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Introduction Chapter 5 Section 1 Copyright © 2016
The McGraw-Hill Companies, Inc. Permission required for presentation or display
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Introduction: Diodes Diodes: two-terminal devices
In semiconductors, made by junctions between two different materials Homojunctions: junctions between two differently doped regions of the same semiconductor material Heterojunctions: junctions between two different types of materials (usually means semiconductor materials) Metal-semiconductor junctions But not all metal-semiconductor junctions are diodes
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Diodes Diodes can act as switches
For one polarity of applied voltage, current flows (nearly a short circuit) For other polarity, current does not flow (nearly an open circuit) Silicon homojunction: turns on at around 0.7V
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Circuit equivalent Open circuit (for V<0.7 V)
Looks like a power supply: 0.7 V and supplies current
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Circuit symbol Ideal diode: More realistic silicon diode:
For V<0, no current flows For V>0, short circuit More realistic silicon diode: For V>≈0.7 V, current flows (current increases exponentially with voltage) For V<0.7 V, current flow is neglible
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Rectifier circuit example
Let input be AC signal of 60 Hz, with 5 volt amplitude Take output voltage across the resistor Let R=2KΩ Let the diode be represented by equivalent circuit shown Plot the input and output waveforms
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Input voltage Current: Is zero until input reaches 0.7 V Then Voltage:
Remains 0 until input reaches 0.7 V Then Vout= Vin-0.7V until Vin returns to 0 (0.7 V dropped across the diode) Vout stays at zero until Vin>0.7 V Current: Is zero until input reaches 0.7 V Then
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How we’ll proceed Operation of semiconductor diode junctions will be explained using energy band diagrams Draw diagrams in electrical neutrality first, lining up vacuum levels Let regions come into contact (in your imagination); charges flow until equilibrium is reached At equilibrium, Ef is a constant Redraw with Fermi levels lined up Then we’ll investigate influence of applied voltages
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Prototype pn homojunctions
Junction between n-type and p-type of same kind of semiconductor Example: start with a p-type substrate Implant phosphorus atoms (donors) in certain region Junction occurs along B-B’ cross section
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Doping profile Background acceptor concentration is constant (NA)
Typical donor concentration profile shown We’ll approximate as step function Junction occurs at x0
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Step junction model Assume doping profile is a step function
On the n-side, ND’=ND-NA (constant) On the p-side, NA’=NA-ND (constant) Assume all impurities are ionized Then n0 on the n side is n0=ND’ Then p0 on the p side is p0=NA’ Neglect impurity-induced band gap narrowing for this model If either side of junction is degenerate, take Ef to be at band edge EC0 or EV0 if ND’ or NA’ is greater than 1018 cm-3
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Key points A pn junction will be a diode
Two-terminal device Diode conducts current in one direction but not the other Will analyze using step junction approximation Will understand using energy band diagrams
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Prototype pn Junctions (Qualitative)- Energy Band Diagrams Part A
Chapter 5 Section 2.1 Copyright © 2016 The McGraw-Hill Companies, Inc. Permission required for presentation or display
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Introduction We’ll assemble the energy band diagram of a pn junction
There will be a built-in voltage There will be an internal electric field Two parts: Part A: Equilibrium, Part B: under bias This section is all qualitative- understand the physics
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Step 1: Electrical neutrality
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Step 2: Join sides When the sides are “joined,” charges flow from one side to the other As electrons flow, they leave behind (positively charged) ionized donors Donors cannot move Holes flow, leaving behind (negatively charged) ionized acceptors Separation of charges sets up electric field
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Currents Drift currents Diffusion currents
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Equilibrium
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The built-in field Near interface, electrons from n-side fill holes on p-side Results in negative charges on p-side (un-neutralized acceptors) Electrons annihilate holes that diffused into n-side Results in positive charges on p-side (un-neutralized donors) Results in a built-in electric field near junction
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The band edges are parallel
Quantities γ, ϕ, and χ are constants of the material Therefore EC is parallel to Evac and EV An electron at EC on n-side must acquire χ energy to escape to vacuum level on n-side- but to escape to vacuum level on p-side must also acquire additional energy qVbi
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The space charge region
On each side is a region of uncompensated charge: space charge region Width on each side depends on impurity concentration on each side and amount of charge needed to equalize the Fermi levels
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The depletion region In the space charge region, there is an electric field Will accelerate mobile charges out of region Mobile charges are depleted here Space charge region also called the “depletion region”
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The built-in voltage Total space charge on either side of the junction is equal (but with opposite sign) There exists a built-in potential energy barrier qVbi Vbi is called the built-in voltage Always taken as positive (by convention)
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The junction width Potential energy barrier is the same in EC as for EV Thus barrier is the same for electrons as for holes If doping concentration is same on both sides (NA’=ND’), width of junction is same on both sides When doping is unequal, width is wider on the more lightly doped side
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The band bending As doping increases, built-in voltage increases
Work functions increase, so bands have to bend more to align Fermi levels
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The electric field Electric field proportional to slope of Evac
Is steepest at metallurgical junction Built-in voltage is mostly on lightly-doped side
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One-sided junction When one side is degenerately doped, most of junction appears on lightly doped side Called “one-sided junction”
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Key points Constructed the energy band diagram under equilibrium
There is a built-in voltage created by uncovered donors and acceptors Space charge region There is an electric field in space charge region that sweeps out electrons and holes, depleting them Space charge region= depletion region Next: what happens when you apply a voltage?
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Prototype pn Junctions (Qualitative)- Energy Band Diagrams Part B
Chapter 5 Section 2.1 Copyright © 2016 The McGraw-Hill Companies, Inc. Permission required for presentation or display
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Introduction We saw how to draw the energy band diagram at equilibrium
What happens when we apply a bias?
