ECSE-6230 Semiconductor Devices and Models I Lecture 10

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ECSE-6230Semiconductor Devices and Models
ILecture 10

• Prof. Shayla Sawyer
• Bldg. CII, Rooms 8225
• Rensselaer Polytechnic Institute
• Troy, NY 12180-3590
• Tel. [phone]
• Fax. [phone]
• e-mail sawyes_at_rpi.edu

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sawyes_at_rpi.edu www.rpi.edu/sawyes/courses.html

May 21, 2014
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Lecture Outline

• Junction Breakdown
• Zener Breakdown
• Avalanche Breakdown
• PN Junction Switching Characteristics
• Charge Control
• Constant Current Turn Off
• Reverse Bias Turn Off
• Midterm notes

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Junction Breakdown

• With the increase in reverse voltage across a pn
  junction, when the voltage reaches the breakdown
  voltage (BV), a large reverse current starts to
  flow.
• Junction breakdown is due to the high electric
  field at the junction.

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Junction Breakdown

• Basically, 2 breakdown mechanisms
• If the BV lt 4 Eg / q ( 4V in Si ), carrier
  tunneling across the junction dominates ( Zener
  breakdown )
• If the BV gt 6 Eg / q ( 6V in Si ), carrier
  multiplication within the depletion region due to
  impact ionization is the major process (
  Avalanche breakdown )

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Junction Breakdown

• Zener breakdown -
• Usually occurs in p/n junctions
• Electrons tunnel from the valence band through
  the bandgap to the conduction band
• Breaking of the covalent bonds due to high
  electric field (called field ionization) is the
  basic mechanism

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Junction Breakdown

• Tunnel barrier is of the triangular shape
• Use WKB (Wentzel-Kramers-Brillouin) approximation
• Put varying conduction band in terms of electric
  field
• Find tunneling probability
• Tunneling current, from either band to empty
  states in the other

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Avalanche Breakdown

• Small initiation current leads
• to large current due to carrier
• multiplication resulting from impact ionization
  caused by the
• high electric field ( gt 105 V/cm ) near the
  metallurgical
• junction.

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Avalanche Breakdown

• Assume Ip0 incident from the left side of the
  depletion region with width WDm.
• With high electric field, e-h pairs are created,
  Ip will increase with distance and reach MpIpo at
  xWDm

• In will increase from In(WDm)0 to In(0)I-Ipo
• The total current is constant at steady state
  (IIpIn)
• The incremental hole current is equal to the
  number of e-h pair generated per second in the
  distance dx

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Avalanche Breakdown

• Boundary condition
• When VR ? BV, Mp ? ?.
• The breakdown condition can be specified as the
  ionization integral
• If the avalanche process is initiated by
  electrons instead of holes

is
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Avalanche Breakdown

• Since avalanche breakdown does not depend on the
  carriers or primary current, either ionization
  integral can be used.
• For semiconductors with equal ionization rates (
  ?n ?p ? ) such as GaP, the ionization
  integral reduces
• Breakdown voltage for one sided abrupt junctions
• Breakdown voltage for linearly graded junctions

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Avalanche Breakdown

• For an abrupt, one-sided pn junction,
• ?max ( 2 q NB VR ) / ?S 1/2
• So, BV ?S ?cr2 / ( 2 q NB )
• ?cr is a weak function of doping because the
  depletion layer width (hence ionization path)
  decreases with increasing doping.

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Avalanche Breakdown Voltage

• With a universal expression, accounting for
  instances where a uniform field over a large
  distance does not exist, Sze has derived
• BV 60 ( Eg / 1.1 )3/2 ( NB / 1016 )-3/4
• for an abrupt junction
• ? NB-3/4
• and
• BV ( Eg / 1.1 )6/5 ( a/(3 x 1020))-2/5
• for a linearly graded junction
• However, these expressions are NOT valid for wide
  bandgap (gt 2 eV) semiconductors, such as SiC and
  GaN. (total voltage must be larger than bandgap)

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Avalanche Breakdown Voltage

• Calculated breakdown voltage as a function of N
  for abrupt junctions
• Dashed line is the upper limit of N for which the
  avalanche breakdown calculation is valid
• Based on criterion 6Eg/q above it tunneling will
  dominate

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Avalanche Breakdown Voltage

• For diffused junctions with a linear gradient
  near the junction and a constant doping on one
  side the BV lies between two limiting cases
• For a large a, the BV is given by the abrupt
  junction results
• For small a, BV is given by the linearly graded
  junction and is independent of NB

