I know this question sounds silly, as if there was a potential difference a current would be created when the terminals are connected together and this would mean energy has come from somewhere.

The reason I ask this though is that from my understanding of the depletion region and built in potential of a diode it seems like if you connected a voltmeter across the whole diode it would show the value of the built in potential.

This is explained in the image below:

pn junction under equilibrium bias

At first, electrons flow from the n type to the p type because there are a higher concentration in the n type, and holes do vise versa. This is called the diffusion current. The first electrons and holes to cross the pn boundary are the ones which are closest to it; these carriers recombine when they meet each other and are then no longer a carrier. This means there is a depletion region of no carriers near the pn boundary. because electrons have left the n type material, and holes have left the p type material, there is a surplus of positive and negative charge on the n and p side of the pn boundary respectively. This causes an electric field that opposes the diffusion current, and so no more electrons or holes cross the boundary and combine. In short, only the electrons and holes near the boundary combine, because after they have done that an electric field is formed that prevents any more carriers from crossing. The current due to this electric field is called drift current, and when in equilibrium this will equal the diffusion current. Because there is an electric field at the boundary (pointing from the positive charge to the negative charge) there is an associated voltage. This is called the built in potential.

If you sample the electric field at each point along the diode from left to right, you would start with 0 in the p region because there are an equal number of protons and electrons. As you approach the depletion region you would see a small electric field pointing back towards the p region, caused by acceptor impurities which now have an extra electron (due to recombination) and therefore now have a net negative charge. This electric field would increase in strength as you get closer to the boundary, and then die away as you get further away.

This electric field means there is a voltage, as shown in graph (d). The p side is at an arbitrary potential, and the n side is at a potential higher than this because there is an electric field between them. This means there is a potential difference across the depletion region; this is known as the built-in potential.

But why when I connect a volt-meter across the whole diode will I not see this built in potential?

enter image description here

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    \$\begingroup\$ I've found an answer on wikipedia but I do not understand it at all. After 3 years of studying EE and classes studying electromagnetism and maxwell's equations I thought I understood what voltage was. Turns out I don't :( \$\endgroup\$ – Blue7 Apr 14 '14 at 5:37
  • \$\begingroup\$ Heck, that's an intimidating Wiki page. I'm going to have to read it again in the morning :) If it makes you feel better, I've been an EE for a decade, and have a good physics background, but I didn't know this... \$\endgroup\$ – bitsmack Apr 14 '14 at 6:47
  • \$\begingroup\$ Because of the law "conservation of energy". Otherwise, we would had had an infinity energy source by simply placing billions of diodes on a silicon chip. \$\endgroup\$ – hkBattousai Apr 14 '14 at 9:05
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    \$\begingroup\$ Consider that a voltmeter does not measure the electric field itself. Ask yourself, "Self, if it doesn't measure the electric field, what does a voltmeter actually measure, and why do we use it rather than a real electric field meter?" \$\endgroup\$ – Adam Davis Apr 14 '14 at 13:48
  • \$\begingroup\$ possible (actually an essential) duplicate of Internal difference in a diode \$\endgroup\$ – Alfred Centauri Apr 14 '14 at 14:13

I think, the answer is relatively simple. Do you know the working principle of a "Schottky diode", which is based on a semiconductor-metal junction? Now - what happens if you connect a voltmeter (or any other load) across the diode? You create two Schottky junctions which exactly compensate the diffusion voltage inside the pn diode. Thus, no voltage can be measured. With other words: You cannot use the diffusion voltage to drive any current through an external load.

  • \$\begingroup\$ It seems like the answers vary on different versions of this question, but I like this answer best. And I don't know the working principle of a schottky diode, can you please explain or link a simple explanation? What happens when you connect a p or n type material to a normal conductor? Another question, is the wikipedia link that I mentioned in the comments nothing to do with the answer? \$\endgroup\$ – Blue7 Apr 14 '14 at 17:53
  • \$\begingroup\$ AS I have mentioned, it is a metal-semiconductor junction. See wikipedia under "Schottky diode". \$\endgroup\$ – LvW Apr 14 '14 at 19:43
  • \$\begingroup\$ I've been reading about metal-semiconductor junctions, and I'm now getting a better understanding of why you can't measure the built in potential. Just to clarify though: Is there an electric field across a metal-semiconductor junction? \$\endgroup\$ – Blue7 Apr 14 '14 at 22:56
  • \$\begingroup\$ Besides Schottky diodes, the same is true of a copper-iron junction, or zinc-acid junctions, etc. A genuine voltage may exist at the junction, but a real-world voltmeter has probes made of metal, and it always forms at least one unwanted junction having opposite voltage! For metals and semiconductors all at the same temp, unwanted junction voltages will exactly cancel the diode voltage, producing an incorrect voltmeter reading of zero. (Heh, for zinc and water you'll detect a voltage, but it will be wrong by over four volts, depending on which type of metal probe touches the water.) \$\endgroup\$ – wbeaty Jun 30 '19 at 8:45

Err, the rest of the answers seem a little dodgy and I just stumbled upon this question so I'll take a shot at it.

I think it's because of the fact that the Fermi level becomes discontinuous under bias. I'm sure you can visualise that what the voltmeter is really measuring is how badly electrons and holes want to cross the junction. At thermal equilibrium, the electrons and holes have no intention of moving across the junction, so the voltage is 0V. IN other words, the voltmeter really only measures the difference in the Fermi levels between the 2 sides.

