Diode AND Gate - Voltage Drop

Hi! There's a schema of a diode AND gate attached. Could anyone of you, please, explain me why the voltage drop across the D₁ diode is so big compared to the voltage drop across the D₂ diode? Thanks a lot in advance!

 

It says that the output voltage (=voltage at the black node) is 0.6V, but I can't really see how they came to this result....

 

Oh I see, so the diode with 4.4V voltage drop is reverse biased then...? shame on me for not noticing that... And btw, when using the constant-voltage-drop model for a diode, does it also mean that the voltage drop across the resistor is also constant since it is connected with the diode in a series..? and to your question - i think i dont know what limits the reverse bias voltage across a diode.

 

I meant the following..: assume A=1 (i.e. 5V) and B=0 (i.e. 0V). when the voltage drop across the diode is constant (say e.g. 0.6V), then - according to Kirchhoff's Voltage Law - the voltage drop across the resistor has to be constant as well (4.4V in this case), independently of its resistance. The current will change with the change of resistance (I = U/R), but the voltage drop (U) will remain the same*. Is this reasoning correct?

* The resistance will of course have to remain in a certain interval, since the constant-voltage-drop model of a diode is restricted to a certain interval of current magnitudes (like 1mA to 10mA).

 

Hope I understand the notion of breakdown voltage correctly - it is the minimal voltage drop across a reverse biased diode that makes the diode conduct.

 

And one more question - if both inputs are 5V, then D₁ and D₂ diodes are reverse biased and the output is equal to (5V - voltage_drop_across_the_resistor) instead of 5V, isn't it?

So when creating these diode gates, should I use resistors with low resistance in case of an AND gate and high resistance in case of an OR gate?