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Avalanche Breakdown Voltage

• It is assumed that the semiconductor layer is
  thick enough to support the maximum
  depletion-layer width WDm at breakdown
• If the semiconductor layer W is smaller than WDm
  the device will be punched through
• Punchthrough breakdown is earlier

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Planar Junctions

• Actual pn junction have junction curvatures that
  are cylindrical or spherical, leading to electric
  field concentrating
• Breakdown voltages are significantly less than
  those of the 1-dim, parallel plane junction that
  we have examined thus far

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Avalanche Breakdown Voltage
With ? ? rj / WDm , Cylindrical
curvature And spherical curvature As the
radius of curvature becomes smaller, so does the
breakdown voltage
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Avalanche Breakdown Voltage
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Transient behavior

• For switching applications the transitions form
  forward bias to reverse bias and vice versa much
  be nearly abrupt and the transient time short
• The response from forward to reverse is limited
  by minority carrier charge storage
• We investigate the switching of a diode from its
  forward state to its reverse state
• Begin with just from on to off

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Large-Signal Charge-Control Model

• Use the time dependent continuity equation.
  Obtain each component of the current at position
  x and time t
• Integrate both sides for instantaneous current
  density
• For injection into a long n region from a p
  region, take current xn0 to be all hole current
  and Jp at xn8 to be zero.

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Large-Signal Charge-Control Model

• Total injected current including time variations
• Hole current injected across the p n junction(
  total diode current) is determined by two storage
  charge effects
• (1) usual recombination term, excess carrier
  distribution is replaced every tp seconds
• (2) charge buildup (or depletion term) carriers
  can be increasing or decreasing in a time
  dependent problem

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Charge Control Equation

• Solve for stored charge as a function of time for
  a given current transient
• Turn off transient, current is suddenly removed
  at t0, leaves the diode with stored charge

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Charge Control Equation

• To solve for v(t) an approximate solution can be
  obtained by assuming an exponential distribution
  for dp at every instant during the decay
• Quasi-steady state approximation neglects
  distortion due to the slope requirement at xn0

Non exponential
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Charge Control Equation

• Not accurate in its details but indicates that
  the voltage across a pn junction cannot be
  changed instantaneously
• Stored charge can present a problem in a diode in
  switching applications
• Problems of stored charge can be reduced by
• Narrow n region (if shorter than hole diffusion
  length, very little charge is stored
• Adding recombination centers such as Au to Si

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Constant Current Turn-Off

• Constant Current Turn-Off Case for a Long-Base
  Diode
• The Charge Control Equation becomes
• dQp / dt Qp(t) / ?p 0
• B.C.s (a) i(t lt 0) I, (b) i(t gt
  0) 0
• Solving using Laplace Transform with Qp(0) I
  ?p
• Qp(s) / ?p s Qp(s) - I ?p 0
• Qp(s) I ?p / ( s 1 / ?p )
• Performing reverse Laplace transform
• Qp(t) I ?p exp ( - t / ?p )
• ?pn(t) pn0 ( exp ( q v(t) / kT ) - 1 )
• What is v(t) ?

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Constant Current Turn-Off

• Assume the quasi-steady state approximation.
• ?pn( x, t ) ?pn( t ) exp ( - ( x xn ) /
  Lp )
• Qp(t) q A ?WN ?pn( x, t ) dx
• ?xn
• q A ?WN ?pn( t ) exp ( - ( x xn ) /
  Lp ) dx
• ?xn
• q A Lp ?pn( t )
• ?pn( t ) pn0 ( exp ( q v(t) / kT ) 1 )
  Qp(t) / ( q A Lp )
• v(t) ( kT/q ) ln ( I ?p ) / ( q A Lp pn0 )
  exp ( - t / ?p ) 1

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Reverse-Biased Turn-Off of pn Junctions

• The switch is flipped from VF at tlt0 to VR tgt0
• Large reverse current occurs first. Why?

