For quite some time I haven't been able to convince myself how reverse bias works using conventional explanations that involve words like 'minority carriers', 'majority carriers' etc. which do not make much sense to me.

Consider the image below: enter image description here Now I am an electron in that depletion region. I was not here initially, I was in the wire (free-moving electron of a copper atom) or perhaps battery (...of an ion?) but now I have been pushed here.

I feel attracted to the positive charges (which are for example phosphorous atoms with four valence electrons, hence the + charge) at the same time I also feel the repulsion from the negative terminal of the battery. I feel tempted (as well as all my electron friends) to move across the border and happily combine with the phosphorous ions, which is not happening. So what stops me there??

I am actually expecting to see answers that require knowledge about solid state physics particularly band gap and band theory, fermi-dirac stats etc., which I do not know enough about and need to get a lot more reading. Can anyone please confirm this so I can move on feeling a little more assured and come back to this after I know enough about solid state physics.

UPDATE: from the same question on physics stack exchange It could be probably due to that, somehow, my friends and I are bound to those impurity trivalent ions i.e. we are immobile in this context. But somehow forward biasing makes us mobile again.

  • \$\begingroup\$ I think you are forgetting about the diffusion currents, Feynamn's explaination is OK. Read 14-4 14-5, feynmanlectures.caltech.edu/III_14.html \$\endgroup\$ – George Herold Mar 27 '18 at 14:44
  • \$\begingroup\$ It seems helpful at first sight, I will read it, thanks. \$\endgroup\$ – Y.JQ Mar 27 '18 at 15:24

Your premise is flawed:

Now I am an electron in that depletion region. I was not here initially, I was in the wire (free-moving electron of a copper atom) or perhaps battery (...of an ion?) but now I have been pushed here.

If you started elsewhere, you have definitely NOT been pushed into the depletion region. If anything, you were pushed AWAY from ever entering it because of the strong electric field that exists there.

In terms of your diagram, electrons from the battery enter the diode from the left. The first thing they encounter is is all the negative ions on the left side of the depletion region, which repels the free electrons.

  • \$\begingroup\$ Ahh, what I mean is that when the diode is in thermal equilibrium and not biased, I was elsewhere outside the diode. Only when the battery is connected, electrons are pushed to the depleted region OR you can say holes are pulled to the positive terminal of the battery (is it ok to assume hole as simply absence of e-?) causing the depleted region to widen. Only when the whole thing reaches equilibrium i.e. the voltage of the battery equals the voltage of the diode(or treated as a capacitor), that electrons are blocked out from getting near. I am among the first batches and not the late comers. \$\endgroup\$ – Y.JQ Mar 27 '18 at 14:57
  • 1
    \$\begingroup\$ It doesn't matter. Even with no external bias, the diode has a depletion region, which is produced because it reduces the total energy of the device to have electrons from the N-side donors "drop into" the holes in the nearby P-side acceptors. This creates the initial electric field. Adding more reverse bias from the battery merely accentuates this effect. \$\endgroup\$ – Dave Tweed Mar 27 '18 at 15:32
  • \$\begingroup\$ Saying this, why doesn't adding reverse bias pushes the electrons that, under no bias, moved into the holes to form depletion region back to n-region? (instead of accentuating the effect) \$\endgroup\$ – Y.JQ Mar 28 '18 at 5:42
  • \$\begingroup\$ No, that's what happens under forward bias, when the external E-field opposes the internal field and eventually cancels it out (at approximately 0.65V for silicon). Under reverse bias, the external voltage causes more electrons to shift sides, with the resulting E-field simply adding to the internal field and increasing the width of the depletion region. \$\endgroup\$ – Dave Tweed Mar 28 '18 at 11:38
  • \$\begingroup\$ You're partially right? Under reverse bias, with the external E-field adding to that internal field, the electrons can't simply shift sides which in the process requires them to travel/diffuse from N to P region, which is going against both the internal and external E-field as both of them want to move electrons from P to N (reverse current). I think it is more appropriate to say that the external E-field puts electrons into P- and pulls electrons out of N-region. Like instead of turning back from living room to kitchen, you go outside, turn to backyard and enter the kitchen from backdoor. \$\endgroup\$ – Y.JQ Mar 28 '18 at 15:06

Lets look at three places in the diode under forward bias and then reverse bias.

Forward bias:-

Anode. Electrons are leaving the anode as they do they leave behind a hole in the p material which is a positive charge carrier this moves under the influence of the electric field towards the junction.

Cathode Electrons arrive at the cathode and move towards the junction.

Junction Electrons and holes meet and cancel preventing a build up of either and keep the current flowing.

Reverse Bias:-

Anode. Anode is negative so it is attracting the charge carriers in p material, holes towards it.

Cathode. Cathode is positive so it is attracting the charge carriers in n material, electrons towards it.

Junction. The holes and electrons have been pulled away from the junction forming a depletion region (depleted of electrons and holes). Hence they cannot meet and cancel once all of the carriers have moved away from the junction current stops.

The answer to your specific question.

Blockquote I feel tempted (as well as all my electron friends) to move across the border and happily combine with the phosphorous ions, which is not happening. So what stops me there??

is very simple the electric field across the junction far exceeds any attraction you have towards the phosphorus ions and you drift off towards the Cathode. Electrons do not feel tempted to do anything, they obey the laws of physics.

  • \$\begingroup\$ That is the common explanation which I find uncomfortable to work with and which does not really make much sense to me without the details and is the reason I'm asking the question. \$\endgroup\$ – Y.JQ Mar 28 '18 at 2:15
  • \$\begingroup\$ @YJQ is you could explain what about this you are not comfortable with and what details you require it would help us answer your question. Have edited the answer to add one clarification. \$\endgroup\$ – RoyC Mar 28 '18 at 8:24
  • \$\begingroup\$ Thank you. I've talked to a prof. and I'll need some time to rework on my understanding about holes and study more about solid state physics. However, if I did not misunderstand your language and you did not make typo, what you added is somewhat insensible in that the electric field across the junction seems to have been generated by the electrons and the phosphorous ions which is the source of attraction that tempts the electrons to drift towards cathode (+ terminal of battery). There' something else going against that electric field, instead of that field going against its own attraction. \$\endgroup\$ – Y.JQ Mar 28 '18 at 14:53
  • \$\begingroup\$ Yes it is the field generated by connecting the anode to a - and the cathode to a +. \$\endgroup\$ – RoyC Mar 29 '18 at 8:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.