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ENE 311 Lecture 9 Junction Breakdown • When a huge reverse voltage is applied to a p-n junction, the junction breaks down and conducts a very large current. • Although, the breakdown process is not naturally destructive, the maximum current must be limited by an external circuit to avoid excessive junction heating. • There are two mechanisms dealing with the breakdown: tunneling effect and avalanche multiplication. Tunneling Effect • If a very high electric field is applied to a p-n junction in the reverse direction, a valence electron can make a transition from the valence band to the conduction band by penetrating through the energy bandgap called tunneling. • The typical field for Si and GaAs is about 106 V/cm or higher. Tunneling Effect • To achieve such a high field, the doping concentration for both p- and n-regions must be very high such as more than 5 x 1017 cm-3. • The breakdown voltage for Si and GaAs junctions about 4Eg/e is the result of the tunneling effect. With the breakdown voltage is more than 6Eg/e, the breakdown mechanism is the result of avalanche multiplication. • As the voltage is in between 4Eg/e and 6Eg/e, the breakdown is due to a mix of both tunneling effect and avalanche multiplication. Avalanche Multiplication • Let consider a p+-n one-sided abrupt junction with a doping concentration of ND 1017 cm-3 or less is under reverse bias. • As an electron in the depletion region gains kinectic energy from a high electric field, the electron gains enough energy and acceleration in order to break the lattice bonds creating an electron-hole pair, when it collides with an atom. Avalanche Multiplication • A new electron also receives a high kinetic energy from the electric field to create another electron-hole pair. This continue the process creating other electronhole pairs and is called avalanche multiplication. Avalanche Multiplication • Assume that In0 is incident current to the depletion region at x = 0. • If the avalanche multiplication occurs, the electron current In will increase with distance through the depletion region to reach a value of Mn In0 at W, where Mn is the multiplication factor. I n (W ) Mn I n0 (1) Avalanche Multiplication • The breakdown voltage VB for one-sided abrupt junctions can be found by EcW s Ec2 1 VB NB 2 2e • The breakdown voltage for linearly graded junctions is expressed as 2 EcW 4 Ec3 / 2 2 s VB 3 3 ea 1/ 2 Avalanche Multiplication The critical field Ec can be calculated for Si and GaAs by using the plot in the above figure. Ex. Calculate the breakdown voltage for a Si onesided p+-n abrupt junction with ND = 5 x 1016 cm-3 Ex. Calculate the breakdown voltage for a Si onesided p+-n abrupt junction with ND = 5 x 1016 cm-3 Soln From the figure, at the given NB, Ec is about 5.7 x 105 V/cm. VB s Ec2 2e NB 1 11.9 8.85 10 21.4 V 14 5.7 10 2 1.6 10 19 5 2 5 10 16 1 Avalanche Multiplication Avalanche Multiplication • Assume the depletion layer reaches the n-n+ interface prior to breakdown. • By increase the reverse bias further, the device will break down. • This is called the punch-through. Avalanche Multiplication • The critical field Ec is the same as the previous case, but the breakdown voltage VB for this punch-through diode is VB W W 2 VB Wm Wm (4) • Punch-through occurs when the doping concentration NB is considerably low as in a p+--n+ or p+--n+ diode, where stands for a lightly doped p-type and stands for a lightly doped n-type. Avalanche Multiplication Breakdown voltage for p+-π-n+ and p+-v-n+ junctions. W is the thickness of the lightly doped region Avalanche Multiplication • Ex. For a GaAs p+-n one-sided abrupt junction with ND = 8 x 1014 cm-3, calculate the depletion width at breakdown. If the n-type region of this structure is reduced to 20 μm, calculate the breakdown voltage if r for GaAs is 12.4. Avalanche Multiplication Soln For the figure, we can find VB is about 500 V (VB >> Vbi) 2 s Vbi V W eN B 2 8.85 1014 12.4 500 1.6 1019 8 1014 2.93 103 29.3 m Avalanche Multiplication Soln When the n-type region is reduced to 20 μm, the punch-through will occur first. VB W W 2 VB Wm Wm 20 20 VB 500 2 449 V 29.3 29.3 Heterojunction A heterojunction is defined as a junction formed by two semiconductors with different energy bandgaps Eg, different dielectric permittivities s, different work function es, and different electron affinities eχ. Heterojunction • The difference energy between two conduction band edges and between two valence band edges are represented by EC and EV, respectively, as EC e 2 1 EV Eg1 e1 Eg 2 e2 Eg EC where Eg is the difference energy bandgap of two semiconductors. Heterojunction • Generally, heterojunction has to be formed between semiconductors with closely matched lattice constants. • For example, the AlxGa1-xAs material is the most important material for heterojunction. • When x = 0, the bandgap of GaAs is 1.42 eV with a lattice constant of 5.6533 Å at 300 K. • When x = 1, the bandgap of AlAs is 2.17 eV with a lattice constant of 5.6605 Å. Heterojunction • We clearly see that the lattice constant is almost constant as x increased. The total builtin potential Vbi can be expressed by Vbi Vb1 Vb 2 2 N 2 Vbi V Vb1 1 N1 2 N 2 1 N1 Vbi V Vb 2 1 N1 2 N 2 (7) where N1 and N2 are the doping concentrations in semiconductor 1 and 2, respectively. Heterojunction • The depletion widths x1 and x2 can be found by x1 x2 21 2 N 2 Vbi V eN1 1 N1 2 N 2 21 2 N1 Vbi V eN 2 1 N1 2 N 2 (8) Heterojunction Ex. Consider an ideal abrupt heterojunction with a built-in potential of 1.6 V. The impurity concentrations in semiconductor 1 and 2 are 1 x 1016 donors/cm3 and 3 x 1019 acceptors/cm3, and the dielectric constants are 12 and 13, respectively. Find the electrostatic potential and depletion width in each material at thermal equilibrium. Heterojunction Soln Vb1 13 3 1019 1.6 12 1 10 1.6 V 13 3 10 12 1 10 1.6 4.9 10 V 12 1 10 13 3 10 2 12 13 8.85 10 3 10 1.6 4.608 10 1.6 10 1 10 12 10 13 3 10 2 12 13 8.85 10 1 10 1.6 1.536 10 1.6 10 3 10 12 10 13 3 10 16 19 16 4 Vb 2 16 19 14 x1 19 5 19 16 16 14 x2 19 19 19 cm 16 16 8 19 Note: Most of the built-in potential is in the semiconductor with a lower doping concentration and also its depletion width is much wider. cm Metal-Semiconductor Junctions • The MS junction is more likely known as the Schottky-barrier diode. • Let’s consider metal band and semiconductor band diagram before the contact. Metal-Semiconductor Junctions • When the metal and semiconductor are joined, electrons from the semiconductor cross over to the metal until the Fermi level is aligned (Thermal equilibrium condition). • This leaves ionized donors as fixed positive charges that produce an internal electric field as the case of one-sided p-n junction. Metal-Semiconductor Junctions • At equilibrium, equal number of electrons across the interface in opposite directions. • Hence, no net transport of charge, electron current Ie equals to zero. The built-in voltage Vbi = m - s. • The barrier for electrons to flow from the metal to semiconductor is given by eb = e(m - χs) or it is called the barrier height of MS contact. Metal-Semiconductor Junctions • When a voltage is applied, the barrier height remains fixed but the built-in voltage changes as increasing when reverse biased and decreasing when forward biased. Metal-Semiconductor Junctions • Reverse bias • Few electrons move across the interface from metal to semiconductor due to a barrier, but it is harder for electrons in the semiconductor to move to the metal. • Hence, net electron transport is caused by electrons moving from metal to semiconductor. Electron current flows from right to left which is a small value. Metal-Semiconductor Junctions • Forward bias • Few electrons move across the interface from metal to semiconductor, but many electrons move across the interface from semiconductor to metal due to the reduced barrier. • Therefore, net transport of charge flows from semiconductor to metal and electron current flows from left to right. Metal-Semiconductor Junctions • Under forward bias, the electrons emitted to the metal have greater energy than that of the metal electrons by about e(m - χs). • These electrons are called hot-carrier since their equivalent temperature is higher than that of electrons in the metal. • Therefore, sometimes, Schottky-diode is called “hot-carrier diode”. Metal-Semiconductor Junctions • This leads to the thermionic emission with thermionic current density under forward bias as J F A**T 2e em s VF / kT where J s A**T 2e J seeVF / kT em s / kT saturation current m* A A. effective Richardson's constant m0 ** * Metal-Semiconductor Junctions • This behavior is referred to a rectification and can be described by an ideal diode equation of J Js e eV / kT 1 (10) where V positive for forward bias and negative for reverse bias. Metal-Semiconductor Junctions • The space-charge region width of Schottky diode is identical to that of a one-sided p-n junction. • Therefore, under reverse bias, they can contain the charges in their depletion region and this is called Schottky diode capacitance. Metal-Semiconductor Junctions Ex. A Schottky junction is formed between Au and n-type semiconductor of ND = 1016 cm-3. Area of junction = 10-3 cm2 and me*= 0.92 m0. Work function of gold is 4.77 eV and eχs = 4.05 eV. Find current at VF = 0.3 volts. Metal-Semiconductor Junctions Soln * m A** A* e 120 0.92 110 A/ cm 2 .K m0 J s A**T 2e em s / kT J J s eeV / kT 1 110 3002 e 4.77 4.05 / 0.0259 . e0.3 / 0.0259 1 0.897 A/(cm 2 ) I A J 103 0.897 0.897 mA Metal-Semiconductor Junctions Ex. Si-Schottky diode of 100 μm diameter has (1/C2) v.s. VR slope of 3 x 1019 F-2V-1. Given r = 11.9 for Si. Find NB for this semiconductor. Metal-Semiconductor Junctions Soln 1 2 Vbi VR ; 2 Cj e s N B slope NB NB Cj C [F/cm 2 ] Area 2 [cm 4 F-2 V -1 ] e s N B 2 slope Area 2 e s 2 2 4 2 19 100 10 19 12 3 10 1.6 10 8.85 10 11.9 2 6.414 1019 cm -3 Ohmic contact • This contact is defined as a junction that will not add a significant parasitic impedance to the structure on which it is used and will not sufficiently change the equilibrium-carrier densities within the semiconductor to affect the device characteristics. • The I-V characteristic of ohmic contact is linear for an ideal case. Ohmic contact • A specific contact resistance RC is given by 1 J RC V V 0 .cm2 1 (11) • A good ohmic contact should have a small specific contact resistance about 10-6 Ω.cm2. Ohmic contact • When the semiconductor is heavily doped with an impurity density of 1019 cm-3 or higher, the depletion layer of the junction becomes very thin so that carriers can tunnel instead of going over the potential barrier.