chapter06sm

chapter06sm - Solutions Chapter 6 6.1 Consider a...

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Unformatted text preview: 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 q m μ λ μ- = - = = - = - E b) Find the electric field in the bulk on the n side. Similarly on the n side, Anderson & Anderson 1 5/12/09 Solutions Chapter 6 ( 29 16 17 0.6 0.3 0.27 6 10 ( ) ln ln 1.8 10 ( ) n D D m x x m N x N x μ λ μ-- = = = - ״ ״ 1 0.026 0.096 / 960 / 0.27 n kT V V m V cm q m μ λ μ- = - = = + =- 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 ) e- x- x ( 29 λ p = 10 18 e- ( x- 0.2) 0.036 N D ( x ) = N D ( x ) e- x- x ( 29 λ 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 5/12/09 Solutions Chapter 6 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), ( 29 ( 29 ( 29 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 F cm V V w cm m qa C cm ε 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....
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chapter06sm - Solutions Chapter 6 6.1 Consider a...

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