Why no current in reverse-biased diode?


by Mårten
Tags: diode, holes, p-n junction
Mårten
Mårten is offline
#1
Apr4-09, 09:18 PM
P: 127
I try to understand why, really, there is no current (except for leakage) when a diode is reverse-biased. I studied several texts, and noone really explains, like for instance this otherwise very pedagogic Wikipedia article.

How I think: The negative terminal connected to the p-region ought to attract the holes there (in other words, when holes are flowing one way, electrons flow the other way, from the negative terminal to the junction). On the other side of the diode, the positive terminal connected to the n-region attracts electrons, so electrons are flowing from the junction to the positive terminal. All in all, this would mean a current is flowing from one end of the diode to the other.

This seems for me to be perfectly symmetrical to what happens when the diode is forward-biased - in the above paragraph, negative terminal becomes positive terminal and vice versa, and attraction becomes repulsion.

I must be missing some of the physics... Please help someone!
Phys.Org News Partner Engineering news on Phys.org
PsiKick's batteryless sensors poised for coming 'Internet of things'
Researcher launches successful tech start-up to help the blind
Researchers propose network-based evaluation tool to assess relief operations feasibility
Bob S
Bob S is offline
#2
Apr4-09, 09:46 PM
P: 4,664
All diodes will break down at some voltage, often destructively. Zener diodes break down at the zener voltage when reverse biased.
chroot
chroot is offline
#3
Apr4-09, 11:05 PM
Emeritus
Sci Advisor
PF Gold
chroot's Avatar
P: 10,424
You are correct with your thoughts on the movement of the carriers. When the carriers are pulled away from the pn junction, the width of the depletion region across the junction (where there are no free carriers) is increased. The wider depletion region presents a much greater obstacle to the movement of carriers across it, and very little current flows.

- Warren

Mårten
Mårten is offline
#4
Apr5-09, 11:57 AM
P: 127

Why no current in reverse-biased diode?


Thanks so far!
Quote Quote by chroot View Post
When the carriers are pulled away from the pn junction, the width of the depletion region across the junction (where there are no free carriers) is increased. The wider depletion region presents a much greater obstacle to the movement of carriers across it, and very little current flows.
But what is this depletion region made up of really? On the p-side, close to the junction, it's made up of holes, now filled with electrons, so we there have full valence shells, but still negatively charged atoms. On the n-side, close to the junction, we have holes where it formerly where free electrons (which were the ninth electron in every valence shell in question), i.e. stable valence shells with eight electrons, but still positively charged atoms. This charge difference ought to start a current of the electrons just to the left of the junction, to flow over just to the right of the junction (and then, pulled by the positive terminal, further on to the right). Or, why not?

(I'm using this pic from WP to more easily imagine what happens.)
cabraham
cabraham is offline
#5
Apr5-09, 12:37 PM
P: 997
The E field vector points the wrong way when a p-n junction is reverse biased. If the cathode is positive wrt the anode, then the E field vector points towards the anode, and away from the cathode.

So, in the anode, electrons are subjected to a force opposite the E field, since the E field by definition points in the direction positive charges flow. Hence the electrons feel a force in a direction towards the cathode. But to reach the cathode, they must travel through the p channel anode. In the p anode, the electrons are *minority* carriers, with low mobility. Very little electron motion occurs since p material has an excess of holes, and electrons immediately tend to recombine with said holes. Hence the electron current from anode to cathode is very low, i.e. the leakage value.

In the n channel cathode, the situation is the same but with polarities reversed. The holes would be directed towards the anode. But the holes travelling through the n cathode are minority carriers, resulting in low mobility, and limited to the leakage value.

Does this explain it?

Claude
Mårten
Mårten is offline
#6
Apr5-09, 01:30 PM
P: 127
Quote Quote by cabraham View Post
So, in the anode, electrons are subjected to a force opposite the E field, since the E field by definition points in the direction positive charges flow. Hence the electrons feel a force in a direction towards the cathode. But to reach the cathode, they must travel through the p channel anode. In the p anode, the electrons are *minority* carriers, with low mobility.
But that in turn is pretty strange also. Electrons are minority carriers, but holes are majority carriers. But holes going one way, is just the same as saying electrons are going the other way. So if holes have high mobility in one direction, it is the same as saying electrons have high mobility in the opposite direction. Or am I wrong, and in that case, why?

Quote Quote by cabraham View Post
Very little electron motion occurs since p material has an excess of holes, and electrons immediately tend to recombine with said holes. Hence the electron current from anode to cathode is very low, i.e. the leakage value.
But this recombining at the same time charges the atoms there negatively. And this charge would feel a Coulomb force directed towards the junction.

