LED blinking with capacitor and schottky diode?

Question

I saw one circuit which connected one 100uF capacitor and one schottky diode as mentioned in the below circuit diagram. It has a negative logic rectangular pulse at cathode of the LED, which is providing ground for the LED.

I tested that circuit but I was not able to understand why they used a capacitor and schottky for this circuit for LED blinking, even that the LED can blink directly if we connected VCC to the LED anode. I attached one screenshot of the oscilloscope:

Node B  is yellow signal 
Node A  is green signal 

• What is the advantage of this circuit?
• Is that giving more voltage variation for the LED to turn on?
• Is that a LED flashing circuit?
• Is it increasing the brightness of the LED?

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After reading the all comments I had one idea, that it might be an LED Brightness circuit.

For that I closed all light of my lab, took a close picture of led, which I am going to share with you. I don't know if it is a right answer but it is quite enough to say it is a LED brightness circuit.

Answer 1

The point of that connection is to make the initial current through the LED high and then turn off, regardless of the low time of the signal. How and when that happens depends on the strength of the generated signal.

If the signal has been high a while, the voltage across the capacitor is very low, in the order of 0.2V or possibly less. (verify for yourself).

Then if the signal goes low, the LED will "see" its anode to be about 3V, so it will conduct what current belongs with 3V across it, as long as the signal can sink that current. The current flowing will charge the capacitor, so the voltage across the capacitor increases. This means the LED voltage decreases, which will make it conduct less and less current until it turns off.

Then when the signal goes high again the diode will very quickly source current into the capacitor the other way, to discharge it.

The time-base at which that happens can easily be a milisecond or even less, depending on the LED, so you might not be looking "fast enough", which is why you see the square wave. The last little slant in the high period is the residual current of the Schottky charging the capacitor a bit more. "Officially" the diode already stopped conducting, but below its Vf at 1mA it will still allow a a trickly current to flow, eventualy all the way down to single uA if you wait long enough.

So this seems like a trick to make a LED quickly pulse brightly on an otherwise slow signal used for something else. Or just to make it pulse brightly in the first place.

Answer 2

I have seen this only a few times as a method to indicate the device (mcu, PLC, PC etc) is still running.

The LED being on (with out the circuit) means just that. But if the mcu has stopped the LED would still be on... not a great "it's alive" indicator.

However with the circuit in place the LED blinks only if it's being driven high and low all the time. The LED should be driven by the main program loop. So if the mcu hangs, dies, is stuck in a loop etc the LED stays out after a short period.

I don't know how watch dog timers work but this is a visual equivalent.