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Fundamentals of Semiconductor Devices

- 1. Introduction Chapter 5 Section 1 Copyright © 2016 The McGraw-Hill Companies, Inc. Permission required for presentation or display
- 2. 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
- 3. 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
- 4. Circuit equivalent • Open circuit (for V<0.7 V) • Looks like a power supply: 0.7 V and supplies current
- 5. Circuit symbol • Ideal 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
- 6. 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
- 7. Input voltage 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 I = Vout R = 5sin 377t( )- 0.7 2 ´103
- 8. 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
- 9. 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
- 10. Doping profile • Background acceptor concentration is constant (NA) • Typical donor concentration profile shown • We’ll approximate as step function • Junction occurs at x0
- 11. 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
- 12. 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
- 13. 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
- 14. 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
- 15. Step 1: Electrical neutrality
- 16. 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
- 17. Currents Jn = qmn nE + qDn dn dx = qmn n + kT q dn dx é ë ê ê ê ê ù û ú ú ú ú Jp = qmp pE - qDp dp dx = qmp pE - kT q dp dx é ë ê ù û ú Drift currents Diffusion currents
- 18. Equilibrium
- 19. 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
- 20. 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
- 21. 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
- 22. 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”
- 23. 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) qVbi = Fp - Fn
- 24. 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
- 25. The band bending • As doping increases, built-in voltage increases – Work functions increase, so bands have to bend more to align Fermi levels
- 26. The electric field • Electric field proportional to slope of Evac • Is steepest at metallurgical junction • Built-in voltage is mostly on lightly-doped side E = 1 q dEvac dx
- 27. One-sided junction • When one side is degenerately doped, most of junction appears on lightly doped side • Called “one- sided junction”
- 28. 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?
- 29. 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
- 30. Introduction • We saw how to draw the energy band diagram at equilibrium • What happens when we apply a bias?
- 31. 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
- 32. 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
- 33. 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
- 34. 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
- 35. The junction voltage • Junction voltage Vj is increased • (increased because recall Va is negative) • Greater field at junction Vj =Vbi -Va
- 36. 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
- 37. Under forward voltage • Now Va is positive • Space charge region gets smaller • Evac not shown-no new information there
- 38. 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?
- 39. 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
- 40. 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
- 42. 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
- 43. 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
- 44. 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
- 45. 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
- 46. 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
- 47. 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
- 48. Forward bias, recombination • Many electrons and holes crossing through depletion region- opportunities to recombine are plentiful • Result is net current to the left
- 49. 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
- 50. 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
- 51. 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
- 52. Tunneling? • Not possible – No empty states in one band at same energy as occupied state in another band
- 53. 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
- 54. 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
- 55. 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
- 56. Tunnel Diodes Chapter 5 Section 2.3 Copyright © 2016 The McGraw-Hill Companies, Inc. Permission required for presentation or display
- 57. 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
- 58. Typical I-V
- 59. Energy Band Diagram ≈10 nm E »106 V /cm
- 60. 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
- 61. I-V characteristic I = Ce qV hkT
- 62. Figure of Merit • Peak-to-Valley ratio • Values of IP/IV ≈ 20 have been observed in GaAs tunnel diodes
- 63. 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)
- 64. 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
- 65. Key points • When both sides of pn junction are heavily doped, can get tunneling • Results in negative resistance region in I-V
- 66. 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
- 67. 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
- 68. Find Vbi • Recall • Express as where on the n-side (non degenerate) • If n-side is degenerate • Take Ef=EC and δn=0 qVbi = Fp - Fn qVbi = Eg - dn +dp( ) dn = EC - Ef = kT ln NC nno = kT ln NC ND '
- 69. Similary for p-type • Non-degenerate • Degenerate: δp=0 dp = Ef - EV = kT ln NV ppo = kT ln NV NA '
- 70. Combine • To get • For one-sided junctions (n+-p or p+-n) qVbi = Eg - dn +dp( ) dn = EC - Ef = kT ln NC nno = kT ln NC ND ' dp = Ef - EV = kT ln NV ppo = kT ln NV NA ' qVbi = Eg - kT ln NC ND ' + ln NV NA ' æ è ç ö ø ÷ é ë ê ê ù û ú ú p-n junction qVbi = Eg - kT ln NV NA ' é ë ê ù û ú n + -p junction qVbi = Eg - kT ln NC ND ' é ë ê ù û ú p + -n junction
- 71. For non-degenerate semiconductor • Recall • Thus ni 2 = NC NV e - Eg kT Eg = kT ln NC NV ni 2 Vbi = kT q ln ND ' NA ' ni 2 pn junction Vbi = kT q ln NV ND ' ni 2 p+ -n junction Vbi = kT q ln NC NA ' ni 2 n+ -p junction And we had qVbi = Eg - kT ln NC ND ' + ln NV NA ' æ è ç ö ø ÷ é ë ê ê ù û ú ú
- 72. 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 Vbi = Eg q - kT q ln NC NV ND ' NA ' p-n junction
- 73. 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 Vbi = kT q ln ND ' NA ' ni 2 pn junction
- 74. 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 Vbi = kT q ln NA ' ND ' ni 2 é ë ê ê ù û ú ú = 0.026eV i 1.6 ´10-19 J / eV 1.6 ´10-19 C ln 1016 1015 1.08 ´1010 ( ) 2 é ë ê ê ê ê ù û ú ú ú ú = 0.026V ln 8.57 ´1010é ëê ù ûú = 0.655V
- 75. 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 Vbi = kT q ln NV ND ' ni 2 p+ -n junction Vbi = kT q ln NC NA ' ni 2 n+ -p junction
- 76. Vbi for one-sided junctions Curves for n+-p and p+-n are indistinguishable
- 77. Note • If impurity-induced band gap narrowing is considered, reduce result from previous plot by amount of apparent band-gap narrowing ΔEg*
- 78. 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
- 79. Copyright © 2016 The McGraw-Hill Companies, Inc. Permission required for presentation or display Energy band diagram with applied voltage: Part A Chapter 5 Section 3.2
- 80. 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
- 82. 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
- 83. 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 dE dx = QV (x) e
- 84. 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 xn < x < xo QV = qND ' x0 £ x £ xp QV = -qNA '
- 85. Integrate to solve • On n-side • On p-side dE o E (x) ò = q ND ' e dx xn x ò dE dx = QV (x) e E (x) = q ND ' e x - xn( ) d E (x) 0 ò = - q NA ' ex xp ò dx E (x) = q NA ' e xp - x( ) xn £ x £ x0 x0 £ x £ xp
- 86. But electric field has to be continuous • We had • Set the solutions equal at x=x0 qND ' (xo - xn ) = qNA ' (xp - xo ) E (x) = q ND ' e x - xn( ) E (x) = q NA ' e xp - x( ) Charge: Field:
- 87. 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 qND ' (xo - xn ) = qNA ' (xp - xo ) wn wp = xo - xn( ) xp - xo( ) = NA ' ND ' Checks out!
