What Causes Electric Field Breakdown in P-N Diodes?

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Electric field breakdown in P-N diodes occurs when a high reverse voltage causes the depletion zone to widen, leading to the breakdown of covalent bonds and the generation of electron-hole pairs. This breakdown allows current to flow due to processes like impact ionization or tunneling. The rectified wave observed on an oscilloscope may not align with the AC source if the oscilloscope is set to AC mode, which centers the waveform by removing the DC component. In full-wave rectification, diodes allow current to flow in both halves of the AC cycle, resulting in a continuous output. Doping determines the concentration of holes and electrons in the P-N junction, influencing the diode's behavior.
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i am not sure if this is considered electrodynamics under classical forum but oh well, here goes

i was reading wiki about p-n diodes and its that for reverse bias diode, because of the anode attracting the holes WITH the cathode attracting the electrons, the depletion zone widens and causes an increase in voltage barrier, so current will be harder to flow.

so when breakdown voltage is reached, the electric field due to the depletion zone breaks down and current flows. so what exactly does the electric field breakdown means?

also regarding rectified ac, i realize that when i rectify an AC source half or full wave, it seems that the oscilloscope shows

ac.png


with the blue part rectified wave, and red the ac transformer source.

so why issnt the blue part directly fitting the red curve? i have to adjust the position of the blue curve upwards on the oscilloscope to fit nicely.

also when full wave rectifying, what gives the extra rectified upward curve? is it the diode causing the original downward ac curve to flip up?

thanks loads sry for the long question
 
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in reverse biasing when we apply high reverse voltage then ,electron in the diode bonded by covalent bond gets free due to the brockage of covalent bond & breakdown ocures.the covalent bond are broke up due to the high electric field across the junction,the break down voltage depend on the doping of the semiconducter...
 
nikhil khatri said:
in reverse biasing when we apply high reverse voltage then ,electron in the diode bonded by covalent bond gets free due to the brockage of covalent bond & breakdown ocures.the covalent bond are broke up due to the high electric field across the junction,the break down voltage depend on the doping of the semiconducter...

why does having the covalent bond breaking up allows for electricity to pass through?

does covalent bonding has anything to do with the holes and electrons in the p-n semiconductor diode?
 
quietrain said:
why does having the covalent bond breaking up allows for electricity to pass through?

does covalent bonding has anything to do with the holes and electrons in the p-n semiconductor diode?

yes ,when covalent bond breaks then electron leave its place from the bond & thus creat a hole there in this manner when one bond break up then a pair of hole& electron is generated.
 
quietrain said:
so when breakdown voltage is reached, the electric field due to the depletion zone breaks down and current flows. so what exactly does the electric field breakdown means?

Breakdown can occur through various processes, but generally carriers are generated by impact ionization (avalanche breakdown), or because they can tunnel through the junction (like in a Zener diode). You can check these terms on wikipedia or google for more details.

so why issnt the blue part directly fitting the red curve? i have to adjust the position of the blue curve upwards on the oscilloscope to fit nicely.

Is your oscilloscope in AC mode by any chance? In AC mode, the oscilloscope will remove the DC component, so the rectified wave is "centered". If you set it to DC mode, the curves should match up.

also when full wave rectifying, what gives the extra rectified upward curve? is it the diode causing the original downward ac curve to flip up?
Yes, sort of. In the half-wave rectifier, the diode stops the current one way. In the full-wave rectifier, there's always a diode that let's the current through.
 
nikhil khatri said:
yes ,when covalent bond breaks then electron leave its place from the bond & thus creat a hole there in this manner when one bond break up then a pair of hole& electron is generated.

so why does the p-n diode already has a high concentrations of holes at one side and another set of high concentration of electrons at another side as explained in wikipedia?

is it due to doping?
 
caffenta said:
Breakdown can occur through various processes, but generally carriers are generated by impact ionization (avalanche breakdown), or because they can tunnel through the junction (like in a Zener diode). You can check these terms on wikipedia or google for more details.



Is your oscilloscope in AC mode by any chance? In AC mode, the oscilloscope will remove the DC component, so the rectified wave is "centered". If you set it to DC mode, the curves should match up.


Yes, sort of. In the half-wave rectifier, the diode stops the current one way. In the full-wave rectifier, there's always a diode that let's the current through.

ah isee thanks!
 
quietrain said:
so why does the p-n diode already has a high concentrations of holes at one side and another set of high concentration of electrons at another side as explained in wikipedia?

is it due to doping?

yes it depend on doping that which region will contain holes & which will contain electron ...
 
nikhil khatri said:
yes it depend on doping that which region will contain holes & which will contain electron ...

ah i see thank you!
 
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