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Reverse-Biased Turn-Off

• Transient time current drops to 10 of initial
  reverse current sum t1 and t2
• t1 is the constant current phase, t2 is the decay
  phase
• During the decay phase, excess charge is being
  removed primarily by recombination
• The device approaches steady state dc in the
  reverse bias condition

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Large-Signal Charge-Control Model

• Assume p/n junction with analysis on the n-type
  side
• Continuity equation
• Boundary conditions initial distribution of
  holes is a steady state solution to the diffusion
  equation and under forward bias the voltage
  across the junction is

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Reverse-Biased Turn-Off

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Reverse-Biased Turn-Off

• Reverse-Biased Turn-Off Case
• When 0 lt t lt tS (or t1 ), Cj can be neglected.
• Charge control equation
• Consider stored charge between 0lttltts or
  t1
• With the initial conditions i( 0 lt t lt tS ) -
  IR, Qp(0) IF?F
• so that
• By setting Qstored0, t1 can be obtained

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Reverse-Biased Turn-Off

• Assume a triangular shape for
• excess hole concentration with
• slopes
• tan(-?1) ? -??pn/?xt0 IF/qADp
• and
• tan(?2) ? -??pn/?xttS -IR/qADp
• giving tan(?1)/tan(?2) IF / IR
• ?pn(tS)/?pn(0)(1 tan(?1)/tan(?2))-1
• 1 / ( 1 (IF / IR))
• Qp(tS)/Qp(0) 1 / ( 1 (IF / IR))

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Reverse-Biased Turn-Off

• Reverse-Biased Turn-Off Case
• After t1 , the hole density starts to decrease
  below its equilibrium value pno. The junction
  voltage tends to reach VR and a new boundary
  condition now holds. This is the decay phase
  with the initial boundary condition
• The solution for t2 is

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Reverse-Biased Turn-Off

• Reverse-Biased Turn-Off Case
• For a plane junction with the length of the
  n-type material W much greater than the diffusion
  length WgtgtLp, for a large IR/IF ratio, the
  transient time can be approximated as
• For a a narrow base junction with WltltLp

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Reverse-Biased Turn-Off

• For IF IR,
• tS() 0.7?p but tS() 0.25?p and
  tS() 0.29?p.
• For t gt tS, the diode gets reverse biased and
  the current flowing through Cj can no longer be
  ignored.
• Define ?R from i(t) - Qp(t) / ?R for
  t gt tS
• Assuming ln ( 1 ( IF / (IF IR )) ? ln ( 1
  ( ?R / ?F )),
• (where ?F / ?R is called the excess charge
  ratio and 4 for a
• short-base diode)
• ?R ?F ( IF / (IF IR ))
• If we assume a resistive load, dv(t) - R
  di(t)
• di(t)/dt ( ?R R Cj ) i(t) ( 1 ( ?R /
  ?F )) 0

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Reverse-Biased Turn-Off

• with B.C. i( t tS ) - IR
• Solution i(t) - IR exp - ( t tS ) / ? ,
  t gt tS
• ? ( ?R R Ct ) / ( 1 ( ?R / ?F ))
• Fall time, tf (or t2 ), is the time taken for the
  reverse current to fall from IR to 0.1IR
• tf ? 2.3 ( ?R RCj ) / (1 ( ?R / ?F ))
  
• If ?R ltlt RCj, tf ? 2.3 RCj
• Cj can be approximated to the average of the
  capacitances at zero and full reverse bias.

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Reverse-Biased Turn-Off

• Ex. A long pn junction diode is forward biased
  to IF 10mA.
• Let a reverse bias of 10V be applied through a
  5k? at t 0.
• Let Cj (V0) 18pF and Cj(V-10V) 6pF, so
  that Cj ? 12pF.
• Assume ?F 0.2?s. Calculate tS and tf.
• Neglecting the voltage drop across the diode,
• IR 10V / 5k? 2mA
• tS (2x10-7s) ln ( 1 (1.5/2) ) - ln ( 1
  (1.5/3.5) ) 4x10-8s
• ?R ?F (1.5/3.5) (3/7) ?F
• tf 2.3 ((3/7)(2x10-7) (5x103)(12x10-12))
  / ( 1 (3/7))
• ? 2.35x10-7s
• tr tS tf ? 2.75x10-7s

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Midterm Short Paper

• 2-3 pages
• Introduction
• Background
• Basics of material (if needed)
• Basics of operation
• How it relates to SDM1
• Technical Relevance, Overall Impact, Applications
• Future Work
• Experimental plan, Milestones
• Equipment and/or simulation program
• Expected Outcomes (optional)
• References (IEEE Style)

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Midterm Short Presentation

• 10 minutes total including questions
• 5-7 slides
• All figures must be referenced in the caption if
  from article or book
• Feedback evaluations from Prof. to provide input
  on speaking and content