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Edited my last question in order to contain the work I've done.

I have to find the output voltage Vo related to the input voltage Vi knowing that the input values range from 0V to 50V. All diodes are ideal.

I started by assuming that D3 is reversed biased. This means that D1 is forward biased and D2 is reversed biased if I am not mistaken.

So, I found VA= 3V (6V/2) using the voltage divider of the 2 5kΩ resistances. I made all this process in order to be able to define the range in which D3 conducts.

So, knowing that VA=3V, when D3 is OFF, I can say that D3 conducts when Ui>3V.

Then I got the range [3V,6V] where D3 and D1 are ON and D2 is OFF and found the output voltage related to the input voltage using the Millman's theorem.

Then I got the range [6V,20V], in which only D3 is ON , so again using a voltage divider on the simplified circuit with the 2,5kΩ and the 5kΩ resistances I found Vo=(2/3)Vi.

Finally I got the range [20V,50V] where both D3 and D2 are ON and D1 is OFF and again using a Millman's theorem found the output voltage.

However I don't know if this is the right way to approach this problem and i am not sure if my process is right. Any help is appreciated! Thanks in advance!

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  • \$\begingroup\$ A wall of text is very hard to read. Put your thoughts in paragraphs. \$\endgroup\$ – StainlessSteelRat Apr 18 '18 at 19:13
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All diodes are assumed to be ideal.

When Vi > 3V, D3 started conducting.

When it started conducting, Vo = VA will be at \$(Vi + 3)/2\$ volts.

Which means, when Vi = 6V, VA = 4.5V. D1 does not stop conducting. It will stop conducting only at VA = 6V, i.e, when Vi = 9V.

After 9V, only D3 will conduct. When it conducts, Vo = VB = \$(2/3)Vi\$ volts.

Which means, Vi should reach 30V, for D2 to start conducting. After that both D2, D3 will conduct upto 50V.

Your ranges [3,6] [6,20], [20,50] therefore are incorrect.

You can learn behavior of these circuits by a simulator.

See here

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