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Apply a voltage Va Applied voltage by convention measured from p to n
When VA is positive, junction is forward-biased When VA is negative, junction is reverse-biased
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Reverse bias Let Va =-1 V Consider pn junction in three regions: n, p, and junction On n-side number of donors is equal to number of electrons (no field)- hence called quasi-neutral region There is another quasi-neutral region on p-side Depletion region contains ions but virtually no free carriers Equilibrium diagram repeated here
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Consider diode as series connection of three resistances of these regions
On the n-side: ρn is inversely proportional to nn0 (subcript n means on n-side) On p-side: ρp inversely proportional to pp0 In transition region, almost no free carriers Resistance therefore high Therefore applied voltage dropped almost entirely across transition region
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To adjust energy band diagram
Use Evac at metallurgical junction as reference Pivot bands around this point The p-side is electrically more negative, so represents a higher potential energy for electrons Thus p-side moves up on diagram The n-side moves down
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The junction voltage Junction voltage Vj is increased
(increased because recall Va is negative) Greater field at junction
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The junction voltage Greater field means more charge on either side
Charges are un-neutralized donors and acceptors Space charge region gets wider (to uncover more of them) Mobile carriers are swept out by field
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Under forward voltage Now Va is positive
Space charge region gets smaller Evac not shown-no new information there
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Key points Learned how to draw energy band diagrams under bias
Applied fields change internal electric fields Applied fields change width of depletion region Next: what about the current?
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Description of Current Flow in a pn Prototype Homojunction
Chapter 5 Section 2.2 Copyright © 2016 The McGraw-Hill Companies, Inc. Permission required for presentation or display
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Introduction We previously saw how to draw the energy band diagram of a pn junction Learned how to modify it for the case of applied bias Now we’ll see how to use the energy band diagram to predict the behavior of the electrons and holes Get a qualitative description of the current
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Start with equilibrium
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Diffusion and drift Electrons diffuse to regions of lower concentration (n to p) Holes diffuse to regions of lower concentration (p to n) Electrons are accelerated by built-in field (toward downhill side) Holes accelerated by built-in field (toward uphill side) At equilibrium, these current cancel out Net current is zero
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Generation and recombination
In the quasi-neutral regions, generation and recombination rates are equal No fields. no concentration gradients, thus no current flow In depletion region, EHPS are generated– and there IS a field Electrons accelerated to left (generation current to right) Holes accelerated to right (generation current to right) Electrons in depletion region can recombine with holes in depletion region Results in regeneration current to the left In any region, net G-R current is zero at equilibrium
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Reverse bias- drift and diffusion
Electrons trying to diffuse to p-side are turned back by the field Same for holes Negligible current Small number of electrons on p-side that wander (diffuse) near junction are swept over Similarly for holes Result is a net current- but small, few carriers available Called minority diffusion current
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Reverse bias, generation/recombination and tunneling
Generation: electron accelerated to left, hole to right- results in net current Neglect recombination in depletion region- number of carriers miniscule Tunneling from VB on p-side directly into CB on n-side can happen Net tunneling current to the right Increases rapidly with reverse bias voltage
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Total reverse current Small minority carrier diffusion current
Small generation current Tunneling current can be large Multiplication current EHPs generated in depletion region can collide Crashing knocks more electrons and holes loose They also get accelerated Will discuss more later
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Forward bias Electrons trying to diffuse to p-side have lowered barrier- lots of them get across Minority electrons on p-side that diffuse into space charge region get swept across- but there aren’t many of them Result is large diffusion current
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Forward bias, recombination
Many electrons and holes crossing through depletion region- opportunities to recombine are plentiful Result is net current to the left
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More on forward bias When electrons are injected into p-side, their charge is neutralized by holes almost instantly (dielectric relaxation time) Holes are supplied ultimately by contact at p-terminal No net charge distribution (very small) so no electric field Hence p-side is quasi-neutral Similarly for n-side
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Excess carriers Excess carriers (Δn and Δp) injected across junction
They recombine as they move, so excess minority carrier concentration decays exponentially with distance away from junction Away from junction, E ≈0, so no drift- current flow is by diffusion
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Away from junction? Excess carriers have recombined, so no diffusion (no concentration gradient) We said no field in quasi-neutral region But there is a small field Resistance is small but not zero, so some voltage dropped across QNR’s Small field but low resistance means current can flow Result: in QNRs current flows by drift
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Tunneling? Not possible
No empty states in one band at same energy as occupied state in another band
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Forward current Primarily diffusion across junction
Recall carrier concentration with energy decreases as energy increases As energy barrier lowered, number of carriers that can cross increases exponentially Thus current increase exponentially with applied voltage
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Forward current summary
When forward voltage applied, electrons and holes are injected across depletion region Become minority carriers on other side Diffuse away from junction, recombine Majority carriers provided for recombination are resupplied from contacts Majority carrier current powered by drift Field small but carrier concentrations large
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Key points Under reverse bias, there is a small leakage current due to minority carriers being swept across junction Small generation current Tunneling possible at higher voltages Under forward bias, minority carriers diffuse across junction (minority carrier diffusion current) Minority carriers replaced by majority carriers Away from junction, current propelled by drift of majority carriers
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Tunnel Diodes Chapter 5 Section 2.3 Copyright © 2016
The McGraw-Hill Companies, Inc. Permission required for presentation or display
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Introduction Tunnel diodes (also known as Esaki diodes) have both side degenerately doped Ef is generally inside the conduction and valence band edges They demonstrate quantum-mechanical tunneling really exists
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Typical I-V
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Energy Band Diagram ≈10 nm
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Assumptions Current is primarily electrons from n going to empty states on p side Electrons cross the forbidden region but remain at same energy (conservation of energy) Electrons occupy states at same K before and after tunneling
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I-V characteristic
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Figure of Merit Peak-to-Valley ratio
Values of IP/IV ≈ 20 have been observed in GaAs tunnel diodes
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Can be used as an oscillator
Negative resistance can be used to make an oscillator Pretty stable Can be used as a switch Valley current tends to increase over time (degradation)
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We said K was conserved Implies direct-gap semiconductors
In Si (indirect), this assumption is violated Negative resistance is still observed Peak-to-valley ratio is much lower (≈4) Probably simultaneous emission or absorption of phonons Ge is indirect but has a relative minimum at K=0 Peak-to-Valley ratios ≈16
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Key points When both sides of pn junction are heavily doped, can get tunneling Results in negative resistance region in I-V
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Prototype pn Homojunctions (Quantitative): Energy Band Diagram at Equilibrium-Step Junction
Chapter 5 Section 3.1 Copyright © 2016 The McGraw-Hill Companies, Inc. Permission required for presentation or display
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Introduction We examined qualitatively how current flows in pn junction Reverse bias Forward bias Considered Majority and minority carrier currents Drift and diffusion Generation and recombination Tunneling Touched on multiplication current Next: quantitative analysis
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Find Vbi Recall Express as where on the n-side (non degenerate)
If n-side is degenerate Take Ef=EC and δn=0
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Similary for p-type Non-degenerate Degenerate: δp=0
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Combine To get For one-sided junctions (n+-p or p+-n)
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For non-degenerate semiconductor
And we had Recall Thus
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For pn junction Can also express Vbi as
This says that as ND’ and NA’ approach NC and NV, Vbi approached the band gap If you include impurity-induced band gap narrowing, could be slightly less than Eg
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Example Find the built-in voltage for a silicon homojunction with ND’=1016 cm-3 and NV’ = 1015 cm-3 Solution: The onset of degenerate doping in Si is ≈4x1018 cm-3, so neither side is degenerately doped Therefore use
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Solution If you have studied diodes in circuits class, you will recognize that this is close to 0.7 V, the typical turn-on voltage for silicon diodes
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What if it were a one-sided junction?
