I am learning the basics of semiconductors. Going through diode equivalent circuits, the material says that Ideal Diodes have a barrier potential of 0 volts. Why is this?

I have understood what a pn junction diode is and how it works, but I cant wrap my head around this.

  • \$\begingroup\$ There is no "why"-- it's a definition. \$\endgroup\$ Oct 31, 2016 at 11:10
  • \$\begingroup\$ An ideal conductor would have zero resistivity. An ideal insulator would have zero conductivity. You can buy them of course just like you can't buy an ideal diode. \$\endgroup\$
    – Andy aka
    Oct 31, 2016 at 11:11
  • \$\begingroup\$ Because ideal diode junctions are formed using lightly doped unobtanium instead of a regular semiconductor material such as silicon. \$\endgroup\$ Oct 31, 2016 at 11:17
  • \$\begingroup\$ I've got some unobtanium but you cannot have any. \$\endgroup\$
    – Andy aka
    Oct 31, 2016 at 11:21
  • \$\begingroup\$ @Andyaka I swap some for my flux capacitor , had to scrap the Delorean :-(. \$\endgroup\$ Oct 31, 2016 at 11:27

4 Answers 4


An Ideal Diode isn't a semiconductor. It isn't a vacuum tube either.

It's the theoretical idea of what a perfect diode would look like if you ever managed to find one. It would be a perfect conductor to currents flowing one way, but a perfect insulator to currents trying to flow the other way.

Such a thing doesn't actually exist, just like you can't find an Ideal Capacitor or Ideal Inductor anywhere.


Because the "ideal diode" is meant to only model the most important characteristic of a diode, which is its rectification effect. The forward voltage is considered a secondary effect that is ignored due to idealization. At higher voltages (24V upwards), which were common in the days electrical circuits started spreading everywhere, the drop can in fact often be ignored.

If you are using a diode as a voltage drop element, and it is always forward biased, you idealize it in a different way, as a voltage source of 0.6 to 0.7 volts. In this case, for theory it doesn't matter that your implementation of the voltage source is a diode, so you call it an (ideal) voltage source instead of an ideal diode.


See the excellent answer by Micheal Karcher, which is spot-on.

I'll add something about modelization, which is a concept that probably you still have to grasp fully, at least judging from what you say in your question.

Keep in mind that the formulas and graphs you use to describe, understand and predict the behavior of a component (and of any physical systems) are just mathematical models which have been validated experimentally to provide some given degree of accuracy.

The Shockley equation of a PN diode is a model, as well as the ideal diode model. And there are more complicated models too (for example those used in SPICE programs to model a real diode). It's just a matter of how accurately you want to describe the behavior of the diode.

As said by Micheal in his answer, the ideal diode model just describes the most basic behavior of a diode when used in applications where it is switched ON and OFF and where its voltage drop, or its dependence on current (and other factors such as temperature), are not worth considering.

  • Ideal Potential for an electron in a semiconductor to escape orbit and create a DC voltage would be zero if the orbit radius was zero. It would conduct perfectly in one direction only at 0V. When reverse biased it acts as a dielectric (capacitance with voltage dependency, ie. Varicap)
    • "All insulators are dielectrics" ( and visa versa.. with limits on breakdown)

The actual zero current diode voltage is the thermal voltage Vt is "approximately" 25.85 mV at 300 'K, or "room temperature" commonly used in device simulations. This is based on the Shockley diode equation ( named after the inventor).

  • However my Rule of Thumb for Si is typically 0.6 ~ 0.7V unless higher currents are involved. For small signals I might estimate 0.65V and expect 1V+ at high currents after memorizing a few datasheets.

  • But when you get down to picoamps of current, you can imagine that diode noise current starts to dominate this DC voltage and then the equivalent series resistance is in the > megaohm range so radiated noise can easily be detected (RF).

Side notes

  • More useful to learn is that the difference between the Base Emitter diode and the Base Collector Diode voltage drop is the junction Vce(sat) at rated current when forward biased.

enter image description here

This difference Vce(sat) is what separates a good "switch", from one that heats up at high current with higher bulk resistance. This is where the diode goes from an exponential curve to a linear property of a resistor from diode saturation which dominates the behavior of all transistors, as a switch.

  • This value Rce is given by Diodes Inc in its switching transistor types in the x to xx milliohm range but typical small signal transistors are related by their case power dissipation and chip size often in the range> 1 Ohm as the incremental V/I slope or ESR. It is always rated at Ic/Ib=10 and for better devices some datasheets may indicate Ic/Ib=20 to 50. But this tends to be 10% of hFE in the linear gain region.

    • But few others if any, give this "bulk resistance" parameter based on the V/I slope, so I just call it Rs or ESR.

    • The same ESR characteristic applies to LED's as well, which of course are also Diodes. Here I plotted the VI curve then computed the slope as ESR for an LED.

    • Here's something you can ask your prof to prove.

    • Prove why is the Rs of the diode at saturated rated current is equal to the inverse of the diode max steady Power Rating. ( +/-50%)
    • This is something I discovered from many years of experience and is due to optimal diode package design with thermal resistance of the package for a given chip size.

enter image description here

See the blue bar above . 
Compute VI power and compare with ESR. What is the Pd rating and ESR?
  • For the advanced reader only,
    • they can read the graph I created from a datasheet VI curve for a Toshiba White power LED
    • @2 Watts, ESR = 0.5Ω and @3W ESR < 1/3Ω . In both examples Pd=1/ESR.
    • This has a wide tolerance, but still useful no less.
    • (I do not know any journal references other than my own to support these findings.
    • Yet it very handy to remember once you understand how to estimate the "knee" of the zener-like curve at 10% Imax.
    • Also Zzt is a same ESR value given for Zeners at rated current and room temp
    • my method applies to all diodes , as a "Figure of Merit" if Pd*ESR<1. enter image description here

"in theory, everything depends on practice."


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