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

F3 Vinanti Engineering 16.02.23 D1

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UGC NET Paper 2: Electronic Science 29 Oct 2022 Shift 1
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  1. \(\rm\frac{1}{2 \pi} \sqrt{\frac{R}{Rs}-1}\)
  2. \(\rm\frac{1}{2 \pi RC} \sqrt{\frac{R}{Rs}−1}\)
  3. \(\rm\frac{1}{2 \pi R C} \sqrt{\frac{R s}{R}}\)
  4. \(\rm\frac{1}{2 \pi R C} \sqrt{\frac{Rs}{R}−1}\)

Answer (Detailed Solution Below)

Option 2 : \(\rm\frac{1}{2 \pi RC} \sqrt{\frac{R}{Rs}−1}\)
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concept:

Tunnel diode:

Tunnel diode has very high doped P-N junction. it has negative resistance.

Equivalent circuit diagram of tunnel diode is:

F3 Vinanti Engineering 16.02.23 D1

where R is negative resistance of tunnel diode.

Rs is resistance of bulk material and contacting plates.

C is  junction capacitance 

L is parasitic series inductance

To define practical frequency range of a tunnel diode,resistive and reactive cutoff frequency are considered.

Resistive cut off frequency : It is calculated when real part of input impedance is equated to zero.

fr =   \(\frac{1}{2\pi R\ C_j}\) \(\sqrt{\frac{R}{R_S}\ -\ 1}\)

Reactive cut off frequencyIt is calculated when imaginary part of input impedance is equated to zero.

fx = \(\frac{1}{2\pi}\) \(\sqrt{\frac{1}{LC_j}\ -\frac{1}{{(RC_j)}^2}}\)

so option 2 is correct.

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