Then we’d use one of these Could happen a lot- let’s plot these to avoid repeating the same calculations over and over
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Vbi for one-sided junctions
Curves for n+-p and p+-n are indistinguishable
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Note If impurity-induced band gap narrowing is considered, reduce result from previous plot by amount of apparent band-gap narrowing ΔEg*
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Key points We derived an expression to find Vbi
The built-in voltage depends on the doping on both sides of the junction Found Vbi≈0.7 V for a typical silicon diode Next: quantitative energy band diagram with applied voltage
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Energy band diagram with applied voltage: Part A
Chapter 5 Section 3.2 Copyright © 2016 The McGraw-Hill Companies, Inc. Permission required for presentation or display
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Introduction Here we will quantitatively determine the features and shapes of the energy band diagram for a pn homojunction We’ll start with variation of charge with position Use that to find the electric field Use the field to calculate the variation of voltage V(x) with position Potential energy is qV(x)– gives us the band shapes
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Coordinate system
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Charges in a step junction
We approximate the charges on either side as constant The n-side is more heavily doped Donors (ND’) are positively charged Acceptors (NA’) are negatively charged No uncompensated charges outside depletion region
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Find the electric field E
Use Poisson’s equation (works for equilibrium and also under bias) Here Qv(x) is the charge per unit volume as a function of position; ε is permittivity Need to know all the charges Electrons, holes, ionized donors, ionized acceptors
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Very few electrons or holes in depletion region
Depletion approximation: n=p=0 Next, write down QV(x) On the n-side On the p-side Outside depletion region, QV =0
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Integrate to solve On n-side On p-side
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But electric field has to be continuous
We had Set the solutions equal at x=x0 Charge: Field:
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Space charge widths Earlier we predicted that the junction would extend further into lightly doped side? Let wn be the width of the depletion region on the n-side, e.g. wn=x0-xn and wp be the width on the p-side From We can write Checks out!
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Next find voltage distribution
We use For n-side, integrate And get For p-side
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Find the total voltage Voltage accumulated on n side: On p-side
Total voltage is
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Voltage has to be continuous
Set solutions equal at junction Take previous results and take ratio Recalling , we get Most of voltage dropped across more lightly doped side
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Plot of voltage
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Key points We started with the charge on either side of the junction
We integrated the charge to get the electric field (Poisson’s Equation) We integrated the field to get the voltage Next episode: We’ll find the widths of the space charge region We’ll calculate the actual shape of the energy band diagram
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Energy band diagram with applied voltage: Part B
Chapter 5 Section 3.2 Copyright © 2016 The McGraw-Hill Companies, Inc. Permission required for presentation or display
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Introduction In Part A we found the charge, field and voltage
Next, junction widths and energy band diagram
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Find expressions for junction widths
Combine: To find
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Similarly for p, and total width
For the p-side And the total junction width is pn junction
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We had Solve for junction voltage Vj
For one-sided junction, e.g. n+-p, w≈wp since in this case ND’>>NA’ and pn junction (n+-p) (p+-n)
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Maximum electric field
Combine To find
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Potential energy related to electric potential
And EC is the potential energy for electrons in the conduction band Thus, EC(x) looks like Vj(x) except inverted And Ev is parallel to EC
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So resulting energy band diagram is
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Junction width for one-sided junction as function of voltage
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Junction width for one-sided junction as function of doping
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Key points We quantitatively derived expressions for the charge, electric field, voltage, potential energy, and energy band diagram for a step junction Charge was constant on either side (step junction) Integrate charge to get the field Integrate the field to get the voltage Potential energy is qV Potential energy is Ec (for electron in CB) Band gap is constant so EV is parallel to EC
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Final result
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Current-Voltage Characteristics of pn Homojunctions (Excess minority carrier concentrations)
Chapter 5 Section 3.3 Part A Copyright © 2016 The McGraw-Hill Companies, Inc. Permission required for presentation or display
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Introduction We’ll obtain I-V characteristics Assumptions:
pn junctions first One-sided junctions later Assumptions: Minority carrier concentration is much less than majority carrier concentration In bulk, majority carrier concentration ≈ equilibrium value (space charge neutrality Δn=Δp and low injection condition)
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More assumptions For minority carriers, can neglect drift in quasi-neutral regions Semiconductor is non-degenerate (can use Boltzmann statistics) Current is defined as positive if Va is positive For positive current, electrons flow from n to p and holes flow from p to n
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Consider long-base diode
Both quasi-neutral regions are much longer than minority carrier diffusion lengths Ln or Lp Minority carriers are recombining in QNR’s, not reaching contacts We’ll do short-base diode later on Most important current mechanism is minority carrier injection or extraction current at junction
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Diffusion current Begin with continuity equations
Quantities Δn and Δp are excess carrier concentrations Thus n=n0+Δn and p=p0+Δp
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Electrons on p-side of junction
Electrons are minority carriers From assumption that we can neglect their drift in QNR Combine with To get
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In steady state We had But in steady state Thus
Where minority carrier diffusion length Ln is
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But np=np0+Δnp We had And np0 is constant, so Solution is
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Solution was Boundary conditions: There are no excess minority carriers away from junction (they have all recombined, so Δnp=0 at x=∞ gives A=0 At x=xp, B.C. is giving And therefore
