Doubts regarding semiconductors

  • #1
What would happen if in reverse bias of a pn junction diode voltage greater than/equal to breakdown voltage is applied?

Consider a zener diode with breakdown voltage 3V(just take as example).If we directly connect it with a voltage supply of 4V what would happen?I read that voltage across zener diode remains constant after breakdown. It means if we connect the 3V zener diode to 4V supply a potential drop of 3V would occur across zener diode, then what about the remaining 1V, where would it drop?If I am wrong and istead of 3V, whole 4V drops across zener then how could it give us constant DC voltage as it would change above breakdown?How could the zener diode be used then as a voltage regulator?

Please explain in detail I am having a lot of confusions regarding semiconductor devices specially in pn junction diodes.

Thank You.
 

Answers and Replies

  • #2
cnh1995
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I'm not an expert but here's what I think will happen..There will be some internal resistance associated with the voltage source. If it is sufficiently high, that remaining 1V will be the drop across this internal resistance. If it is very low, then you'll have almost 4V across the zener. For a practical zener, the v-i graph is not a vertical straight line after the breakdown region, but a curve. So, on that curve, whatever the current is at V=4V(almost), will flow through the circuit. It will be very very high and can damage the diode.
images?q=tbn:ANd9GcRSdwb9MvgQuhhdtXh3GD_aEr52amC1HDzKgsPrf4REyTl12mEKkUmIb7zW.jpg
 
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  • #3
I'm not an expert but here's what I think will happen..There will be some internal resistance associated with the voltage source. If it is sufficiently high, that remaining 1V will be the drop across this internal resistance. If it is very low, then you'll have almost 4V across the zener. For a practical zener, the v-i graph is not a vertical straight line after the breakdown region, but a curve. So, on that curve, whatever the current is at V=4V(almost), will flow through the circuit. It will be very very high and can damage the diode.
images?q=tbn:ANd9GcRSdwb9MvgQuhhdtXh3GD_aEr52amC1HDzKgsPrf4REyTl12mEKkUmIb7zW.jpg
Thanks for the help.
But then how could zener diode be used as voltage regulator because for different input voltage the output would be different?
 
  • #4
cnh1995
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Thanks for the help.
But then how could zener diode be used as voltage regulator because for different input voltage the output would be different?
In that case, there is a resistane in series with the source.

upload_2016-2-21_13-9-16.png


images?q=tbn:ANd9GcTHM0PskEHdv-Fg4Jzb0-JrVpfjZ1Kt0RsKMFdeDV-tClVxvipp.png
 

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  • #5
In that case, there is a resistane in series with the source.

View attachment 96229

images?q=tbn:ANd9GcTHM0PskEHdv-Fg4Jzb0-JrVpfjZ1Kt0RsKMFdeDV-tClVxvipp.png
Thank You very much.
Could you please answer the first question?thats only the doubt remaining in my mind now.
"What would happen if in reverse bias of a pn junction diode voltage greater than/equal to breakdown voltage is applied?"
 
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  • #6
cnh1995
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"What would happen if in reverse bias of a pn junction diode voltage greater than/equal to breakdown voltage is applied?"
It will behave almost in the same manner as the zener does. Current will be determined by the reverse characteristic curve. The only difference is a pn juction diode is more likely to be damaged in reverse breakdown than the zener(which is "designed" to operate in the reverse breakdown mode).
 
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  • #7
sophiecentaur
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One problem associated with Zener or Avalanche breakdown of a diode is that the voltage drop is high and P = IV so the power dissipation limit (which must be relevant) will dictate a much lower safe current than for forward biassed operation.
 

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