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It's well-known that if a diode is forward biased at a sufficient voltage, there is a voltage drop across the diode, i.e. it dissipates power.

However, I was reading Wikipedia's article on the depletion region of the diode (aka PN junction), and it occurred to me the electric field, E, as labeled below, is set up by immobilized ions and causes the voltage, V, as indicated to be positive:

enter image description here

I've also drawn the symbol of the diode in the above.

This result is definitely what I would expect, because if a battery (and resistor) is connected to the diode in the usual, forward-biased way, there should be voltage drop across the diode and the voltage V (as labeled) should be negative.

Where did I go wrong?

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    \$\begingroup\$ It isn't clear what you think is "wrong". Yes, that internal field exists; it's the reason that it requires about 0.65V to get significant current flowing in the forward direction. The current flows once the externally-applied field cancels out that internal field, eliminating the depleted region at the junction. \$\endgroup\$ – Dave Tweed Jun 22 '16 at 11:15
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    \$\begingroup\$ @DaveTweed I believe I was also taught that external field needs to "overcome" the internal one for current to flow, though the OP raises a good point: had the external and internal voltage cancel, there would be 0 V, and that's clearly not the case - there is a voltage drop. \$\endgroup\$ – LCW Jun 22 '16 at 11:24
  • \$\begingroup\$ The external field (measureable) overcomes the internal field (inaccessible to measurement) \$\endgroup\$ – Chu Jun 22 '16 at 14:06
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You understanding is not wrong, but incomplete. You basically showed how a diode is "built", but not how it is used.

schematic

simulate this circuit – Schematic created using CircuitLab

If you have a circuit like the one above, for instance, the 3 V supply must overcome the built-in voltage of your diode (the one you drew in your picture) to generate an electric field (directed from anode to cathode) strong enough to push current from anode to cathode.

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The drawing you have is correct, this is the equilibrium electric field (voltage) caused by carriers diffusing over the junction. This process of diffusion continues until the the rate of recombination in the P- and P-sides of the junction is equal to the diffusion rate. The level of doping used on the P and N sides of the junction determines the width of this region at equilibrium. This is why in a loose sense diodes won't conduct (much) until the applied electric field (voltage) exceeds the built in electric field.

To answer the question, where did I go wrong? I think you may just be misunderstanding that this is the diode's equilibrium state with no externally applied electric field and that the diagram is showing the built-in potential that has to be overcome in order for the diode to conduct.

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  • \$\begingroup\$ Well...are you saying those labeled + and - ions somehow will cease to exist when a battery is connected (external field applied)? I don't think those fixed ions in the lattice are going anywhere when the ext. field is applied. I'm just trying to understand your Answer, since I don't have one myself. \$\endgroup\$ – LCW Jun 22 '16 at 11:29
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    \$\begingroup\$ @LCW: There are no "fixed ions" in the lattice, just donor and receptor dopant atoms, which are electrically neutral. They only become ionized (when no external field is applied) because it reduces the total energy of the system slightly when electrons migrate across the junction from the donor atoms to the receptor atoms, creating the ions in the process. This also depletes the region near the junction of any "free" charge carriers, which is why no current can flow until this situation is reversed. \$\endgroup\$ – Dave Tweed Jun 22 '16 at 11:36
  • \$\begingroup\$ @LCW The ions in the drawing are not fixed, carriers are continuously diffusing across the junction. The equilibrium shows the state of the PN junction at which the rate of diffusion across the junction is equal to the rate of recombination in the second material. Take the P-side for example, the electrons that diffuse across the junction take some time to recombine with a hole this is why there appear to be static electrons on that side of the junction. \$\endgroup\$ – Captainj2001 Jun 22 '16 at 11:38
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    \$\begingroup\$ What I meant was that, for example, if phosphorus is introduced as a dopant donor atom, then almost always it will become a positive ion having given up the electron; it is this positive ion that is not free to travel (but the electron is) - this is what I meant by "fixed". But what's special in the pn junction is that the electron has moved elsewhere, so the + ion is "exposed" and no longer neutral (locally). But of course when the situation is reversed the Ph +ion is no longer exposed and I think this is what both you meant by no stationary lattice ions, right? If so, then I'm in agreement. \$\endgroup\$ – LCW Jun 22 '16 at 11:56
  • \$\begingroup\$ Getting back to OP's question...with the situation (ie ++ and -- separated) reversed, the pn junction will be neutral everywhere, there is no accumulation of free charges anywhere, and the voltage across it will be 0 V. How would you reconcile this with the 0.65 V drop in forward bias? \$\endgroup\$ – LCW Jun 22 '16 at 11:58
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According to Sedra/Smith: Microelectronic Circuits:

The voltage across the depletion region will be exactly canceled by the voltages at the (metal) contacts and therefore will not appear between the terminals.

Therefore \$V _0\$ will not create a voltage drop. Also, as others have pointed out, your drawing is for the case that there is no forward current. Once you have forward current, your depletion region changes, which then results in the voltage drop.

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From an energy standpoint, the battery would provide an electromotive force that does work against this internal field, and hence energy is dissipated in the internal field as you illustrate, causing the voltage drop.

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