Maybe it all comes down to that the Coulomb force attracting charges, is less than the "force" holding down the electrons in stable full valence shells (even though the atoms then become non-neutral)?
cabraham
cabraham is offline
#7
Apr5-09, 01:54 PM
P: 997
Quote Quote by Mårten View Post
But that in turn is pretty strange also. Electrons are minority carriers, but holes are majority carriers. But holes going one way, is just the same as saying electrons are going the other way. So if holes have high mobility in one direction, it is the same as saying electrons have high mobility in the opposite direction. Or am I wrong, and in that case, why?


But this recombining at the same time charges the atoms there negatively. And this charge would feel a Coulomb force directed towards the junction.

Maybe it all comes down to that the Coulomb force attracting charges, is less than the "force" holding down the electrons in stable full valence shells (even though the atoms then become non-neutral)?
Holes having high mobility in p material does not mean that electrons also have high mobility in any direction. In p material, holes have high mobility because they do not recombine with atoms having a hole, or missing an electron. Electrons, OTOH, do just that. A hole can easily transit through the entire p region unscathed, not recombining with atoms, since an atom missing an electron will capture an electron in recombination, not a hole. Vice versa for n material.

The recombining action does not charge the atoms negatively. An atom with a hole is positive. When an electron recombines with said atom, charge neutrality is attained.

A good peer-reviewed university level text on semiconductor physics will dive into the details. The explanantion I gave is just a sketchy overview. It's correct, but not knowing how deep your knowledge in semi physics is, that is what I can tell you in a couple of paragraphs. BR.

Claude
Mårten
Mårten is offline
#8
Apr5-09, 02:45 PM
P: 127
Quote Quote by cabraham View Post
The recombining action does not charge the atoms negatively. An atom with a hole is positive. When an electron recombines with said atom, charge neutrality is attained.
Wait... Young-Freedman's University Physics, 9ed, says on p. 1364:

Caution: Saying that a material is a p-type semiconductor does *not* mean that the material has a positive charge; ordinarly, it would be neutral. Rather, it means that its majority carriers of current are positive holes (and therefore its minority carriers are negative electrons).

An atom with a hole, is e.g. a boron atom surrounded by Si-atoms, making the boron atom be surrounded by 7 valence electrons (4 from Si, 3 from boron itself). This is perfectly neutral as I understood it - but it's not a full valence shell, and that's a different matter. Or have I misunderstood something here?

(Btw, this is the textbook I have, but it doesn't explain the things I wonder in this thread so deep.)
cabraham
cabraham is offline
#9
Apr5-09, 08:58 PM
P: 997
Quote Quote by Mårten View Post
Wait... Young-Freedman's University Physics, 9ed, says on p. 1364:

Caution: Saying that a material is a p-type semiconductor does *not* mean that the material has a positive charge; ordinarly, it would be neutral. Rather, it means that its majority carriers of current are positive holes (and therefore its minority carriers are negative electrons).

An atom with a hole, is e.g. a boron atom surrounded by Si-atoms, making the boron atom be surrounded by 7 valence electrons (4 from Si, 3 from boron itself). This is perfectly neutral as I understood it - but it's not a full valence shell, and that's a different matter. Or have I misunderstood something here?

(Btw, this is the textbook I have, but it doesn't explain the things I wonder in this thread so deep.)
You ask a question, I give an answer. You dispute me, so I elaborate. Now you present a straw man argument by putting words into my mouth. I never said a p material has a positive charge. Holes and electrons are generated thermally so ions are formed. Then when recombination takes place, neutrality is restored.

You are obviously not interested in anything I have to say, so why do you bother to ask? Good day sir.

Claude
Mårten
Mårten is offline
#10
Apr6-09, 06:31 AM
P: 127
I'm very, very sorry to have insulted you, that was absolutely not my meaning. I appreciate very much the help I get from you and all others here. My mother tounge is not english, so maybe I have unintentionally written things that signals tones I didn't mean. I'm very sorry for that.

Now to the point. You said earlier:
An atom with a hole is positive. When an electron recombines with said atom, charge neutrality is attained.
1) How could an atom with a hole be positive? If you mean a boron atom as I described above, I thought then one has an atom with seven valence electrons, but still it's uncharged. Or am I wrong there, and if so, why?

2) If the atoms with holes are positive, how come the whole p-area being not positive? I thought that must be the same thing. Why is it not the same thing?

