How this shortcut indicator works?

[amateur_25]

Hi
I thought a short circuit indicator would be nice to have on my new psu. My question is why in the attached circuit does the polarity of D2 get reversed when there's a short? I thought when there's a short it means there a path of negligible resistance so the circuit draws more power then it should so the components then burn out. 

[Simon]

It is not clear what you are asking, why are you under the impression that the polarity reverses. Are you basing this from examining the diagram or an actual measurement ?

[Nerull]

During normal operation, current is passing through D2, keeping T2 on, which then keeps T1 off. When the output is shorted, voltage drops too low to conduct through D2, which allows T2 to turn off and T1 to turn on, activating the buzzer.

The polarity never reverses.

[owiecc]

My question is why in the attached circuit does the polarity of D2 get reversed when there's a short?

When there is no short the anode of diode D2 is connected to positive supply voltage minus voltage drop on D1 (Vin–V(D1)).
During short the voltage on the output is close to zero therefore the diode is not forward biased any more.

I thought when there's a short it means there a path of negligible resistance so the circuit draws more power then it should so the components then burn out. 

Depends on the power supply. During short current will rise due to Ohm's law. This increased current can damage the power supply, cables, connectors, PCB traces and so on. However the current cannot rise to infinity in practical circuit.

Sometimes the power supply may be current limited and the circuit may be designed to handle this current. If you take the formula for power P=I²R if the current is limited to a reasonable value it will not destroy cables, connectors and traces because R in those components is very small.

BTW. The circuit is not good. During short the input voltage will be close to a single volt so the red LED and the buzzer will not work. D1 will probably burn. It should work ok if you replace D1 with a fuse.

[ciccio]

I agree with the above two posts, that were written when I was writing mine....
This is my humble opinion: It will not work
Le's examine it:
When the output is not shorted, there is some output voltage.
The green LED is on, D2 drives T2 base through R2, T2 collector is low, T1 base is low and so T1 does not conduct and the buzzer and the red LED are off.
Theoretically, if the output is shorted, T2 will not conduct (it's base will be low), so it's collector should go high and will drive T1 base.
The buzzer and the red LED will be ON.
The fact is that, if the ouput is shorted, the circuit will be powered by only 1 to 0.5 V (D1 direct drop) and so it will not operate (The supply voltage is not sufficient for the operation.
 
Maybe the circuit will work if it's supply (R1 and buzzer) is connected  not to the ouptput voltage, but to the unregulated voltage feeding the regulator transistor or IC.

Best regards

[Fank1]

Pay attention to Neurell

[amateur_25]

oh www.electroschematics.com/5487/short-circuit-indicator/ said D2 reverse biased. I confused it reverse polarity because that's what I've used diodes for.  So this should work if the buzzer and led had an isolated supply right? or still no?

Perhaps there's a good reason way all the bench supplies I have seen don't have a output shorted led?

[Clear as mud]

Perhaps there's a good reason way all the bench supplies I have seen don't have a output shorted led?

Actually, most bench supplies have a maximum current they will supply if the output is shorted, and have built in current-limiting to keep the supply from destroying itself when an overcurrent is present.  Some supplies can be set to regulate both current and voltage, with both being settable, and an indicator to indicate which is being regulated.  So in that sense, there are bench supplies with an "output shorted" LED.  The LED only indicates that the set current limit has been reached; the reason can be a short, an incorrect setting, or some component drawing more current than you anticipated.  When that happens, the output voltage falls to whatever level it takes to maintain the current at that level.

Some of the recent EEVblog videos have shown Dave using such a supply.  I think one of the recent "precision 1-amp current supply" videos shows it.