The equivalent circuit of a Tunnel diode is shown in the following Figure. The resistive cut off frequency is given by:

Electronics Problem: Tunnel Diode

3) The equivalent circuit of a Tunnel diode is shown in the following Figure. The resistive cut off frequency is given by:

(1)   f_r = [1 / 2π] √[ (R / R_s) - 1 ]

(2)   f_r = [1 / 2πRC] √[ (R / R_s) - 1 ]

(3)   f_r = [1 / 2πRC] √[ R_s / R ]

(4)   f_r = [1 / 2πRC] √[ (R_s / R) - 1 ]

Step-by-Step Solution:

1. The total input impedance Z_in is given by the sum of the series components and the parallel combination of the diode's junction capacitance and negative resistance:

Z_in = R_s + jωL_s + [ (-R) || (1 / jωC) ]

2. Solve the parallel impedance part:

Z_parallel = -R / (1 - jωRC) = [ -R(1 + jωRC) ] / [ 1 + (ωRC)² ]

3. Separate the real part of the total impedance:

Re(Z_in) = R_s - R / [ 1 + (ωRC)² ]

4. The resistive cutoff frequency (f_r) is the frequency at which the real part of the input impedance becomes zero (the point where the device stops exhibiting negative resistance behavior):

R_s - R / [ 1 + (2πf_rRC)² ] = 0

5. Solve for f_r:

1 + (2πf_rRC)² = R / R_s

(2πf_rRC)² = (R / R_s) - 1

2πf_rRC = √[ (R / R_s) - 1 ]

f_r = [ 1 / 2πRC ] √[ (R / R_s) - 1 ]

Conclusion: Option (2) is the correct expression.

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