Test: Diode Resistance & Diode Capacitances - Electrical Engineering (EE) MCQ

 According to Ohms law the electric current in the circuit is directly proportion to voltage and inversely proportional to resistance so, R=V/I.

In the diode’s volt ampere characteristics, the line joining the operating point and the origin, at any point of the line is equal to the conductance so, it is reciprocal of the resistance.

 We know that R= (n*V_T) /I, by substituting the value of n, V_T, I we get R= 260 ohms, (1*26)/100*10^-3 = 260 ohms.

We know that, in volt ampere characteristics the resistance is equal to the reciprocal of the line joining the origin and operating point, R = dV/dI, by substituting the value of dV and dI we get R= 10ohms.

In piecewise linear characteristics the forward resistance will be equal to reciprocal of the slope so, R_F = 1/slope, R_F = 1/0.5 which is equal to 2 ohms.

The diode made up of semiconductor has a certain threshold voltage only after which it behave as closed circuit in the sense it performs some operation if the threshold voltage is greater than the voltage in circuit.

 We know that the thermal voltage of diode is approximately equal to room temperature which is 300K then for all practical purpose the thermal voltage of diode is taken as 25mV so it will be 25mV at 300K.

The thermal voltage of the diode is given by, V_T = KT/q, by substituting the values of T, K which is Boltzmann constant and q which is the charge of the electron we get V_T = (300*1.38*10^-23)/ (1.602*10^-19), V_T= 25.8mV.

 When the current in the circuit is zero there will be no flow of charges to resist hence the diode resistance will be zero.

 The diode acts as an ideal diode when it is a perfect conductor and has zero voltage across it during forward bias, a perfect insulator and zero current through it during reverse bias.

We know, C_T=Aε/W and
W ∝ (1/N_A+1/N_D) ^1/2. So, C_T ∝ (1/N_A+1/N_D)^-1/2
So when N_A and N_D increases, depletion capacitance C_T increases.

In reverse bias condition, depletion region increases and acts as an insulator or dielectric medium. So, the transition capacitance increases. In forward bias condition, due to stored charge of minority carriers, diffusion capacitance increases.

C_jo is the capacitance at zero bias, that is V_R=0V, C_jo=C_j for V_R=0V. We know, C_j = C_jo/(1+(V_R/V_bi))m , m=1/2 for abrupt. So, putting C_j=0.1nF/cm² where, V_R=99V and V_bi=1V we get, C_jo= 0.1(1+99)^1/2 = 0.1nF/cm².

We know, C_j1/ C_j2=[(V₀+V_R2)/(V₀+V_R2)]^1/2
So, C_j2=C_j1/ {(0.75+7.25)/(0.75+1.25)}^1/2 we get C_j2=C_j1 /2 =5/2=2.5Pf.

We know, C_j1/ C_j2=[(V₀+V_R2)/(V₀+V_R1)]^1/2
C_j2=C_j1(1/4)^1/2=1/2 .
We get C_j2=1/2=0.5pF.

We know, C_T =Aε₀ε_r /W
C_T/A= (8.85×10^-12)(11.7)/10
=10
By putting the values we get 10µF/m².

C_T = K/(V₀+V_B)^1/2
As it’s having uniform doping on both sides, the voltage V₀ will be zero. So, C_T=K/(VB)^1/2. The variation of transition capacitance with built in capacitance is (V_B )^-1/2.

For an abrupt PN junction diode, C_T = K/(V₀+V_B)ⁿ. Here, n=1/2 for abrupt PN junction diode and 1/3 for linear PN junction diode. When the doping concentration of a diode varies within a small scale of area, then the diode is called as an abrupt diode.

C_D =τ I /n₀ V_T
Where, I is the current and VT is temperature factor. The diffusion capacitance is directly proportional to current and indirectly proportional to the temperature.

Transition capacitance occurs in reverse bias. We obtain a depletion layer in that case. Hence it’s also called as depletion capacitance. The diffusion capacitance occurs in forward bias.