ECE331_Wi06_hw6_sol

ECE331_Wi06_hw6_sol - m is 2 7 15 19 / 1600 10 10 10 6 . 1...

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1/4 pages ECE331 Homework #6 Solution 1. Streetman 3.12
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2/4 pages 2. Streetman, 3.13 (a)
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3/4 pages 3. Streetman, 3.14 (a) (b)
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4/4 pages 4. Streetman 3.15 (i) For an applied voltage of 2.5 V across its length, 5 µ m cm V cm V / 5000 10 5 / 5 . 2 4 = × = ε , which is below 10 4 V/cm, and is not in the velocity saturation region. 1 1 15 19 0 ) ( 24 . 0 1500 10 10 6 . 1 = = × × × = = ρ µ σ cm n q n cm = 167 . 4 A L R / = L VA R V I / / = = The current density for an applied voltage of 2.5 V across 5 µ m is 2 4 / 1200 ) 10 5 167 . 4 /( 5 . 2 / / cm A cm cm V L V A I J = × × = = = (ii) How about a voltage of 2500V? For an applied voltage of 2500 V across its length, 5 µ m cm V cm V / 10 5 10 5 / 2500 6 4 × = × = , which is in the velocity saturation region Thus, v s =10 7 cm/s. The current density for an applied voltage of 2500 V across 5
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Unformatted text preview: m is 2 7 15 19 / 1600 10 10 10 6 . 1 cm A v qn J s = = = 5. Streetman, 3.17 where notice we were careful to use mks units under the radical. This is the velocity between collision, which we expect to be much greater than the drift velocity, since the electron collisions many times and bounces in random directions, make a slow net progress. We check our conclusion by looking at Figure 3-24, which shows the drift velocity is only about 2x10 5 cm/s, or roughly 100 times slower. =2.3 10 5 m/s=2.3 10 7 cm/s...
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ECE331_Wi06_hw6_sol - m is 2 7 15 19 / 1600 10 10 10 6 . 1...

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