ECE 3455: Electronics. Diode Physics: A Brief Tour. CurrentVoltage Characteristics. Figure 3.7 The i– v characteristic of a silicon junction diode. We’ll try to justify this…. …but we’ll leave this alone until Solid State Devices.
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Diode Physics: A Brief Tour
CurrentVoltage Characteristics
Figure 3.7 The ivcharacteristic of a silicon junction diode.
Well try to justify this
but well leave this alone until Solid State Devices.
Figure 3.8 The diode iv relationship with some scales expanded and others compressed in order to reveal details.
Semiconductors of practical interest are generally crystalline, meaning their atoms are arranged in periodic arrays.
This figure shows the diamond lattice: if the atoms are carbon, this is diamond; if they are Si or Ge we get the corresponding semiconductors (silicon or germanium).
If the atoms alternate between Ga and As, we get GaAs (gallium arsenide), one of the important optically sensitive semiconductors (LEDs, photodetectors).
Conduction (i.e., current flow) can take place via movement of either electrons or holes when a voltage (i.e., and electric field) is applied. A hole is the absence of an electron and moves when successive electrons take its place.
Often, either electrons or holes dominate the current; the other current component is then very small.
We can add either electrons or holes to increase the conductivity. This process is called doping and involves intentional addition of impurities to the semiconductor.
Electrons are added to Si by doping with a Group V (periodic table) element, typically P but also Sb. These have a fifth valence electron which is very loosely bound to the lattice and can move if a voltage is applied. The P or Sb atoms are donors.
Holes are added to Si by doping with a Group III element, typically B but also Al. The B or Al atoms are acceptors.
ptype
ntype
Lets make a diode!!
Ingredients: mostly holes
Ingredients: mostly electrons
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ntype
Electrons have been added via doping with donors.
Holes have been added via doping with acceptors.
Here is a pn junction diode
and the corresponding circuit symbol.
Why do electrons (or holes, for that matter) move in a semiconductor in the first place? There are two possibilities
Diffusion: if there are more electrons (or holes) in one place than another, they will tend to diffuse to even out the concentration.
Due to the concentration gradient, electrons will tend to move toward the right.
Drift: if an electric field is applied (by applying a voltage), electrons will move toward more positive potential.
e

+
ptype
ntype
ptype

ntype
+
So when p and n meet
holes here diffuse to nside
electrons here diffuse to pside
But that sets up a positively charged region on the ntype side and a negatively charged region on the ptype side... (but see note below)
eo
which generates an electric field that opposes further diffusion.
The charge shown in the lower figure is not due to electrons and holes. Once the electrons get to the ptype side, they disappear when they find holes there (the electrons and holes recombine); the same goes for holes on the ntype side. The charge is due to donors (on the nside), which become positively charged when they lose their electrons (because they move to the other side!!), and to acceptors (pside), which become negatively charged when they lose their holes. The donors and acceptors do not move; they are bound to the lattice, and are responsible for the charged regions shown.
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In fact, electrons and holes never stop moving to the other side via diffusion, and the electric field never stops pulling them back to where they started. But eventually we reach a balance between diffusion and drift; this is equilibrium. (We are assuming there is no applied voltage here.)
eo
Were here
Electron drift
Electron diffusion
Hole diffusion
Hole drift
e
e
Another way to look at this is the following: at equilibrium, there is an energy barrier keeping more (net) electrons from getting to the ptype side. But we could get another electron to diffuse over if we add energy somehow.
Electron energy (Joules)
diffusion
If we add energy, this electron can diffuse to the pside.
extra energy
weak electron
p
n
Instead of adding energy, could we reduce the energy barrier?
Now it takes less energy to get the electron over the barrier. The number of electrons that get over the hill increases exponentially with voltage.
Electron energy (Joules)
Similar arguments can be made for holes.
e
ef
Va
How can we reduce the barrier? We can apply a voltage!!!
The increase in the number of electrons that can get over the barrier is exponential: the current goes as eVa/VT, where VT is the thermal voltage
Note that Va generates an electric field that is opposite in sign to eo. So the net field reduces to ef, and the barrier to diffusion drops.
What if we reverse the sign of Va?
e
er
Va
Now Va generates an electric field that is in the same direction as eo. So the net field increases to er, and the barrier gets big. Theres no diffusion now.
But whats been happening to the drift current?
ACME
Red Arrows
(Box of 25)
Wile E. Coyote Theorem
small field
ef
big field
er
The drift current doesnt change very much with field, because of the Wile E. Coyote Theorem: regardless of how big the cliff is (no matter what voltage is applied), Wile E. will either fall, or not. Electrons that do fall contribute to the drift current just the same, whether the field is big or small.
e
Now we put this all together:
e
er
ef
v > 0
v < 0
Forward bias: the diffusion current, which is positive, increases exponentially with voltage. The drift current is negligible.
Reverse bias: Diffusion is off: the only current is the drift current, which is small and negative.
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What about that threshold voltage thing?*
eo
It turns out that the equilibrium electric field corresponds to a voltage of about 0.8 V or so. When the applied voltage is about 0.6 0.7 V, we start to see a significant current**. So we can approximate a threshold voltage Vth as being about 0.7 V in many cases.
*Dont confuse the threshold voltage (Vth or Vf in your textbook) with the thermal voltage VT; these are not the same thing!!
**In fact, the applied voltage does not all go to changing eo. Some of it is dropped across the equivalent resistance of the n and ptype regions. So at Va = 0.7 V, something less than that is appearing across the barrier and reducing it enough to generate a noticeable current.