Hello!

So i have two diodes together and volatage 1V, the upper diode should be open and fully conduct current, but the second diode (reverse-biased) acts like a resistor with very high resitance? There should be some current in circuit, but how to calculate it, is this current only the leaking current from the reverse biased diode? And if I would make the source voltage something like 0,2V then the upper diode would not be opened, because it takes ~0,7V to open a diode, so then there would be even less current in circuit?

napos

 

For the circuit you have drawn, the current would be restricted by the saturation current of the reverse-biased diode; i.e. Is in the Ebers-Moll equation. Recall this current is of the order of pA, so a reduction of your input voltage to 0.2V would mean that current around the circuit changes very little.

In essence, your circuit is completely useless!! But it is interesting to use this as a thought exercise.

For more information on diodes, refer to: AAC: Volume 3 Chapter 3

Dave

 

Originally posted by napos@Mar 22 2006, 06:59 PM
Hello!

So i have two diodes together and volatage 1V, the upper diode should be open and fully conduct current, but the second diode (reverse-biased) acts like a resistor with very high resitance? There should be some current in circuit, but how to calculate it, is this current only the leaking current from the reverse biased diode? And if I would make the source voltage something like 0,2V then the upper diode would not be opened, because it takes ~0,7V to open a diode, so then there would be even less current in circuit?

napos

[post=15322]Quoted post[/post]​

Hi you should not say the upper diode should be open and fully conducting but rather it is forward biased and fully conducting however no current will flow since thesecond diode is reverse biased (open circuit). (An open diode is non conducting)