- 88. Next find voltage distribution • We use • For n-side, integrate • And get • For p-side E = - dV dx dV V (xn ) V (x) ò = - E (x)dx xn x ò = - qND ' e (x - xn )dx xn x ò V(x) -V(xn ) = - qND ' 2e x - xn( ) 2 V(xp )-V(x) = - qNA ' 2e xp - x( ) 2 xn £ x £ xo xo £ x £ xp
- 89. Find the total voltage • Voltage accumulated on n side: • On p-side • Total voltage is V(xn ) -V(xo ) =Vj n = qND ' 2e xo - xn( ) 2 = qND ' 2e wn 2 V xo( )-V xp( )=Vj p = qNA ' 2e xp - xo( ) 2 = qNA ' 2e wp 2 Vj =Vj n +Vj p = q 2e ND ' (xo - xn )2 + NA ' (xp - xo )2 é ë ù û
- 90. Voltage has to be continuous • Set solutions equal at junction • Take previous results and take ratio • Recalling , we get V(xo ) =V(xn ) - qND ' 2e xo - xn( ) 2 =V(xp )+ qNA ' 2e xp - xo( ) 2 V(xn )-V(x0 ) V(x0 )-V(xp ) = Vj n Vj p = ND ' wn 2 NA ' wp 2 Vj n Vj p = NA ' ND ' wn wp = NA ' ND ' Most of voltage dropped across more lightly doped side
- 91. Plot of voltage
- 92. 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
- 93. Copyright © 2016 The McGraw-Hill Companies, Inc. Permission required for presentation or display Energy band diagram with applied voltage: Part B Chapter 5 Section 3.2
- 94. Introduction • In Part A we found the charge, field and voltage • Next, junction widths and energy band diagram
- 95. Find expressions for junction widths • Combine: • To find V(xn ) -V(xo ) =Vj n = qND ' 2e xo - xn( ) 2 = qND ' 2e wn 2 V(xn )-V(x0 ) V(x0 )-V(xp ) = Vj n Vj p = ND ' wn 2 NA ' wp 2 V(xo ) =V(xn ) - qND ' 2e xo - xn( ) 2 =V(xp )+ qNA ' 2e xp - xo( ) 2 wn wp = NA ' ND ' wn = xo - xn( )= 2eVj n qND ' é ë ê ê ù û ú ú 1 2 = 2eVj qND ' 1+ ND ' NA ' æ è ç ö ø ÷ é ë ê ê ê ê ê ù û ú ú ú ú ú 1 2
- 96. Similarly for p, and total width • For the p-side • And the total junction width is wp = xp - x0( )= 2eVj p qNA ' é ë ê ê ù û ú ú 1 2 = 2eVj qNA ' 1+ NA ' ND ' æ è ç ö ø ÷ é ë ê ê ê ê ê ù û ú ú ú ú ú 1 2 w = wn + wp = 2eVj NA ' + ND ' ( ) qNA ' ND ' é ë ê ê ù û ú ú 1 2 pn junction
- 97. We had • Solve for junction voltage Vj • For one-sided junction, e.g. n+-p, w≈wp since in this case ND’>>NA’ and w = wn + wp = 2eVj NA ' + ND ' ( ) qNA ' ND ' é ë ê ê ù û ú ú 1 2 Vj = qND ' NA ' w2 2e ND ' + NA ' ( ) pn junction w = 2eVj qNA ' é ë ê ê ù û ú ú 1 2 (n+-p) w = 2eVj qND ' é ë ê ê ù û ú ú 1 2 (p+-n)
- 98. Maximum electric field • Combine • To find E (x) = q ND ' e x - xn( ) w = wn + wp = 2eVj NA ' + ND ' ( ) qNA ' ND ' é ë ê ê ù û ú ú 1 2 Vj = qND ' NA ' w2 2e ND ' + NA ' ( ) Emax = qND ' e wn = qNA ' e wp = 2qVj ND ' NA ' e ND ' + NA '( ) é ë ê ê ù û ú ú 1/2 = 2Vj w
- 99. 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 dEP dx = -q dV dx = dEC dx = dEV dx
- 100. So resulting energy band diagram is
- 101. Junction width for one-sided junction as function of voltage
- 102. Junction width for one-sided junction as function of doping
- 103. 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
- 104. Final result
- 105. Copyright © 2016 The McGraw-Hill Companies, Inc. Permission required for presentation or display Current-Voltage Characteristics of pn Homojunctions (Excess minority carrier concentrations) Chapter 5 Section 3.3 Part A
- 106. Introduction • We’ll obtain I-V characteristics – 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) np pp @ NA ' pn nn @ ND ' nn @ nno = ND ' pp @ ppo = NA '
