Let's try to wrap this up

January 20, 2013 4:21:58 PM CST

All right.. last time I went through a series of contortions that produced the following equation to describe the current in a diode's depletion region:

  • B = A * e^-(C/k)

The actual diode equation is:

  • pn = pp * e^-Vbi/Vt

where:

  • pn is the hole concentration in the N-type material
  • pp is the hole concentration in the P-type material
  • e is the 'natural logarithm'.. roughly 2.71828
  • Vbi is the 'built-in voltage' of the depletion zone.. usually about .6v
  • Vt is the 'thermal voltage' for silicon.. roughly 26mV

About those voltages:

Vbi you already know.. it's the voltage that comes into existence as diffusion current pulls carriers out of their native material and leaves the dopant atoms behind as fixed charges. It depends on the dopant concentrations in the P-type and N-type materials, but in most semiconductors it runs about .6v.

Vt takes a bit more explaining because it's only counts as a voltage due to a technicality.

The expanded version is:

  • Vt = kT/q

where:

  • k is Boltzmann's constant.. 1.38 * 10^-23 Joules per degree Kelvin
  • T is the temperature of the silicon in degrees Kelvin
  • q is the charge of an electron.. 1.6 * 10^-19 Coulombs

It's basically a way of describing a carrier's thermal energy in terms of voltage. The temperature tells you how much energy is in the system, Boltzmann's constant tells you how much of that energy goes to a single carrier, and the electron charge is there so our answer ends up having the units of "Joules per Coulomb", which is another way of saying "Volts".

In physical terms, voltage is the amount of energy a charged particle gains gains or loses when it moves in an electrical field. If the particle moves with the field, it receives a certain amount of energy per unit distance. To push the particle against the field, you have to exert a certain amount of energy per unit distance.. which is a lot like what we're trying to with the whole diffusion-and-drift thing.

Vt says that one degree Kelvin worth of thermal energy gives a particle as much energy as an 87 microvolt electric field.

Degrees Kelvin are the same size as degrees Centigrade, but '0' on the Kelvin scale is at absolute zero. Most components are rated for performance in cold (-40C, 235K), room temperature (25C, 300K), and hot (125C, 400K) conditions.

We include Vt because the extra thermal energy makes it easier to bump electrons up into the conduction band. That matters because semiconductors have resistance, and resistance generates heat. The more current you push through a semiconductor, the hotter it gets. The heat changes the way the device behaves, and under certain conditions can destroy it. That's why the CPU in your computer has a heatsink and/or fan.

At 25C, the thermal voltage is about 25.8mV, which gets rounded to 26mV.

Vt controls the diffusion current, Vbi controls the drift current. The ratio between them represents the balance of currents in the depletion zone.

Adding an external voltage:

As we've already seen, an external voltage doesn't have any effect on the diffusion current, but adds to or subtracts from the drift current.

Since the most interesting diode behavior happens when the external voltage works against the built-in voltage, we write the equation like so:

  • pn = pp * e^(Vf-Vbi/Vt)

where Vf is the external (forward) voltage.

An aside about exponentials:

Mathematicians call the exponential a 'fast-moving' function.

When you compare the speed of two functions you see which one gets big faster. I'm not talking "five times bigger" either.. I'm talking "so much bigger you can't even measure the gap in terms of the smaller function." Compared to a fast function, you can safely round a slow one down to zero.

The exponential e^x is 'fast' compared to almost any other function you're likely to know. The only common function that grows faster than the exponential is the factorial, and most people forgot what that was as soon as they left 8th grade algebra. To get really fast functions, you need special machinery like Knuth's up-arrow notation.

Since e^x gets large faster than almost everything else, 1/e^x (usually written e^-x) gets close to zero faster than almost everything else.

And just as a reminder, the basic rules of exponents say e^0 equals 1.

To show the exponential changing from positive to negative I have to use an extra variable, so the function I'll graph is e^ax, for the variable 'a' having values of 1, 0, and -1. I tried drawing it by hand but eventually gave up and did it in software:

plots of exponentials

  • The red trace is e^ax for a=1: e^x
  • The purple trace is e^ax for a=0: e^0
  • The blue trace is e^ax for a=-1: e^-x

The graph shows e^x for values of x between -3 and 3. For reference, e^1 is about 2.7, e^2 is about 7.4, e^3 is about 20, and it just keeps getting steeper from there.. e^14 is a bit north of a million.

How does this relate to diodes?

That sudden change from "gets small really fast" to "gets large really fast" is what makes diodes work.

  • If Vf is smaller than Vbi, Vf-Vbi will be negative, so the exponential will stay in the "gets small really fast" zone.
  • When Vf equals Vbi, the exponential reduces to e^0.
  • When Vf is larger than Vbi, Vf-Vbi becomes positive, moving the exponential into the "gets large really fast" zone.

So what happens?

The exponential represents the number of holes that diffuse all the way through the depletion zone to the edge in the N-type material, which are called 'injected minority carriers'. As the drift current (which holds them back) falls, more and more of them can escape the depletion zone and recombine with free electrons in the N-type material's bulk.

That creates another concentration imbalance, which causes diffusion, but this time there are no uncovered fixed charges or drift current to oppose the diffusion current. Instead, there's a general increase in free holes throughout the bulk, and new carriers diffuse in from the wire at the end of the bulk.

There's a similar equation for electrons passing through the depletion zone and recombining with free carriers in the P-type material, and the mechanics of carrier motion work the same.

