XOR Gate Circuit From Diodes

The diagram I marked with yellow and blue shows A at V+ and B at 0V / ground.

If the inputs are reversed, so B at V+ and A at 0V, the upper right diode is forward biased because the transistor base at about 1.2V less than V+ and A is at 0V.
As I just explained above..

It's just the same principle as the lower right diode pulling the base to ground as it is drawn, with B at 0V; the blue line.

Either input can supply power to the emitter via the left side diodes.
Either input can pull the base resistor down via the right side diodes.

It has to have one of each for both power to be available and the transistor to turn on.

Thanks for the answer.
Acording to my knowledge the only thing that determines if a diode is forward biased or reverse biased is :the silver mark on it -is the negative side and the oposite side is the positive side.
So if you aplly positive voltage to the silver mark side and zero voltage to the other side it will not conduct current because it is in reverse biased condition.
And if you aplly poitive voltage to the positive side of the diode and 0 vols to the negative side
the current will flow fro the positie side t the negative side.
thanks a lot

rjenkinsgb said:

As I just explained above..

Click to expand...

Seems like maybe your talking, explaining to a bot.

shortbus= said:

Seems like maybe your talking, explaining to a bot.

Click to expand...

My thoughts are pretty much unrepeatable [DK?].. It's a good test of patience & keeping calm.

sxy said:

the silver mark on it -is the negative side and the oposite side is the positive side.

Click to expand...

Correct. The bar is the cathode, both on a device and on the symbol.
That is how they all work in the circuits & descriptions above.