- 107. 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 Jnp = qDn dnp dx Jpn = -qDp dpn dx no = NC e - EC -Ef kT æ è ç ö ø ÷ po = NV e - Ef -EV kT æ è ç ö ø ÷
- 108. 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
- 109. Diffusion current • Begin with continuity equations • Quantities Δn and Δp are excess carrier concentrations – Thus n=n0+Δn and p=p0+Δp ¶n ¶t = ¶Dn ¶t = 1 q ¶Jn ¶x æ èç ö ø÷ + Gop - Dn tn æ è ç ö ø ÷ ¶p ¶x = ¶Dp ¶x = - 1 q ¶Jp ¶x æ è ç ö ø ÷ + Gop - Dp t p æ è ç ö ø ÷
- 110. Electrons on p-side of junction • Electrons are minority carriers • From assumption that we can neglect their drift in QNR • Combine with • To get Jn = Jndiff = qDn dn dx ¶n ¶t = ¶Dn ¶t = 1 q ¶Jn ¶x æ èç ö ø÷ + Gop - Dn tn æ è ç ö ø ÷ ¶n ¶t = ¶Dn ¶t = Dn ¶2 n ¶x2 - Dn tn
- 111. In steady state • We had • But in steady state • Thus • Where minority carrier diffusion length Ln is ¶n ¶t = ¶Dn ¶t = Dn ¶2 n ¶x2 - Dn tn ¶n/ ¶t = 0 ¶2 n ¶x2 = d2 n dx2 = Dn Dn tn = Dn Ln 2 Ln = Dn tn
- 112. But np=np0+Δnp • We had • And np0 is constant, so • Solution is ¶2 np ¶x2 = d2 Dnp dx2 = Dnp Ln 2 ¶n ¶t = ¶Dn ¶t = Dn ¶2 n ¶x2 - Dn tn Dnp = Ae x Ln + Be -x Ln
- 113. 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 Dnp = Ae x Ln + Be -x Ln Dn(xp ) = Be - x-xp( ) Ln B = Dn(xp )e xp Ln And therefore Dn(x) = Dn(xp )e -(x-xp ) Ln
- 114. 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 Dn(x) = Dn(xp )e -(x-xp ) Ln Jn = qDn dn dx = qDn dDn dx Jn = qDn Ln Dn(xp )e -(x-xp ) Ln
- 115. 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
- 116. 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
- 117. Copyright © 2016 The McGraw-Hill Companies, Inc. Permission required for presentation or display Current-Voltage Characteristics of pn Homojunctions (Diffusion current, forward bias) Chapter 5 Section 3.3 Part B
- 118. 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
- 119. Energy band diagram at equilibrium • On the n side: • On the p side • Barrier for both electrons and hole is qVbi nno = ND ' pno = ni 2 ND ' ppo = NA ' npo = ni 2 NA '
- 120. Let’s relate Vbi to doping • Neutral p-region • Multiply by to obtain • But barrier qVbi=(Ecp-Ecn), so npo = NC e - ECp -Ef kT æ è ç ö ø ÷ e - ECn kT e + ECn kT npo = NC e - ECn-Ef kT æ è ç ö ø ÷ e - ECp -ECn kT æ è ç ö ø ÷ npo = NC e - ECn-Ef kT æ è ç ö ø ÷ é ë ê ê ù û ú ú e - qVbi kT æ è ç ö ø ÷
- 121. 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 npo = NC e - ECn-Ef kT æ è ç ö ø ÷ é ë ê ê ù û ú ú e - qVbi kT æ è ç ö ø ÷ np (xp ) = npo = ND ' e - qVbi kT = nn0 e - qVbi kT pn (xn ) = pno = NA ' e - qVbi kT = pp0 e - qVbi kT
- 122. 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
- 123. Now consider forward bias • At edges of depletion region np (xp ) = ND ' e _ q Vbi -Va( ) kT pn (xn ) = NA ' e - q Vbi -Va( ) kT or np (xp ) = npo e qVa kT pn (xn ) = pno e qVa kT
- 124. 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
- 125. 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 Dnp (xp ) = np (xp )- npo Dpn (xn ) = pn (xn ) - pno np (xp ) = npo e qVa kT Dnp (xp ) = npo e qVa kT -1 æ è ç ö ø ÷ Dpn (xn ) = pno e qVa kT -1 æ è ç ö ø ÷
- 126. 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 Dnp (x) = Dnp (xp )e - (x-xp ) Ln Dpn (x) = Dpn (xn )e - (xn-x) Lp
