I have wondered myself why is that value fixed to around 0.7 V (0.3 Ge). I have researched about this topic over and over again, but I always find the same answer. They say "Because the voltage for Silicon Diodes is 0.7". That is just like saying that the sky is blue because blue is the color of the sky.

I am familiar with the Shockley diode equation, but I don't see the connection with the threshold voltage (I'm saying this because people have given me a link to its Wikipedia page).

I have also read something about the concentration of impurities near the junction being related to the voltage barrier (I am hoping to get an answer related to that, and the manufacture process).

Another answer I have been given is that that is silicons nature (I kind of hate this answer, because what I get from it is that the voltage is an intensive property, instead of extensive - which would make materials more "workable").

So the question per se is: Why 0.7 and not 0.4, 0.11, 1.2 (for Silicon)?

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    \$\begingroup\$ it is not always 0.7 V. It can be 0.62 or 0.82 too. It depends on the doping concentration. The magnitude of the built-in voltage can be defined based on energy level diagram or fermi levels. electronics.stackexchange.com/questions/252702/… \$\endgroup\$
    – User323693
    Feb 16, 2017 at 3:42
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    \$\begingroup\$ I think you are talking about the equilibrium voltage. Study this page: wiki.analog.com/university/courses/electronics/text/chapter-5 And then, if you want a deeper understanding, get a book on micro-electronics and read the first few chapters. Still deeper? Get a physics book. Also, I think Feynman's lecture series is on the web -- see chapter 14 of volume 3. Also, a quantitative potential computation depends on physical construction details and temperature. \$\endgroup\$
    – jonk
    Feb 16, 2017 at 3:48
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    \$\begingroup\$ It is even worse. The forward voltage of a real diode strongly depends on forward current AND temperature. Examine, say, technical data on the very common diode 1N914 / 1N4148, fairchildsemi.com/datasheets/1N/1N914.pdf , Figures 3 to 6. You will find out that the Vf changes from 300 mV at 2 uA to 1.4 V at 800 mA. :-) :-) \$\endgroup\$ Feb 16, 2017 at 4:00
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    \$\begingroup\$ Useful values: if 0.7v at 1mA, then will be 3*0.06v=0.18v less at 1/1000th the current; thus 0.52v at 1uA; 0.34v at 1nA; 0.16v at 1picoAmp; that value of 0.06v for 10:1 change in current is temperature dependent. \$\endgroup\$ Feb 16, 2017 at 4:54
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    \$\begingroup\$ Unlike resistors, silicon conduction in a diode is exponential in current as a function of voltage. So, only a narrow band of forward voltages corresponds to measurable-but-not-destructive currents (four orders of magnitude in current corresponds to about 0.24V change). The lower limit (zero) and upper limit (bandgap voltage) leave that as a small range around 0.6 to 0.7V. \$\endgroup\$
    – Whit3rd
    Feb 16, 2017 at 10:21

3 Answers 3


A slightly-more ELI5 answer:

When we touch any two different metals together, they charge up, one becoming positive, the other negative. They form a self-charging capacitor, or something like a low-voltage battery. This effect was detected in the early days of physics, discovered during sensitive measurements of electrostatic charge. It behaved much like contact-charging of silk rubbed against rubber. But with metals, no friction was needed. Later on it became clear that two different metals always produce the same voltage between them. (Well, same at room temp. The voltage changes slightly with temperature.)

But this voltage can never be detected by normal voltmeters. We can build our circuits out of copper, aluminum, iron, etc., and for every copper-aluminum junction, there will always be an aluminum-copper junction somewhere else. The metals-charging effect might be very large, yet it sums to exactly zero around a closed circuit. The neg terminal of one "battery" always faces the neg terminal of another. It's not an energy source (not the perpetual motion machine that Alessandro Volta thought he'd discovered!)

