Regarding diodes

[zms]

1). what would happen if d.c current was supplied to the input terminals of a bridge rectifier ? will this damage the circuit in anyway ?

2). Does a LED rectify a.c current ?

 

[Herschel Peeler]

1). what would happen if d.c current was supplied to the input terminals of a bridge rectifier ? will this damage the circuit in anyway ?

2). Does a LED rectify a.c current ?

1) Nothing special. The output is still the same.
2) Yes, it would but would only work at a low voltage and current. Like 4 V and 20 mA. And be a high loss rectifier dropping far more than 600 mV a diode would.

 

[davenn]

1). what would happen if d.c current was supplied to the input terminals of a bridge rectifier ? will this damage the circuit in anyway ?

This is a good and common thing to do to provide a bit of gear with reverse polarity protection. As it means the + and - leads from the power supply can be connected either way and the correct polarity comes out of the bridge for the gear.
You just have to keep in mind the voltage drop through the 2 diodes that will be conducting. around 0.7V each


Dave

 

[hevans1944]

Some folks use this as a method of "reverse polarity" protection. It doesn't matter how you connect the + and - DC supply to the input, the output is always between a positive cathode and and negative anode. Disadvantage is you add two forward-biased diode voltage drops in series with your DC power, about a 1.4 volt loss if any significant current flows.

Sure it does. It's a light-emitting DIODE. But why would you want to reverse-bias an LED by applying AC to it? Well, some pairs of LEDs in the same package are designed for this, so forward biasing one yields one color, forward biasing the other yields another color, and if you apply AC it alternately lights both LEDs and yields a mixture of the two colors. However, LEDs are not intended to be used as rectifiers. If the reverse bias voltage is high enough they will break down and destroy themselves.

 

[davenn]

Some folks use this as a method of "reverse polarity" protection. It doesn't matter how you connect the + and - DC supply to the input, the output is always between a positive cathode and and negative anode. Disadvantage is you add two forward-biased diode voltage drops in series with your DC power, about a 1.4 volt loss if any significant current flows.

giggle ... somehow I don't think you read my response

 

[hevans1944]

giggle ... somehow I don't think you read my response

I did read it, but after I had already posted. Isn't it wunnerful how great minds think along similar tracks?