Download Semiconductor p-n junction diodes

Semiconductor p-n junction diodes
p
n
p-n junction formation
p-type material
n-type material
Semiconductor material
doped with acceptors.
Semiconductor material
doped with donors.
Material has high hole
concentration
Material has high
concentration of free
electrons.
Concentration of free
electrons in p-type material
is very low.
Concentration of holes in
n-type material is very
low.
p-n junction formation
p-type material
n-type material
Contains
NEGATIVELY
charged acceptors
(immovable) and
POSITIVELY charged
holes (free).
Contains
POSITIVELY charged
donors (immovable)
and NEGATIVELY
charged free electrons.
Total charge = 0
Total charge = 0
p-n junction formation
What happens if n- and p-type materials are in close contact?
p-type material
n-type material
Contains
NEGATIVELY
charged acceptors
(immovable) and
POSITIVELY charged
holes (free).
Contains
POSITIVELY charged
donors (immovable)
and NEGATIVELY
charged free electrons.
Total charge = 0
Total charge = 0
p- n junction formation
What happens if n- and p-type materials are in close contact?
Being free particles, electrons start diffusing from n-type material into p-material
Being free particles, holes, too, start diffusing from p-type material into n-material
Have they been NEUTRAL particles, eventually all the free electrons
and holes had uniformly distributed over the entire compound crystal.
However, every electrons transfers a negative charge (-q) onto the pside and also leaves an uncompensated (+q) charge of the donor on the
n-side.
Every hole creates one positive charge (q) on the n-side and (-q) on the
p-side
p- n junction formation
What happens if n- and p-type materials are in close contact?
p-type
n-type
Electrons and holes remain staying close to the p-n junction because
negative and positive charges attract each other.
Negative charge stops electrons from further diffusion
Positive charge stops holes from further diffusion
The diffusion forms a dipole charge layer at the p-n junction interface.
There is a “built-in” VOLTAGE at the p-n junction interface that prevents
penetration of electrons into the p-side and holes into the n-side.
p- n junction current – voltage characteristics
What happens when the voltage is applied to a p-n junction?
p-type
n-type
The polarity shown, attracts holes to the left and electrons to the right.
According to the current continuity law, the current can only flow if all
the charged particles move forming a closed loop
However, there are very few holes in n-type material and there are
very few electrons in the p-type material.
There are very few carriers available to support the current through the
junction plane
For the voltage polarity shown, the current is nearly zero
p- n junction current – voltage characteristics
What happens if voltage of opposite polarity is applied to a p-n junction?
p-type
n-type
The polarity shown, attracts electrons to the left and holes to the right.
There are plenty of electrons in the n-type material and plenty of holes in
the p-type material.
There are a lot of carriers available to cross the junction.
When the voltage applied is lower than the built-in voltage,
the current is still nearly zero
When the voltage exceeds the built-in voltage, the current can flow through
the p-n junction
Diode current – voltage (I-V) characteristics
Semiconductor diode consists of a p-n junction with two
contacts attached to the p- and n- sides
p
n
V
0

 qV  
I = I S  exp 
 − 1
 kT  

k – Boltzmann constant
T – junction temperature (K)
Note that at room temperature,
(kT/q) ≈ 0.026 V
IS is usually a very small current, IS ≈ 10-17 …10-13 A
When the voltage V is negative (“reverse” polarity) the exponential term ≈ -1;
The diode current is ≈ IS ( very small).
When the voltage V is positive (“forward” polarity) the exponential term
increases rapidly with V and the current is high.
The I-V characteristic of the diode

I = I S  exp

 qV  

 − 1
 kT  
IS
At room temperature, a simplified form of the I-V characteristic can be used:


 V

I = I S  exp 
−
1


0.026




where V is in [Volts] and
the term kT/q was substituted with 0.026V
The experimental I-V characteristic of a Si diode
p- n diode circuit notation
p
n
When “plus” is applied to the p-side,
 qV

exp 
− 1
 kT

the current is high. This voltage
polarity is called FORWARD.
When “plus” is applied to the n-side,
the current is nearly zero. This voltage
polarity is called REVERSE.
IS
p- n diode applications:
current rectifiers
 qV

