pn junction section 2.2-2.3. tentative schedule #datedaytopicsection 1 1/14tuesdaydiagnostic test l...
TRANSCRIPT
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PN Junction
Section 2.2-2.3
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Tentative Schedule
# Date Day Topic Section1 1/14 Tuesday Diagnostic Test L 1/14 Tuesday Lab protocol, cleaning procedure,
Linus/Cadence intro
2 1/16 Thursday Fundamental concepts from Electric Circuits
3 1/21 Tuesday Basic device physics 2.1L 1/21 Tuesday I-V characteristics of a diode
(Simulation)
4 1/23 Thursday Drift/Diffusion current
5 1/28 Tuesday Physics of PN junction diode 2.2-2.3
L 1/28 Tuesday Diode logic circuit 6 1/30 Thursday Applications of diodes: diode
logic/Review7 2/4 Tuesday Test #1 L 2/4 Tuesday Diode Logic 8 2/6 Thursday Class Canceled!
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Important Dates
• 2/4: Test #1• 2/6: Class Canceled!
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Review
• Intrinsic Semiconductor• Extrinsic Semiconductor• Currents
• Drift Current• Diffusion Current
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At T=0K Electrons gain thermal energy and break away from the bonds. They begin to act as “free
charge carriers”—free electron.
Not A Whole Lot of Free Electrons at Room Temperature
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Phosphorus has 5 valence electrons. The 5th electron is “unattached”.This electron is free to move and serves as a charge carrier.
Add Phosphorous to Silicon to Create an silicon
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if we dope silicon with an atom that provides an insufficientnumber of electrons, then we may obtain many incomplete covalent bonds.
A boron has only 3 valence electrons and can form only 3 covalent bonds. Therefore, it contains a hole and is ready to absorb a free electron.
Add Boron to Silicon to Create a p-type Silicon
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Two Ways to Produce Currents
Mechanism: Electric Field Mechanism: ConcentrationGradient
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Drift Current
Drift current is composed of the drift current due to holes and the drift current due to electrons.Drift current is caused by the presence of an electric field.
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if charge carriers are “dropped” (injected) into a semiconductor so as to create a nonuniform density. Even in the absence of an electric field, the carriers move toward regions of low concentration, thereby carrying an electric current so long as the nonuniformity is sustained.
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Diffusion current due to Holes
Where does the – sign come from?
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Diffusion Current Due to Electron
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What do we get by introducing n-type and p-type dopants into two adjacent sections of a piece of silicon?
(p-type)(n-type)
Cathode Anode
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IS=Reverse Saturation=leakage current
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P side is suddenlyjoined with the n side
Each e- that departsfrom the n side leavesbehind a positive ion.
Electrons enter the P side and create neg.ion.
The immediatevincinity of the junctionis depleted of freecarriers.
Creation of Depletion Region
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Electric field within the depletion region points from the left to the right.The direction of the electric field make it difficult for more free electronsto move from the n side to the p side.
Equilibrium does not mean that there is no movement of carriers, but instead We have the gradient to push holes to the left.
E is there to push the drift current to the right.
PN Junction Without Bias Voltage
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Electric Field/Voltage
Definition of Voltage: The work done in moving a unitpositive charge in an electric field.Alternative definition:
-+ Vo
Caution:You can’t useVo as a battery!!!
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Net charge =0 Net charge =0
E depends on the net chargeincluded in the imaginary surface.
Extra Credit:Derive Built in Voltage
(P is neutral, even though it carries 5 electrons, one of them being a free electron.)
(B is neutral, even though it carries 3 electrons. )
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Different ways of Crossing PN Junction
np=ni2
DiffusionDiffusion
Majority carriers cross the pn junction via diffusion (because you have the gradient)Minority carriers cross the pn junction via drift( because you have the E, not the gradient)
Drift Drift
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PN Junction under Reverse Bias
Reverse: Connect the + terminal to then side.
Depletion region widens.Therefore, stronger E.
Minority carrier to cross the PN junction easilythrough drift.
Current is composed mostly of drift current contributedby minority carriers.
np to the left and pn to the right.
Current from n side to p side,the current is negative.
E
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PN Junction as a capacitor
As the reverse bias increases, the width of the depletion region increases.
Smaller capacitance.(More charge separation)
Large capacitance.(Less charge separation)
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c02f25
Bias dependent capacitance.Useful in cell phone applications.
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Forward Bias Diode
Depletion region shrinks due to charges from the battery.The electric field is weaker.Majority carrier can cross via diffusion;Greater diffusion current.Current flows from P side to N side
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Equilibrium Forward Biased Diode
Majority carriers cross the junction via diffusion.Minority carriers increased on both sides of the junction.
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nn,f enters the p side as minority carriers (np,f). np,f will recombinewith the pp,f, which are abundant.
(gradient of minority carriers)
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In the vincinity of depletion region, the current consists mostly of minority carriers because you have the gradient!
Away from the depletion region, the current consists mostlyOf majority carriers.At each point along the x-axis, the two components add upTo Itot. (This is the bottom line)
ID must be constant at all points along x
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IS=Reverse Saturation=leakage current
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Measure Forward Biased Diode Current
Listed R1=330 Ohms, Measured R1=327.8 Ohms, % error=-0.66 %
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Measured Value (Forward Bias)
VF
(V)
IF (Computed)
0.455 30.50 uA
0.509 0.10 mA
0.551 0.26 mA
0.603 0.77 mA
0.650 2.10 mA
0.70 5.74 mA
0.748 13.8 mA
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Measured Diode Voltage
400 440 480 520 560 600 640 680 720 760 8000
3
6
9
12
15
D
iod
e C
urr
en
t (m
A)
Diode Voltage
Measured DataBarrier Potential is ~ 665 mV
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Reverse Biased Diode
IS=Reverse Saturation=leakage current
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Dynamic Resistance
VF
(V)
IF (Computed)
0.70 5.74 mA
0.748 13.8 mA
Dynamic Resistance from the measurement:(0.748-0.70)/(13.8 mA-5.74 mA)= 48 mV/8.06 mA =5.95 Ohms
From the manufacture’s specification=8.33 Ohms, using data from 0.7V and 0.725 V in Figure 4.
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If VD is less than VD, On, the diode behaves like an open circuit.The diode will behave like an open circuit for VD=VD,on
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Reverse Bias
Measured R2 is 0.997 MOhms. % Error is about -0.3 %
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Reverse Bias
VS
(Measured)
IR
(Computed)
5 3 nA
10 3 nA
15 3 nA