- 127. Diffusion currents • On p-side: • Now combine with and To obtain Similarly Jn = qDn dn dx = qDn dDn dx Dnp (x) = Dnp (xp )e - (x-xp ) Ln Dnp (xp ) = npo e qVa kT -1 æ è ç ö ø ÷ Jp (xn ) = q Dp Lp pno e qVa kT -1 æ è ç ö ø ÷ Jn (xp ) = q Dn Ln npo e qVa kT -1 æ è ç ö ø ÷
- 128. 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
- 129. Similarly for holes • Thus total current across junction is sum of minority carrier diffusion currents
- 130. 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
- 131. Total current density • We know • And we had • Thus • Or where J = Jn(xp)+ Jp(xn) Jp (xn ) = q Dp Lp pno e qVa kT -1 æ è ç ö ø ÷ Jn (xp ) = q Dn Ln npo e qVa kT -1 æ è ç ö ø ÷ J = q Dn npo Ln + Dp pno Lp æ è ç ö ø ÷ e qVa kT -1 æ è ç ö ø ÷ J = Jo e qVa kT -1 æ è ç ö ø ÷ Jo = q Dn npo Ln + Dp pno Lp æ è ç ö ø ÷
- 132. Repeating previous results • These yield • Point: J0 is proportional to ni 2 Jo = q Dn npo Ln + Dp pno Lp æ è ç ö ø ÷ Ln = Dn tn , Lp = Dp t p , np0 = ni 2 NA ' and pn0 = ni 2 ND ' J0 = qni 2 Dn tn × 1 NA ' + Dp t p × 1 ND ' æ è ç ç ö ø ÷ ÷
- 133. 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(!) J = Jo e qVa kT -1 æ è ç ö ø ÷
- 134. 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 Jn (xp ) Jp (xn ) = Dn npo Ln i Lp Dp pno = Dn Dp i Lp Ln i ND ' NA '
- 135. 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 J = Jo e qVa kT -1 æ è ç ö ø ÷ I = Io e qVa kT -1 æ è ç ö ø ÷ I » I0 e qVa kT Va = kT q ln I I0 = kT q ln I - ln I0( )
- 136. 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 I = Io e qVa kT -1 æ è ç ö ø ÷
- 137. Copyright © 2016 The McGraw-Hill Companies, Inc. Permission required for presentation or display Current-Voltage Characteristics of pn Homojunctions (Reverse Bias) Chapter 5 Section 3.3 Part C
- 138. Introduction • We had derived expressions for current through a diode for forward bias • We found • Now let’s look at reverse bias J = Jo e qVa kT -1 æ è ç ö ø ÷ Jo = q Dn npo Ln + Dp pno Lp æ è ç ö ø ÷
- 139. 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
- 140. At edge of depletion region, minority carrier concentration ≈ 0 • Excess carrier concentration: • p-side: • n-side: • Now net diffusion is toward junction Dnp (xp ) = np (xp )- np0 = 0- np0 = -np0 Dpn (xn ) = pn (xn )- pn0 = -pn0
- 141. These equations still valid • We derived (forward bias) • For Va<-3kT, exponential term 0 and • Quantity J0 called “reverse leakage current” J = Jo e qVa kT -1 æ è ç ö ø ÷ Jo = q Dn npo Ln + Dp pno Lp æ è ç ö ø ÷ J » -J0
- 142. 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
- 143. Short base diode, continued • Diffusion current is still • But now concentration gradient nearly linear, or • So, from (we had before) • We now have Jn = qDn dn dx = qDn dDn dx dDn dx = Dnp (xp ) WB Jn (xp ) = q Dn Ln npo e qVa kT -1 æ è ç ö ø ÷ Jn (xp ) = q Dn WB npo e qVa kT -1 æ è ç ö ø ÷ Jo = q Dn npo WB + Dp pno Lp æ è ç ö ø ÷
- 144. 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, Jn (xp ) = q Dn WB npo e qVa kT -1 æ è ç ö ø ÷ Jo = q Dn npo WB + Dp pno Lp æ è ç ö ø ÷ Jo = q Dn npo WB( p) + Dp pno W(n) æ è ç ö ø ÷
- 145. Current ratios • If both sides short, electron-to-hole current ratio is Jn Jp = Dn Dp i W(n) WB( p) i ND ' NA '
- 146. 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!
