# The Diode Model

## by Jamie Honan

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Essential to the running any sort of simulation are the
questions:

What am I trying to model?

How well does my model fit the actual device?

For what range of what parameters is my model valid?

It is difficult to get any meaning from a simulation unless you
have some understanding of the model used. These notes attempt
to help understand the diode model used in the `analog' component
of the Chipmunk package.

The equation used to model the diode is:

Id = Is * (exp(Vd * lamda) - 1) * BLEND +
(1 - BLEND) * ((Is * Vd) / Vcrow) * exp(lamda * Vcrow)

where
BLEND = 1 / (1 + exp(lamda * Vd) * exp(-lamda * Vcrow))
or     = 1 / (1 + exp(lamda * (Vd - Vcrow)))

lamda = 1 / (n * (k * T)/ q)

Id is the current flowing through the diode
Vd is the voltage difference between diode terminals
The value  (k * T)/q is a constant, generally taken to be

This leaves the variables n, Vcrow and Is to be explained.

Before I do that, the model itself needs some explanation,
particularly in the light of simple models such as

Id = Is * (exp(Vd * lamda) - 1)

which is the traditional model used.

The traditional model does not take into account two aspects
of static diode behaviour.

The exponential rise predicted by exp(lamda * Vd) covers diode
behaviour up until large voltages, which in our model are
voltages above Vcrow. The exponential function also does not
cover the reverse breakdown voltage.

The large forward behaviour is modeled by our model somewhat
better. Instead of exp(lambda * Vd) rising ever onwards and
upwards, there is a kink in diode behaviour at Vcrow. Beyond
this voltage, the current flowing through the diode behaves
in a more resistor like manner. This voltage, Vcrow, is fixed
for a particular diode.

This kink is achieved in our model by the BLEND function.
There are two distinct parts to the model, the part multiplied
by BLEND and the part multiplied by (1 - BLEND).  BLEND is
controlled by exp(Vd - Vcrow). When Vd equals Vcrow, BLEND is
0.5. When Vd is much larger than Vcrow, the exp(Vd - Vcrow) becomes
exponentially larger, making BLEND tend towards 0. In this case
the first part of the model becomes insignificant, and the model
tends towards ((Is * Vd) / Vcrow) * exp(lamda * Vcrow).  Vcrow is
a fixed value, as is Is and lambda, so this part of the model is
simply a constant times Vd.

When Vd is very much smaller than Vcrow, the exp(Vd - Vcrow)
becomes exponentially smaller, with BLEND tending towards 1.
In this case, the first part of the model dominates, with the
second part dropping out, our model returning to the traditional
model described above.

The reverse breakdown voltage is not modeled by our system, (in
common with the traditional model) so the model is only valid up
until this reverse breakdown voltage.

This still leaves us with the variables n and Is.

The variable n is the `emission coefficient' or `ideality factor',
with values around 2. This is related to the doping levels of the
materials used in the pn junction. Later we'll see how to derive
this number.

`Is' is the leakage current, or reverse saturation current.

How do we tie these variables to a particular diode's
characteristics? I have a table listing VR, IF, VF @ IF, IR @ VR
for different diodes. Crowbar voltages, n, and Is seem to
be missing.

For our model, the important values are the VF @ IF values, and
the IR @ VR values.  The IR gives the reverse saturation current,
while the reverse voltage has not yet reached reverse breakdown.
This IR value becomes our Is value directly.

The VF @ IF values give the crowbar voltage (Vcrow) and we can
derive the n variable using the current given.

To derive n, at Vd = Vcrow our model becomes:

Id = Is * (exp(lambda * Vcrow) - 1)

(At this voltage, the -1 could be dropped. The C code has the
dropped for the part of the model controlled by (1 - BLEND).)

This becomes

lambda = ln(Id / Is + 1) / Vcrow

n * (k * T)/ q = Vcrow / ln (Id / Is + 1)
n = 40 * Vcrow / ln (Id / Is + 1)

In my table for an 1N4148, IR is 0.025 uA, IF is 1 volt and IF is
10 mA.

Id = 10 * 10 ^ -3
= 10 ^ -2
Is = 0.025 * 10 ^ -6
= 25 * 10 ^ -9
Id / Is = (10 ^ -2) * (4 * 10 ^ 7)
= 4 * 10 ^ 5
n = 40 / ln ((4 * 10 ^ 5) + 1)
= 40 / 12.9
= 3.1

The value requested for nkT_q then would be 3.1 * 0.025 = 0.0775.
(crowbar voltage would be 1, Is is 0.025 uA, as indicated above).

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Email
lazzaro@cs.berkeley.edu
Phone
(510) 643 4005
SMail
UC Berkeley / CS Division / 387 Soda Hall / Berkeley CA 94720