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 electron-hole 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. (1)

Avalanche Multiplication The breakdown voltage VB for one-sided abrupt junctions can be found by The breakdown voltage for linearly graded junctions is expressed as

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 one-sided p+-n abrupt junction with ND = 5 x 1016 cm-3

Ex. Calculate the breakdown voltage for a Si one-sided 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.

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 (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)

Avalanche Multiplication Soln When the n-type region is reduced to 20 μm, the punch-through will occur first.

Heterojunction A heterojunction is defined as a junction formed by two semiconductors with different energy bandgaps Eg, different dielectric permittivities s, different work function es, 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 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 built-in potential Vbi can be expressed by (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 (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 Note: Most of the built-in potential is in the semiconductor with a lower doping concentration and also its depletion width is much wider.

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 eb = 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

Metal-Semiconductor Junctions This behavior is referred to a rectification and can be described by an ideal diode equation of (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

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

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 (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.