There will be current when diode is reverse biased

[PainterGuy]

hi everyone,

please see fig 42.25 a and b on this google doc:---

https://docs.google.com/viewer?a=v&...EyZGYtM2RmMTA3MDQxMTlm&hl=en&authkey=CKP-7OsD

i can understand the fig 42.25a and why the current flows. but i have questions about fig 42.25b.

what does the word 'biased' in "forward-biased" and "reverse-baised" mean as far as english meaning is concerned?

in fig 42.25b, positive terminal of battery will such electrons out of n-side which will leave behind 'positive' holes. negative terminal of battery will feed the electrons to holes in p-side. as n-side is deficient of electrons because positive terminal of battery has sucked them so this n-side will attract electrons out p-side which is now rich in electrons because negative terminal of battery has supplied electrons to p-side, so this will result in electric current. so there would also be current when battery terminals are reversed as in fig 14.25b. so where is i am wrong? please show me the light. thank you.

 

[Studiot]

The answer is best illustrated with reference to the red curve in fig 42.26c of your link.

This is a graph or plot of the current that flows at any given voltage across the diode (PN junction).

The graph can be divided into two zones by the current axis.
The right had half where the voltage across the diode is positive, we call this forward biased.
The left hand half where the voltage is negative, we call this reverse biased.

Taking the positive right hand or forward biased half first, notice that the current near zero voltage is very small and increases slowly as we increase the (forward) voltage. But at about half a volt the current starts increasing rapidly until it is going nearly straight up.
This means that we require just over half a volt to 'turn the diode on', but then almost no extra voltage to drive as much current as the diode will support.

A diode in this condition is working in the forward or pass condition and has a forward bias of 0.5 volts.

Now turning to the left hand or reverse biased zone of the graph we see that if we reverse the supply polarity very little current passes, no matter what the voltage across the diode.

The diode is said to be in the reverse or blocking condition. In an ideal diode no current passes in this condition. The small amount that does pass in real world diodes is called the leakage current.

To summarise

Diodes have a forward bias of 0.5 to 0.6 volts - fixed we can't impress say 6 volts forward on the diode.

Diodes can have any reverse bias we like, so long as we remain within their voltage rating.
This voltage rating is called the peak inverse voltage or PIV and may range from 50 volts to many thousands of volts, depending upon diode type.

go well

 

[PainterGuy]

many thanks, Studiot. what do you mean by "fixed we can't impress say 6 volts forward on the diode."

sorry to say this but i am still having the same difficulty understand the concept. which means i still think current can flow even when diode is reverse biased. if you read my previous post you will see my reason (no matter how stupid it could sound to you but at least you would be able to see gaps in my understand and where i go wrong)...a bundle of thanks for this assistance.

 

[Studiot]

Are you trying to understand diode action from a circuit point of view (what diodes do and why we use them) or from a physics point of view (how a PN junction works)?

These are quite different questions and many become confused if they are mixed, especially as knowledge of one is very little help with the other.

Let us try a thought experiment.

Let us take a variable voltage DC supply, set it to zero, and connect it across the diode so the negative (black) terminal is connected to the anode and the positive (red) terminal is connected to the cathode.

If we now slowly increase the voltage at the red terminal and monitor the current we find that almost no current flows as we increase the voltage. Say our ouput meter reaches 10 volts.

The diode has 10 volts reverse bias.

Now say we reset the power supply to zero disconnect the wires and reconnect in the opposite order.
Now the cathode is connected to the black and the anode is connected to the red.

Again we slowly increase the voltage.
This time a current large current flow when we reach 0.5 volts. If we try to increase the voltage above this value the current increases rapidly, but the voltage does not. We may reach 0.6 volts or even 1.0 volts but no more.
If our power supply is large enough it will supply enough current to destroy the diode, but the voltage will still not rise beyond 0.5 - 1.0 volts.

The diode is said to be forward biased in this condition.

 

[PainterGuy]

Are you trying to understand diode action from a circuit point of view (what diodes do and why we use them) or from a physics point of view (how a PN junction works)?

big thanks. I'm trying to understand from physics point of view, how a PN junction works. in my post i have reasoned why I'm unable to understand the functioning of PN junction under reverse-biased condition. please lead me to the right path. I'm very much grateful to you.

