In an open-circuited step-graded junction, the left-half of the bar is p-type with a constant concentration NA, whereas the right-half is n-type- with a uniform density ND. In this type of doping, the density changes abruptly from p-type to n-type. What is the contact difference of potential V0?

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UPSC ESE (Prelims) Electronics and Telecommunication Engineering 19 Feb 2023 Official Paper
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  1. 1.6021 × 10-19 J
  2. \(\rm V_{n_0}-V_{i_0}=V_{n_i} \)
  3. V21 = V0 ln \(\rm (p_{p_0} / p_{n_0})\)
  4. V0 = VT ln \(\rm (N_A N_D / n_i^2)\)

Answer (Detailed Solution Below)

Option 4 : V0 = VT ln \(\rm (N_A N_D / n_i^2)\)
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Detailed Solution

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Contact potential difference:

The contact potential difference, Vo, in an open-circuited step-graded junction is given by:

V0 = \(\frac {k.T}{q}.ln(\frac{N_A.N_D}{n^2_i})\)

where, 

k is the Boltzmann constant.

T is the temperature.

q is the elementary charge.

NA is the acceptor concentration.

ND is the donor concentration.

ni is the intrinsic carrier concentration.

\(V_0 = V_T.ln(\frac{N_A.N_D}{n^2_i})\)

Where,\(V_T = \frac {k.T}{q}\) is the thermal voltage.

Here, option 4 is correct.

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