Sample Exam # 2 ECEN 3320 Fall 2013 Semiconductor Devices October 28, 2013 Due November 4, 2013

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1 Sample Exam # 2 ECEN 3320 Fall 203 Semiconductor Devices October 28, 203 Due November 4, 203. Below is the capacitance-voltage curve measured from a Schottky contact made on GaAs at T 300 K. Figure : Capacitance voltage curve for problem 2. Some relations: The maximum field in the metal to n-type Schottky is given by E max qn dx d ɛ s. The built-in potential is the integral of this or φ i V a qn dx 2 d 2ɛ s. The built-in potential is also ressible as φ i φ b q ln Nc N d the relation that defines φ i as a quasi Fermi level. The depletion width can be found from the built-in potential by x d φ i V a ). qn d This x d can be used to find a well defined ression for the E max E max φ i V a ). qn d

2 From the x d, we can also find the capacitance per unit area by C A ɛ s qɛs N d x d 2φ i V a ). This is the ression that we really need here, for this first problem. Actually, we need the inverse square of this relation that is Calculate: a) the built-in potential, φ i, A C ) 2 2 φ i V a ) qɛ s N d Solution: The built-in potential is evidently the intercept or b) the doping density, N d, φ i 0.9 V Solution: We have that the slope, s, of the A/C) 2 curve is given by s 2 qɛ s N d or that N d 2 sqɛ s. and The slope can be read from the curve as -3/ giving N d c) the barrier height, φ B cm 3 Solution: The barrier height is the distance from the metal Fermi level to the unperturbed conduction band edge. The built in potential is the distance from the metal Fermi level to the unperturbed semiconductor Fermi level. The difference is the distance from the Fermi level to the conduction band edge in the semiconductor or φ b φ i + q ln Nc ln N d Consider a Schottky contact between gold and n-type GaAs with a cross-sectional area of 0 4 cm 2. 2

3 Solution: If we consider only the thermionic and diffusion current in the Schottky diode that is, we ignore the quantum tunneling current), we find that J J t + J d A T 2 + qv d N c ) qφ ) [ ] b qva J t0 + J d0 ) qφ ) [ ] b qva. The relation can also be written as J J t0 + J d0 ) qφ ) [ b qva qv R + µ n E max )N c qφ ) [ b where the v R is given by We note here that v R 2πm. m { GaAs m 0.08 Si, ] qva ) ] N c { cm 3 GaAs cm 3 Si and from Table 3.2. in the book) { 0.9 ev Au-GaAs φ b. 0.8 ev Au-Si The N c is signifcantly smaller for GaAs than Si. The µ n is a factor 6 larger qφ b for GaAs, and the v R due to the /m is larger for GaAs by) a factor of roughly 4. These factors are insignifcant with respect to the N c qφ b that is two orders of magnitude larger for silicon. a) Plot the current for forward bias voltage range of 0 V a 0.5 V. ). Solution: The plot follows the shape of qva b) Do the same for a Schottky contact made with gold and n-type silicon. ). Solution: The plot follows the shape of qva c) Discuss the differences between the two results in a) and b). Solution: The GaAs curve will lie more than an order of magnitude below the silicon curve. The exact number depends on the ratio of the thermionic current that can be calculated from the above) to the diffusion current that cannot be calculated exactly from the above as there is not enough information to calculate the E max ). The difference is not especially great, though, as the diffusion current and thermionic currents have differ by nearly the same factors in GaAs versus silicon. 3

