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a question I don't understand about the diode

circuit975

New Member
Hello everyone, friends, I am facing a problem and I could not get out of it as a result of long efforts

1719514640660.png



D3 Si Diode
D4 Ge Diode
has a semiconductor structure.



Asks us to determine and draw the change in the output voltage Vo when the input is given to this circuit. But I could not find how to solve this question. I have knowledge about diodes in general, but I could not find how to solve this question. I would be grateful if anyone can solve it by explaining it.
 
Both diodes are Zener diodes with a breakdown voltage of +5.1 VDC. The silicon diode has a forward voltage drop of about 0.7 VDC and the Germanium diode has a forward voltage drop of 0.3 VDC. It will be a non-symmetric clipper circuit.
 
Assuming Ge zeners exist...

View attachment 146288
Both diodes are Zener diodes with a breakdown voltage of +5.1 VDC. The silicon diode has a forward voltage drop of about 0.7 VDC and the Germanium diode has a forward voltage drop of 0.3 VDC. It will be a non-symmetric clipper circuit.
friends thank you very much for your answers, if we try to express it mathematically, si diode and germanium diode will only provide vout as much as the diode's own voltage, right? I hope I understand correctly.
 
friends thank you very much for your answers, if we try to express it mathematically, si diode and germanium diode will only provide vout as much as the diode's own voltage, right? I hope I understand correctly.
Zener diodes are normally used reverse biased and that is the condition in which their voltage rating applies.

When a zener diode is reverse biased, the voltage drop is approximately the rated voltage. For the 5.1 V zener diodes, when reverse biased the voltage will be about 5.1 V

When forward biased, silicon diodes, whether zeners or not, have a voltage drop of about 0.6 V.

When forward biased, germanium diodes, whether zeners or not, have a voltage drop of about 0.3 V.

Diodes do not really provide a voltage. They have a voltage drop when current is applied to them.
 

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