# Help with reverse engineering an LED driving circuit

I am trying to reverse engineer an LED driving circuit. The schematic I generated from analyzing the board looks as follows:

The main circuit is this one: https://www.monolithicpower.com/pub/media/document/MP3202_r0.91.pdf

From the datasheet I understand that the FB pin is regulated to 104mV. The currently mounted LED needs a max. current of 140mA. However I am not sure if R7 is the only resistor influencing the max. current through the LED? If yes, this would result in 138mA? The input signal on the left is generated by a Microcontroller and only used for dimming?

Now I need to modify the circuit so the current is max. 60mA only. So is the only modification I have to do to change R7 to 1.75 Ohms?

I am sorry if this seems to be a silly simple question...

• Your assumptions are correct and raising D2 fakes a high Vfb so it attenuated which can be PWM controlled – Tony Stewart Sunnyskyguy EE75 May 24 '18 at 16:29

The lower input port dims (or turns off) the LED when high.

The higher current (assuming the input is less than 0.5V or so) is set by R7 only. Whatever that port is used for, it can only dim the LED.

So if you want the nominal maximum current to be 60mA, you need about 1.75 ohms for R7.

(there is a small, about +4% effect from the resistor divider, which I am ignoring, so pick a bit higher value for R7 - the tolerance of the chip Vref is about +/-10% )

Now I need to modify the circuit so the current is max. 60mA only. So is the only modification I have to do to change R7 to 1.75 Ohms?

The FB pin reference voltage is 104 mV +/- 10 mV so, if 60 mA is a maximum limit then you should use 114 mV as the reference voltage and hence the reference resistance will be 1.90 ohms. This assumes no R3, R4 and R5.

The other resistors (R3, R4 and R5) will tend to lower the voltage fed to the FB pin by 4.1% hence with R7 at 1.9 ohms you will get 4.1% more current than 60 mA so, R7 should be increased by 4.1% from 1.9 ohms to 1.98 ohms.

I'd use a 2 ohm 1% resistor if 60 mA is an absolute maximum value that should not be exceeded.

A positive signal present on D2's anode will tend to fool the chip into thinking that the LED current is higher than it actually is and therefore the chip will drive less current through the LED.