Why is there no current in a reversed biased diode?

  1. kmm

    kmm 133
    Gold Member

    I understand there is some "leakage" current in a reversed biased diode but I don't understand why there is significantly less current in a reversed biased diode. I've seen the band diagrams and if all the electrons were only on the n-side of the diode, I would understand why the electrons can't flow through the diode since there would be a big hill to climb. However, I see there being electrons coming from the battery or whatever source, being injected into the p-side and then being attracted to the positive potential placed on the n-side.

    Here's how I imagine things. So the depletion region was at some width before the reverse bias was applied, then when the reverse bias is applied, I see the electrons on the n-side get pulled away increasing the number of positively charged ions on the n-side. During this, the electrons injected into the p-side I assume combine with some holes increasing the number of negatively charged ions on the p-side. That's how I see the depletion region forming. After that here's where I get stuck, an electron from the wire enters the p-side Adding up the forces on the electron, we have the negative and positive side of the battery plus the positive ions on the n-side all pushing and pulling the electron to the n-side and only the negative ions on the p-side pushing aopposite of that. So what prevents the electron from making it all the way across?
     
  2. jcsd
  3. Simon Bridge

    Simon Bridge 15,261
    Science Advisor
    Homework Helper
    Gold Member

    When reverse biased, the applied field draws electrons and holes away from each other.
     
  4. kmm

    kmm 133
    Gold Member

    I'm not sure how this addresses the specifics I gave in my question.
     
  5. Drakkith

    Staff: Mentor

    My understanding:

    When you apply a reverse bias voltage, electrons are pushed into the P-side and pulled off of the N-side. The electrons combine with holes on the P-side and are pulled off of the N-side. This results in a decreased number of carriers on both sides. (Holes are carriers on the P-side, while electrons are carriers on the N-side) Since there are very few carriers available, the electrons can't easily get from the P-side to the N-side and holes can't get from the N-side to the P-side, so the entire diode acts like an insulator. Thus a voltage builds up until it matches the applied voltage.

    When the diode is forward biased, the reverse is true. Electrons are pulled from the N-side to the P-side, and holes are pulled from the P-side to the N-side, but are replaced by the voltage source, resulting in current flow.

    Someone correct me if I'm wrong.
     
  6. kmm

    kmm 133
    Gold Member

    This may be the root of my problem. That I was thinking about the electrons as being able to move freely within the diode, as though it was a conductor. So on the N-side, we have a semiconductor doped to increase it's conductivity, meaning the electrons are near the conduction band. If we reverse bias the diode, the electrons pumped into the P-side are combined with holes so are not near the conduction band. Perhaps a small number make it to the depletion region and are swept across giving the "leakage" current.

    However, I guess I would run into that same problem with it forward biased. I'm still not so sure.
     
  7. Not on P and N “sides” in the volume, of course, but only in the space adjacent to the transition. Reverse bias causes electric potential to have a sharp change near the transition. There are very few conduction electrons on N side near the transition; it causes this region to have positive charge (due to electronic structure of N-type semiconductor). Likewise, there are very few holes on P side near the transition; it causes this region to have negative charge (due to electronic structure of P-type semiconductor). But total charge is almost zero. It’s this double layer that provides abrupt change in electric potential.
     
  8. kmm

    kmm 133
    Gold Member

    A thought I had that maybe someone could clarify for me. If I have some circuit with a battery and then hook a another battery that opposes the first, no current will flow. Is this essentially what's happening with a reverse biased diode?

    With the batteries however, the section of wire close to the negative terminal seems to be isolated from the positive terminal. What I mean is, we don't extend the electric field from the positive through the negative terminal. If we did, then the battery wouldn't be doing it's job of maintaining the potential difference. However, this is what I want to do with the diode. The depletion region has it's positive side on the same side as the positive terminal of the battery, dido the negative side. Within the diode however, I want to extend the electric field from the positive side through the negative side and add the total electric fields on the P-side just to apply the superposition principle(same with the n-side). By doing this it seems that the total field across the diode points from the n-side to p-side. Therefore, electrons should be flowing from the p-side to n-side. What's wrong with this?
     
  9. kmm

    kmm 133
    Gold Member

    I think I may have figured it out. When I referred to two batteries that are connected in opposition to each other, current won't flow only if the potential differences of each battery is the same. When I look at the total potential on the P-side and N-side without considering the battery, there will be a potential difference since each side is opposite charge. Now if we connect the battery to reverse bias the diode, there will actually be current for a brief period, but since the depletions regions grow, the potential difference between each side grows. This should continue until the potential difference within the diode equals that of the battery, and then current should stop.
     
  10. Diode resembles an (electrolytic) capacitor, not a battery. BTW, you can’t understand anything about a chemical cell thinking of electrostatic fields, whereas about diodes and capacitors you can. Both in capacitor and reverse-biased diode, there is a strong field where we haven’t carriers. And there is a plenty of carriers… where electric potential doesn’t change. So nothing happens, in average.

    But there is also a difference. In an electrolytic capacitor, electrolyte has carriers of both signs, so you must have some layer of insulator that deters [STRIKE]cathions[/STRIKE] anions from rushing towards metallic [STRIKE]cathode[/STRIKE] anode. In a reverse-biased diode, no insulator is needed because carriers on both sides avoid vicinity of the transition.
     
    Last edited: Sep 6, 2014
  11. Drakkith

    Staff: Mentor

    They do flow from the P-side to the N-side when in reverse bias. But the number of available carriers in the depletion region is so low that very little current can flow. When reversed bias, both holes and electrons flow away from the junction. When the holes flow from the junction towards the negative terminal, the negatively charged region of the P-side near the junction widens, and when electrons flow towards the positive terminal the positively charged region of the N-side near the junction widens.

