20085ee2_1_20085ee2_1_ps4

20085ee2_1_20085ee2_1_ps4 - kT E E C D D C F e N N N n / )...

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EE 2 Problem Set 4 Solutions I. E e E E m h E V E f E Z p kT E E V v E E V V V V + - - = - = - - - ) 1 1 1 ( ) ( ) 2 ( 4 )) ( 1 ( ) ( / ) ( 2 / 1 2 / 3 3 π kT E E V kT E E v V F V e N e kT m h p / ) ( / ) ( 2 / 3 2 / 3 3 2 ) ( ) 2 ( 4 - - - = × = II. kT E V C i g e N N n 2 / - = From the given information at the set temperature, 3 19 10 51 . 2 - × = cm N N V C Plugging this back in the main equation for different temperatures we get: k T cm n k T cm n i i 150 @ 5 . 3 200 @ 10 99 . 1 3 3 5 = = = × = - - III. Assuming full ionization, it is an n-type material and thus 2 4 ) ( 2 2 0 i D D n n N N n + + = + + . However, since N D is much larger than n i , we can approximate the electron density to be the same as that of the donors, 3 4 15 2 3 15 10 5 10 2 10 2 0 0 - - × = × = × = cm n p cm n i n n IV. (a) 3 15 2 3 5 10 , 10 - - = = = cm n n np cm p i
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(b) eV E E N N kT E C C C D F 263 . 0 ) ln( - = + = + Assuming fully ionization. Notice since N D is less than N C , material is non-degenerate and the logarithm is negative. V. Assume full ionization: (a) Boron is acceptor (p): 3 15 10 - = cm N A Arsenic is donor (n): 3 15 10 4 - × = cm N D > A D N N material is n-type. (b) eV E N N N kT E E C C A D C F 234 . 0 ) ln( - = - + = VI. Assuming donor dominance over thermally generated:
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Unformatted text preview: kT E E C D D C F e N N N n / ) ( 2 2-= = kT E E C D C F e N N / ) ( 2-= Taking log from both sides and simplifying ) 2 ln( ) 2 ln( ) 2 ln( C D C F C D C F C F C D N N kT E E N N kT E E kT E E N N + = =--= VII. n D n p N n + = + from 2 n p = we get: + + + + = = = = = + = D i i D i n n i D n n D n N n n N n n n n n p N n n N n 2 2 ) 2 ( ) 2 ( , 2 2 2 2 2 VIII. ( ) p n p n q + = 1 . Electron is the majority carrier here so we can simplify the above to n qn 1 From question 4, and figure 17 in the reader: cm cm n n = =-91 . 3 10 6 . 1 , 10 3 3 15...
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20085ee2_1_20085ee2_1_ps4 - kT E E C D D C F e N N N n / )...

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