4 3. Consider a silicon pin photodetector under a back bias loaded with a resistor R L. a) Draw the circuit diagram labeling all relevant voltages and currents. Solution: The p contact connects to the high side of a battery and the n contact to the load resistor that in turn connects to the low side ground) of the battery. b) Calculate the junction capacitance assuming that x i µm is much larger than the x n and x p due to doping. Use A 0 4 cm 2. Solution: We have that C ɛa x d pf c) Sketch the RC time constant and the circuit receiver) bandwidth /RC) as a function of R L. Solution: A straight line with slope C. d) Sketch the output voltage as a function of R L for a mw input power and a responsivity of R A/W. Solution: A straight line with slope ma. e) Find an ression and a value for a gain bandwidth product for the circuit. Solution: The bandwidth is /R L C and the transfer characteristic is V RL /P in RR L. The transfer characteristic bandwidth product is then R/C and is independent of the circuit, that is, only dependent on the properties of the diode. 4. Consider a step pn junction made of GaAs. The doping densities and carrier lifetimes are given as N d 0 7 cm 3, N a cm 3 and τ n 0 7 s, τ p 0 5 s. Assume long diode, i.e. long quasi-neutral regions. a) Find the built-in potential, Solution: We have that φ i q ln Na N d n 2 i ) q ln ev. b) Find the depletion region widths on the n- and p-sides, Solution: The depletion width is given by x d + ) φ i q Nd Na cm Given the disparity in doping densities, the lowest approximation is that x n 0 and x p x d. c) the capacitance per unit area, 4

5 Solution: We have that and C A ɛ s x d F/cm 2. d) Sketch the energy band diagram indicating the energy levels, E c, E f and E v, the built-in potential φ i and the depletion widths x n and x p. Solution: There is nothing tricky here. e) Find the ratio of the electron current to the total current at the edge of the p-side depletion region. Solution: We have that yielding that J tot J n + J p qn 2 i [ N a D n + τ n N d ] Dp τ p J n J tot Dn τ n Dn τ n + Na Dp N d τ p + Na Dpτn N d D nτ p

6 Useful Relations g c E) 2π 2 F E) 2m ) 3/2 e E h 2 Ec g v E) 2π 2 [E E f )/ ] + 2m ) 3/2 Ev h h 2 E Ei E c n i N c Ef E c n 0 N c Ev E i n i N v Ev E f p 0 N v n 0 p 0 n 2 i Fn E c Ev F p n 0 + δn N c p 0 + δp N v n t J n q x + G n R n n J n qd n x + qnµ ne I opt G n α n E ph R n δn τ n p t J p q x + G p R p p J p qd p x + qpµ ne I opt G p α p E ph R p δp τ p x n N ax d N a + N d q J J s φ i V a qn dx 2 n 2ɛ s + qn dx 2 p 2ɛ s N a φ i V a ) x p N dx d N d N d + N a N a + N d q + ) φ i V a ) Na Nd x d x n + x p q C A ɛ s x d N a N d n 2 φi V a i N a N d n 2 φi V a i qva J s J n + J p qn 2 i ) N a D n + τ n N d 6 ) Dp τ p N d φ i V a ) N a N d + N a

7 J J t + J n qv R + v d )N c [ v R 2πm v d µ n E max J J t + J n A T 2 + qv d N c ) A 4πqm k 2 h 3 Si Material Parameters Band gap energy at 300 K: E g.24 ev Relative permittivity: ɛ s.7 Intrinsic carrier concentration at 300 K: n i 0 0 cm 3 Effective density of states at 300 K: N c cm 3, N v cm 3 Effective mass: m e.08m 0, m h 0.8m 0 Mobility: µ n 400 cm 2 /V s, µ p 450 cm 2 /V s Diffusion coefficients: D n 32.cm 2 /s, D p cm 2 /s qφ b [ qφ b ) ) )] qva )] qva GaAs Material Parameters Band gap energy at 300 K: E g.424 ev Relative permittivity: ɛ s 2.9 Intrinsic carrier concentration at 300 K: n i cm 3 Effective density of states at 300 K: N c cm 3, N v cm 3 Effective mass: m e 0.067m 0, m h 0.45m 0 Mobility: µ n 8400 cm 2 /V s, µ p 400 cm 2 /V s Diffusion coefficients: D n 20cm 2 /s, D p 0. cm 2 /s 7

8 Physical Constants Permittivity of vaccum: ɛ F/cm Planck s constant: h J s Speed of light: c cm/s Electronic charge: q C Electron rest mass: m kg Boltzmann constant: k J/ K Thermal energy at 300 k: ev Energy conversion: ev J 8