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When a voltage(neglecting sign as I am considering reverse bias region) more than the reverse breakdown is applied then why diodes adjust themselves to reverse breakdown voltage? Please provide an intuitive answer. What is happening physically? What happens to the remaining voltage applied?

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    \$\begingroup\$ When you apply a voltage, irrespective of the diode's reverse voltage being exceeded, the same voltage occurs across the diode terminals even if 1,000,000 amps is caused to flow through the diode. \$\endgroup\$
    – Andy aka
    Jul 15, 2019 at 14:58
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    \$\begingroup\$ Do you understand what a depletion region is and how it gets wider under reverse bias? \$\endgroup\$
    – DKNguyen
    Jul 15, 2019 at 14:58
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    \$\begingroup\$ @Andyaka I think the OP is asking about the actual physics behind the zener IV curve, not a constant voltage model. \$\endgroup\$
    – DKNguyen
    Jul 15, 2019 at 14:59
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    \$\begingroup\$ If 1000000 amps flows through the diode, I don't think there would be a diode, only smoke. \$\endgroup\$
    – Voltage Spike
    Jul 15, 2019 at 15:02
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    \$\begingroup\$ It takes more energy for electrons to cross (or rather, mess up) the wider depletion region during avalanche breakdown which means more voltage drop, but this voltage change is very small with respect to the change in current which is why a zener explodes if you directly apply even a moderately higher voltage than breakdown directly to it. It requires too much current to increase voltage drop to eat up the extra voltage so you use a resistor to offload the extra voltage elsewhere. This is what lets you approximate the voltage drop in breakdown as constant. \$\endgroup\$
    – DKNguyen
    Jul 15, 2019 at 15:11

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