To understand why it does this, you have to know how a voltmeter works. Rather than literally measuring the difference in the energy level of an electron at both ends of the diode (which would be awesome), it just measures the current flowing through its high resistance. In a diode at thermal equilibrium, there's no net movement of any charge carriers and so there's no current. No current means no voltmeter reading.

  • \$\begingroup\$ Sorry to chime in here, but your answer would imply that if I had some device to measure voltage without driving any current, that I would in fact measure a voltage. I'm being hypothetical here of course, but you're saying that the fact that you connected metal to silicon is what causes the voltage to disappear; correct? \$\endgroup\$ – user2662833 May 24 '18 at 1:52
  • \$\begingroup\$ That is correct. I tried to focus my answer on why specifically a voltmeter would not register a reading without having to go into the behaviour of metal-semiconductor contacts. Of course one line of reasoning would be that there is simply no potential difference between the anode and cathode of a diode once metal contacts are installed since there would be voltages at the contacts that when summed are equal and opposite to the voltage across the depletion region. \$\endgroup\$ – Dr Coconut May 25 '18 at 2:27
  • \$\begingroup\$ Awesome :) I really appreciate you taking the time to respond to me. Have an awesome one! \$\endgroup\$ – user2662833 May 26 '18 at 13:18
  • \$\begingroup\$ "Dodgy" is putting it mildly. You've got it right. Heh, just make your voltmeter leads out of long strands of p- and n-type semiconductor, so there are no junctions formed at the probe tips! Oops, the voltmeter still must contain an internal pn junction between its leads, and that junction is oriented opposite to the diode being measured. Hence, the voltmeter reads zero, even though hundreds of mV may actually exist between its leads! So, must use an electrometer-type voltmeter, a fieldmill voltmeter (or, rotate the diode at high RPM, to remotely measure its e-field capacitively.) \$\endgroup\$ – wbeaty Jun 30 '19 at 8:45

It is a very nice curiosity question! Same question came up to me when I was in my second year. But until I came across the Threshold voltages in Transistors and PN junction voltage drops, the picture became little clear.

You are absolutely right (last paragraph), because there is a change in potential due to electric field in the depletion region, there is higher potential from n-type side and negative potential from p-type side, making the intrinsic potential difference built up. That is why, to allow the current to flow through diode (PN junction) you would need higher potential from P-type and n-Type such that their difference is larger than the intrinsic potential difference which is in opposite direction to applied voltage across diode. This is what we call forward biased diode! I am sure you know this basics. Now lets go to the real question ->

If you were to probe your virtual Digital voltmeter exactly at the two depletion boundaries then I am sure you would see the voltage difference there, but its quite impossible to do with the regular multimeter. I am sure there are ways that semiconductor companies have special probes to sense these voltage differences. But if you were to measure the disconnected diode from your regular multimeter (same this is taken in consideration when you simulate it in LTSPICE that the probing is done at the ends of the diode not internally). Basically, your Graph (D) has this answer it self. Graph shows that both ends of diode have no electric field present. since the Electric field is conservative, and two diode ends (ends of P and N type materials) have no charge and electric fields at the ends are cancelled due to diffusion, as a results there is no electric field present at after the diffusion region ends, that means their difference is also 0 and measured voltage difference is 0 V as well. Hope this helps!


If you had an electrostatic voltmeter with a resistance much higher than your D.U.T. Series Resistance, which is possible, but the diode leakage would have to be equally high to prevent discharging the Static Potential.


Giving this question a shot. There are two types of currents at a PN junction. Diffusion currents are caused by carriers moving down a carrier density gradient. Drift currents are caused by carriers moving down an electric field. When no bias is applied to an isolated pn junction, the diffusion current moves carriers across the depletion region, building up charges on each side of the depletion region. The accumulated charges create an electric field across the depletion region, and this electric field induces a current in the opposite direction. The process naturally tends toward an equilibrium in which the diffusion current is exactly canceled by the drift current. One could model this as two equal valued current sources connected in an anti-parallel fashion. If one were to connect a volt-meter across such an anti-parallel connected pair of current sources, one would measure 0 volts.


The answer is quite simple.The barrier potential exist across the depletion region not across the diode, so the region of existence of electric field lines is limited to the depletion region only.

The multi-meter used is connected across the terminals of the diode. And there exist n and p regions between multi-meter probe and depletion region.The unbiased n and p region acts as an insulator so as result no field lines is received at the probes so no voltage is shown in multi-meter.


The answer is quiet simple: You confuse the electrostatic potential with the electric potential. What you measure with a voltmeter is a difference in the electric potential.

The electric potential however does include the chemical potential of charge carriers. Note: The chemical potential µ, or more precisely the gradient -grad(µ) of the chemical potential, is the "driving force" behind diffusion.

In the case of a PN junction, net diffusion of carriers occurs until the difference of the electrostatic potential between the two conductors equals the difference of the chemical potential between the two conductors in magnitude. Since both potential differences have opposite signs, their sum is zero -> there is no electric potential difference to measure, despite a non-vanishing difference in the electrostatic potential!


Though there is a potential barrier across the pn junction point,it is unable to send any current in output circuit.as there is no other sources are present, the wire must be heated.experimets show that it never happens.otherwise,the junction should be cool as there is no external source.so there will be created a thermal instability.so the current must be zero.the contact potential of metal and semiconductor neutralizes the potential barrier.so the above type of case happens.


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