Hi

I recently finished my first project which was a lab power supply made from an old computer power supply. But something remains unclear to me.
I used two diodes to indicate its state - a red one connected to the violet 5v standby line (it lights up then the mains switch is on) and a green one which is connected to the gray power good cable in order for it to flash when i flip the turn on switch (connected to green wire). These diodes have 330 ohm resistors connected in series to them to limit the brightness.
When i turn the mains power on, the red LED obviously turns on. But when i turn it off, the diode dimms slowly before it turns off completly.
Does that mean that these big and dangerous capacitors in the suppy discharge through the red diode when the main power switch is turned off?

 

Hi
Does that mean that these big and dangerous capacitors in the suppy discharge through the red diode when the main power switch is turned off?

It means that the standby supply continues to run from the charge stored in those capacitors until their voltage is too low for it run any more. There will still be voltage across those capacitors.

 

Why do you think the capacitors are "big and dangerous"?

 

No the big capacitors are for the mains side to feed the oscillator, it's a good sign if it takes longer to go out, put a 5 ohms 10W resistor across the red and black wires.

 

My suppy has two 330uF capacitors rated for 200V and as an amateur i rather to act careful around them.
Is there a way for them to automatically discharge (in a controlled and safe manner) after the main switch goes off?

As a dummy load, i use two 10 ohm 10W resistors connected in parallel

 

There is, or should be, a resistor, or probably two, across them which will discharge them completely. It might take something like 10 minutes for them to get to really safe voltage. See R2 and R3 in the diagram below:

 

I looked through the schematic and my supply circuit and I think, sadly, that there is no much room to put anything.
What is the estimated time for them to discharge themselves to a safe level through leakage current ?

 

I would be surprised if there aren't resistors in there already.
Have you measured the voltage left on them when the red LED goes off? It may already be quite low.

 

In my case, if im not wrong, it is not possible to mesaure their voltage without disassemling the whole thing which i dont want to do.

 

Does that mean that these big and dangerous capacitors in the suppy discharge through the red diode when the main power switch is turned off?

Yes, but not with a Direct Electrical connection.

There is a auxiliary switched mode unit that supplies the Standby +5 V for the Main SMPS IC to power on. This auxiliary unit draws its supply from the rectified mains input supply (C1/C2 combination in the generic schematic) and discharges the main capacitors to a point where the voltage is insufficient for the Aux smps to operate.

There is no need for additional resistors to discharge the Main Filter Capacitors.