Physics 120 - David Kleinfeld - Spring 2016
1
1.1
Notes on diode analysis
Background
Mass action says that it is more probable for charge to move from a region with high carrier density
to one with lower density. This gives rise to the notion of the diode, where current can flow more
strongly in one direction than the other. The semiconductor diode is built by taking blocks of pure
semiconductor, like Si, and doping one boock so that it has a net excess of negative carriers (n-side)
and doping the other block so that it has a net excess of positive carriers (p-side) and then bringing
the blocks in contact (Fig.)
Let Nn be the number of negative change carriers (electrons) and Np be the number of positive charge
carriers (holes). Right at the junction, the extra holes in the p-side will be swept toward the n-side and
the extra electrons on the n-side will be swept onto the p-side. This results in a steady-state situation
with a junction that has no free charge and a smooth transition in potential from one side to the other.
The total potential drop, φ, can be expressed in terms of carrier density
Np (n − side)
= e−eφ/kB T
Np (p − side)
(1)
Now the current of holes that flows is
Current
∝ Hole from (p − side) moving to (n − side) − Hole from (n − side) moving to (p − side)
Np (p − side) e−eφ/kB T
∝
=
−
Np (n − side)
(2)
0
Suppose that we now apply a potential, denoted VD , to the n-side that lowers the potential; tho is
called the forward-biased condition as the applied potential is higher on the p-side than the n-side.
The current is now
Np (p − side) e−e(φ−VD )/kB T − Np (n − side)
∝
Current
∝ Np (p − side) e−e(φ−VD )/kB T − Np (p − side) e−φ/kB T
h
∝
Np (p − side) e−eφ/kB T
ih
i
eVD /kB T − 1 (3)
We can repeat this argument in terms of electrons
Nn (p − side)
= e−eφ/kB T
Nn (n − side)
(4)
The current of electrons that flows with VD > 0 is again
Current
=
Electron from (n − side) moving to (p − side) − Electron from (p − side) moving to (n − side)
∝
Nn (n − side) e−(φ−VD )/kB T
−
Nn (p − side)
∝
−(φ−VD )/kB T
−
Nn (n − side)e−eφ/kB T
∝
Nn (n − side) e
h
ih
i
Nn (n − side) e−eφ/kB T eVD /kB T − 1
1
(5)
We can add the two currents up to get
h
ih
i
I ∝ Np (p − side) e−eφ/kB T + Nn (n − side) e−eφ/kB T eVD /kB T − 1
h
i
= I0 eVD /kB T − 1
(6)
The prefactor depends on the band gap energy, Egap . The product of theNn Np can be estimated as
follows. In equilibrium, the numbers of holes and electrons are constant. There is a chance for an
electron to leave the valence band and join the conduction band, thus forming an electron-hole pair.
Thus the production of electron-hole pairs is given by the Boltzmann factor exp(Egap /kB T ) where
Egap is the gap energy, about 1.1 eV for silicon. These charge pairs will not accumulate, but reach a
steady state through the recombination of electron and hole pairs. The probability of decay depends
on the product Nn Np ,
Np Nn ≈ e−Egap /kB T
(7)
The next order to business to to use the diode in a circuit. Let’s see how we handle the nonlinearity!
2
Load line analysis for a diode circuit (I versus VD)
A plot of the current, I, versus the voltage drop across the diode, VD, will yield a straight
line, called the "Load Line", of possible values of current flow in the circuit. We form the
line by noting that I = 0 at VD = VS and that I = VD/(RS + RL) at VS = 0, as shown below:
The so called "Operating Point" is found through the second equation that relates I and
VD, namely the diode equation I = I0[exp(VD/KBT) - 1]. The I-V relation for the diode will
cross the Load Line at the Operation Point (open circle above). This provides a
graphical solution for the currents and voltages in a circuit with a diode.
What happens when the source voltage changes with time, i.e., VS = VS(t)? Here the
Load Line varies with time; the slope is constant at dI/dV = -1/(RS + RL) while the
intercept shifts, as shown below. When VS(t) varies symmetrically around zero, as with
the AC line, we see that the maximum positive value of VS(t) leads to the maximum
current flow, while the maximum negative value of VS(t) leads to a minimal current, so
that asymptotically I(VS → -∞) → -I0. The rhythmic change in Operating Point (open
circles below) is the basis of the half-wave rectifier that you constructed.
Load line analysis for a diode circuit (VD versus I)
Just to emphasize what we learned, let's look at a second slightly more complicated
load-line situation in which we have two supplies, VS and VB. We ignore the resistance
of the diode, rD, calculated as rD = (dVD/dI)-1, in writing Kirchhoff's voltage law:
For a value of RB that is too large, the operating or Q-point moves to the left and, if it
moves too close to the origin, the diode conducts poorly (and rD cannot be neglected)