Now say we reset the power supply to zero disconnect the wires and reconnect in the opposite order.
Now the cathode is connected to the black and the anode is connected to the red.

Again we slowly increase the voltage.
This time a current large current flow when we reach 0.5 volts. If we try to increase the voltage above this value the current increases rapidly, but the voltage does not. We may reach 0.6 volts or even 1.0 volts but no more.
If our power supply is large enough it will supply enough current to destroy the diode, but the voltage will still not rise beyond 0.5 - 1.0 volts.

The diode is said to be forward biased in this condition.

do you mean that voltage drop across diode won't exceed 0.6v-1.0v. otherwise voltage supply can supply any range of voltage. i think you you mean voltage drop across a diode. could you please confirm it.

cheers

 

[Studiot]

do you mean that voltage drop across diode won't exceed 0.6v-1.0v

Yes that's right.
Look again at the graphic in your link.

You need to get firmly hold of this idea that you cannot have (say) 10 volts across a diode in the forward direction, in order to progress. You can have anything in the range 0 - 1 volt
This fact is a consequence of the diode action.

It is also a consequence of diode action that you can have any reverse voltage 0.1, 1 10, 100, 1000 volts if the diode will stand it.

A diode is limited by the current it can safely pass in the forward direction. Its voltage is constant.
A diode is limited by the voltage it can withstand in the reverse direction. It does not pass (appreciable) current.

 

[PainterGuy]

many thanks, Studiot. I'm afraid you have missed one question. could you please help me how PN junction works from physics point of view? I'm still under the impression that current can flow even in reverse biased condition as i explain the first post above.

once again, many thanks for all the information.

 

[Studiot]

I do believe that I have said in every post in this thread words to the effect that there is a small reverse current which is unintentional and would be designed out if diode fabricators possibly could.

Understanding how these things work is a big leap and doesn't come to most people with a single reading so keep re-reading.
The workings of a PN junction are sufficient for a thread all of their own, indeed ther are several already here and available for the searching.

 

[granpa]

when forward biased holes and electrons recombine at the junction
when reverse biased holes and electrons are formed at and move away from the junction

 

[webberfolds]

First of all the information I give is based on what I read and if I'm wrong on anything please say. So the holes and electrons tend to stay in the semiconductor. In the first diagram the electrons would move counterclockwise and the electrons originally in the semiconductor would be repelled by the electrons moving towards them making the electrons move inwards. The holes would move inwards too and that would make a smaller depletion region. That would make it easier for current to flow across the "gap" (depletion region).That situation would be reverse biased. In the second diagram the depletion region would become wider and so it is reverse biased. As for atoms and that sort of stuff: http://books.google.ca/books?id=Bs6...g#v=onepage&q=examples varactor diode&f=false. There's a section on that there. And yes there'll probably be some current moving when reverse bias but I haven't heard people say that there'd be an appreciable amount of it although I wonder if some diodes would conduct in reverse bias closer to how well they conduct forward bias.

hi everyone,

please see fig 42.25 a and b on this google doc:---

https://docs.google.com/viewer?a=v&...EyZGYtM2RmMTA3MDQxMTlm&hl=en&authkey=CKP-7OsD

i can understand the fig 42.25a and why the current flows. but i have questions about fig 42.25b.

what does the word 'biased' in "forward-biased" and "reverse-baised" mean as far as english meaning is concerned?

in fig 42.25b, positive terminal of battery will such electrons out of n-side which will leave behind 'positive' holes. negative terminal of battery will feed the electrons to holes in p-side. as n-side is deficient of electrons because positive terminal of battery has sucked them so this n-side will attract electrons out p-side which is now rich in electrons because negative terminal of battery has supplied electrons to p-side, so this will result in electric current. so there would also be current when battery terminals are reversed as in fig 14.25b. so where is i am wrong? please show me the light. thank you.

 

[Studiot]

Have you checked the date on this thread?

 

[webberfolds]

I didn't check before but it's from 2011 I think, I guess I should comment on new ones instead.

Have you checked the date on this thread?