I don't like asking duplicate questions but I couldn't find one that was quite the same. I apologize if it is similar to another.

TL;DR: How can a charge neutral substance have a potential? Why would measuring the voltage across a pn-block read 0 despite the built in voltage being present? Feel free to get as technical as necessary.

My question is one that I have been unable to find a satisfying answer for even in my microelectronics class. I understand the basics of the PN-junction. P-doped material connected to N-doped material through some process that maintains their crystalline structure and allows electrons to freely flow between the two. In equilibrium there is a balance between drift and diffusion currents which results in a depletion region of negative charge on the Pside and positive charge on the Nside. This creates an electric field which is the main idea behind the PN diode.

Now I have two questions... we were taught that at some level of doping, a semiconductor has a potential. It can be found using a logarithmic relationship which I don't want to type here since my question is more conceptual than anything. How can there be an inherent potential in a doped semi-conductor if it is charge neutral? The way I understand potential is that it requires some sort of charge. The equation I mentioned before comes from the Boltzmann relations between concentrations of particles at different potentials. If a piece of silicon is placed in a potential then an imbalance in the number of holes vs electrons is created based on the strength of that potential. So artificially creating the imbalance of holes and electrons creates a natural potential. However, it is still supposed to be charge neutral... but it has a potential... does my confusion make sense yet?

Ok! I now understand this! I misunderstood something I read and that was causing my sadness. Basically, the concentration gradient of charge carriers between the Pside and Nside has to be continuous (no weird deltas or steps). The levels concentrations of these carriers are entirely determined by the doping of the Pside and Nside via the law of mass action. \$ n_i^2 = n_0 \times p_0 \$ where n_i is the intrinsic amount of charge carriers in a substance at room temperature. This quantity is conserved even under doping so we can see that the concentrations of minority carriers can be determined easily thank to this law. Now, we use boltzmann relations to determine the potential difference that the Nside and Pside will have relative to each other. The equation for that can be found here. We use the intrinsic concentration and 0 potential as our reference voltage. This allows us to determine the potential difference due to the concentration gradient.

My second question is about the PN junction as a whole. We know in equilibrium there is no current flow... but there is a built in potential. That's fine... diffusion current opposing drift current explains that. However, I don't understand why we can't measure the voltage drop across the PN block with a voltmeter. I've read that, again because the whole thing is charge neutral, as far as the universe is concerned it doesn't have a potential associated with it... but there is one INSIDE it? I don't understand.

I think I have isolated the issue with my understanding here... something weird happens with contact potential... can someone explain exactly what? When metal makes contact with the PN-Junction there must be some sort of "infinite" electric field

If you took the time to read this wall of text, thanks! I would love an answer to this question.


4 Answers 4


How can a charge neutral substance have a potential?

To say that a PN junction has built-in potential isn't to say that the PN junction has a potential relative to ground or infinity etc.

How can there be an inherent potential in a doped semi-conductor if it is charge neutral?

Charge has been separated within the PN junction and, thus, there is an electric field across the depletion region and an associated potential difference. A charged capacitor is neutral but there is a potential difference (voltage) across the dielectric. There are some similarities but...

However, I don't understand why we can't measure the voltage drop across the PN block with a voltmeter.

As explained, for example, here, the built-in potential is not readily measured with, e.g., a voltmeter. In other words, this question has been asked here several times (which means you are not the only one perplexed by this - most are at first) and there are good answers already available.

  • \$\begingroup\$ -Why ISN'T it the same as saying that? There is a potential difference between them, yet the universe views them as the same from outside the substance. -I don't mean in the PN junction, unless i've misunderstood what that equation tells us, which I've been thinking might be the case. I meant a single doped bit of semi-conductor. It doesn't make sense to me that it has a potential, so I suppose it doesn't. \$\endgroup\$ Commented Jun 5, 2014 at 22:47
  • \$\begingroup\$ And to follow up on the last point, we were shown a graph of the potentials for a PN-Junction with metal contacts in my microelectronics class. It said the contact potential was 0 on both sides meaning there was a near instantaneous drop between the semi-conductor and metal contact point. Doesn't that imply a "delta" or infinite electric field somewhere at the contact points? \$\endgroup\$ Commented Jun 5, 2014 at 22:50
  • \$\begingroup\$ @XavierHubbardAnderson, a doped semiconductor, whether n-type or p-type, should not have a built-in potential. Where did you read that? Do you have a link? \$\endgroup\$ Commented Jun 5, 2014 at 22:58
  • \$\begingroup\$ I think the notes I read were correct and I just misunderstood them. So what I understand is that the Boltzmann relation idea is only used to find a potential due to doping when it is actually part of a PN junction. The reference voltage we used was ground and the ground reference concentration was the intrinsic concentration of the material. Is that correct? \$\endgroup\$ Commented Jun 6, 2014 at 1:09

Why can't we measure the junction potential?