- 147. 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
- 148. Copyright © 2016 The McGraw-Hill Companies, Inc. Permission required for presentation or display Current-Voltage Characteristics of pn Homojunctions (Generation and recombination currents, reverse bias) Chapter 5 Section 3.3 Part D
- 149. 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
- 150. 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=ni 2 and R=G=0 as expected✔ tn = t p = to R - G = np - ni 2 to (n+ ni + p + ni )
- 151. 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 n(x) = nnoe -[EC (x)-EC (xn)] kT p(x) = ppoe [EV (x)-EV (xp )] kT
- 152. 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
- 153. 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 R = 0, G = ni 2to
- 154. Current density • Generation current density is • And depletion region width w for step junction is • Thus (reverse bias) JG = -qGw = - qni w 2to w = 2e ND ' + NA ' ( ) Vbi -Va( ) qND ' NA ' é ë ê ê ù û ú ú 1 2 JG = - ni to qe ND ' + NA ' ( ) Vbi -Va( ) 2ND ' NA ' é ë ê ê ù û ú ú 1 2
- 155. 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 NA ' = 1017 cm-3 ND ' = 1017 cm-3 tn @ t p = to @ 6 ´10-6 s Vj = Vbi -Va( )= 5V Dn = 20 cm2 /s Dp = 11 cm2 /s
- 156. Example, continued • Also find L’s from graph • Minority carrier concentrations are np0=pn0=ni 2/1017=1.16 x103 cm-3 • Built-in voltage is Lp Ln Lp = 102 mm Ln = 73mm Vbi = kT q ln ND ' NA ' ni 2 = 0.026V ln 1017 ´1017 1.08´1010 ( ) 2 æ è ç ç ö ø ÷ ÷ = 0.83V
- 157. 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 Jo = q Dn npo Ln + Dp pno Lp æ è ç ö ø ÷ = 1.6 ´10-19 C 20 cm2 s æ èç ö ø÷ 1.16 ´103 cm-3 ( ) 73´10-4 cm + 11 cm2 s æ èç ö ø÷ 1.16 ´103 cm-3 ( ) 102 ´10-4 cm é ë ê ê ê ê ê ù û ú ú ú ú ú = 7.1´10-13 A/ cm2
- 158. Plug result into • For generation current, use J = Jo e qVa kT -1 æ è ç ö ø ÷ Jdiff = J0 e qVa kT -1 æ è ç ö ø ÷ = 7.1´10-13 A/ cm2 e 1.6´10-19 C´(-4.17V ) 0.026×1.6´10-19 J /eV -1 æ è ç ö ø ÷ @ -J0 = -7.1´10-13 A/ cm2 = -7.1´10-21 A/ mm2 JG = - qni w 2to = - 1.6 ´10-19 C 1.08 ´1010 cm-3 ( ) 0.36 ´10-4 cm( ) 2 6´10=6 s( ) = -5.2 ´10-9 A/ cm2 = 5.2 ´10-17 A/ mm2
- 159. Ratio • We find Jdiff<<JG so under reverse bias, we can safely neglect diffusion current JG Jdiff @ 5.2 ´10-17 7.1´10-21 = 7.3´103
- 160. 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?
- 161. Copyright © 2016 The McGraw-Hill Companies, Inc. Permission required for presentation or display Current-Voltage Characteristics of pn Homojunctions (Generation and recombination currents, forward bias) Chapter 5 Section 3.3 Part E
- 162. 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
- 163. 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 n(x) = nn0 e - EC (x)-ECn( ) kT and p(x) = pp0 e - EVp -EV (x)( ) kT
- 164. Then the np product • We had • And we recall • Can show that n(x) = nn0 e - EC (x)-ECn( ) kT and p(x) = pp0 e - EVp -EV (x)( ) kT np = ni 2 e qVa kT Vbi = kT q ln ND ' NA ' ni 2
- 165. Then the np product • We had • And we recall n(x) = nn0 e - EC (x)-ECn( ) kT and p(x) = pp0 e - EVp -EV (x)( ) kT np = ni 2 e qVa kT Vbi = kT q ln ND ' NA ' ni 2
- 166. 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 np = ni 2 e qVa kT
- 167. Forward bias • Since under forward bias np>>ni 2 , from • We can write • And max recombination rate is where dR/dn=0 – More convenient: d(1/R)/dn=0 R - G = np - ni 2 to (n+ ni + p + ni ) R = np to n+ p( ) = ni 2 e qVa kT to n+ p( )
- 168. Where is recombination rate greatest? • We had • Set slope=0, use • Result is where or R = ni 2 e qVa kT to n + p( ) p = ni 2 n e qVa kT n = p = ni e qVa 2kT Rmax = ni e qVa 2kT 2to
- 169. 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 Vbi -Va 2 æ èç ö ø÷ JGR = JGR0 e qVa 2kT -1 æ è ç ö ø ÷
- 170. 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 JGR0 = - ni to qe ND ' + NA ' ( ) Vbi -Va( ) 2ND ' NA ' é ë ê ê ù û ú ú 1 2 J0(diff ) = qni 2 Dn tn × 1 NA ' + Dp t p × 1 ND ' æ è ç ç ö ø ÷ ÷ JGR0 µ ni J0(diff ) µ ni 2
- 171. 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 J = JGR + Jdiff = JGR0 e qVa 2kT -1 æ è ç ö ø ÷ + J0 e qVa kT -1 æ è ç ö ø ÷ J = JGRo e qVa 2kT + Jo e qVa kT
- 172. Compare • Over some range of current, generally approximated as • Where n is the “quality factor” and usually varies between 1 and 2. J = Js (e qVa nkT -1)
- 173. 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
- 174. Next up • Tunneling under reverse bias • Current amplification under reverse bias (avalanche)
- 175. Copyright © 2016 The McGraw-Hill Companies, Inc. Permission required for presentation or display Current-Voltage Characteristics of pn Homojunctions (Tunneling and current amplification, reverse bias) Chapter 5 Section 3.3 Part F
- 176. Introduction • We have considered diffusion and drift • We have examined generation and recombination • Two more effects: tunneling and current amplification under reverse bias
- 177. 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
- 178. 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 T = e -2 a dxò a = 2m* 2 EP * (x)- E( )é ë ê ù û ú 1 2 M* is some average of effective masses for both bands EP* is effective potential energy
- 179. 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
- 180. 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 EP * (x)- E( )= EC (x) - E( ) E - EV (x)( ) EC (x)- E( )+ E - EV (x)( ) = EC (x)- E( ) E - EV (x)( ) Eg
- 181. 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 T = e - pWT m* Eg 2 3 2 æ è ç ç ç ö ø ÷ ÷ ÷ WT = Eg qE
- 182. Current-voltage characteristics • We’ll discuss avalanche multiplication, a different effect, next
- 183. Carrier multiplication 1. EHP generated thermally 2. Electron accelerated to left (high field under reverse bias) 3. Electron collides 1. Loses energy 2. Knocks another electron loose, generating a another EHP (impact ionization) 3. Now there are three carriers
- 184. Carrier multiplication, continued 4. Both electrons accelerated to left, new collision • Two electrons arrived at left side of junction while only one entered- multiplication of 2Note: 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.
