Question

In figure, a battery of emf 2V is used. The length of the block is 0.1m and the area is 1 x 10^–4m². If the block is of intrinsic silicon at 300 k, find the electron and hole currents.

What will be the magnitude of the total current? What will be the magnitude of total current if germanium is used instead of silicon?
Given-For silicon:
Mobility of electrons = 0.135 SI units
Mobility of holes = 0.048 SI units
Number density of electrons holes
= 1.5 x 10¹⁶ per unit volume
For germanium:
Mobility of electrons = 0.39 SI units
Mobility of holes = 0.19 SI units
Number density of electrons holes = 2.4 x 10¹⁹ per unit volume

Solution

We know,
Current i due to charge carriers of density 'n' and having charge 'e' through a block of cross-section area A is given by,

i = n A e v

...(i)
where, v is the average drift speed of the charge carriers.

The mobility

μ

of charge carriers is velocity attained per unit external field E.

i.e,

μ = \frac{v}{E}

\Rightarrow

v = μE

....(ii)

The electric field E due to the applied potential V,

E = \frac{V}{l}

....(iii)

From equations (i) and (ii) and (iii), we have

i = n A eμ (\frac{V}{l})

Electron current for silicon,
E.m.f of the battery = 2 V
Length of the bloack = 0.1 m
Area of the block, A = 1 x 10^–4m²Temperature, T = 300 K
Mobility, μ₁ = 0.135
Number density of elcetrons/holes, n = 1.5 x 10¹⁶ per unit volume

For electron current,

i_{1} = {An}_{1} {eμ}_{1} (\frac{V}{l})

= \frac{10^{- 4} \times 1.5 \times 10^{16} \times 1.6 \times 10^{- 19} \times 0.135 \times 2}{0.1} = 6.45 \times 10^{- 7} A = 0.645 μA

Similarly, for holes

μ_{2}

= 0.048 SI units.

Therefore,
Hole current,

i_{2} = A n_{2} {eμ}_{2} (\frac{V}{I})

\Rightarrow

i_{2} = \frac{10^{- 4} \times 1.5 \times 10^{16} \times 1.6 \times 10^{- 19} \times 0.048 \times 2}{0.1} = 2.304 \times 10^{- 7} = 0.2304 μA

Thus, total current is,

i = i_{1} + i_{2}

= 0.645 + 0.2304 = 0.8754 μA

If germanium is used instead of silicon then,

Number density,

n = 2.4 \times 10^{19} per unit volume

μ_{1}' = mobility of electrons = 0.39 SI units

∴

Electronic current,

i_{1}'

= \frac{2.4 \times 10^{19} \times 10^{- 4} \times 1.6 \times 10^{- 19} \times 0.39 \times 2}{0.1} = 2.99 \times 10^{- 3} A = 2.99 mA

The mobility μ₂’ of holes in germanium = 0.19 SI units

Hole current, i'_{2} = 1.49 \times 10^{- 3} A = 1.49 mA

∴

The total current i' is

i' = i'_{1} + i'_{2} = 2.99 + 1.49 = 4.48 mA .

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