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Substituting Substitute into And obtain the electron current
So electron (diffusion) current is decreasing as you move away form the junction, but total current has to be constant (Kirchhoff current law) Difference is made p by increasing hole current Hole current is drift caused by the small electric field in the quasi-neutral region
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Key points That was a general derivation of the excess minority carrier concentration as a function of position- when there are excess carriers present (non-equilibrium) Excess carriers injected across junction decay exponentially away from junction
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Next up It would be useful to obtain expressions for the minority carrier concentrations at either end of the transition region We’ll do this in Part B of this section
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Current-Voltage Characteristics of pn Homojunctions (Diffusion current, forward bias)
Chapter 5 Section 3.3 Part B Copyright © 2016 The McGraw-Hill Companies, Inc. Permission required for presentation or display
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Introduction We had derived expressions for the excess minority carrier concentration Δn(xp) and the functional form away from the junction (exponential decay) How does Δn(xp) vary with applied voltage? Start with equilibrium case
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Energy band diagram at equilibrium
On the n side: On the p side Barrier for both electrons and hole is qVbi
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Let’s relate Vbi to doping
Neutral p-region Multiply by to obtain But barrier qVbi=(Ecp-Ecn), so
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Result from last slide But part in square brackets is nn0, so
And similarly the minority carrier density on p-side is Minority carrier concentrations at edges of depletion region are functions of the doping on the other side Don’t forget Vbi contains information about doping on both sides
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Now consider forward bias
Barrier has now changed to ECp-Ecn=q(Vbi-Va) More electrons have enough energy to diffuse to p-side (injection of minority carriers) More holes have enough energy to diffuse to n-side (injection of minority carriers) Result is net current flow
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Now consider forward bias
At edges of depletion region or
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Under forward bias Excess electrons are injected into p side
They become minority carriers They diffuse to the right They recombine as they go The excess carrier concentration decays exponentially with distance Holes injected to the n-side, diffuse to the left, same thing
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Excess carrier concentrations
Let Δnp be excess carrier concentration on p-side Let Δpn be excess carrier concentration on p-side Then from We can write And similarly
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The excess carriers diffuse
They diffuse to regions of lower concentration (away from junction) We already solved for the distribution And on n-side, for holes Note that varying distribution means diffusion currents
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Diffusion currents On p-side: Now combine with and To obtain Similarly
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Electron injection Recall we’re neglecting generation, recombination
Thus all excess electrons on p-side had to come from n-side They had to cross xn as well as xp Recall we also assumed negligible drift in the junction Therefore, all minority current across junction is due to diffusion
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Similarly for holes Thus total current across junction is sum of minority carrier diffusion currents
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Total current picture (fwd. bias)
Outside the junction, total current is still constant Minority carrier diffusion current decreasing Thus majority carrier current must increase Majority carrier current is by drift Electric field is small, but there are lots of majority carriers
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Total current density We know And we had Thus Or where
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Repeating previous results
These yield Point: J0 is proportional to ni2
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Diode equation We derived
This is diode current equation used in circuits courses Current flow is due to lowering of potential barriers Current flow is due to diffusion of minority carriers(!)
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Ratio of electron to hole current
Ratio depends on ratio of doping on n-side (ND’) to doping on p-side (NA’) This result important to bipolar junction transistors
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Diode characteristic, forward bias
We had (current density J=I/A) Thus When Va>> kT/q, For current to change by factor of 10: Then ln(I)=ln(10)=2.3, so Va varies by factor of 2.3
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Key points We found expressions for excess carrier concentrations on either side of junction under forward bias We derived the diode equation Next: look at reverse bias
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Current-Voltage Characteristics of pn Homojunctions (Reverse Bias)
Chapter 5 Section 3.3 Part C Copyright © 2016 The McGraw-Hill Companies, Inc. Permission required for presentation or display
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Introduction We had derived expressions for current through a diode for forward bias We found Now let’s look at reverse bias
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Energy band diagram Minority carriers diffuse
If they wander to close to the depletion region, they feel electric field Minority carriers get swept over Extraction
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At edge of depletion region, minority carrier concentration ≈ 0
Excess carrier concentration: p-side: n-side: Now net diffusion is toward junction
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These equations still valid
We derived (forward bias) For Va<-3kT, exponential term and Quantity J0 called “reverse leakage current”
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But what if it’s a short-base diode?
What if one side is shorter than a diffusion length? Happens in “base” of transistor, hence the name Forward bias: carrier injection process is the same But, on short side, carriers don’t have a chance to recombine “naturally” Their excess concentrations forced to zero by presence of contact
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Short base diode, continued
Diffusion current is still But now concentration gradient nearly linear, or So, from (we had before) We now have
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Short-base diode result
We had (last slide) (short side) Total current is (long (n) side is normal, p-side short:) If both sides are short,
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Current ratios If both sides short, electron-to-hole current ratio is
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Key points Under reverse bias, carriers are extracted
Resulting current is very small We also looked at short-base diodes One side or both so short that carriers don’t have time to recombine (forward bias) or be generated (reverse bias) between depletion region and contact All results so far are for non-degenerate materials!