- 185. 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
- 186. 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 nin (1+ P+ P2 + P3 + )
- 187. Multiplication, continued • We had • Which can be expressed as • Thus multiplication factor M is • If P=1, M=∞, and avalanche occurs nin (1+ P+ P2 + P3 + ) nin (1- P) M = 1 (1- P)
- 188. 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
- 189. 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
- 190. Key points • Under reverse bias, diodes can break down – Not destructive, means the curve breaks down • Two mechanisms – Tunneling – Carrier multiplication • If multiplication large, can have avalanche
- 191. Copyright © 2016 The McGraw-Hill Companies, Inc. Permission required for presentation or display Reverse Bias Breakdown Chapter 5 Section 3.4
- 192. 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
- 193. Breakdown vs doping
- 194. 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
- 195. Avalanche breakdown • For one-sided junctions Vbr avalanche( )= CN - 3 4 Semiconductor Bandgap (eV) C 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
- 196. 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
- 197. 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
- 198. Find breakdown voltage • Vbr =4 V (Va=4V) 4V
- 199. Find maximum field and junction width • Junction width: ( we used Vj=Vbi-Va=1.04-(- 4)=5.04V) • Maximum electric field: w = 2eVj qND ' é ë ê ê ù û ú ú 1 2 = 2 ´11.8 ´ 8.85´10-12 F / m( )´ 5.04V( ) 1.6 ´10-19 C( )´ 8´1023 m-3 ( ) é ë ê ê ù û ú ú 1 2 = 9.1´10-8 m = 91nm Emax = 2Vj w = 2´5.04V 9.1´10-8 m =1.1´108 V / m
- 200. Now to find the tunneling width WT = Eg (eV ) qEmax = 1.12eV 1.6 ´10-19 C( ) 1.1´108 V / m( ) 1eV 1.6 ´10-19 V æ èç ö ø÷ = 1.0 ´10-8 m = 10nm. Thus, for silicon, an appreciable tunneling current flows for WT<10 nm
- 201. 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
- 202. Copyright © 2016 The McGraw-Hill Companies, Inc. Permission required for presentation or display Chapter 5 Section 4.1 Small signal impedance of prototype homojunctions- Junction Resistance
- 203. 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
- 204. 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
- 205. 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 GP = dI dVa
- 206. Junction resistance • Resistance is reciprocal of conductance, or RP = 1 GP = dI dVa é ë ê ù û ú -1
- 207. 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 RP = dI dVa é ë ê ù û ú -1 I = I0 e qVa kT -1 æ è ç ö ø ÷ = I0 e qVa kT - I0 @ I0 e qVa kT e qVa kT >>1 dI dVa = q kT I0 e qVa kT = qI kT = 1 RP
- 208. Example, continued • We had • Thus • Since kT/q=0.026V: – At I=1 mA, RP=26 Ω – At I=1 μA, RP=26 KΩ dI dVa = q kT I0 e qVa kT = qI kT RP = kT qI .
- 209. 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
- 210. Copyright © 2016 The McGraw-Hill Companies, Inc. Permission required for presentation or display Chapter 5 Section 4.2 Small signal impedance of prototype homojunctions- Junction Capacitance
- 211. 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
- 212. 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
- 213. 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
- 214. • 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 Differential junction capacitance Cj º dQ dVa Cj = e A w A=junction area ε=permittivity of material w=junction width
- 215. But, w depends on √(voltage) • Thus Cj is not linear with voltage • Expression for w (pn junction): • So (pn junction) w = wn + wp = 2eVj NA ' + ND ' ( ) qNA ' ND ' é ë ê ê ù û ú ú 1 2 Cj = e A w = A qeND ' NA ' 2 ND ' + NA ' ( ) Vbi -Va( ) é ë ê ê ù û ú ú 1 2 = A qeND ' NA ' 2 ND ' + NA ' ( )Vj é ë ê ê ù û ú ú 1 2
- 216. Results • For pn junction (repeated) • For one-sided step junction Cj = A qeN' 2 Vbi -Va( ) é ë ê ê ù û ú ú 1 2 = A qeN' 2Vj é ë ê ê ù û ú ú 1 2 Cj = A qeND ' NA ' 2 ND ' + NA ' ( )Vj é ë ê ê ù û ú ú 1 2 N’ is the net doping concentration on the lightly doped side
- 217. Junction capacitance per unit area for one-sided junction • Reverse bias shown
- 218. 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 Cj = A qeN' 2 Vbi -Va( ) é ë ê ê ù û ú ú 1 2 = A qeN' 2Vj é ë ê ê ù û ú ú 1 2 Vj =Vbi - Va - IRS( )
- 219. 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
- 220. Copyright © 2016 The McGraw-Hill Companies, Inc. Permission required for presentation or display Chapter 5 Section 4.3 Small signal impedance of prototype homojunctions- Stored Charge Capacitance
