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In a forward bias Diode, why is it that when current gets really large :

1) voltage drop across the diode does not increase a lot ( its because of logarithmic nature? and if so why is it logarithmic)

2) The gradient would represent 1/R hence as current goes up and V remains relatively the same (0.7-1v) this would mean the resistance would tend towards 0? why does this happen and how is it that the resistance is able to change like this. I would appreciate your explanations. Many thanks.

enter image description here I would appreciate your explanations. Many thanks.

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    \$\begingroup\$ With infinite current through the diode, you do get infinite voltage across it. It's just that the voltage grows a lot slower than the current. Precise details of why would require an understanding of solid-state physics. \$\endgroup\$
    – Hearth
    Nov 3, 2018 at 20:34
  • \$\begingroup\$ you sort of have it backward ... the current is dependent of the voltage because voltage is the parameter that is being controlled...... as forward voltage passes Vf, the current suddenly starts to approach infinity \$\endgroup\$
    – jsotola
    Nov 3, 2018 at 22:19

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Any Diode has its own barrier potential in forward bias that when the voltage on the two sides of the diode is less than that potenial, diode doesn't let the current to flow. This potential depends on diode material. For Silicon this potential is about 0.7 volts and for Germanium is about 0.3 volts. After that voltage applied to the sides of diode is greater than barrier potential, Diode acts like a wire and only voltage drop at the diode is just about 0.7 volts(or about 0.3 volts) and approximately doesn't change.

As i said above, if Voltage applied to the sides of diode is greater than barrier potential, then diode acts like a wire and then it's Resistance tends to about zero Ohm.

You can check the link below for more information about P-N junctions:

https://www.electronicshub.org/characteristics-and-working-of-p-n-junction-diode/

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  • \$\begingroup\$ The potential also depends on doping concentration, just not enough to really matter very much. \$\endgroup\$
    – Hearth
    Nov 3, 2018 at 21:13
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The junction is only a small region of the PN semiconductor.

To reach the junction, current flows thru doped silicon that also has resistance.

Think of the diode like this

schematic

simulate this circuit – Schematic created using CircuitLab

where the two resistors may be 0.1 ohm or even smaller, if the diode is rated to handle 100 amps (I have some of those, with 1cm-diameter multi-stranded wire entering the anode side, in my junk box).

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