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The built-in potential of a P N junction diode is 0.7 V at room temperature. What will be the approximate value of built-in potential if the doping concentrations on both sides are doubled?
Assume In (2) = 0.7
Assume In (2) = 0.7
- a)0.9846 V
- b)0.73604 V
- c)0.7182V
- d)0.49 V
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
The built-in potential of a P N junction diode is 0.7 V at room temper...
The built-in potential of a P-N junction diode is determined by the difference in the work functions of the P and N regions. It is given by the formula:
\(V_{bi} = \frac{kT}{q} \ln \left( \frac{N_AN_D}{n_i^2} \right)\)
where:
- \(V_{bi}\) is the built-in potential
- \(k\) is the Boltzmann constant
- \(T\) is the temperature in Kelvin
- \(q\) is the elementary charge
- \(N_A\) and \(N_D\) are the doping concentrations of the P and N regions, respectively
- \(n_i\) is the intrinsic carrier concentration
In this question, we are given that the built-in potential is 0.7 V at room temperature. Let's assume the room temperature is 300 K.
To find the approximate value of the built-in potential if the doping concentrations on both sides are doubled, we need to calculate the new value of \(V_{bi}\) using the formula above.
Doubling the doping concentrations means that \(N_A\) and \(N_D\) are both multiplied by 2.
Let's plug in the values into the formula:
\(V_{bi2} = \frac{kT}{q} \ln \left( \frac{2N_AN_D}{n_i^2} \right)\)
Since we are only interested in the approximate value, we can use the approximation \(\ln(2x) \approx \ln(x) + \ln(2)\).
\(V_{bi2} = \frac{kT}{q} \left( \ln \left( \frac{N_AN_D}{n_i^2} \right) + \ln(2) \right)\)
The term \(\ln \left( \frac{N_AN_D}{n_i^2} \right)\) is the same as the term in the original formula, which gives the built-in potential at room temperature. Let's call this term \(V_{bi1}\):
\(V_{bi1} = \frac{kT}{q} \ln \left( \frac{N_AN_D}{n_i^2} \right)\)
Therefore, we can rewrite the equation as:
\(V_{bi2} = V_{bi1} + \frac{kT}{q} \ln(2)\)
Substituting the values, we get:
\(V_{bi2} = 0.7 + \frac{(1.38 \times 10^{-23} \times 300)}{(1.6 \times 10^{-19})} \times \ln(2)\)
\(V_{bi2} \approx 0.73604\) V
Therefore, the approximate value of the built-in potential if the doping concentrations on both sides are doubled is 0.73604 V.
\(V_{bi} = \frac{kT}{q} \ln \left( \frac{N_AN_D}{n_i^2} \right)\)
where:
- \(V_{bi}\) is the built-in potential
- \(k\) is the Boltzmann constant
- \(T\) is the temperature in Kelvin
- \(q\) is the elementary charge
- \(N_A\) and \(N_D\) are the doping concentrations of the P and N regions, respectively
- \(n_i\) is the intrinsic carrier concentration
In this question, we are given that the built-in potential is 0.7 V at room temperature. Let's assume the room temperature is 300 K.
To find the approximate value of the built-in potential if the doping concentrations on both sides are doubled, we need to calculate the new value of \(V_{bi}\) using the formula above.
Doubling the doping concentrations means that \(N_A\) and \(N_D\) are both multiplied by 2.
Let's plug in the values into the formula:
\(V_{bi2} = \frac{kT}{q} \ln \left( \frac{2N_AN_D}{n_i^2} \right)\)
Since we are only interested in the approximate value, we can use the approximation \(\ln(2x) \approx \ln(x) + \ln(2)\).
\(V_{bi2} = \frac{kT}{q} \left( \ln \left( \frac{N_AN_D}{n_i^2} \right) + \ln(2) \right)\)
The term \(\ln \left( \frac{N_AN_D}{n_i^2} \right)\) is the same as the term in the original formula, which gives the built-in potential at room temperature. Let's call this term \(V_{bi1}\):
\(V_{bi1} = \frac{kT}{q} \ln \left( \frac{N_AN_D}{n_i^2} \right)\)
Therefore, we can rewrite the equation as:
\(V_{bi2} = V_{bi1} + \frac{kT}{q} \ln(2)\)
Substituting the values, we get:
\(V_{bi2} = 0.7 + \frac{(1.38 \times 10^{-23} \times 300)}{(1.6 \times 10^{-19})} \times \ln(2)\)
\(V_{bi2} \approx 0.73604\) V
Therefore, the approximate value of the built-in potential if the doping concentrations on both sides are doubled is 0.73604 V.
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Community Answer
The built-in potential of a P N junction diode is 0.7 V at room temper...
Concept:
- As the N-type material has lost electrons and the P-type has lost holes, the N-type material has become positive with respect to the P-type.
- Then the presence of impurity ions on both sides of the junction causes an electric field to be established across this region with the N-side at a positive voltage relative to the P-side.
- The problem now is that a free charge requires some extra energy to overcome the barrier that now exists for it to be able to cross the depletion region junction.
- This electric field created by the diffusion process has created a “built-in potential difference” across the junction with an open-circuit (zero bias) potential
Calculation:
Given built-in voltage of the diode is 0.7 V
Now doping concentrations on both sides are doubled. i.e, NA’ = 2NA and ND’ = 2ND
Newly built-in potential is:
Given built-in voltage of the diode is 0.7 V
Now doping concentrations on both sides are doubled. i.e, NA’ = 2NA and ND’ = 2ND
Newly built-in potential is:
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The built-in potential of a P N junction diode is 0.7 V at room temperature. What will be the approximate value of built-in potential if the doping concentrations on both sides are doubled?Assume In (2) = 0.7a)0.9846 Vb)0.73604Vc)0.7182Vd)0.49VCorrect answer is option 'B'. Can you explain this answer?
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