- 221. 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
- 222. 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
- 223. 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 Dnp (x) = Dnp (xp )e - x-xp Ln
- 224. 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 Qs = -qA Dnp (x)dx xp ¥ ò = qADnp (xp )Ln
- 225. Current is by diffusion of minority carriers (primarily) • So current across x=xp is • Combine with • Result I xp = qADn dDnp (x) dx xp = -qADn Dnp (xp ) Ln Qs = -qA Dnp (x)dx xp ¥ ò = qADnp (xp )Ln Qs = ILn 2 Dn = Itn Ln = Dn tn
- 226. 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 • Since • 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 Dn xp( )µ e qVa kT Jn = qDn dn dx Referred to as reclaimable stored charge
- 227. 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) Csc = dQsr dVa = d dQs dVa = q kT d Itn I xp = -qADn Dnp (xp ) Ln = -qADn np0 e qVa kT Ln Qs = Itn
- 228. 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 µ µ Csc = q kT d Itn = q kT dnp0 e qVa kT tn Cj = A qeNA 2 Vbi -Va( )
- 229. 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: Cj = A qeNA 2 Vbi -Va( ) é ë ê ê ù û ú ú 1 2
- 230. First find Vbi • Find Vbi from chart for one-sided junction Vbi=0.98V
- 231. Calculate Cj’s • For Va=-5V • and for Va=0.75V Cj (-5) = 100mm2 10-8 cm2 1mm2 æ èç ö ø÷ i i (1.6 ´10-19 C) 11.8( ) 8.85´10-14 F cm æ èç ö ø÷ 1017 cm-3 ( ) 2 0.98+5( ) é ë ê ê ê ê ù û ú ú ú ú 1 2 = 0.053pF Cj (0.75) = 0.27 pF x x Cj = A qeNA 2 Vbi -Va( )
- 232. For stored-charge capacitance • Need to find I – Use the diffusion current – Generation/recombination current does not contributed to stored-charge capacitance • We find I = I0 e qVa kT -1 æ è ç ö ø ÷ = qA Dn np0 Ln æ è ç ö ø ÷ e qVa kT -1 æ è ç ö ø ÷ np0 = ni 2 NA ' = (1.08´1010 cm-3 )2 1017 cm-3 =1.17 ´103 cm-3 Csc = dQsr dVa = d dQs dVa = q kT d Itn
- 233. • Look up Dn (minority carriers on p-side) I = I0 e qVa kT -1 æ è ç ö ø ÷ = qA Dn np0 Ln æ è ç ö ø ÷ e qVa kT -1 æ è ç ö ø ÷ Dn=20 cm2/s
- 235. Find τn τn=3 µs
- 236. So I0 becomes • and I at Va=-5V is -5.3 x 10-19 A • At Va=+0.75V, I0 = qA Dn np0 Ln æ è ç ö ø ÷ = 1.6 ´10-19 C ×100 ´10-8 cm2 ´ 20 cm2 s ×1.17 ´103 cm-3 70 ´10-4 cm æ è ç ç ç ö ø ÷ ÷ ÷ = 5.3´10-19 A I = I0 e qVa kT -1 æ è ç ö ø ÷ = 5.3´10-19 A e 0.75 0.026 -1 æ èç ö ø÷ =1.8mA
- 237. Stored-charge capacitances • At Va=-5V: • At Va=+0.75V Csc (-5) = q kT d Itn = 1 0.026V æ èç ö ø÷ (0.5)(5.3´10-19 A)(3´10-6 s) = 3.1´10-23 F @ 0 Csc (0.75) = q kT d Itn = 1 0.026V æ èç ö ø÷ (0.5)(1.8´10-6 A)(3´10-6 s) = 100 pF
- 238. Compare results Cj Csc Va= -5 V 0.053 pF ≈0 Va= +0.75 V 0.27 pF 100 pF
- 239. 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
- 240. Copyright © 2016 The McGraw-Hill Companies, Inc. Permission required for presentation or display Chapter 5 Section 5.1 Turn-off transient
- 241. 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
- 242. 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)
- 243. Take an n+-p junction • Then can ignore stored hole charge on n-side • At time t=0, the diode is on Dnp (x,t = 0) = Dnp (xp ,0)e - x-xp Ln • Vj(on)<Vbi • VF>>Vbi>Vj • Thus forward current IF is (up to t=0) IF (0) @ VF R1
- 244. 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
- 245. 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) IR @ VR R1
- 246. 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 ts @ tn ln 1+ IF IR æ è ç ö ø ÷ ts = tn ln2 = 0.68tn
- 247. 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 IF IR
- 248. 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
- 249. Copyright © 2016 The McGraw-Hill Companies, Inc. Permission required for presentation or display Chapter 5 Section 5.2 Turn-on transient
- 250. 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
- 251. Now must build up stored charge • Slopes constant again • Current IF constant • Reach steady state when injection rate equals recombination rate
- 252. 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
- 253. 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 VOLTAGE DIODE CURRENT
- 254. 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 VOLTAGE DIODE CURRENT
- 255. 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 VOLTAGE DIODE CURRENT
- 256. 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
- 257. For short-base diode • WB<<Ln • Recombination mostly happens at contact (neglect recombination in p-QNR) • and Dnp (x) = Dnp (0) 1- x WB æ è ç ö ø ÷ In = qADn dDn dx = - qADn Dn(0) WB
