The doping concentrations on the p-side and n-side of a silicon d
The doping concentrations on the p-side and n-side of a silicon diode are 1 × 1016 cm-3 and 1 × 1017 cm-3, respectively. The forward bias of 0.3 V is applied to the diode. At T = 300 K, the intrinsic carrier concentration of silicon ni = 1.5 × 1010 and \(\frac{{kT}}{q} = 26\;mV\). The electron concentration at the edge of the depletion region p-side is
A. 2.3 × 10<sup>9</sup> cm<sup>-3</sup>
B. 1 × 10<sup>16</sup> cm<sup>-3</sup>
C. 1 × 10<sup>17</sup> cm<sup>-3</sup>
D. 2.25 × 10<sup>6</sup> cm<sup>-3</sup>
Please scroll down to see the correct answer and solution guide.
Right Answer is: A
SOLUTION
Concept:
The electron concentration at the edge of the depletion region on the p-side is given by:
\({n_p} = {n_{{p_0}}} \cdot {e^{\left[ {{V_f}/{V_T}} \right]}}\)
np0 = minority carrier electron concentration in the p-type semiconductor under thermal equilibrium and is given by mass action law, i.e.
\({n_{{p_0}}} = \frac{{n_i^2}}{{{N_A}}}\)
ni = Intrinsic carrier concentration
NA = Acceptor concentration at p-side
Vf = Forward bias voltage
VT = Thermal Voltage
\({V_T} = \frac{{kT}}{q}\)
Calculation:
Given:
\({n_i} = 1.5 \times {10^{10}}c{m^{ - 3}}\)
\({N_A} = 1 \times {10^{16}}\;c{m^{ - 3}}\)
\(T = 300\;K,\;\;{V_f} = 0.3\;V,\;\frac{{kT}}{q} = 26\;mV\)
\({n_{{p_0}}} = \frac{{\left( {1.5 \times {{10}^{10}}} \right)}}{{{{10}^{16}}}}\)
\({n_{{p_0}}} = 2.25 \times {10^4}\;c{m^{ - 3}}\)
Now, we can write:
\({n_p} = {n_{{p_0}\;}}{e^{\left[ {q{V_f}/kT} \right]}}\)
\({n_p} = 2.25 \times {10^4}\;{e^{\left[ {0.3/.026} \right]}}\)
\({n_p} = 2.25 \times {10^4} \times 1.025 \times {10^5}\)
\({n_p} = 2.306 \times {10^9}\;c{m^{ - 3}}\)