What if we bump a slab of p-type silicon against a slab of n-type silicon? That's a self-charging capacitor, and it produces roughly 0.7V between the silicon slabs. One slab steals electrons from the other, but just until the difference in orbit-energies of the mobile carriers is cancelled out. Note that diodes needn't be formed at the contact point. Instead we could use high-doped ++p and --n "metallic" silicon which cannot form diodes, yet when touched together, the slabs still produce that spontaneous charging and that same potential-diff. We could even solder the p and n silicon together (first silver-plate the ends, so solder will wet them,) and still that same 0.7V potential appears.

Why do diodes turn on at 0.7V, rather than at zero volts? It's because the depletion layer of the diode always contains that spontaneous "differing-metals-contact" 0.7 volts inside. The voltage keeps the diode turned off. On a disconnected diode this is not a measurable voltage (you'll never detect it directly, not unless you start measuring the e-fields surrounding the diode's terminals.) Heh, if we could form diodes from iron and copper, then instead of 0.7V, those diodes would turn on at the natural iron-copper potential-difference that all iron-copper junctions exhibit.

When we apply an external voltage to forward-bias the diode junction, the diode turns on when the external voltage cancels out the constant built-in invisible voltage. In other words, diodes only turn on when we've reduced the "invisible" junction-voltage to near zero: shorted it out by applying an opposing potential-diff.

All of this connects to many other physics effects. If we make a closed ?metal? no, semiconductor ring, a half-ring of p-type connected to a half-ring of n-type, then heat one of the junctions, many mA or perhaps amps will flow, since the two "invisible" voltages are no longer the same, and the small difference produces a large current in the circuit. In other words, PN junctions' "thermocouple" voltages are just a tiny remainder of this magical "invisible voltage," the thermo-voltage only arising because of an imbalance. We only detect the imbalance, but not the original potential-difference which always appears between two materials. [EDIT: metal junction's built-in potential changes little, while much larger thermo-voltages appear in the metal legs of the junction, rather than in the junction itself. For all-metal thermocouples, the metal, not the junction, becomes the "dc generator." ]

We can create a "source of cold:" a semiconductor refrigerator. If we solder any p-type silicon against n-type, then force a reverse current through the junction, where holes flow away from electrons, then the p-to-n connection becomes cold, and the metal contacts elsewhere become equally warm. Note that no diode was formed, since two separate silicon blocks were connected by solder. Swap the polarity and instead the pn-soldered junction heats up, while the metal contacts become equally cool.

Also, this means that solar-cells don't work as most people imagine. Inside the dark solar cell, the pn junction has a natural 0.7V potential difference. Elsewhere in the circuit we find opposite differences (probably found mostly at the metal contacts to the semiconductor.) The other connections all add up to the same 0.7V, which cancels the 0.7V of the pn junction. So, when the light hits the junction, carriers flood across it, and the junction-potential ...gets shorted out! Then, all the other potential-differences from other parts of the circuit will provide the e-fields which then force the charges to flow around the circuit. An illuminated pn junction in a solar cell doesn't provide the drive voltage. Weird! Instead, the metal contacts of the wires provide the drive-voltage, and the illuminated pn junction provides a missing voltage which normally would halt any current. Missing junction-potentials are an oddity which isn't found in any normal circuit. When a voltmeter (made of junctions of copper, solder, silicon, etc.) is connected to an illuminated solar cell, the missing junction-potential of the pn junction lets us measure the total potential of all the other conductor junctions present. (Or, instead we could take the micro-view, and say that the absorbed photons are elevating the energy-level of mobile charges in the junction, allowing them to cross it, regardless of the strong e-field of the natural 0.7V trying to repel them back again.) The flood of high-energy mobile carriers have shorted out the junction, discharging the self-charged capacitor. But this also means that the V-out of a solar cell will NOT be related to the photon energy (won't give higher outupt voltage for UV, or for x-rays.) Instead, the maximum V-out of the solar-cell is just the (now missing) potential-barrier of the pn junction.

But why do two different metals charge up when touched together?

