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Course by: Bijay_Kumar Sharma. E-mail the author

# Tutorial 3_supplementary_AE Lecture No.3.

Module by: Bijay_Kumar Sharma. E-mail the author

Summary: This gives problems on diffusion capacitances and junction capacitances.

Tutorial 3_supplementary_AE Lecture No.3.

Problems on parasitic capacitances associated with Diode at high frequencies.

Problem 1.Given a diode under reverse bias with a depletion width 4µm.

1. If cross sectional area is 1mm square, determine CjD given

Absolute permittivity ε0 = 8.854×10-12F/m and relative permittivity or dielectric constant (k) = εr= 11.6 [Answer: 26pF]

1. If CjD= 15pF then determine the cross sectional area. [Ans.0.6 mm2 ]

Problem 2. A diode is under forward bias condition at VD = 0.7V. Reverse Saturation current IS=16nA at 300K. Minority carrier life time are τn = τp = 25µsec.

1. Find Cd = diffusion capacitance ; [1/(rdCd) = 1/ τn +1/ τp ; [Answer 5.74µF]
2. Find VD which gives Cd = 10µF; [Answer VD = 0.729V]

Problem 3. A silicon PN junction has doping concentrations NA= 1019/cc & ND= 1015/cc. Cross sectional area A= 0.001cm2. Since it is one sided step junction major role is played by holes injected in N-type region. Hence the life-time of holes is given τp(P region)= 0.3µsec.

Under forward bias condition of VD = 0.6V , determine:

1. Diffusion Capacitance Cd ;
2. Transition or Junction Capacitance CjD;

Solution: Here ID is not given nor the saturation current IS is given. We can use the doping to find either of the two.

ID = Qpp and

Qp= charge stored in N region under forward biased condition =

q{pn(0)-pn(thermal equilibrium)}LpA;

and Lp=√(Dpτp) and Dp = µ×VT = 450 cm2/(V-sec)×26mV= 11.7 cm2/sec;

Therefore Lp= 1.94×10-3 cm.

Therefore ID= q{pn(0)-pn(thermal equilibrium)}LpA/τp = 2.453mA;

Therefore rd = 26mV/2.453mA =10.59 Ω.

rd×Cd= τp therefore Cd =28,000pF.

Calculation of CjD:

CjD= εA/dn where dn = the depletion width on N-side = √(qεND/2)√(øB0-VF);

In a one sided step junction d~ dn ( depletion width on the lightly doped side);

Therefore CjD= εA/[√(qεND/2)√(øB0-VF)] = √(2ε/(qND))×(1/√( øB0-VF));

øB0 = Built in barrier potential under zero bias condition= VTln(NAND/ni2)=817.056mV;

Therefore CjD= 19.8 pF.

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