3) I though that when an electron "settles itself" in such a hole, then you actually have a negatively charged atom (or maybe you can call it an ion). I understand it as being like a chloride atom - a chloride atom would like to have an extra electron to fill its valence shell with 8 electrons, and therefore it gets negatively charged, Cl-. Isn't it the same thing here?

Sorry again for being a little bit confused here...
Mårten
Mårten is offline
#11
Apr6-09, 05:37 PM
P: 127
I think this thread lost focus a little bit... Back to where we started - I think the following things is what puzzles me the most:

1) There seems to be a force from atoms with non-full valence shells, attracting electrons flying by. Even though we have an electric field from Coulomb forces (i.e. from a battery), this force seems to be greater. That's for instance the case in the depletion area, the electrons like to stick to their holes. Is that correct? And, what kind of force is this willingness to have a full valence shell? Is it some kind of Coulomb force?

2) I have still problems understanding major/minor carriers and why they have different mobility. In a p material, doped with boron, the major carrier are holes. If a hole there moves from one atom to another, an electron must move from that other atom to the first. Am I right? So why would the movement of the hole be more easily done? Isn't the phenomenon of a hole moving, the same thing as saying an electron moved the other way?

I hope someone will help...
Redbelly98
Redbelly98 is offline
#12
Apr27-09, 07:06 PM
Mentor
Redbelly98's Avatar
P: 11,984
Hi Mårten, I'll respond with what I think is going on, but I invite others to critique what I say. Understanding of diode and semiconductor physics has never come easy to me.

Quote Quote by Mårten View Post
Thanks so far!

But what is this depletion region made up of really? On the p-side, close to the junction, it's made up of holes, now filled with electrons, so we there have full valence shells, but still negatively charged atoms.
Pretty much. What had been holes--in reality the absence of many valence-band electrons--is now essentially filled in with electrons in the valence band. Note: electrons in the valence band are not mobile charges. While there is a net negative charge due to the extra electrons, they occupy the valence band and not the conduction band.

On the n-side, close to the junction, we have holes where it formerly where free electrons (which were the ninth electron in every valence shell in question), i.e. stable valence shells with eight electrons, but still positively charged atoms.
I would say it differently. We formerly had conduction-band (or "free") electrons, but those have been removed. Result: essentially no mobile charges are present. Also, the removal of the conduction electrons leaves a net positive charge, but does not produce holes.

So, the situation is essentially no mobile charges in the semiconductor, which means a relatively high resistance and insignificant current flow.

This charge difference ought to start a current of the electrons just to the left of the junction, to flow over just to the right of the junction (and then, pulled by the positive terminal, further on to the right). Or, why not?
Again, I repeat, these electrons do not occupy the conduction band, but rather are localized around individual atoms and will not take part in the flow of current.

(I'm using this pic from WP to more easily imagine what happens.)
Mårten
Mårten is offline
#13
Apr29-09, 05:52 AM
P: 127
Thanks for the reply! This thing with valence-band electrons cleared up a little bit in my confusingness... But still:
Quote Quote by Redbelly98 View Post
What had been holes--in reality the absence of many valence-band electrons--is now essentially filled in with electrons in the valence band. Note: electrons in the valence band are not mobile charges. While there is a net negative charge due to the extra electrons, they occupy the valence band and not the conduction band.
But why, really, in the first place, would these electrons from the n-region like to flow over the p-n junction to the holes in the p-region? I read something about a diffusion force, but what is that really? I've learned in physics that we have four fundamental forces (gravity, EM, strong/weak nuclear). But what is this diffusion force then really? How could this force move electrons over the p-n junction though there is in the beginning (when the p- and n-material where put together) neutrality over the junction? (If there is neutrality, there could not be any coulomb/EM force that makes the particles move.)
Redbelly98
Redbelly98 is offline
#14
Apr29-09, 06:07 AM
Mentor
Redbelly98's Avatar
P: 11,984
Use of the word force here is something of a misnomer. But there is diffusion, or a tendency towards a state of higher entropy. It's the same process that makes a gas fill the entire volume available to it.
miva
miva is offline
#15
May1-09, 10:05 AM
P: 10
if you need anymore information on this topic pm me and i have some lecture slides that are perfect for this.
jsgruszynski
jsgruszynski is offline
#16
May1-09, 07:37 PM
P: 272
Quote Quote by Redbelly98 View Post
Use of the word force here is something of a misnomer. But there is diffusion, or a tendency towards a state of higher entropy. It's the same process that makes a gas fill the entire volume available to it.
This is the key. Diffusion happens. Most electrical currents are diffusion-based at the meso-atomic level. This is true for metals and semiconductors. There is an overlying drift component on top of diffusion due to E-fields/Voltage but under most conditions diffusion equations are largely dominant.