    So while this results in an electric field within the diode that points in the same direction as the field from the applied reverse bias voltage, the majority carriers in the depletion region are depleted, leaving only the minority carriers available to carry current. This effectively increases the resistance of the junction, resulting in very little current flow. The higher the reverse bias voltage, the larger the depletion region and the resistance of the depletion region becomes. This continues until the breakdown voltage of the diode is reached.
     
  12. kmm

    kmm 133
    Gold Member

    I think I understand what is going on with the electrons within the diode. I'm trying to understand what is going on with the electrons outside of the diode. When the diode is first turned on, there is a brief time where current flows, but then this quickly stops. My thought is that there simply must be a potential difference that builds up inside the diode which stops the current flow. I was thinking of a circuit with two batteries as an analogy of why current would stop. I'm not sure if there is something else you were getting at.

    I knew that there was some small amount of current flow and I understand why the depletion region would widen, I just wasn't sure why electrons from outside the diode didn't continue to flow through mostly unimpeded. Using my battery analogy from #8, when the breakdown voltage is reached, the potential difference inside the diode mostly preventing current from flowing is broken down so current begins to flow significantly.
     
  13. Drakkith

    Staff: Mentor

    As far as I know, the electric field inside the diode isn't impeding the current when in reverse bias. The region near the barrier is simply depleted of charge carriers, forming an insulator.

    When the breakdown voltage is reached, electrons are ripped away from their ions, forming electron-hole pairs that move in opposite directions, allowing current to flow. Electrons from outside the diode don't flow through because they can't. The resistance is too high.
     
  14. kmm

    kmm 133
    Gold Member

    OK so then it was as I thought earlier in #5 where I suspected my problem was looking at the diode as a conductor. I understand how the depletion region is formed, but I've been trying to understand what is happening based on the resulting electric field created by it. It sounds like the bigger issue is actually that any electron from outside the diode injected into the p-side runs into the depletion region which has become essentially a poorly conducting material(insulator) and so it is stopped, regardless of the electric field that resulted. Am I thinking correctly?
     
  15. Drakkith

    Staff: Mentor

    That looks correct to me, but I can't say with 100% certainty that it's right.
     
  16. kmm

    kmm 133
    Gold Member

    The problem I see now though is that if the depletion region is seen as an insulator and the resistance is too high for electrons to pass through, then why are electrons able to pass through the "insulator" when forward biased?
     
  17. kmm

    kmm 133
    Gold Member

    For a forward biased diode of sufficiently applied voltage, we say that the electrons are able to push through the electric field barrier created by the depletion region. In this case we are talking about how the electrons interact with the field then. So I'm tempted to go back to my original analogy and say that this is what happens for a reverse biased diode. That the electrons are prevented to flow by the electric field created by the depletion region unless the breakdown voltage is reached.
     
  18. It is the electric field that keeps (most of) carriers out of the transition. All voltage drop is concentrated in a vicinity of the transition.

    It is depleted of carriers, but forms a charged double layer. It is not infinitely thin, of course. Its depth depends on the voltage, but the dependence might be complicated because in realistic semiconductors transitions are not absolutely sharp themselves.
     
    Last edited: Sep 6, 2014
  19. Drakkith

    Staff: Mentor

    When forward biased, the depletion region shrinks and the voltage barrier drops, enabling electrons and holes to flow more easily.

    From wiki: http://en.wikipedia.org/wiki/P–n_junction#Forward_bias

    Only majority carriers (electrons in N-type material or holes in P-type) can flow through a semiconductor for a macroscopic length. With this in mind, consider the flow of electrons across the junction. The forward bias causes a force on the electrons pushing them from the N side toward the P side. With forward bias, the depletion region is narrow enough that electrons can cross the junction and inject into the P-type material. However, they do not continue to flow through the P-type material indefinitely, because it is energetically favorable for them to recombine with holes. The average length an electron travels through the P-type material before recombining is called the diffusion length, and it is typically on the order of micrometers.[2]

    Although the electrons penetrate only a short distance into the P-type material, the electric current continues uninterrupted, because holes (the majority carriers) begin to flow in the opposite direction. The total current (the sum of the electron and hole currents) is constant in space, because any variation would cause charge buildup over time (this is Kirchhoff's current law). The flow of holes from the P-type region into the N-type region is exactly analogous to the flow of electrons from N to P (electrons and holes swap roles and the signs of all currents and voltages are reversed).

    Therefore, the macroscopic picture of the current flow through the diode involves electrons flowing through the N-type region toward the junction, holes flowing through the P-type region in the opposite direction toward the junction, and the two species of carriers constantly recombining in the vicinity of the junction. The electrons and holes travel in opposite directions, but they also have opposite charges, so the overall current is in the same direction on both sides of the diode, as required.
     
    Last edited: Sep 6, 2014
  20. Drakkith

    Staff: Mentor

    I see. Can you expand on what happens to the carriers at the junction when the diode is reversed biased? I'm afraid I'm definitely no expert and I don't want to give the wrong info to someone.
     
  21. Ī can reiterate that the depletion region is not an insulator; it would be a misconception to think so. There is the same energy gap between conduction and valence bands as a regular semiconductor has; that’s why some carriers are created there and cause a noticeable current (although a very small one because the volume of depletion region is tiny). Although it is, technically, a semiconductor, it does not behave as a semiconductor because a strong electrostatic field quickly wipes carriers of both signs away.
     
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