Semiconductor p-n junction diodes p n
p-n junction formation p-type material Semiconductor material doped with acceptors. Material has high hole concentration Concentration of free electrons in p-type material is very low. n-type material Semiconductor material doped with donors. Material has high concentration of free electrons. Concentration of holes in n-type material is very low.
p-n junction formation p-type material Contains NEGATIVELY charged acceptors (immovable) and POSITIVELY charged holes (free). Total charge = 0 n-type material Contains POSITIVELY charged donors (immovable) and NEGATIVELY charged free electrons. Total charge = 0
p-n junction formation What happens if n- and p-type materials are in close contact? p-type material Contains NEGATIVELY charged acceptors (immovable) and POSITIVELY charged holes (free). Total charge = 0 n-type material Contains POSITIVELY charged donors (immovable) and NEGATIVELY charged free electrons. 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 p- side 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 V p n 0 qv I = I S 1 exp kt k Boltzmann constant T junction temperature (K) Note that at room temperature, (kt/q) 0.026 V I S is usually a very small current, I S 10-17 10-13 A When the voltage V is negative ( reverse polarity) the exponential term -1; The diode current is I S ( 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 qv I = I S exp 1 kt I S At room temperature, a simplified form of the I-V characteristic can be used: I V = 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, the current is high. This voltage exp qv kt 1 polarity is called FORWARD. I S When plus is applied to the n-side, the current is nearly zero. This voltage polarity is called REVERSE.
p- n diode applications: current rectifiers + - + - exp qv kt I S 1 Voltage Current Time Time
A flux of water rotates the coils of the AC generators and produces AC voltage Why do we need rectifiers? Hydroelectric Power Stations produce 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 E PH [ ev ] = λ[ µ m]
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 R B V R I 0.03 0.025 0.02 I qv I D = S exp 1 kt E V D 0.015 0.01 0.005 0 0 0.2 0.4 0.6 0.8 Typical LED bias circuit (forward bias applied) V D P-n diode I-V characteristic The p-n diode I-V does not follow the Ohm s law. The KVL: E = V R + V D = I R + V D qv I I D = S exp 1 kt
E R B V R Solving a circuit with a p-n diode method 1: varying V D V D Assuming we know the diode voltage V D, we can calculate the diode current. Knowing the current we can find the V R. Then we can find V R +V D The sum must be equal to E. If it is not, then the assumed V D was wrong. We need to change the V D and try again. E = V R + V D = I R + V D I qv I D = S exp 1 kt
Solving a circuit with a p-n diode method 1: varying V D 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
R B V R Solving a circuit with a p-n diode method 2: Load line Considering V D as a variable we can calculate the diode current I D. Assuming we know V D, we can find the resistor current I R using KVL. E I D V D qv I D = S exp 1 kt I R V = R = R B E V R In the load line approach two currents: I D and I R are plotted as functions of V D B D In the series circuit, I R = I D 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 0.2 Current, A 0.15 0.1 0.05 0 0 0.5 1 1.5 2 Voltage, V