The equivalent circuit of a tunnel diode shown below has the total input impedance

i) \(\left[ {{R_s} + \frac{{ - R}}{{1 + {{\left( {\omega R{C_j}} \right)}^2}}}} \right] + j\left[ {\omega {L_s} + \frac{{ - \omega {C_j}{R^2}}}{{1 + {{\left( {\omega R{C_j}} \right)}^2}}}} \right]\)

ii) R_s + ωRC_j

iii) \(\left[ {{R_s} + \frac{{ - R}}{{\left( {1 + \omega R{C_j}} \right)}}} \right] + \left[ {\omega {L_s} + \frac{{\omega {C_j}{R^2}}}{{1 - {{\left( {\omega R{C_j}} \right)}^2}}}} \right]\)

iv) \(\omega {L_s} + \frac{{\omega {C_j}{R^2}}}{{1 + {{\left( {\omega R{C_j}} \right)}^2}}}\)

Code:

Concept:

The passive elements and their impedances are shown below table:

Calculation:

The parallel equivalent impedance is:

\(Z = \frac{{R \times \frac{1}{{s{C_j}}}}}{{R + \frac{1}{{s{C_j}}}}}\)

Converting into the ‘ω’ form we get

\(Z = \frac{R}{{Rj\omega {C_j} + 1}}\)

\(Z = \frac{{R\left( {1 - j\omega R{C_j}} \right)}}{{\left( {1 + j\omega R{C_j}} \right)\left( {1 - j\omega R{C_j}} \right)}}\)

\(Z = \frac{{R - j\omega {R^2}{C_j}}}{{1 + {{\left( {\omega R{C_j}} \right)}^2}}}\)

The input impedance will be

\({Z_{in}} = {R_s} + s{L_s} + Z\)

\({Z_{in}} = {R_s} + j\omega {L_s} + Z\)

\({Z_{in}} = {R_s} + j\omega {L_s} + \frac{{R - j\omega {R^2}{C_j}}}{{1 + {{\left( {\omega R{C_j}} \right)}^2}}}\)

Separating the real and imaginary parts

\({Z_{in}} = {R_s} + \frac{R}{{1 + {{\left( {\omega R{C_j}} \right)}^2}}} + j\left[ {\omega {L_s} - \frac{{\omega {R^2}{C_j}}}{{1 + {{\left( {\omega R{C_j}} \right)}^2}}}} \right]\)