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Next up We still need to consider effects of regeneration and recombination: Parts D&E There are also more considerations about reverse bias, like tunneling and carrier multiplication: Part F
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Current-Voltage Characteristics of pn Homojunctions (Generation and recombination currents, reverse bias) Chapter 5 Section 3.3 Part D Copyright © 2016 The McGraw-Hill Companies, Inc. Permission required for presentation or display
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Introduction We had derived expressions for current through a diode
We considered diffusion of minority carriers across junction Forward bias: injection Reverse bias: extraction Now we consider effects of generation and recombination
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Recall G-R Primarily happens via interband (trap) states ET near Ei
Suppose there are trap states in our diode For simplicity, assume lifetimes Then net recombination rate is At equilibrium, np=ni2 and R=G=0 as expected✔
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Net recombination rate
Depends on numbers of available electrons and holes At position x inside the transition region Thus n decreases rapidly as you approach the p-side, and p decreases rapidly as you approach the n-side
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Energy band diagram under reverse bias
Recall transition region widens (compared to equilibrium) Means more opportunities for G-R in depletion region But, not many carriers there because electric field sweeps them out
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The depletion approximation
Depletion approximation: assume n and p can be neglected for xn<x<xp If no carriers, no recombination, so neglect that (reverse bias only) Implies generation rate constant in depletion region
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Current density Generation current density is
And depletion region width w for step junction is Thus (reverse bias)
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Solution: assume a protype junction with
Example: Estimate the value of generation current relative to the diffusion current for a typical Si pn junction under reverse bias conditions. Solution: assume a protype junction with From graph find D’s
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Example, continued Also find L’s from graph
Lp Also find L’s from graph Minority carrier concentrations are np0=pn0=ni2/1017=1.16 x103 cm-3 Built-in voltage is Ln
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We said we’d assume Vj=5V
Vj=5V=Vbi-Va=0.83-Va or Va=-4.17V Next, find diffusion current- find leakage current first
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Plug result into For generation current, use
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Ratio We find Jdiff<<JG so under reverse bias, we can safely neglect diffusion current
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Key points Under reverse bias, there are essentially no free carriers in depletion region (they are swept out by the field) Therefore negligible recombination There is generation current in depletion region Diffusion current negligible under reverse bias Next: what happens under forward bias?
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Current-Voltage Characteristics of pn Homojunctions (Generation and recombination currents, forward bias) Chapter 5 Section 3.3 Part E Copyright © 2016 The McGraw-Hill Companies, Inc. Permission required for presentation or display
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Introduction We are examining generation and recombination currents in pn homojunctions Under reverse bias, found that diffusion current was negliglible Negligible recombination current There was some generation current (small) Here we will consider forward bias
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GR current under forward bias
Here, electrons in transition region are in thermal equilibrium with n-side Holes in transition region in thermal equilibrium with p-side Can show
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Then the np product We had And we recall Can show that
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Then the np product We had And we recall
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For reasonable forward bias…
We had When Va≥3kT/q, recombination current is large compared to generation current Lots of electrons crossing the transition regions Lots of holes crossing the other way Same place, same time, recombination likely
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Forward bias Since under forward bias np>>ni2 , from
We can write And max recombination rate is where dR/dn=0 More convenient: d(1/R)/dn=0
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Where is recombination rate greatest?
We had Set slope=0, use Result is where or
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More on recombination in junction
Most of the recombination occurs where R≈Rmax Therefore “barrier” to recombination current looks like half the potential barrier for diffusion G-R current often approximated by where JGR0 is slightly voltage dependent
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Leakage currents Compare JGR0 to J0(diff) (and recall that under reverse bias J=-J0 From reverse bias discussions we had So we see that while
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Total current density And usually JGR0>>J0
For forward bias (Va>3kT/q), At small Va, recombination current predominates because JGR0>>J0 but at larger Va diffusion dominates
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Compare Over some range of current, generally approximated as
Where n is the “quality factor” and usually varies between 1 and 2.
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Key points Under forward bias, recombination current occurs in junction due to high numbers of carriers crossing the depletion region Dominates under small forward bias Diffusion current also significant Dominates under “normal” forward bias
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Next up Tunneling under reverse bias
Current amplification under reverse bias (avalanche)
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Current-Voltage Characteristics of pn Homojunctions (Tunneling and current amplification, reverse bias) Chapter 5 Section 3.3 Part F Copyright © 2016 The McGraw-Hill Companies, Inc. Permission required for presentation or display
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Introduction We have considered diffusion and drift
We have examined generation and recombination Two more effects: tunneling and current amplification under reverse bias
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Reverse bias tunneling
Remember electrons tunnel to empty states at the same energy If forbidden region is not too thick Here “barrier” is the forbidden gap
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Tunneling There is a finite probability that electrons will cross the forbidden gap Also called “Zener tunneling” Tunneling probability for electron at energy E normally incident on forbidden region is M* is some average of effective masses for both bands EP* is effective potential energy
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Effective potential energy
Recall potential energy for electron in conduction band is EC Potential energy for hole in valence band is EV Both are varying with position in this case
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While electron is tunneling..
While electron is in forbidden gap, it’s affected by both potential energies Effective potential energy analogous to resistances in parallel
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Can show Tunneling probability is Where WT is the tunneling distance
Tunneling probability depends on band gap and reverse bias voltage, and doping and effective mass
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Current-voltage characteristics
We’ll discuss avalanche multiplication, a different effect, next
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Carrier multiplication
EHP generated thermally Electron accelerated to left (high field under reverse bias) Electron collides Loses energy Knocks another electron loose, generating a another EHP (impact ionization) Now there are three carriers
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Carrier multiplication, continued
Both electrons accelerated to left, new collision Two electrons arrived at left side of junction while only one entered- multiplication of 2 Note: for multiplication, electron must gain enough energy to exceed the band gap to generate a new EHP. This doesn’t happen at steps 4 and 5 here so there is no further electron multiplication.