− 1
exp 
 kT

+
-
+
-
IS
Voltage
Current
Time
Time
Why do we need rectifiers?
Hydroelectric Power Stations produce AC voltage
A flux of water
rotates the coils of
the AC generators
and produces AC
voltage
Many appliances need DC voltage sources
Diode Rectifiers
p- n diode applications:
Light emitters
P-n junction can emit the
light when forward biased
+
p-type
n-type
Electrons drift into p-material and find plenty of holes there. They
“RECOMBINE” by filling up the “empty” positions.
Holes drift into n-material and find plenty of electrons there. They also
“RECOMBINE” by filling up the “empty” positions.
The energy released in the process of “annihilation” produces
PHOTONS – the particles of light
Photon Energy – Light wavelength – Light Color
Visible light - that which is detectable by the human eye - consists
of wavelengths ranging from approximately
0.780 micrometer down to 0.390 micrometer.
The photon energy - wavelength relation:
1.24
EPH [eV ] =
λ [ µm]
Semiconductor Bandgap – Light Color
Lighting the Future by Solid-State Lighting
p- n diode applications:
Photodetectors
+
-
P-n junction can detect light
when reverse biased
p-type
n-type
When the light illuminates the p-n junction, the photons energy RELEASES free
electrons and holes.
They are referred to as PHOTO-ELECTRONS and PHOTO-HOLES
The applied voltage separates the photo-carriers attracting electrons toward
“plus” and holes toward “minus”
As long as the light is ON, there is a current flowing through the p-n junction
Photodetector applications- Optical
communications
Photodetector applications- Security systems
Solar Cells
“Solar revolution" is the idea that one day we will all
use free electricity from the sun.
On a bright, sunny day, the sun shines approximately
1,000 watts of energy per square meter of the planet's surface
Solar cells (contined)
1967 - Soyuz 1 is the first manned spacecraft to be powered by solar
cells
The spacecraft crashed during its return to Earth. One of
the reasons: the left solar panel deployment failure
Electrical circuit with p-n diodes
0.035
RB

 qVD  
− 1
I = I S exp 

 kT  

0.03
I
0.025
VR
0.02
0.015
E
VD
0.01
0.005
0
Typical LED bias circuit
(forward bias applied)
0
0.2
0.4
0.6
VD
P-n diode I-V characteristic
The p-n diode I-V does not follow the Ohm’s law. The KVL:
E = VR + VD = I R + VD

 qVD  
− 1
I = I S exp 

 kT  

0.8
Solving a circuit with a p-n diode
method 1: varying VD
Assuming we know the diode voltage VD,
RB
we can calculate the diode current.
Knowing the current we can find the VR.
VR
E
Then we can find VR +VD
VD
The sum must be equal to E.
If it is not, then the assumed VD was wrong.
We need to change the VD and try again.
E = VR + VD = I R + VD

 qV  
I = I S exp  D  − 1
 kT  

Solving a circuit with a p-n diode
method 1: varying VD
MATLAB code and results
% Circuit with p-n Diode
% Method 1: varying VD
clear all
close all
%Device and circuit parameters
E=10;
RB=400;
Is=2e-12;
kTq=0.026;
%Solution
Vmax=0.75;
VD=0:0.001:Vmax;
I=Is*(exp(VD/kTq)-1);
VR=I*RB;
Vtot=VD+VR;
dE=abs(E-Vtot);
[Eop,m]=min(dE);
%Results
VDop=VD(m)
Iop=I(m)
VRop=I(m)*RB
%Validation
Eop=VRop+VDop
Results:
VDop = 0.6030
Iop = 0.0236
VRop = 9.4489
Eop = 10.0519
Solving a circuit with a p-n diode
method 2: Load line
Considering VD as a variable
RB
we can calculate the diode current ID.
Assuming we know VD, we can find the
resistor current IR using KVL.
VR
E
VD

 qV  
I D = I S exp  D  − 1
 kT  

VR E −VD
IR =
=
RB
RB
In the load line approach two currents: ID and
IR are plotted as functions of VD
In the series circuit, IR = ID
The intercept of the two plots provides the
operating current.
Solving a circuit with a p-n diode
method 2: Load line
MATLAB code
% Circuit with p-n Diode
% Method 2: Load Line
clear all
close all
%Device and circuit parameters
E=2;
RB=20;
Is=2e-12;
kTq=0.026;
Vmax=E;
VD=0:0.001:Vmax;
ID=Is*(exp(VD/kTq)-1);
% Load-line
IR=(E-VD)/RB;
dI=abs(IR-ID);
[dImin,m]=min(dI);
IDop=ID(m)
IRop=IR(m)
VDop=VD(m)
plot(VD,ID,VD,IR,VDop,IDop,'ro')
xlabel('Voltage, V')
ylabel('Current, A')
axis([0 E 0 0.25])
grid on
Solving a circuit with a p-n diode
method 2: Load line
MATLAB results
0.25
Current, A
0.2
0.15
0.1
0.05
0
0
0.5
1
Voltage, V
1.5
2