Is This Diode Circuit Analysis Correct?

[Drao92]

Hello everyone.
Can you tell me if i solve right this circuit?
D2 is on off mode so I2 is approximately 0,D1 is the same and D3 is on ON mode so the equivalent circuit is like AFTER.
Is it ok what i did?
http://postimage.org/image/mblq1dr7n/

 

[gneill]

If you model the diodes with a "typical" 0.7V turn-on potential, then you might want to begin by removing all the diodes and calculating the potential across R6. Will there be enough potential "across the rails" to turn on any of the diodes?

If a more realistic model for the diodes is to be used, then D3 is the only one that has the possibility of conducting (and it can pass a very small current prior to the "agreed" turn-on threshold).

 

[Drao92]

E is 10 V in the real circuit, that values are only the default values put by the software. So if E was 1V ( big chance of not opening one diode) i use the after circuit, i change the D3 diode with a wire and i calculate the voltage across R10 or R11, parallel - same voltage? Or i have to use the initial circuit, change every diode with a wire and i calculate every node potential then i check if the D3 diode opens?
Thanks very much for the answer. One of my friend tried to solve the circuit and he didnt took in consiederation the R6 (R11) resistor so i was a bit confused.

 

[gneill]

E is 10 V in the real circuit, that values are only the default values put by the software. So if E was 1V ( big chance of not opening one diode) i use the after circuit, i change the D3 diode with a wire and i calculate the voltage across R10 or R11, parallel - same voltage? Or i have to use the initial circuit, change every diode with a wire and i calculate every node potential then i check if the D3 diode opens?
Thanks very much for the answer. One of my friend tried to solve the circuit and he didnt took in consiederation the R6 (R11) resistor so i was a bit confused.

You would be better off removing all the diodes first and finding the various potentials to see what diodes might conduct. Add back the diodes that might conduct one at a time and recalculate as you go. Remember that a conducting diode still drops its turn-on voltage across it, so don't just replace conducting diodes with a "wire". A better choice (for analysis) is to replace them with a voltage source equal to the turn-on potential.