The corner cases are:
  • When you have a PN junction you have "minority carrier transport", where recombination limits the depth of conduction exponentially from a PN junction boundary. This why bipolar transistors have narrow base widths and how you get more gain with narrower bases. Minority carrier transport is diffusive but it's limited in conduction distance.
  • When the drift component shifts the diffusion equilibrium to result in an effective temperature that is above room temperature (diffusion is a statistical distribution effect) then you have "hot carrier transport" which is central to the primary reliability degradation mechanism of MOS devices: hot carrier injection (into the oxide interface).
  • When transit distances is less than the mean-free path (which is a diffusion concept) then you have "ballistic transport" which is a faster current than diffusion current. Many nanoelectronic devices rely on ballistic transport.
  • When the thickness of an insulator goes below a certain limit defined by quantum mechanics then you have "quantum tunneling transport" through the insulator which comes in a number of forms but is largely the same basic result, conduction through insulators. Flash memory absolutely depends on this. Some types of conduction for magnetoresistive devices rely on the magnetic field applied changing the probability of tunneling through resistive grain boundaries which results in the change in resistance due to B-field.
  • When the number of electrons involved in the current are "countable small" like 1 to dozens, etc., "quantum conduction transport" dominates which is described by Schoedinger's equations rather than Ohm's law or conventional circuits or quasi-classical diffusion conduction. Those nanoelectronic devices that don't rely on ballistic transport usually rely on quantum transport. For example "Single Electron Transistors".

Some stupid trivia: tin is the only metal that conducts electricity by hole currents rather than electron currents. Strictly speaking it's a semiconductor or semimetal. The differences between insulator, semiconductor and metal are simply different points on a continuum.
Mårten
Mårten is offline
#17
May7-09, 10:22 AM
P: 127
Sorry for this late reply, took some days vacation...
Quote Quote by Redbelly98 View Post
Use of the word force here is something of a misnomer. But there is diffusion, or a tendency towards a state of higher entropy. It's the same process that makes a gas fill the entire volume available to it.
Okey, I'll accept that. I'll guess I have to start a new thread about what temperature and diffusion really is, and their connection with the four fundamental forces... I've just discovered I don't really know what temperature is on a fundamental level, that would be interesting to dwell into...

Back to where this thread started. Why no current in reverse-biased diode. I've understood it has something to do with different mobility for majority/minority carriers. Okey, electrons are minority carriers with low mobility in the p region, because they would recombine with the holes there, so if the diode is revese-biased, no current will flow. But now think of the forward-biased diode (see pic below). The electrons get a push from the battery, making them leave the n region and going over to the p region. While in the p region, they are now minority carriers, so they will recombine with the holes there. So why do we have a current in this forward-biased diode? Okey, someone would say: because in the p region the holes are majority carriers, so the holes will easily flow towards the n region. But, if the holes flow towards the n region, that is the same as saying that the electrons are flowing in the opposite direction. But, we just said that electrons as minority carriers hardly didn't move in that direction! How does it make sense?



P.s. Thanks jsgruszynski for the list of different types of transport, that was good to have.
Defennder
Defennder is offline
#18
May7-09, 12:29 PM
HW Helper
P: 2,618
Sorry if my reply has been brought up before; I didn't read all the posts to date before posting.

I'll draw a band diagram for better understanding. The following is from my notes:

By defennder

As you can see, applying a reverse bias voltage would increase the potential barrier majority carriers on either side of the pn junction have to surmount before they can cross over. At the same time, minority carriers on either side now require less energy than before, so there is a slight but negligible reverse bias current due to flow of minority carriers from both sides.

For clarification: The top diagram shows the situation under zero applied voltage, bottom shows one under reverse bias. The arrows denote flow of both majority and minority carriers on both sides; the size of the arrow roughly proportionate to the percentage increase/decrease of current due to carrier diffusion.

Hope this helps. I took this course nearly two years ago, so I might have forgotten a fair bit of stuff.


Register to reply

Related Discussions
How come when you reverse a diode the current/voltage? is zero? General Physics 4
How come when you reverse a diode the current/voltage? is zero? Introductory Physics Homework 5
Forward biased diode Atomic, Solid State, Comp. Physics 2
Finding reverse sat current of pn junction diode Engineering, Comp Sci, & Technology Homework 2
Reverse Breakdown of a Diode Engineering, Comp Sci, & Technology Homework 7