chapter06sm - Solutions Chapter 6 6.1. Consider a...

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Solutions Chapter 6 6.1. Consider a base-collector junction of a silicon BJT (bipolar junction transistor) like that if Figure 6.1. Assuming a linearly graded junction with a =1.2x10 18 cm -3 /um, find V bi This can be done iteratively from Equation (6.11) V bi = 2 kT q ln a 2 n i 12 ε V bi qa 1 3 We choose a starting guess of 1V, and find: Vbi cube root term logarithm new Vbi 1 0.000402474 10.0150146 5.21E-01 5.21E-01 0.00032381 9.79753926 5.09E-01 5.09E-01 0.000321449 9.7902212 5.09E-01 The built-in voltage is 0.51 V. 6.2. A silicon pn homojunction has a doping profile as indicated in Figure P6.1. a) Find the value of the electric field in the bulk on the p side. From Equation (4.16), From Equation (4.22), 1 0.026 0.72 / 7.2 / 0.036 p kT V V m kV cm qm µ λµ =− = E b) Find the electric field in the bulk on the n side. Similarly on the n side, Anderson & Anderson 1 1/19/05 Solutions Chapter 6
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() 0 16 0 17 0.6 0.3 0.27 61 0 ln ln 1.8 10 n D D m xx m Nx µ λ == = ⎛⎞ ⎡ ⎤ × ⎜⎟ ⎢⎥ × ⎣⎦ ⎝⎠ 1 0.026 0.096 / 960 / 0.27 n Vm V c m qm λµ =− = =+ = kT V E c) Plot N A - N D as a function of position [ Hint : see Equation (4.16)], and find the slope a . We first find the grading constant a. One way to obtain it is express the quantity N A - N D as a function of x , which we do using Equation (4.16): N A ( x ) = N A ( x 0 ) e x x 0 p = 10 18 e ( x 0.2 ) 0.036 N D ( x ) = N D ( x 0 ) e x x 0 p = 6 × 10 16 e ( x .3) 0.27 and the difference plotted below: The slope is approximately –10 18 cm -3 / µ m, or a= 10 18 cm -3 / µ m (10 22 cm -4 ). d) Find the built-in voltage. This can be done iteratively from Equation (6.11) Anderson & Anderson 2 1/19/05 Solutions Chapter 6
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V bi = 2 kT q ln a 2 n i 12 ε V bi qa 1 3 Picking a starting voltage of 1V, this converges rapidly to 0.831V: Vbi cube root term logarithm new Vbi 1 1.98517E-05 16.0336938 8.34E-01 8.34E-01 1.86843E-05 15.9730874 8.31E-01 8.31E-01 1.86608E-05 15.9718251 8.31E-01 e) Find the junction width at equilibrium From Equation (6.7), () ( ) 1 1 14 3 3 5 19 22 4 12 11.8 8.85 10 / 0.83 12 1.9 10 0.19 1.6 10 10 bi Fcm V V wc qa Cc m m m µ −− ⎡⎤ ⋅× ⎢⎥ == = × = × ⎣⎦ f) For the junction width you found in (e), comment on the validity of the linear approximation used over this distance. From the figure above in part (c) we find that over a distance of 0.2 µ m on either side of the junction, the approximation is poor. 6.3. In section 6.2.2, it was claimed that hyper-abrupt junctions exhibit a large fractional change in junction capacitance with applied voltage. Explain physically why we should expect this to be the case. In a hyper-abrupt junction, the doping concentration decreases as one gets further from the junction. Let us consider the reverse bias regime. As the bias voltage gets larger, the junction width must get large to uncover the appropriate number of ionized dopants. Since the dopants get more scarce away from the junction, as the bias increase the junction width increases more and more rapidly. This changes the distance between the capacitive plates more and more rapidly, thus changing the capacitance more rapidly.
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This note was uploaded on 06/30/2010 for the course EECE 332 taught by Professor Constable during the Spring '08 term at Binghamton University.

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chapter06sm - Solutions Chapter 6 6.1. Consider a...

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