It's because even two lone metal atoms also charge up when touched together. The energy-levels of different metal atom's orbitals are not the same. If touched together, one atom tends to steal electrons from the other ...but just enough to cancel the difference in orbital levels. Rather than single atoms, if instead we used two long chains of metal atoms, one of copper and one of iron, then when their ends touched, one chain would steal electrons from the other, until the magic invisible voltage-value appeared between the chains. It's a self-charging 2-plate capacitor. Works for metals, works for semiconductors. Search term: work-function of metals, and work-function difference of metal junctions (and, Volta or Galvani potentials in electrochemistry.)

[Beware, this is a first-approximation gradeschool ELI5 answer. As mentioned elswhere here, diode's turn-on potentials are only proportional to the work-function difference, not equal to it. Disconnected diodes don't actually have zero junction current, instead they have carrier-mobility effects, equal and opposite carrier diffusion currents, etc.]

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    \$\begingroup\$ Excellent answer, wish my profs started with something like this instead of just math! \$\endgroup\$
    – mbrig
    Feb 17, 2017 at 5:52
  • \$\begingroup\$ "What if we bump a slab of p-type silicon against a slab of n-type silicon? That's a self-charging capacitor, and it produces roughly 0.7V between the silicon slabs." This statement implies that any off-shelf silicon diode would produce 0.7V standalone, if measured say, with a hi-impedance voltmeter. This doesn't sound right. \$\endgroup\$ Feb 17, 2017 at 22:48
  • \$\begingroup\$ When the metal terminals are connected to the silicon, one junction charges to ~+0.35V, the other to ~-0.35V. These exactly cancel the PN junction-potential (so if the leads are touched together, zero picoamps.) A PN diode is like one PN junction plus two Shottky metal/silicon junctions in series. Yes, an eletrometer-voltmeter should detect the 0.7V 'charge' on large silicon pieces. Make an insulated Faraday Cup, connect electrometer to the cup and to earth. Briefly stick a charged object into the cup and observe electrometer reading. Or just measure mV with a field-mill instead. \$\endgroup\$
    – wbeaty
    Feb 17, 2017 at 23:30
  • \$\begingroup\$ That's the issue. 99% of educational materials omit this circumstance that two other metal-semiconductor junctions must be present in the actual electronic device. Then the distinction between rectifying Shottky junction and non-rectifying Ohmic contact gets lost, without any explanations how the "heavy doping" (for making the contact "ohmic") is blended with presumably "normal" doping levels at the p-n junction. All this makes all web-based pictures of zero-biased diodes brutally misleading, starting with Wikipedia. \$\endgroup\$ Feb 20, 2017 at 8:14
  • \$\begingroup\$ Why does the voltage change with temperature? Separately, this description is drastically different than how the Seebeck effect is described in various places. I'm trying to noodle on whether one description is wrong or it's two completely different ways to describe the same macro effect with both still being valid. Thoughts? \$\endgroup\$
    – horta
    Aug 16, 2017 at 22:27

The voltage drop varies with temperature and you can make a good temperature sensor from a diode or transistor by measuring the drop. Calibrate with ice water and boiling water.

In the materials used for LEDs, band gap energy is also the energy of photons produced by a current. A red LED has a band gap of around 1.8 volts and the red light has an energy of around 1.8 electron volts, or a wavelength of around 700nm. You can test this with a voltmeter and a spectroscope. Likewise for IR, green, blue, and UV LEDs. The voltage drop across the diode increases as you move toward the UV, which has more energetic photons.

(Remarks about silicon deleted.)

