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When you have a diode with a certain barrier voltage (e.g., 0.7 V for Si) and you apply a voltage higher than this barrier potential, why does the voltage across the diode remain at 0.7V?

I understand that the output voltage across the diode will increase as a sinusoidal input is applied until it reaches the 0.7 mark, I don't seem to understand why it remains constant after that point however.

It makes sense to me that any potential greater than this barrier potential will allow current to pass, and correspondingly, the potential across the diode should be the applied voltage minus the 0.7 V.

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    \$\begingroup\$ Who told you it was constant? \$\endgroup\$
    – Dmitry Grigoryev
    Mar 27, 2017 at 7:35
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    \$\begingroup\$ "Why is the diode forward voltage constant?" It's not, so the rest of the question is rather pointless. \$\endgroup\$
    – Olin Lathrop
    Mar 29, 2017 at 11:37
  • \$\begingroup\$ @DmitryGrigoryev in intro to electronics courses at my university at least, all diodes in homework problems and exams are constant forward-voltage diodes. \$\endgroup\$
    – taylor swift
    Mar 30, 2017 at 19:32
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    \$\begingroup\$ @taylorswift We used ideal diodes for that purpose. The advantage of an ideal diode is that you know it's ideal, so there's no room for questions like this one. \$\endgroup\$
    – Dmitry Grigoryev
    Mar 31, 2017 at 6:12
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    \$\begingroup\$ Upvoted just because it was a question I asked me years ago when having electronics courses : it is a legit question and the answers are very instructive for beginners. You should accept one of the massively upvoted answers. \$\endgroup\$
    – Benj
    Mar 31, 2017 at 9:20

8 Answers 8

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The voltage across the diode does not remain at about 0.7 V. When you increase the current, the forward voltage also increases (here: 1N400x):

1N4001 forward voltage vs. forward current

And when you increase the current even further, the power dissipation becomes too large, and the diode eventually becomes a LED (light-emitting diode) and shortly afterwards a SED (smoke-emitting diode). So a larger forward voltage cannot happen in practice.

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    \$\begingroup\$ massive upvote for the smoke emitting diode \$\endgroup\$
    – bobflux
    Mar 26, 2017 at 21:20
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    \$\begingroup\$ N.E.D. = Noise Emitting Diode. ;-) \$\endgroup\$
    – Mike Waters
    Mar 27, 2017 at 1:07
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    \$\begingroup\$ Joined just to upvote this for SED. \$\endgroup\$
    – TheValyreanGroup
    Mar 27, 2017 at 2:24
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    \$\begingroup\$ lol. it should be noted that the plot above is log current vs. linear voltage. so a straight line (at the left) is actually an exponential curve. that means the current is increasing much faster than the voltage is. so the voltation moves a little from 0.7 v, but not much before you getta SED. \$\endgroup\$
    – robert bristow-johnson
    Mar 27, 2017 at 3:23
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    \$\begingroup\$ Back in my college days ('70s) I had a roommate that bought surplus computer boards that had tons of glass diodes on them. He would clip the ends of an AC power cord across each diode in turn, place a shot glass over the diode and then plug the cord into the outlet. There was sound and light but basically no smoke as the diode vaporized. The hot glass spatter would deposit on the inside of the shot glass. After 100s of diodes there was a considerable layer built up in his shot glass. (Please avoid doing this at home it was a silly and potentially dangerous activity). \$\endgroup\$
    – Michael Karas
    Mar 28, 2017 at 11:16
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Voltage is what we can observe and measure, but what is also changing is resistance.

A diode starts out as a large resistance, as you apply voltage to it that resistance remains fairly constant until you approach the forward breakdown voltage. At that point the resistance starts to drop.

enter image description here

Past the knee the resistance is very low. Any further increase after knee causes little change in the resistance.

Since R has gone down, in order to maintain that voltage you have to increase the current... a lot. The diode has become a small resistor "switch" and can therefor be referred to as ON.

The full voltage current relationship of a diode looks like this.

enter image description here

The slope before the knee is the forward off conductance (1/R), the slope past the knee is the forward ON conductance.

The actual math is of course a lot more complicated than that, but I find this description helps folks understand.

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    \$\begingroup\$ "Past the knee the resistance is very low. Any further increase after knee causes little change in the resistance" - true, but most diodes are not operated much past the knee because it causes excessive voltage drop (and power dissipation). \$\endgroup\$
    – Bruce Abbott
    Mar 27, 2017 at 1:14
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    \$\begingroup\$ Re, "what is actually changing is resistance" Beware saying "actually." Ask a physicist what is "actually" happening, and you'll get an ear full of quantum field theory. The word "resistance" comes from Georg Ohm's model of how electicity flows in conductors. A PN diode does not really fit that model, but if it helps you to think of diodes as having variable resistance, then it's part of your model. If it works for you, then Hey! It works for you. So long as we all agree on the same I/V curve, then all is cool. \$\endgroup\$
    – Solomon Slow
    Mar 27, 2017 at 16:54
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    \$\begingroup\$ @sdpatel, Sorry, I don't know solid-state physics. I am only a software geek who sometimes tinkers with simple electronic circuits. My understanding of semiconductor diodes is limited to the idea that, so long as you don't let the magic smoke out, then the operating point will be somewhere on that fixed curve. And really, most of the time, I go by an even simpler model: The one that says, "the forward voltage will be somewhere close to N volts" (where N depends on whether it's some particular color of LED, a Schottkey diode, or a 1N400_x_.) \$\endgroup\$
    – Solomon Slow
    Mar 28, 2017 at 2:43
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    \$\begingroup\$ The V-I graph is simply wrong. It is an "artist's impression" of what changing the scale from positive (mA) to negative (uA) current would look like. And the artist got it very wrong. There are no inflection points near the origin. The curve is basically an exponential translated to pass through the origin. If you scale it right, it would appear to have a discontinuity near the origin. The artist wanted to make a pretty curve and joined the two sides with what must have seemed the prettiest wiggly line. Result: a wrong graph that is propagated to confuse students all over the galaxy. \$\endgroup\$
    – Sredni Vashtar
    Mar 28, 2017 at 17:32
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    \$\begingroup\$ Found it: electronics-tutorials.ws/diode/diode_3.html , I also tried to contact the author via electronics-tutorials.ws/contact but all I got is "Failed to send your message. Please try later or contact the administrator by another method." I hate wasting my time like this. \$\endgroup\$
    – Sredni Vashtar
    Mar 28, 2017 at 18:10
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why does the voltage across the diode remain at 0.7V?