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Carrier multiplication, holes
Holes can also create impact ionization (5 here) In this case total multiplication is three (if no more impacts) Above some critical field, process can avalanche
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Derivation of multiplication
Let P be probability of either electron or hole creating EHP Let nin be number of electrons entering depletion region from p-side There will be Pnin ionizing collisions Result is nin(1+P) electrons reaching n-side Also generates Pnin holes Result is P(Pnin)=P2nin pairs Total number of carriers crossing junction is
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Multiplication, continued
We had Which can be expressed as Thus multiplication factor M is If P=1, M=∞, and avalanche occurs
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Breakdown When current exceeds some value, diode is in “breakdown”
Means the I-V curve has broken downward on the graph, Device is not harmed
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Reverse breakdown in silicon
For Si, if Vbr is greater than about 8 V, breakdown mechanism is primarily avalanche If Vbr is less than around 6V, is Zener tunneling When you buy a “Zener diode,” not necessarily breaking down via tunneling mechanism- in fact, probably not
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Key points Under reverse bias, diodes can break down Two mechanisms
Not destructive, means the curve breaks down Two mechanisms Tunneling Carrier multiplication If multiplication large, can have avalanche
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Reverse Bias Breakdown
Chapter 5 Section 3.4 Copyright © 2016 The McGraw-Hill Companies, Inc. Permission required for presentation or display
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Introduction We saw that a diode doesn’t conduct much for reverse voltages- up to a point At some voltage, current starts to flow and the I-V characteristics turns downward Called reverse breakdown Does not mean the device is damaged Breakdown mechanism can be via: Tunneling (Zener breakdown) Avalanche
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Breakdown vs doping
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Explanation: avalanche
For avalanche to occur, need a big enough field that electrons gain Eg or more to create ionizing impacts Takes a larger field (reverse voltage) to get there With increasing doping, junction width gets smaller so field increases for a given voltage
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Avalanche breakdown For one-sided junctions Semiconductor Bandgap (eV)
Ge 0.67 2.4 x 1013 Si 1.12 5.3 x 1013 GaAs 1.43 7.0 x 1013 4H-SiC 3.26 3.0 x 1015 GaN 3.44 6.1 x 1015
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Explanation: tunneling
Tunneling probability increases with decreasing width Gap narrows with decreasing band gap, so probability increases and tunneling happens at lower voltages Increased doping means increased field; higher slope decreases WT
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One-sided junction, so use plot to find Vbi
Example: Estimate the tunneling distance for appreciable tunnel current. Consider a p+-n Si junction with ND’=8.0 x 1017 cm-3 (8.0 x 1023 m-3). Tunneling begins at breakdown, so find the junction voltage at breakdown. One-sided junction, so use plot to find Vbi Vbi=1.04 V
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Find breakdown voltage
Vbr =4 V (Va=4V) 4V
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Find maximum field and junction width
Junction width: ( we used Vj=Vbi-Va=1.04-(-4)=5.04V) Maximum electric field:
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Now to find the tunneling width
Thus, for silicon, an appreciable tunneling current flows for WT<10 nm
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Key points Breakdown voltage increases with increasing band gap
Important for power devices that must withstand large voltages- wide band gap materials advantageous Breakdown voltage decreases with increasing doping Provides a way to “engineer” reverse breakdown devices
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Small signal impedance of prototype homojunctions- Junction Resistance
Chapter 5 Section 4.1 Copyright © 2016 The McGraw-Hill Companies, Inc. Permission required for presentation or display
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Introduction We have looked at I-V characteristics in detail
Now look at small-signal ac response Bias at some DC voltage Vary voltage by small amount around that point
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Small signal equivalent
Series resistance RS (constant with voltage, contact resistance plus resistance of QNR) Differential resistance RP varies with voltage Two capacitances: Junction capacitance Cj Stored charge capacitance Csc Both associated with the junction
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Junction conductance Small signal conductance GP:
Is slope of I-V curve at a given voltage For small variation of input voltage, can determine output current Small so slope is a constant in the vicinity
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Junction resistance Resistance is reciprocal of conductance, or
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And we know for reasonable forward bias
Example: Find the junction resistance of a diode at forward currents of 1 mA and 1 μA. Assume the ideality factor is unity and RS = 0. We had And we know for reasonable forward bias “Reasonable” means Then
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Example, continued We had Thus Since kT/q=0.026V: At I=1 mA, RP=26 Ω
At I=1 μA, RP=26 KΩ
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Key points Small signal resistance is reciprocal of the local slope at the bias point There is also a series resistance due to contacts and quasi-neutral regions Next up: junction capacitance
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Small signal impedance of prototype homojunctions- Junction Capacitance
Chapter 5 Section 4.2 Copyright © 2016 The McGraw-Hill Companies, Inc. Permission required for presentation or display
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Introduction We were looking at the small-signal model of a prototype homojunction We looked at the junction resistance (and mentioned the series resistance) Two capacitances to take into account Junction capacitance Stored charge capacitance
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Recall there are charges on either side of the transition region
Charges are due to ionized impurities Number of charges depends on depletion width Varies with voltage Change in charge on either side of a dielectric region with voltage is capacitance
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As voltage changes… Suppose applied voltage changes by dVA
Charge on one side changes by dQ Charge on other side changes by –dQ As applied voltage changes, mobile charges move They move out of junction if reverse bias They move into junction if forward bias Electrons and holes moving=current Must be an equal displacement current flowing across the junction
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Differential junction capacitance
Also called small-signal junction capacitance Already some charge on either side of junction at equilibrium When voltage changes, charge is added or subtracted in sheets at edges of depletion region Looks like parallel plates A=junction area ε=permittivity of material w=junction width
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But, w depends on √(voltage)
Thus Cj is not linear with voltage Expression for w (pn junction): So (pn junction)
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Results For pn junction (repeated) For one-sided step junction
N’ is the net doping concentration on the lightly doped side
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Junction capacitance per unit area for one-sided junction
Reverse bias shown
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What if Va approaches Vbi?