Easy: it's for the same reason we cannot measure the potential which appears when copper touches iron, or when salt-water touches zinc. This has nothing to do with diodes, with rectifying junctions. Even with Ge and Si, any connection produces a built-in voltage, and no diode-junction need be present: the same problem appears when a dirty hunk of p-type semiconductor is placed against an uncleaned n-type, or even touches against any metal contact. The built-in potential appears with all contacts between all possible conductors, not just diodes. Touch two metals together and they magically become charged.

The classical-physics answer goes back to the nature of thermocouples and the 'work function' voltages of metals. For example, whenever we touch copper against steel, one metal steals electrons from the other. One charges positive and the other charges equally negative. A genuine voltage appears across the junction, and an e-field surrounds the metal pieces. (This happens at constant temperature, with no thermocouple voltages produced. It's something like a self-charging capacitor.)

However, we cannot use a normal voltmeter to measure this steel/copper potential difference. Unfortunately our meter leads are made of metal! If they're copper, then we can touch our copper meter-probe against the copper piece, but when we touch the second copper meter probe against the steel piece, it produces a backwards potential which cancels out the potential we wanted to measure. The total voltage seen by the voltmeter, adding up the contributions of all metal junctions around the loop, is always zero! (If it weren't, then we'd have a perpetual motion machine.) The voltages are really there, but we can't get at them when using metal meter probes.

Think about it: all the metals near you in your environment, whenever they touch each other, are producing DC e-fields and voltages ...voltages which we cannot measure. Solder against copper, steel against chromium, stainless screws in iron plates. They're really there, but we can't use them to power anything. All these built-in potential differences are small; they're typically below 1/10Vdc but for semiconductors they're higher, and for metal/water junctions they can be as high as 6vdc.

OK, back to the diodes. When we connect a p-type and n-type semiconductor together, it always creates one of these built-in voltages of roughly 0.6v, even if no rectifying junction is formed. But then when we add copper contacts to the ends of the n- and p-type blocks, we create two new built-in voltages at those contact points. And, these metal-semiconductor voltages will exactly cancel the PN junction's built-in potential, just like with the thermocouples. The PN voltage is really in there, but we can't get at it with normal meters, since it's being canceled by the metal-semiconductor contacts.

Obviously the built-in voltage in the PN junction has real-world effects: our diodes turn on at around 0.6Vdc. All PN junctions are automatically biased by this built-in voltage. When a diode is sitting on a shelf in storage, the voltage-force sweeps carriers away from the junction, turning the diode off. If the weird potential wasn't in there, then diodes would turn on at 0Vdc, and would only turn off if reverse-biased.

Weird trivia: solar cells put out a max voltage which is determined by the built-in junction potential. But in the case of solar cells, the potential only appears on the output wires because the sunlight has shorted-out the PN junction! In other words, the output voltage is coming from the metal contacts on the solar cell, not from the PN junction. The PN junction voltage has been removed; reduced to near zero, because the junction is shorted out by the flood of electron-hole carrier pairs. The missing PN junction voltage allows all the other built-in circuit voltages to pump charges through the loop.

More weird trivia: built-in voltages of metal pairs were detected in the early 1800s with electrostatic meters. When Volta invented his electric pile, he was convinced that the battery potentials were actually these built-in potentials, rather than created by chemical reactions. In other words, Volta was an early Free Energy nutcase! He was convinced that he'd discovered a perpetual motion machine. (His great secret was to avoid copper/tin/copper/tin, but instead to stack up copper/tin/saltwater/copper/tin/saltwater.) Arguments with colleagues broke out, and eventually the scientific community determined that the battery-voltage came from active charge-pumping action. Chemical energy was being removed from the metals as they corroded, and this supplied the work provided by such a "battery." The famous electrostatic-motor clocks of DuLuc and Zamboni wouldn't run forever, but only for a few centuries until their metal layers corroded away. The Oxford Electric Bell will eventually stop.

  • 1
    \$\begingroup\$ So then how do thermocouples work given this? \$\endgroup\$ Commented Oct 11, 2017 at 20:49
  • \$\begingroup\$ @XavierHubbardAnderson noticed comment. Metal thermocouples don't use Contact Potential, since its value remains constant with temperature. Instead, a Seebeck potential appears in each metal wire having a temperature gradient. Different metals have different Seebeck voltage, so connecting them gives a measurable output. But this only applies to metals. If we make thermocouples out of semiconductor, then the Contact Potential changes hugely with temperature. In other words, Peltier cooling modules are not based on Seebeck potentials, they're actually based on Volta/Galvani potentials. \$\endgroup\$
    – wbeaty
    Commented Nov 17, 2018 at 9:07

The key concept you need to think about is the Fermi potential. In thermal equilibrium the Fermi potential throughout both materials must be equal. Contacting the n-type and p-type regions together causes a transition region to be formed - " the depletion region" this is caused by the necessity of charge balance (no external fields nor charge generation in the bulk). This simply means that there will be as many holes on one side as electrons on the other and the doping levels determines the extent (size) of the region. Far away for the depletion region the fermi level is close to the band edges and this "built-in voltage" is shielded. Close to the depletion region the bands bend as they transition from P-type to n-type. The barrier is formed from the band edge displacement.