It doesn't. Most of the time, a constant 0.7 V is good enough, just as flat earth is good enough for driving around town.

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Diodes have a logarithmic relation between current through the diode and the voltage across the diode. A ten:1 increase in current causes 0.058 volts increase across the diode. (the 0.058 V depends on several parameters, but you can see that number in lots of on-chip-silicon bandgap voltage-references].

What if the current changes 1,000:1, either increasing or decreasing? You should expect to see (at least) 3 * 0.058 volts change in Vdiode.

What if the current changes 10,000:1? Expect at least 4 * 0.058 volts.

At high currents (1 mA or higher), the bulk resistance of the silicon starts to affect the logarithmic behavior, and you get more of a straight line relation between Idiode and Vdiode.

The standard equation for this behavior involves "e", 2.718, thusly

$$Idiode = Is * [e^-(q*Vdiode/K*T*n) - 1]$$ and at room temperature and ideal doping profiles (n=1) $$Idiode= Is *[e^-Vdiode/0.026 -1]$$

By the way, this same behavior exists for bipolar transistor emitter-base diodes. Assuming 0.60000000 volts at 1 mA, at 1 µA, expect 3 * 0.058 V = 0.174 V less. At 1 nanoampere, expect 6 * 0.058 V = 0.348 V less. At 1 picoampere, expect 9 * 0.058 volts = 0.522 volts less (ending up with only 78 millivolts across the diode); perhaps this pure-log behavior ceases to be an accurate tool, near zero volts Vdiode.

Here is Vbe plot over 3 decades of Ic; we expect at least 3*0.058 volts or 0.174 volts; reality for this bipolar transistor is 0.23 volts. enter image description here

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As the other answers have explained, the voltage is not constant at 0.7V, but based on the reference to barrier potential in your question, I suppose you realize this and are asking more about the semiconductor physics behind why this happens.

The reason is that the depletion region of a diode (with zero voltage applies) creates the barrier potential, as you already noted, of about 0.7V (assuming a typical Silicon diode). As you apply forward voltage, the depletion region becomes smaller. With low voltage the larger depletion region restricts most current, and as the voltage increases, the reduced depletion region results in a reduction in resistance (and therefore increased current). This continues until approaching ~0.7V where the depletion region is very small as well as the resistance. This causes the exponential V-I relationship.

This article has some good diagrams and explanations, as does the Wiki page.

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The point is that you can't "apply a voltage higher than this barrier potential", the diode doesn't let you.

That is, the marginal impedance of the diode in conduction mode is less than the source impedance of your voltage supply: your voltage source can't drive more than "0.7V" across a 0.7V diode, so "the voltage across the diode remain[s] at 0.7V".

Of course, the marginal impedance of a diode in conduction mode is not exactly zero, so there will be some rise in voltage if your voltage supply attempts to supply more than zero current. And the marginal impedance of your voltage supply may be very low, comparable to a diode, so it may be able to boost the diode voltage up quite high before the diode fails. Those are the second-order effects. The simple model of a diode, conducting above 0.7V, is a device that limits voltage by accepting infinite current.

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Once diode is switched ON with sufficient biasing, it acts a voltage source of 0.7 or 0.6(depends on material) with a small series resistor.

So if we increase the input voltage, current across the small resistor will also increase. So as input voltage increases there is variation across output taken across diode.

Usually diode is considered to be ideal, so there is no resistor in series. So o/p voltage across diode remains constant.

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No,it will increase. But it will remain constant if ideal nature is assumed.This concept is generally taught in high schools and it is of no practical use.

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    \$\begingroup\$ Can you provide evidence to support your claim that an assumption of constant forward voltage is "of no practical use"? And it's not enough to provide one example of where this assumption is not useful, you need to show that this assumption is never of practical use, in any situation. Or retract your last statement. \$\endgroup\$
    – Elliot Alderson
    Nov 12, 2020 at 20:59
  • \$\begingroup\$ In reality, the $V-I$ graph of the diode(Forward Biased) is an increasing graph which looks like an exponential function. This is because when when the diode is forward biased, the barrier potential reduces and the drift current increases. But this does not happen instantaneously. It takes some time for this buildup. But if we assume the ideal nature(and ideal means perfect, right? ) we will get a straight line having a infinite(undefined) slope.(Remember that in reality nothing is perfect) \$\endgroup\$
    – user254781
    Nov 13, 2020 at 16:11
  • \$\begingroup\$ (EDIT): An Ideal diode is a perfect conductor when forward biased and acts as a perfect insulator when reversed biased \$\endgroup\$
    – user254781
    Nov 13, 2020 at 17:11

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