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Solutions to class activity 16 questions related to the physical origins, values, and differences between junction and diffusion capacitance in p-n diodes. It includes calculations for reverse bias conductance, junction capacitance, and diffusion admittance.
Typology: Assignments
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ECSE-2210 Microelectronics Technology Class Activity 16 – Solution
= 1.52 × 10-4^ cm.
= 10 -12^ x 1/1.52 × 10 - = 6.57 nF. At zero bias, the junction capacitance will be larger than the above value since the depletion layer width will be smaller.
equivalent circuit for the diode. I = I 0 exp ( qV / kT ) I ( V = 0.6 V) = 1.15 × 10 -4^ A
Diffusion conductance: G D = qI / ( kT ) = 1.15 × 10-4^ / 0. = 4.44 × 10-3^ S
= 4.44 nF
Æ Y D = GD + jω CD
The small signal equivalent circuit under forward bias is as shown in the figure above.
Of the three diodes I will have the least depletion region width W and III will have the largest width. As the junction capacitance and the depletion width are inversely proportional to each other, diode III will have the least capacitance and diode I will have the largest capacitance.
p+^ N D=10 15 cm-3^ p +^ N D=10 17 cm- p+ intrinsic
N D=10 15 cm-