The static resistance R of the diode is given by __________
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 volt ampere characteristics of the diode, the slope of the line joining the operating point to the origin at any point is equal to reciprocal of the _________
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.
. At room temperature (VT = 26) what will be the approximate value of r when n=1 and I=100mA?
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.
In the diode volt ampere characteristics what will be the resistance if a slope is drawn between the voltages 50 to 100 and corresponding current 5 to 10?
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 what will be the RF value if the slope is 0.5?
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.
A diode will behave as an open circuit if the voltage in the circuit is less than __________
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.
What will be the approximate value of thermal voltage of diode?
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.
What will be the thermal voltage of the diode if the temperature is 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.
What will be the diode resistance if the current in the circuit is zero?
When the current in the circuit is zero there will be no flow of charges to resist hence the diode resistance will be zero.
Which of these following is not a characteristic of an ideal diode?
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.
Compared to a PN junction with N_{A}=10^{14}/CM^{3}, which one of the following is true for N_{A}=N_{D}= 10^{20}/CM^{3}?
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.
If CT is the transition capacitance, which of the following are true?
1) in forward bias, CT dominates
2) in reverse bias, CT dominates
3) in forward bias, diffusion capacitance dominates
4) in reverse bias, diffusion capacitance dominates
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.
For an abrupt PN junction diode, small signal capacitance is 1nF/cm^{2} at zero bias condition.If the built in voltage, V_{bi} is 1V, the capacitance at reverse bias of 99V is?
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/cm^{2} where, V_{R}=99V and V_{bi}=1V we get, C_{jo}= 0.1(1+99)^{1/2} = 0.1nF/cm^{2}.
The built in capacitance V_{0} for a step graded PN junction is 0.75V. Junction capacitance C_{j}at reverse bias when V_{R}=1.25V is 5pF. The value of C_{j} when V_{R}=7.25V is?
We know, C_{j1}/ C_{j2}=[(V_{0}+V_{R2})/(V_{0}+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.
Consider an abrupt PN junction. Let V0 be the built in potential of this junction and VR be the reverse bias voltage applied. If the junction capacitance C_{j} is 1pF for V_{0}+V_{R} =1V, then for V_{0}+V_{R} =4V what will be the value of C_{j}?
We know, C_{j1}/ C_{j2}=[(V_{0}+V_{R2})/(V_{0}+V_{R1})]^{1/2}
C_{j2}=C_{j1}(1/4)^{1/2}=1/2 .
We get C_{j2}=1/2=0.5pF.
A silicon PN junction diode under revers bias has depletion width of 10µm, relative permittivity is 11.7 and permittivity, ε0 =8.85×10^{12}F/m. Then depletion capacitance /m^{2} =?
We know, C_{T} =Aε_{0}ε_{r} /W
C_{T}/A= (8.85×10^{12})(11.7)/10
=10
By putting the values we get 10µF/m^{2}.
The transition capacitance, C_{T} of a PN junction having uniform doping in both sides, varies with junction voltage as ________
C_{T} = K/(V_{0}+V_{B})^{1/2}
As it’s having uniform doping on both sides, the voltage V_{0} will be zero. So, C_{T}=K/(VB)^{1/2}. The variation of transition capacitance with built in capacitance is (V_{B} )^{1/2}.
The CT for an abrupt PN junction diode is ________
For an abrupt PN junction diode, C_{T} = K/(V_{0}+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.
The diffusion capacitance of a PN junction _______
C_{D} =τ I /n_{0} 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 is also called as _______
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.
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