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why diode voltage drop is almost always equal in reverse and forward bias

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matrixofdynamism

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The voltage drop in the resistors seems to be varying depending on the power supply and the other impedances in the circuit. But the one across he diode seems to be constant. A 10 ohm resistor can have a varying voltage drop across it but a zenner diode always has the same voltage drop across it always, HOW?
Also, in the reverse breakdown according to the graph an infinite amount of current shall flow through the diode right? This means that the diode shall be destroyed. But no such breakdown and huge current flows in the resistor.
I just need answers in terms of the internal working of the diode vs resistor. Thanks :).
 

In practical electrical terms, a diode is a self-adjusting resistance.

Its range of behavior encompasses a narrow volt range (a little to either side of its threshold voltage).

To pass a wide range of current, a diode adopts an enormous range of resistance values.

A particular volt drop goes with a particular current goes with a particular resistance. More or less.

So the diode assumes a value of megohms when dropping 0.3 V at 0.1 uA.

Then it assumes a value below 1 ohm when dropping 0.8 V at 2 A.

--------------------------------------------------------

Reverse breakdown does not take place instantaneously. As with forward biasing, reverse biasing develops gradually.

At the 50 volt level, it may span several volts, during which current flow goes from 1 uA to damaging levels.

When is it a damaging reverse current? When it goes above 1mA? 100mA? If it overheats the device? Don't know the answer to that.

And is a 50 V zener diode really different from an ordinary diode which has a reverse rating of 50 V? Don't know the answer to that one.
 
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When we speak about the "resistance" of a diode we always have to remember that such a semiconductor is a strong non-linear device.
That means we always have to distinguish between the STATIC resistance (ratio of dc values) and the DYNAMIC (i.e. differential) resistance (ratio of small changes around the bias point, slope of the V-I characteristic). Then, the function and the properties of a Z-diode can be understood very easily.
 

This is a very simplified sight (neglecting the series resistance) and does NOT apply to Z-diodes in the breakdown area.
I doubt, Zener diode is nothing but Diode. And exponential fenomina does apply to Z-diodes too.
Difference is Zener diodes are fabricated on more minority doped substrate to create reverse bias threshold.
By this minority diffusion , z-diodes reverse bias threshold are adjusted, and it acts as short (very low resistance path) after threshold, as normal diodes in forward bias. ( resistance may vary)
All the PN junction whether it is forward or reverse, have to follow exponential relation. [That is my understanding]
 


...z-diodes ... acts as short (very low resistance path) after threshold, as normal diodes in forward bias. ( resistance may vary)
All the PN junction whether it is forward or reverse, have to follow exponential relation. [That is my understanding]

Example: Breakdown voltage Vz=6 volts; current Iz=1mA : Static resistance Rz= 6 kohms. Short?
Avalanche Breakdown with exponential law? I don't think so.
 

Yes that depends on doping ( i already mention this), as if Vz =50V and static resistance Rz = 100kohm. {depends on application)
Now Point:try to provide more than 6 volts (reverse voltage) on Vz of 6V z-diode. It should not vary that voltage. What is that mean? Current increased with considerable amount.
 

As mentioned in posting #4 already: You have to distinguish between static and dynamic resistance.
 

may be I am not clear... I need to read more, trying to find reverse biased small signal resistance relation.
 

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