Conceptually, a diode's depletion region is sort of an electrical black hole. Carriers that get close to it just sort of disappear, creating a vacuum that pulls in more carriers from the rest of the circuit. In reality, electrons from the conduction band in the P-type material move to the valence band in the N-type material, but it's more fun to think of it as electron current and hole current annihilating each other.

Getting down to hard numbers:

If you recall, the exponent in the current equation was -(Vf-Vbi)/Vt, and that Vt is about 26mV at room temperature. Thing is, Vt describes the thermal energy in the lattice, and the thermal energy also produces reverse saturation current by causing spontaneous electron-hole pair generation inside the depletion zone.

...

Now there's a sentence I never expected to write for a casual audience.

...

Anyway, the depletion zone is a small target, so the reverse saturation current lives somewhere in the 10^-13 amps neighborhood.

That gives us one diode property that's really small, and another that gets big really fast. With some appropriate manipulation of the equations for diffusion current in the bulk, we can treat the exponential equation as a scaling factor for the reverse saturation current, which gives us the "Ideal Diode Equation":

  • Id = Is * (e^Vf/Vt - 1)

where

  • Id is the current through the diode
  • Is is the reverse saturation current
  • Vf is the forward voltage applied to the diode
  • Vt is the thermal voltage

and I've always thought the '- 1' was someone's idea of a joke.

Let me demonstrate with an example.. if we set Vf to .6v and Is to 10^-13 amps, the numbers crunch out as follows:

  • Vf/Vt = .6v/.026v =~ 23
  • e^23 =~ 9744803446
  • e^23 - 1 =~ 9744803445

which just makes all the difference in the world, dunnit?

To finish the job:

  • 10^-13A * 9744803445 =~ .000974, or about 1mA.

Now, for the sake of comparison, let's say our estimate of Is was 1% off:

  • .99 * 10^-13 * 9744803445 =~ .000984

which is still about 1mA, but the difference due to a 1% change in Is is several orders of magnitude larger than the error from ignoring the '- 1' would be.

The joke is that in the real world, Is can vary by a factor of 5 from one diode to the next. It's not that the fabs aren't doing a good job -- in point of fact they control their processes incredibly well -- it's just that the quantities they're trying to control are incredibly small. A few more atoms of dopant in one place or another can make a significant difference in the final device.

The good news is that even large changes in Is have a manageable effect on a diode's overall performance.

Let's say we have a diode whose Is is ten times as large as the one in the previous example:

  • 10^-12 * 9744803445 =~ .009748, about 10mA

Pretty damning, right?

Thing is, that's only the difference in current when Vf is .6v. To get back down to 1mA, we just have to lower Vf.

Offset compensation in diodes:

To compensate for Is being ten times as large, we have to make the exponential multiplier ten times smaller. Turns out we can do that just by finding the exponent for e^n = .1, then multiplying that by Vt.

Crunching those numbers gives us a value of about 60mV.

Let's test that:

  • Vf/Vt = .54v/.026v =~ 20.7
  • e^20.7 =~ 977002725
  • 977002725 / 9744803446 =~ .1
  • 10^-12 * 977002725 = .000977, about 1mA again.

so compensating for a 10-to-1 variation in reverse saturation current costs 60mV of forward voltage. Yes, that's a 10% change from the original .6v, but you could flip that reasoning around and say that the forward voltage necessary to get 1mA from any diode will probably be within 10% of .6v.

We can extend the idea even farther though..

60mV is nothing more than ln(10) * .026mV. Adding and subtracting exponents ends up multiplying or dividing the final result, so adding or subtracting 60mV to Vf will always change the diode current by a factor of 10.

Again, let's test that:

  • Vf/Vt = .72v/.026v =~ 27.7
  • e^27.7 =~ 1071413585016
  • 10^-13 * 1071413585016 = .10714, about 100mA.

So adding 120mV increases the current about 100 times.

  • Vf/Vt = .78v/.026v =~ 30
  • e^30 =~ 10686474581524
  • 10^-13 * 10686474581524 = 1.068, about 1A.

And adding 180mV increases the current 1000 times.. just as predicted.

No matter what Vf and Id happen to be for a specific diode, increasing Vf by 60mV will always multiply Id by 10 and lowering Vf by 60mV will always reduce Id by a factor of 10.

At long last, the payoff:

By playing with the numbers, we can generalize that idea to some useful rules of thumb:

  • Adding 60mV to Vf will multiply the current by 10.
  • Adding 42mV to Vf will multiply the current by 5.
  • Adding 18mV to Vf will multiply the current by 2.
  • Adding 10mV to Vf will increase the current by 50%.
  • Adding 5mV to Vf will increase the current by 20%.
  • Adding 1mV to Vf will increase the current by 4%.

and you can combine those to get surprisingly good back-of-the-envelope calculations. Want to triple the current? Double it (18mV) then increase it by 50% (10mv) for a total of 28mV.

Checking it:

  • .028/.026 =~ 1.08
  • e^1.08 =~ 2.944, which comes pretty close.

Just to be sure:

  • .628/.026 =~ 24.15
  • e^24.15 =~ 30775969150
  • 10^-13 * 30775969150 = .00307, about 3mA

yep, it works.

The Grand Finally:

(old theatre joke.. it's the part of an opera where the audience says, "whew.. finally!")

Trying to explain that table of rules, dear friends, is what led me down this rabbit hole however many months ago.

Henceforth, anyone who dares to ask, "well yeah, but where did you get those numbers?" will be forced to read the whole damn saga until they can recite it from memory.

Random brain cookies:

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