chp 9 hw solutions mse 120

8/12/2019 Chp 9 HW Solutions MSE 120

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Solutions

9.1 From [eqn 8.17],

where E Fn , E Fp are the Fermi levels in the n- and p-type materials

respectively; N c is a constant given by [eqn 8.18].

Divide the two equations to get

However, the 'built-in' voltage = difference in Fermi level/e

Hence,

Similarly, for holes, from [eqn 8.20]

9.2 At thermal equilibrium, J e = 0

(1)

Integrate across the transition region.

R.S. of eqn (1) gives where N en , N ep are the electron

densities beyond the transition region in the n-type and p-type

materials respectively.

Note that dx across the transition region is the built-in

voltage.

Thus

Similarly, by taking J h = 0, we obtain


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By comparing the results with that in Ex 9.1, we get

e h

e h

D D kT=

e which is the Einstein relationship. (Q.E.D.)

Eliminate from the expressions of J e and J h , and note that J e and J h

are separately zero,

But D e / e = D h/ h

9.3 The density gradient of impurities implies a spatially varying

electron density and a resulting diffusion current. Therefore =

Ne e e . At thermal equilibrium, conduction current = diffusion current,

Integrate across the semiconductor,

the built-in voltage

where N e1 , N e2 are the

electron densities at the low and high impurity ends respectively.

Rearrange to get

or

9.4 In the semiconductor (see Fig S9.1)


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Integrate once,

where is the electric field at

the insulator-semiconductor boundary.

Suppose the width of the depletion region is x n. Then at x = x n , is

zero because there is no charge imbalance to the right of x n .

Fig. S9.1 The metal-insulator-semiconductor junction.

Hence

The potential difference across the semiconductor is

The electric field in the insulator is

the potential drop across it

So the total voltage drop

giving

9.5 (i) When d < d 0 , the space charge density is given by

the electric field

Thus the 'built-in' voltage ( by symmetry,

the voltage drop from d/2 to 0 is the same as that from 0 to d/2)

Hence

(ii) When d > d 0 ,


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Note that the integration constants are determined by

(i) the electric field at x = d/2 is zero

(ii) matching the electric field at x = d 0/2

or

9.6 Since (1)

and from the expression given in Ex 9.1,

Substitute the values into the expression,

9.7 The density of holes will increase from the equilibrium density

by a factor of exp eV/kT when a forward bias voltage V is applied,

i.e.


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Similarly,

9.8 (i)

Now +D DN = N

or

(ii)

But

or

(iii) the 'built-in' voltage = difference in the Fermi levels

giving

9.9 From [eqns 9.5 and 9.6],


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or and

From [eqn 9.8],

(1)

9.10 The continuity equation is

where J is the current density vector.

Now the diffusion current dominates, therefore

Assume that

yielding

where A, B are constants.

When therefore B = 0

At x = 0 (chosen to be in the n-type material, at the end of the

transition region), N h = injected hole density = N hn exp(eU 1/kT).

Hence

9.11 From the result of Ex 9.10, the injected hole density is reduced

by a factor of e when

9.12 Neglect the conduction current,

chp 9 hw solutions mse 120

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