While an electron transiting the system would experience this, you cannot directly experience it as your probe will affect the Fermi level, changing the system.

It's probably best to think in terms of the space charge, the electric that is generated and the voltage is the integral of the field.

Qualitatively the built in voltage arises because the fermi level far away from the depletion region must be close to the doping level. Since one doping is close to the valence edge and the other doping is close to the conduction band edge there must be a potential difference.

  • \$\begingroup\$ Can you suggest good readings on Fermi Potential? I understand everything in terms of space charge EXCEPT for what happens when metal makes contact with the edges of the PN-block, ie: when you try and measure the voltage across it. Thanks for this explanation! \$\endgroup\$ Commented Jun 6, 2014 at 23:27

Allow me to make several unsubstantiated assumptions that might or might not be correct (this is my disclaimer).

I think this problem is a lot easier to think of from a chemistry point of view (see the picture below). Just some background information about how I think of a p/n junction and the standing electric field that is created. A molecule has electrons & protons & when these are not equal there is an overall charge. Molecules generally like to have 8 electrons in their valence shells (for atoms of lower atomic weight) which is discussed here. Since Si-P (common n-type material) has 9 total shared e- it is not at its lowest energy level (most stable configuration). Since Si-B (p-type) has 7 shared e- it is not at its lowest energy level either. So, Si-P transfers 1 e- to Si-B (creating an electric potential in the process) and everyone is happy. Other Si-P molecules can continue transferring e- to other Si-B molecules until the E field opposing their transfer gets too big.

Your question as I understand it is: why or why not is there a potential & voltage drop?

More of my background understanding: voltage is just a measure of how much e- wants to go from one place to another.

My answer (with a question): why would e- want to leave a full valence shell thus recreating 2 molecules with unstable valence shells? Yes it is true that e- wants to go to molecules with positive charges but my totally unfounded belief is that the desire to remain in a stable molecule is a stronger force than the desire to go to a positively charged molecule.

Since no e- wants to flow, no external voltage is created. I believe the reason you could still say that it has a potential is due to the standing electric field that is created across the juction boundary (depletion region).

pn junc drawing

On a mostly unrelated side note, the way solar cells create an external voltage is that the sun rays dislodge e- around the doped interface and the e- is swept across the boundary by the standing E field. Since there are an excess of e- in the Si-P region now, if you connect a wire between the 2 sides, the e- would really like to flow back towards the Si-B region (that is now positively charged since its e- was previously swept across the interface).

Hope it helps, I welcome any criticism/correction of my explanation.

  • \$\begingroup\$ The chemistry perspective is helpful!! I've only ever seen this from an EE point of view. However, I understand conceptually why a space charge region exists in the PN-Junction. The bit I don't understand is why you cannot measure this built in voltage with an external meter! There was an answer on another question, but I wasn't satisfied with it. \$\endgroup\$ Commented Jun 6, 2014 at 23:32
  • \$\begingroup\$ The quantities of electrons are almost identical in both the SI-B & Si-P regions after the p/n junction has stabilized so no voltage would show up on a voltmeter. The electric field across the p/n junction remains b/c their are 9 protons (& 8 e-) in Si-P & 7 protons (& 8 e-) in Si-B setting up the charge differential (& E field along with it). I think it is important to understand the difference b/t the quantity of electrons & the charges of the molecules. \$\endgroup\$ Commented Jun 8, 2014 at 17:35
  • \$\begingroup\$ Is that true? the Law of mass action says that there should be less electrons in the P-doped region than in the N-doped region by several orders of magnitude. p*n = n_i^2 has to be conserved, right? so if we are introducing holes, we are removing electrons and vice-versa. Anyhow, I don't think the answer lies with the concentrations of charge carriers. I think it has to do with ohmic contact and schottky barriers. I'm just not sure what it is specifically yet. \$\endgroup\$ Commented Jun 8, 2014 at 20:50
  • \$\begingroup\$ That stuff is above my head, sorry I can't help anymore. Also, you should probably accept answers to some of your questions. \$\endgroup\$ Commented Jun 9, 2014 at 15:09
  • \$\begingroup\$ oh! i didn't know to do that! thanks for telling me! \$\endgroup\$ Commented Jun 9, 2014 at 18:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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