driverledover-voltage-protection
I designed a small circuit to drive LEDs with a classical constant current source (transistor and opamp). Then I realized, that a colleague using this circuit might do something wrong and tries to drive a super high current through the LEDs (SW problems we all now). The LEDs would die. To avoid this I want to limit the input signal (the voltage at the non-inverting input). What do you think about this protection circuitry?
I think a one or two channel op-amp and a precision voltage shunt would be your friend here. for the shutoff, simply make a voltage divider that at 13.0V, would output an equal voltage as your voltage shunt. Connect that voltage into the one input of the op-amp and the voltage regulated by the voltage shunt to the other. Connect the op-amp output to the PWM dimmer. That will take care of the low-voltage drop out.
The "work" on this circuit will be primarily in finding a particular op-amp and resistor combination to create a dimming profile that you like. (I.E. does it dim the right amount at 13.2V, 13.3V, etc.) To be clear, I am using the term "op-amp" loosely to include comparators and similar op-amp variants.
Additionally, if you want to ensure that everything is 100% on at >13.5V you can use a 2nd op-amp and a second voltage divider similar to above but with reverse the hookups to the op-amp so that it switches on at higher voltages and off at lower.
The SC4509 feedback pin requires 0.2 volts across it to set the LED current. The current that flows thru the LEDs also flow thru a resistor (R1) that sets the 0.2 volts depending on the current flow.
You say your design needs 40mA and this is well within the limits for the device so, alter the resistor value to set 40mA instead of 30mA.
So is it correct assumption that the IC can be used in this design since the total power remain with in the limit?
No, this is not correct. If R1 has a value that produces 30mA then it will produce 30mA and not 40mA. It's a constant current device not a constant power device.
Best Answer
Per The Photon's comment, you might find something like this simpler
simulate this circuit – Schematic created using CircuitLab
where R3 and R2 are chosen to give the desired current at the maximum input voltage which your colleagues can produce. This, of course, means that you'll have to supply a higher input voltage to produce the same current level, but that may or may not be a problem. Certainly, providing voltages as low as you do seems a bit unusual.