Excercise 1: emiconductors / n-junctions Carrier concentrations 1. Consider undoed GaAs (N c=4.7x10 17 cm -3, N v=7.0x10 18 cm -3, E g=1.424 ev) and i (N c=2.8x10 19 cm -3, N v=1.04x10 19, cm -3, E g=1.12 ev). At what temerature is the intrinsic carrier density of GaAs equal to that of i at 200K? Ans: The intrinsic carrier concentration is n i n N c N v Eg ex 2kT which gives n i(i, 200K) = 1.33e5 cm -3. olving for T gives T ln E N g 2 k n i c N v and inserting the result for n i(i, 200K) gives T= 273 K. Due the larger band ga of GaAs comared to i there are fewer carriers excited to the conduction band (i.e. smaller art of the tail of the Fermi-Dirac distribution has to be considered) at a secific temerature. Raising the temerature allows more carriers to be excited. Transort 2. Find the resistivity at 300 K for a i samle doed with Phoshorus (P): 1.0x10 14 cm -3 Arsenic (As): 8.5x10 12 cm -3 Boron (B): 1.2x10 13 cm -3 The electron mobility of i is µ n=1500 cm 2 /Vs and the hole mobility µ =500 cm 2 /Vs. Hint: the doants can be either accetors or donors deending on the grou in the eriodic table of elements. Ans: P and As are in grou V i.e. they have one extra valence electron comared to i and therefore act as donors of electrons which gives a total donor concentration of N D=1x10 14 + 8.5x10 12 = 1.085x10 14 cm -3. B is in grou with one fewer electron in the valence band comared to i i.e. it s an accetor giving a density N A=1.2x10 13 cm -3. The donors and accetors comensate (cancel out) each other giving an electron concentration of n=n D-N A=9.65x10 13 cm -3 2. According the mass action law n i = n* where the intrinsic carrier concentration is given by Eg ni n NcNv ex = 6.67e10 9 cm-3 2kT which gives a hole concentration 1
n n 2 i 9 2 ( 6.67 10 ) 5 3 4.6110 cm 13 9.65 10 Both electrons and hole contribute to the conductivity q( µ n µ n ) 1.6022 10 19 (1500 9.65 10 13 5 500 4.6110 ) 0.0232 ( cm) 1 Giving a resistivity of 1 43. 1 cm As can be seen, the hole concentration is so small (and the mobility is lower as well) so it can safely be ignored and the resistivity is set by the electron transort. 3. Calculate the electron and hole drift velocities through a 10-µm thick layer of intrinsic silicon across which a voltage of 1V is alied. Let µ n = 1350 cm 2 /Vs and µ = 480 cm 2 /Vs. Ans: The electric field across the layer is ε=v / L = 1 (V) / 10-5 (m) = 10 5 V/m. The electron velocity is given by v n= ε µ n = 10 5 (V/m) 1350 10-4 (m 2 /Vs) = 13500 m/s The hole velocity is given by v = ε µ = 10 5 (V/m) 480 10-4 (m 2 /Vs) = 4800 m/s Note that the electric field is lower than the critical field for the onset of velocity saturation (see lecture 1). f the voltage would be increased >100V the carrier velocity would saturate and become indeendent of bias. 4. Holes (minority carriers) are steadily injected into a iece of n-doed i (N D=10 16 cm -3 ) at x=0 and extracted at x=w resulting in the hole concentration shown in the image below, where n0 is the hole carrier concentration without injection. Use an intrinsic carrier concentration of n i=1.5*10 10 cm -3 and W=5 µm. The hole mobility is µ = 480 cm 2 /Vs. Calculate the current density that flows in the x- direction. 2
Ans: The electron concentration is determined by the doing i.e. n N D. The equilibrium (without injection) hole (minority carriers) concentration is then given by the mass action law: no= n i2 /N D = (1.5 10 10 ) 2 / 10 16 = 22500 cm -3. The current density (due to diffusion of holes from 0 to W) is determined by the concentration gradient as J qd 1.38 10 d( x) kt q µ dx q 23 d( x) ktµ dx 2 3 J / K300 K 480 cm / Vs (1 1000) 22500 cm 0 510 4 cm n0 1000 0 W n0 8.9310 The situation in the roblem resembles that of the n-side of a forward biased njunction where minority carriers are steadily injected at the deletion region edge (x=0) and extracted at the contact (x=w). ince the minority carrier concentration decreases linearly there is no recombination in the region and the current is constant at each oint. 8 A/ cm 2 3
n-junctions / diodes 5a (roblem 1.1 in book). Calculate the built-in otential, deletion layer deths and maximum field in an abrut (not graded doing) n junction in i with doing densities N A=8*10 15 cm -3 and N D=10 17 cm -3 for three biases: - reverse bias of 5 V - zero bias - forward bias of 0.3 V For Zero bias only the built-in otential of 0.75 V is used in the calculation. 4
5b (roblem 1.2 in book). Calculate the junction caacitance at a bias of 0, -3 and 0.5V. Assume a junction area of 2*10-5 cm 2 5
6. For the circuits shown below using ideal diodes, find the values of the voltages and currents indicated. Ans a) The diode is forward biased i.e. its resistance is negligible and it can be treated as a short between the middle and lower terminals i.e. V = -5 V. The current is =U/R= 5-(-5) [V] / 10 [kohm] = 1 ma b) The diode is reverse biases i.e. its resistance is very high and it can be treated as an oen circuit between the middle and lower terminals i.e. V = +5 V. The current = 0 A. c) Diode is forward biased. V=5 V and = 1 ma. d) Diode is reverse biased V=-5 V and = 0 A. 7. a) At what forward voltage does a diode for which n=2 conduct a current equal to 1000 s ( s= saturation current at reverse bias)? b) Exressed in terms of s, what current flows in the same diode when its forward voltage is 0.7V? Ans: a) Use the ideal diode equation qv / nkt 1) et =1000 and solve for V to get V 1000 ln 2kT 1 q 0.355 V b) The ideal diode equation gives qv / nkt q0.7 / nkt 1) 1) 749 10 3 s 6
8. A diode for which the forward voltage dro is 0.7 V at 1.0 ma and for which n=1 is oerated at 0.5V. What is the value of the current? Ans: Use the ideal diode equation qv / nkt 1) and solve for to get qv 3 10 / nkt 19 23 1.78 10 1.602210 0.7 /1.3810 300 1) 1) 12 ma nserted in the diode equation this gives qv / nkt 1) 1.78 10 15 1.602210 19 0.5/1.3810 23 300 1) 4.40 10 4 ma Note the very large difference in current of four orders of magnitude in the forward direction for only a change of 0.2 V. Equivalent circuits 9. Find the Thevenin and Norton equivalent circuits for the circuit shown below. Take care that you orient the olarity of the voltage source and the direction of the current source correctly relative to the terminals a and b. 7
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