Cj appears to go infinite But as Va increases, current increases Have IRs drop across series resistance Junction voltage is So in practice Vj is always greater than zero
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Key points Junction capacitance arises because of ionized donors and acceptors in transition region As voltage is applied, depletion region gets wider (or narrower) Sheets of charge are added or removed Sheets of charge look like a parallel plate capacitor Junction capacitance is smallish for large reverse bias, increases nonlinearly as voltage becomes more positive Next: stored charge capacitance
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Small signal impedance of prototype homojunctions- Stored Charge Capacitance
Chapter 5 Section 4.3 Copyright © 2016 The McGraw-Hill Companies, Inc. Permission required for presentation or display
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Introduction There are two kinds of capacitance associated with pn junctions We already looked at junction capacitance Caused by ionized donors and acceptors on each side of junction Now we’ll investigate stored-charge capacitance Caused by change in minority carrier density
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Consider an n+p junction
Take the case of forward bias For n+p, have injection of electrons primarily When you change the bias, the amount of injected carriers changes Change in charge caused capacitance
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First, steady state We already know the density of excess electrons on p-side is In steady state, this concentration remains constant in time (still varies with distance) Electrons are constantly diffusing across junction Electrons are constantly diffusing into p-region and recombining There is a constant “pile” of charge on p-side
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When the voltage changes
The amount of injection changes The size of the “pile” changes Change in charge “stored” in the pile is capacitance Total charge stored is integral of injection pile Will relate this to current
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Current is by diffusion of minority carriers (primarily)
So current across x=xp is Combine with Result
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Now change voltage by dVa abruptly
Size of pile changes Since then Δn(xp) changes abruptly But the rest of the distribution can’t change instantaneously Now peak concentration is not at xp but to right of it the stored electrons will diffuse to left and right of peak The ones that diffuse to the right recombine and disappear Only the ones that diffuse to the left flow into external circuit and contribute to the capacitance Referred to as reclaimable stored charge
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Let δ be the fraction of charge that is reclaimable
Then Qsr is reclaimable stored charge and recalling that and We obtain For prototype step junction, δ=0.5 For graded doping, δ depends on the profile We’ll visit in more detail when we talk about bipolar junction transistors (Chapter 10)
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Stored-charge capacitance diffusion current
Thus Csc increases exponentially with Va But junction capacitance goes as square root of Va Therefore, for reverse bias and small forward bias, junction capacitance dominates For large forward bias, stored-charge capacitance dominates
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Junction capacitance:
Example: Compare the junction capacitance and stored charge capacitance under reverse (Va=-5V) and forward bias (Va =+0.75V). Consider a prototype n+-p junction with NA’=NA=1017 cm-3, with junction area A=100 μm2 and fractional reclaimable charge δ =0.5 Junction capacitance:
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First find Vbi Find Vbi from chart for one-sided junction Vbi=0.98V
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Calculate Cj’s For Va=-5V and for Va=0.75V x x
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For stored-charge capacitance
Need to find I Use the diffusion current Generation/recombination current does not contributed to stored-charge capacitance We find
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Look up Dn (minority carriers on p-side)
Dn=20 cm2/s
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Find Ln Lp Ln=70 μm Ln
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Find τn τn=3 µs
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So I0 becomes and I at Va=-5V is -5.3 x A At Va=+0.75V,
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Stored-charge capacitances
At Va=-5V: At Va=+0.75V
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Compare results Cj Csc Va= -5 V 0.053 pF ≈0 Va= V 0.27 pF 100 pF
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Key points Under forward bias, as voltage changes, the amount of charge “stored” in the injection changes, leading to capacitance When voltage is changed, not all the “stored” charge is recovered- reclaimable fraction is δ=1/2 for step junction Under reverse bias, junction capacitance dominates Under large forward bias, stored-charge capacitance dominates Next: we’ll look at transient behavior of diodes
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Turn-off transient Chapter 5 Section 5.1 Copyright © 2016
The McGraw-Hill Companies, Inc. Permission required for presentation or display
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Introduction A pn junction is often used as a switch
Change state from “on” (forward bias, VF) to “off” (reverse bias, VR) Junction has capacitance Cannot change state instantly There will be turn-on and turn-off transients Investigate turn-off transient first
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Here is the situation We’ll assume that both VF and VR are larger than Vbi Assume R1>>Rs (so we can neglect internal series resistance of diode)
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Take an n+-p junction Then can ignore stored hole charge on n-side
At time t=0, the diode is on Vj(on)<Vbi VF>>Vbi>Vj Thus forward current IF is (up to t=0)
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At t=0 switch voltage to VR
Excess carrier concentration goes from injection to extraction Some carriers diffuse to the right (not all are reclaimed) Notice slope of Δnp is constant near junction during transition
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Find the current Recall current crossing the plane xp is same as the current anywhere else in the device But it’s easy to evaluate here- it’s entirely due to diffusion Since slope is constant during the switch, current is constant until t=ts (ts is storage time) After that, slope flattens to steady state Current decays to steady state value (reverse current ≈0)
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The derivation is involved
But approximate solution is For VF=VR, |IF|≈|IR| and Note IR is zero in steady state, only reaches |IF| during transition
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To shorten turn-off time, reduce charge in p-region
Can do this by decreasing (adjust voltages in circuit) Can do this by reducing electron lifetime By doping with traps such as Au or Cu Traps, however, increase G current in the off state Increases power consumption Shortening lifetime also decreases diffusion length Ln Can also reduce thickness of more lightly doped side Can also grade doping- more in Chapter 9
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Key points When switching from “on” to “off”, initial condition is injected charge stored on lightly doped side of one-sided junction When voltage is reversed, takes time to deplete the excess charge (source of capacitance) On turn-off, current is constant for storage time ts, then decays To speed up turn-off: shorten lifetime via traps states, adjust on/off voltages
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Turn-on transient Chapter 5 Section 5.2 Copyright © 2016
The McGraw-Hill Companies, Inc. Permission required for presentation or display
