf = 2.9 V and has maximum current of 350 mA.

Circuit input power is 5 V@2 A (J1 and J4 on schematic).

There is a situation that the LED output port (LED_P in the schematic, J2) can be short to LED return port (GND, J3).

Also there is an option that only single LED will be connected to the output port (LED_P).

Issue #1: Short output = the circuit fails, the input current is over 2 A and drop the voltage bellow 3 V.

Issue #2: Connecting single LED, the current on that LED is over the designed limit (350 mA) and the actual current is 1.2 A.

I need to prevent those situations.

A basic solution for single LED connection can be adding a diode with Vf = 2.2 V (PN: RFV5BM6STL) between J3 and GND, so when connecting a single LED, the diode acts as additional LED.

This solution will not work on short output because the output act like there is a single LED.

Thank you for your help.

Bellow the actual circuit: Circuit Image

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  • \$\begingroup\$ What's the voltage at IADJx pins? If you leave them as shown in the schematic (i.e. IADJ1 = IADJ2 = 5V) the output current should be less than 350mA because the ref voltage for output current adjustment will be taken as 100mV. \$\endgroup\$ Apr 19, 2022 at 7:15
  • \$\begingroup\$ J6 is a header pins for potentiometer, the voltage range is 880mV to 0V \$\endgroup\$
    – Gal Magen
    Apr 19, 2022 at 8:42

1 Answer 1

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The circuit is a constant-current boost converter with OVP.

Issue #1: Short output = the circuit fails, the input current is over 2 A and drop the voltage bellow 3 V.

This should normally be detected by the IC as it has cycle-by-cycle current limit. Make sure you selected correct sense resistors for the required limits. Sorry, I won't re-calculate them for you. The datasheet should provide enough info for this.

Issue #2: Connecting single LED, the current on that LED is over the designed limit (350 mA) and the actual current is 1.2 A.

This can't be prevented by the current design. Placing series diode(s), as you suggested, might be a solution but brings extra, unnecessary loss. Although the current exceeds the set threshold it's used to stop the switching. But there'll be no switching whatsoever because the output voltage is lower than the input (i.e. boost is stopped). So the single LED will be driven by the supply (5V) directly.

The simplest solution is to use a buck-boost regulator which can operate as a buck or a boost depending on the input and output. Or you may want to use an external "smart" circuit to detect the load voltage and switch to CV driving (i.e. driving directly from the supply with a series resistor) or a buck regulator (way efficient) when the load voltage is lower than some threshold. Remember that this will bring extra complexity, size occupation and cost. So the better option is to use a buck-boost.

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