Hello! Here are the answers to your questions:


1. Why does the PWM signal come from the cathode to the anode?


Since the anode is pulled up to the supply voltage, the cathode must be pulled down by the PWM signal for the LED to turn on. When the LED is on, it drives the transistor and pulls it down to the level of the emitter (low). Otherwise, when the PWM input to the cathode is high, the voltage across the LED is not enough to turn it on, preventing it from driving the transistor. When this happens, the output remains pulled up to the level of the collector (high).


To summarize, the circuit logic is as follows:


PWM input = high ; LED = off ; Transistor output = high
PWM input = low ; LED = on ; Transistor output = low
Circuit logic: Active High


This circuit could also work if the anodes are connected to the PWM signals, and the cathodes are connected to ground. This would however, result in the following circuit logic:


PWM input = high ; LED = on ; Transistor output = low
PWM input = low ; LED = off ; Transistor output = high
Circuit logic: Active Low


The PWM signal was probably connected to the cathode instead of the anode in order to operate with an active high logic.


2. Why is a capacitor placed between the cathode and the anode?


Having a capacitor across the LED prevents the applied voltage from changing instantaneously. This prevents noise from falsely triggering the LED.


3. Why 12V is given externally to the anode?


I may need to ask for the datasheet of the optocoupler used in this design to answer this question. The required supply voltage for the LED is dependent on its forward voltage. Any voltage level (within the max ratings) that is greater than the forward voltage would be okay to use as long as the level (logic high) of the PWM signal matches that of the supply. A series resistor may be necessary to limit the current going into the LED.


Hope this helps! :)