This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Solutions Chapter 6 6.1. Consider a basecollector junction of a silicon BJT (bipolar junction transistor) like that if Figure 6.1. Assuming a linearly graded junction with a =1.2x10 18 cm3 /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 5.21E01 5.21E01 0.00032381 9.79753926 5.09E01 5.09E01 0.000321449 9.7902212 5.09E01 The builtin 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 11/19/07 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 cm3 / μ m, or a= 10 18 cm3 / μ m (10 22 cm4 ). d) Find the builtin voltage. This can be done iteratively from Equation (6.11) Anderson & Anderson 2 11/19/07 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 1.98517E05 16.0336938 8.34E01 8.34E01 1.86843E05 15.9730874 8.31E01 8.31E01 1.86608E05 15.9718251 8.31E01 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. 6.3. In section 6.2.2, it was claimed that hyperabrupt junctions exhibit a large fractional change in junction capacitance with applied voltage. Explain physically why we should expect this to be the case....
View
Full Document
 Anderson & Anderson Solutions, Anderson Solutions Chapter

Click to edit the document details