hw_4_sol

# hw_4_sol - HOMEWORK 4 SOLUTIONS CHAPTER 4 4.7 Draw the...

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HOMEWORK 4 - SOLUTIONS CHAPTER 4 4 C (0.8 eV/&m) x for 0< x <0.5 &m. a) Find the effective electric field .7. Draw the energy band diagram for GaAs that is doped such that E - E f =0.2+ for electrons The electric field is 19 * 19 4 1 1 0.4 (1.6 10 / ) 8/ 1.6 10 0.5 10 C e dE eV J eV kV cm qd x C cm −− × == = ×× E b) Indicate the directions of J n(diff) , J n(drfit) , J p(diff) , and J p(drift) . (below) c) Indicate the direction of the electric field. CHAPTER 5

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5.5. A step pn junction diode is made in silicon with the n-side having ' D N =2&10 16 -3 cm and on the p-side the net doping is ' A N 15 -3 a) Draw, to scale, the energy band diagram of the junction at equilibrium. We begin by finding the locations of the Fermi levels on each side of the junction. On the n side, =5&10 cm . E C E f =− kT ln n 0 N C kT ln N D ' N C 0.026 eV ln 2 × 10 16 2.89 × 10 19 = 0.19 eV This assumes complete ionization and nondegeneracy. On the p side, we have '1 5 0 19 ln ln 0.026 ln 0.25 3.1 10 A fV 51 0 VV pN E Ek T k T e V e V NN −= = = = ⎜⎟ × ⎝⎠ ⎛⎞ × From these we can construct the entire energy band diagram. The built-in voltage is then 1.12 0.19 0.25 eV −− =0.68eV 0.19eV 1.12eV 0.70eV 0.25eV 0.31eV 0.37eV
b) Find the built-in voltage, and compare to the value measured off your drawing in part (a). () '' 1 6 3 1 5 3 2 2 10 3 (2 10 ) (5 10 ln 0.026 ln 0.71 1.08 10 DA bi i NN kT cm cm VV qn cm −− ⎛⎞ ×⋅ × ⎜⎟ == = × ⎝⎠ V which agrees reasonably well with the result above. The difference results from round-off errors. c) Find the junction width. We use 14 16 3 ' ' 19 16 3 15 3 6 2( ) 2(11.8)(8.85 10 / )(0.71 )(2.5 10 ) (1.6 10 )(2 10 )(5 10 ) 48 10 0.48 bi A D AD VN N Fcm V cm w qN N C cm cm cm m ε μ + ×× × = d) Find the width of the n-side of the depletion region and the p-side of the depletion region, and the voltage dropped across each side of the transition region. 14 16 3 ' 19 16 3 ' 15 3 ' 6 2 2(11.8)(8.85 10 / )(0.71 ) 21 0 1.6 10 (2 10 ) 1 1 51 0 9.6 10 0.096 j n D D A V V w cm N Cc m qN cm N cm m × × + + × = 14 15 3 ' 19 15 3 ' 16 3 ' 6 2 2(11.8)(8.85 10 / )(0.71 ) 0 1.6 10 (5 10 ) 1 1 0 38 10 0.38 j p A A C V V w cm N m qN cm N cm m × × + + × = For the voltages, we have

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'1 9 1 6 3 26 14 1.6 10 2 10 () ( 9 .
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## This note was uploaded on 01/11/2012 for the course ECE 432 taught by Professor Lu during the Fall '08 term at Ohio State.

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hw_4_sol - HOMEWORK 4 SOLUTIONS CHAPTER 4 4.7 Draw the...

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