chapter04sm

# chapter04sm - Chapter 4 Solutions 4.1. Consider a p-type Si...

This preview shows pages 1–4. Sign up to view the full content.

Chapter 4 Solutions 4.1. Consider a p-type Si sample of length 0.1 µ m in which the net doping varies exponentially from 5 × 10 17 to 5 × 10 15 . Find the electric field. Following the example of Section 4.2.2, x x 0 λ = ln N A ' x 0 ( ) N A ' x () and ( ) 4 0 6 ' 0 ' 0.1 10 2.2 10 ln 100 ln A A cm xx cm Nx × == = × ⎛⎞ ⎜⎟ ⎝⎠ From Equation (4.22), 6 11 0.026 12 / 2.2 10 kT Vk qc m =− × V c m E The field points in the negative x-direction. 4.2. Consider the graded-composition Si:Ge alloy discussed in Section 4.3. The hole concentration is assumed constant ( is essentially constant and all dopants are ionized). There is no electric field or diffusion for holes at equilibrium but there is a field for electrons in the conduction band. Since at equilibrium there is no net current, the drift electron current must be offset by an opposing diffusion current. Identify and explain the source of the varying electron concentration that produces the diffusion current. N A ' Since the bandgap is changing, n i changes. Thus even though p 0 is constant, n 0 = n i 2 p 0 is not. At the end where the band gap is smaller, n i is Solutions Chapter 4 1 1/19/05 Anderson & Anderson

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
larger, and thus n 0 is larger. This produces an electron diffusion flux to the left that offsets the drift flux gong to the right. 4.3. Find the time required for an electron to traverse the p-region of Example 4.2 due to drift alone. The electric field is 20 kV/cm. at this field the electrons travel near their saturation velocity. From Figure 3.9, 6 sat 8.5x10 / . vc m s =≅ v Then 4 6 0.14 10 1.6 8.5x10 / xc m tp m s ∆× == = s . 4.4. A graded alloy is manufactured in the AlGaAs system. At x =0, the material is pure GaAs ( χ =4.07, E g =1.43), and over a distance of 2 µ m the composition changes to Ga 0.6 Al 0.4 As ( χ =3.4 eV and E g =1.92 eV). The material is intrinsic. Find the effective electric fields for electrons and holes, and find the true electric field. We first draw the energy band diagram by drawing the Fermi level as a constant, and then constructing the energy band diagram of the two materials around it assuming the intrinsic level to be at midgap. At the GaAs end, E C 1.43/2=0.715 eV above E f , and E vac is 4.07 eV above E C . At the AlGaAs end, the band edges are each 1.92/2=0.96 eV from the Fermi level and E vac is 3.4 eV above E c . Solutions Chapter 4 2 1/19/05 Anderson & Anderson
2.0 1.5 1.0 0.5 0.0 E C V f vac E E E distance (um) 0.715 eV 4.07 eV 3.4 eV 0.96 eV The effective electric field for electrons is, from Equation (4.27), ( ) ( ) 19 19 4 0.96 0.715 1.6 10 / 11 1.6 10 2 10 1.2 / C eV J eV dE qd x C cm kV cm −− −⋅ × == ×× = * e E For holes, we have ( ) ( ) 19 19 4 0.96 0.715 1.6 10 / 1.6 10 2 10 1.2 / V eV J eV dE x C kV cm ⋅× =− * p

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 06/30/2010 for the course EECE 332 taught by Professor Constable during the Spring '08 term at Binghamton.

### Page1 / 12

chapter04sm - Chapter 4 Solutions 4.1. Consider a p-type Si...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online