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Introduction We looked at the turn-off transient for a step-junction diode Now we’ll consider turn-on Transient associated with going from VR back to VF
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Now must build up stored charge
Slopes constant again Current IF constant Reach steady state when injection rate equals recombination rate
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Effects of the two capacitances
For turn-on, beginning state is reverse bias Junction capacitance important there Amount of charge needed to discharge the junction capacitance small compared to amount needed to fill stored charge pile Stored charge established by diffusion (slow) Thus turn-on time determined mostly by time needed to set up the minority carrier steady-state distribution small large
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When switching from high to low
Current switches from VF/R1 to VR/R1 During storage time, diode voltage constant (current is constant) Real diode: series resistance Rs not really 0 VOLTAGE IN DIODE CURRENT DIODE VOLTAGE
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Voltage drop across series resistance of diode
Diode voltage is junction voltage plus voltage across Rs Voltage across diode terminals goes from (Vj(on)+IFRS) to (Vj(on)+IRRS) Remember IF is positive, IR is negative Diode model DIODE CURRENT DIODE VOLTAGE
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After storage time… Current flow decays to “off” value
Voltage across terminals of diode decays to VR Notice turn-on time very short compared to turn-off Maximum switching frequency limited by turn-off VOLTAGE IN DIODE CURRENT DIODE VOLTAGE
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Real life Usually use V=0 for off voltage (not a negative voltage)
Junction capacitance and stored-charge capacitance vary with voltage Messy Simulate with SPICE or equivalent
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For short-base diode WB<<Ln
Recombination mostly happens at contact (neglect recombination in p-QNR) and
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Excess charge in short base is
where tT is transit time across the short QNR (“base transit time”) Base transit time given by
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Solution: We had So we can write
Example: Compare the amount of minority carrier stored charge in a forward-biased short-base diode with that in a long-base diode. Solution: We had So we can write x short long x
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Let WB≈0.1 μm, Ln=31 μm This value of WB typical for the base of a transistor Can speed up switching time by about 5 million by using a short-based diode x
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Key points Turn-off time is much slower than turn-on time
Switching frequency limited by turn-off Reduce time, increase frequency by shortening the width of the lightly doped side Can also introduce trap states to speed up recombination but that incurs a power penalty
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Effects of temperature
Chapter 5 Section 6 Copyright © 2016 The McGraw-Hill Companies, Inc. Permission required for presentation or display
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Introduction When designing a diode (or any component) have to consider the temperature range over which it has to operate Commercial grade products might have to operate over 0 C to 70 C Military grade: -55 C up to 125 C
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Current in an n+-p diode
Quantities Dn , τn , NC , NV , and Eg all vary with temperature Setting those aside, have kT in exponent Compare fractional change in J0 to J0
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Fractional change for Si
or J0 varies by 14% per °C!
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What about JGR0? That depends on ni rather than ni2 (J0 depends on ni2) Recall So JGR0 changes with temperature at about half the rate of J0
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And the voltage Suppose a diode is biased at a constant forward current in typical range 0.1 to 1 mA Diode voltage increases about 2 mV per degree C
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Key points Diode current and voltages vary with temperature
Have to take into account when designing devices
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Summary Chapter 5 Section 7 Copyright © 2016
The McGraw-Hill Companies, Inc. Permission required for presentation or display
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Introduction We considered prototype pn junctions
Real diodes have complex doping gradients Have to solve equations numerically- provides little insight We chose step junction to illustrate physical processes We considered pn junctions (neither side degenerate) one-sided junctions (one side is degenerate) n+-p p+-n
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Fermi levels Non-degenerate n and p regions:
Degenerate: can often assume that
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Built-in voltage in general:
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Junction width Width of region where there are uncompensated ions
No free carriers Depletion region Transition region Space charge region
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Junction width equations
pn n+-p p+-n Vj I the junction voltage Vj=Vbi-Va Va is the applied voltage
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Junction width Most of the junction width appears on the lightly doped side Most of the voltage is dropped across the lightly doped side
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Three current mechanisms
Drift (caused by electric field) Diffusion (caused by gradients in carrier concentrations) Generation/recombination Total current density is if evaluated at edge of transition region where you can neglect drift
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Diode current density J0=J0diff+JGR0 J0 increases by about 14% per °C
J increases by factor of 10 for Va increase of 60 mV (room temperature) Total current is I=AJ where A is the cross-sectional area of the junction
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Leakage currents Transition width w goes as √Vj Long base diode
Short base diode (p-side short) Both sides short
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Leakage current comments
Usually JGR0>>J0 JGR0 important for reverse bias and small forward bias J0 important for larger forward bias
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Junction breakdown Two mechanisms Tunneling (Zener breakdown)
Avalanche multiplication Zener is softer breakdown, occurs in heavily doped junctions Avalanche sharper breakdown, comes with lightly doped junctions
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Capacitance Two kinds Junction capacitance (variation in ionized charges in depletion region) Stored charge capacitance (variation in charge in excess carriers near junction)
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Junction Capacitance Junction capacitance (variation in ionized charges in depletion region) Goes as Va-1/2 Predominates under reverse bias and low forward bias
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Stored Charge Capacitance
Stored charge capacitance (variation in charge in excess carriers near junction) Dominates under large forward bias
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Transient effects Switching time determined by how quickly you can charge/discharge internal capacitances Turn-on time small compared to turn-off Operating frequency limited by turn-off Use short-base diode to improve
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Key points Prototype (step) junction greatly simplified
Does provide great physical insight Will refine the model in Chapter 6 See also Supplement to Part II which has additional information
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