  • \$\begingroup\$ I think you are mixing things up: Band gap is an energy not a voltage and it's numerical value for Si is 1.1eV. It is not the forward voltage of a diode. \$\endgroup\$
    – Curd
    Feb 16, 2017 at 8:51
  • \$\begingroup\$ An electron accelerated by a potential of 1 volt will have an energy of 1eV. That is why we call it an electron volt. You must be thinking of pure silicon. The band gap in a junction transistor is adjusted with the impurities used to make P-type and N-type semiconductors. Check here under the "Physics" heading en.wikipedia.org/wiki/Light-emitting_diode \$\endgroup\$ Feb 16, 2017 at 9:02
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    \$\begingroup\$ Forward voltage and band gap energy are completely different quantities. There is a relationship between the two but it's not as easy as "\$V_f\$ in volts is the same as \$E_{gap}\$ in eV". See physics.stackexchange.com/questions/177910/… \$\endgroup\$
    – Curd
    Feb 16, 2017 at 9:25
  • \$\begingroup\$ I'm not sure why this was voted down, but I'm giving it an up vote. For direct gap semiconductors (LED's and such) it's a pretty good rule. (though can be complicated by shallow dopants.) I think it fails a bit for Si because it has an indirect gap. The direct gap in Si is ~1.1 eV.. I think the indirect gap is about 0.6V.. but I could not find a reference for that. \$\endgroup\$ Feb 16, 2017 at 18:47
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    \$\begingroup\$ My mistake. I was confusing lab write-ups in the American Journal of Physics 25 or 30 years ago for LEDs and photon energy, with how I use transistors or diodes as temperature sensors. I have been through the whole thing with full Ebers-Moll for IR photo-diodes and fempto-amp currents a couple of times. Ironically, it was to read temperature of silicon wafers in RTP. \$\endgroup\$ Feb 17, 2017 at 5:37

Why [Diode Threshold is] 0.7 and not 0.4, 0.11, 1.2 (for Silicon)?

Here is an answer from electrical engineering standpoint (since this is a EE site):

There is no actual "threshold" in a forward-biased diode. The I-V curve for a forward-biased diode is a strong exponential function. The "knee voltage" (also known as "contact potential" or "built-in voltage") of 0.7V is a characteristic point in a piece-wise LINEAR APPROXIMATION of the actual I-V curve for a forward-biased P-N junction of typical silicon material with typical dopants. This is the simplest linear model, see Section 5.4 of the link suggested by "jonk". It reads:

The linear model of the diode approximates the exponential I - V characteristics by a straight line that is tangent to the actual curve at the DC bias point. Figure 5.8 shows the curve with the tangent line at the point (VD, ID). The curve intersects the horizontal axis at the voltage VD0. For small changes in VDand ID about the tangent point, the tangent line gives a good approximation to the actual curve.

This is a good first-approximation large-signal model for silicon diodes, which is widely used in EE ballpark estimations. For more accurate modeling more complex models are used as SPICE model.

The next question would be, why the I-V curve for a silicon-based diode has this particular exponential shape, such that its "knee" is located near the value of 0.7 V? The answer needs to be searched in the physics of semiconductors, in the theory of P-N junctions and transistors, and the answer will likely take several lectures. At the bottom, the property of current flow is determined by intrinsic atomic structure of the particular semiconductor with its particular band gap, (see electronic band structure), and quantum dynamics of electron-hole interactions with its crystalline structure across two differently doped regions (p and n). For a different intrinsic semiconductor material (like Germanium) with a different band parameters, the resulting linear approximation of I-V curve would yield a different knee value of about 0.3V.

The explanation of how the "contact potential" is related to bandgap voltage can be found on local Physics site. It says that typically the "contact potential" is about 0.3V less than the corresponding bandgap voltage.

  • \$\begingroup\$ I would add that there is more of a way to define the threshold voltage. Millman and Halkias, in their 1967 milestone "Electronic Devices and Circuits" call it [start of M&H Quote] "cut-in, offset, breakpoint or threshold voltage" and define it as the voltage V_gamma below which the current is very small (say, less than 1 percent of maximum rated value). (p. 128). [end of M&H quote]. If you scale your exponential graph to fit that much current range, you end up with a curve that basically rises from the V axis in correspondence of V_gamma. Like you would see on a curve tracer. \$\endgroup\$ Feb 16, 2017 at 22:24
  • \$\begingroup\$ @SredniVashtar, I totally agree that the task to approximate a strongly non-linear exponential function by a simple straight line is indeed challenging, and it is very subjective how to draw that single line. The criteria of 1% is also quite subjective. 1% of which rated value? Relative to constant DC rating, or to an impulse rating that can be 100X of that? \$\endgroup\$ Feb 17, 2017 at 2:57

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