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I am facing difficulties in understanding the concept of zener diode and problems related to it.

My doubt is, In many texts I have read that if the input voltage is less than zener voltage, the diode behaves as a normal diode. Now doesn't that means that it is a normal pn diode and now across the load the voltage drop must be 0.7V. But in many problems we find the Voc i.e. the open circuit voltage, such that we remove the zener diode completely in that case.

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the diode behaves as a normal diode. Now doesn't that means that it is a normal pn diode and now across the load the voltage drop must be 0.7V.

The drop across a "normal" diode is only 0.7 if it is conducting current in the forward direction.

If it isn't conducting current, the voltage across it can be anything less than 0.7 V.

The same is true for a zener that isn't conducting in zener mode. Its voltage can be anything from $-V_z$ to $V_f$ and it won't conduct current in either direction. If the voltage gets above $V_f$ it will start to conduct strongly in the "forward" direction. If the voltage gets below $-V_z$, it will start to conduct strongly in the "reverse" direction. If you google "zener diode I-V curve" you should find many examples of graphs that show this behavior.

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  • $\begingroup$ does that means say input voltage must be greater than 0.7 or even 1V to conduct as a pn diode $\endgroup$ Feb 2, 2023 at 7:09
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    $\begingroup$ @AnshulSharma, it depends on how much current and what kind of diode. There are many diodes where the forward voltage will go to 1 V if you pass, say, 1 A through them. If you want to conduct only ~1 mA, Vf is not likely going to be that high for an ordinary silicon diode. $\endgroup$
    – The Photon
    Feb 2, 2023 at 16:29
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    $\begingroup$ But the simple model we often use for hand calculations is that the forward voltage is 0.7 for any amount of forward current. $\endgroup$
    – The Photon
    Feb 2, 2023 at 16:30

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