EE105 Spring 2006 Discussion 1

1. Semiconductors - Semiconductors are a group of materials, which have electrical conductivities intermediate between metals and insulators. The conductivity of these materials can be varied over orders of magnitude by changes in temperature, optical excitation and impurity content. This property makes semiconductor materials natural choices for electronic devices. - Si is in Group IV and has a diamond lattice.

2. States of an Atom - From quantum mechanics, electrons exist in certain stable, circular orbits about the nucleus. These circular orbits correspond to discrete energy levels.

- As isolated atoms are brought together to form a solid, various interactions occur between neighboring atoms and these energy levels split. Now because of the large number of atoms in close proximity, the discrete energy levels become bands. - The conductivity of the material is determined by the spacing between the conduction and valence band. In the case of semiconductors, the energy gap is approximately 1ev. When a bond is broken, a free electron leaves behind positive charge: a hole. - Recombination describes the process by which a hole meets a conduction band electron. They cancel each other, which decreases the amount of free carriers.

3. Intrinsic Carrier Concentration - A perfect semiconductor crystal with no impurities or lattice defects is called an intrinsic semiconductor. Since the electrons and holes are created in pairs, the conduction band electron concentration n (electrons per cm cubed) is equal to the concentration of holes in the valence band p (holes per cm cubed). Each of these intrinsic carrier concentrations is commonly referred to as ni, thus n = p = ni.

4. Doping by Ion Implantation In addition to the intrinsic carriers generated thermally, it is possible to create carriers in semiconductors by purposely introducing impurities into the crystal. This process is called doping. a. Donors These are group V elements (P, As). Each ionized donor will contribute an extra free electron. b. Acceptors These are group III elements (Boron). This leads to one free hole. c. Mass Action Law There is a balance between generation and recombination po*no = ni2. For n type material, we can approximate no = Nd-Na For p type material, we get po = Na-Nd. However the material is charge neutral so the total charge concentration must sum up to zero so : no + Na = po + Nd.

5. Drift and Currents The net flow of electrons and holes in a semiconductor will generate currents. The process by which these charged particles move is called transport. The two basic transport mechanisms in a semiconductor crystal are: - Drift: the movement of charge due to electric fields. - Diffusion: the flow of charge due to density gradients. Note: Temperature gradients in a semiconductor can also lead to carrier movement. However this can be ignored because of the small size of the semiconductor devices.

a. Drift currents In the case of holes, the drift velocity is in the same direction at the E.

The hole is:

In the case of electrons, the drift velocity is in the opposite direction at the electric field E and the electron drift current is given by:

This gives us a total drift current of:

Example: Calculate the drift in a semiconductor for a given electric field. Consider a germanium sample at T = 300K with doping concentrations of Nd = 0 and Na = 1016 cm-3. Assume complete ionization and electron and hole mobilities are 3900 cm2/V.sec and 1900cm2/V.sec. The applied electric field is E = 50V/cm.

Solution: Since Na>Nd, the semiconductor is p type and the majority carrier hold concentration po is approx. Na = 1016 cm-3. The minority carrier electron concentration is : no = ni2/po = 5.76x1010 cm-3. For this extrinsic ptype semiconductor, the drift current density is: dr dr dr J = Jp + Jn = q(pµp + nµn)E which can be approximated by qNaµpE Therefore, Jdr =152A/cm2.

Comment: Significant drift current densities can be obtained in a semiconductor applying relatively small electric fields. The drift current will be due primarily to the majority carrier in an extrinsic semiconductor. b. Diffusion Current The diffusion currents are expressed in terms of the electron or hole diffusion coefficient Dn or Dp, as shown below: diff Jn = -(-q)Dn dn/dx = qDn dn/dx diff Jp = -(+q)Dp dp/dx = qDp dp/dx

Einstein Relation: Dn = (kT/q) µn. c. Resistivity - Resistivity is defined by the following:

- Ohm’s Law states:

Now a new parameter is defined by eliminating the thickness variable, the sheet resistance, with Rsheet = ρ /t in units of ohms per square.

Example: a) Given that a does Qd = 1012 cm-2 of arsenic is implanted into p-type silicon, with a junction depth of xj = 400nm, estimate the sheet resistance of this n-type layer.

SOLUTION: The average arsenic concentration in the implanted layer is the thickness of the layer (in cm) divided into the dose (per cm2)

12 Qd 1"10 16 #3 Nd = = #5 = 2.5 "10 cm x j 4 "10 From Fig. 2.8 of the textbook (page 36), the electron mobility for the average concentration is µn = 1000 cm2/Vs. Substituting into the definition of sheet resistance we get: ! 1 1 Rsheet = = #19 16 #5 $ 6k% # per # square qµn Nd x j (1.6 "10 )(1000)(2.5 "10 )(4 "10 )

Note: This is an approximation for reasons, which will be discussed when covering pn junctions. ! b) For this n-type layer, find the length L for a straight resistor (no corners) of value R = 56kΩ. The minimum width W=2.5µm and the two contact areas have a “dogbone” contact area.

SOLUTION: First the total number of squares needed is:

R 56k" N = = = 9.3 Rsheet 6k"/square

Subtracting off the contributions of the contacts, the rectangular portion of the resistor has: ! L W = N " 2(0.65) = 9.3"1.3 = 8. Therefore, the length L=8x2.5µm=20µm.

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