- 258. Excess charge in short base is • where tT is transit time across the short QNR (“base transit time”) • Base transit time given by Qs = -qADn(0)i WB 2 = In tT tT = WB( ) 2 2Dn x
- 259. 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 Qs = -qADn(0)i WB 2 = In tT tT = WB( ) 2 2Dn Qs = ILn 2 Dn = Itn Ln 2 = Dn tn Qs (short - base) Qs (long - base) = 1 2 i WB Ln æ è ç ö ø ÷ 2 x x short long
- 260. 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 Qs (Short-Base) Qs (Long Base) @ 1 2 i 0.1 31 æ èç ö ø÷ 2 @ 5´10-6x
- 261. 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
- 262. Copyright © 2016 The McGraw-Hill Companies, Inc. Permission required for presentation or display Chapter 5 Section 6 Effects of temperature
- 263. 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
- 264. 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 J0 = q NA ' Dn tn ni 2 = q NA ' Dn tn NC NV e - Eg kT dJ0 dT æ èç ö ø÷ J0 = de - Eg kT dT æ è ç ç ç ö ø ÷ ÷ ÷ e - Eg kT = Eg kT 2 æ è ç ö ø ÷ = Eg kT × 1 T
- 265. Fractional change for Si • For Si, • or J0 varies by 14% per °C! dJ0 dT æ èç ö ø÷ J0 = de - Eg kT dT æ è ç ç ç ö ø ÷ ÷ ÷ e - Eg kT = Eg kT 2 æ è ç ö ø ÷ = Eg kT × 1 T 1.12 0.026 × 1 300 = 0.14
- 266. What about JGR0? • That depends on ni rather than ni 2 (J0 depends on ni 2) • Recall • So JGR0 changes with temperature at about half the rate of J0 ni 2 = NC NV e - Eg kT
- 267. 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
- 268. Key points • Diode current and voltages vary with temperature • Have to take into account when designing devices
- 269. Copyright © 2016 The McGraw-Hill Companies, Inc. Permission required for presentation or display Chapter 5 Section 7 Summary
- 270. 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
- 271. Fermi levels • Non-degenerate n and p regions: • Degenerate: can often assume that Ef = EC - kT ln NC ND ' = Ei + kT ln ND ' ni n-region Ef = EV + kT ln NV NA ' = Ei - kT ln NA ' ni p-region Ef @ EC n+ -region Ef @ EV p+ -region
- 272. Built-in voltage Vbi = 1 q Fp - Fn( ) Vbi = 1 q Eg - kT ln NC NV ND ' NA ' é ë ê ù û ú = kT q ln ND ' NA ' ni 2 prototype pn junction Vbi = 1 q Eg - kT ln NV NA ' é ë ê ù û ú = kT q ln NV NA ' ni 2 prototype n+ -p junction Vbi = 1 q Eg - kT ln NC ND ' é ë ê ù û ú = kT q ln NC ND ' ni 2 prototype p+ -n junction in general:
- 273. Junction width • Width of region where there are uncompensated ions • No free carriers – Depletion region – Transition region – Space charge region
- 274. Junction width equations • pn • n+-p • p+-n w = wn + wp = 2eVj NA ' + ND ' ( ) qNA ' ND ' é ë ê ê ù û ú ú 1 2 w = 2eVj qNA ' é ë ê ê ù û ú ú 1 2 w = 2eVj qND ' é ë ê ê ù û ú ú 1 2 Vj I the junction voltage Vj=Vbi-Va Va is the applied voltage
- 275. Junction width • Most of the junction width appears on the lightly doped side • Most of the voltage is dropped across the lightly doped side wn wp = xo - xn( ) xp - xo( ) = NA ' ND ' Vj n Vj p = NA ' ND '
- 276. 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 J = JGR + Jdiff @ JGRo e qVa 2kT -1 æ è ç ö ø ÷ + Jo e qVa kT -1 æ è ç ö ø ÷
- 277. 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 Jdiff = Jo e qVa kT -1 æ è ç ö ø ÷
- 278. Leakage currents Transition width w goes as √VjJGRo @ qni w 2to Jo = q Dn npo Ln + Dp pno Lp æ è ç ö ø ÷ Long base diode Short base diode (p-side short)Jo = q Dn npo WB + Dp pno Lp æ è ç ö ø ÷ J0 = q Dn np0 WB( p) + Dp pn0 W(n) æ è ç ö ø ÷ Both sides short
- 279. Leakage current comments • Usually JGR0>>J0 – JGR0 important for reverse bias and small forward bias – J0 important for larger forward bias
- 280. 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 J e - pwT m* Eg 2 3 2 æ è ç ç ç ö ø ÷ ÷ ÷ M = 1 (1- P)
- 281. Capacitance • Two kinds – Junction capacitance (variation in ionized charges in depletion region) – Stored charge capacitance (variation in charge in excess carriers near junction)
- 282. Junction Capacitance • Junction capacitance (variation in ionized charges in depletion region) • Goes as Va -1/2 • Predominates under reverse bias and low forward bias Cj = A qeND ' NA ' 2 ND ' + NA ' ( ) Vbi -Va( ) é ë ê ê ù û ú ú 1 2
- 283. Stored Charge Capacitance • Stored charge capacitance (variation in charge in excess carriers near junction) • Dominates under large forward bias Csc = q kT d Itn (longbase) Csc = q kT d ItT = q kT d I WB 2 2Dn æ è ç ö ø ÷ (shortbase)
- 284. 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
- 285. 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