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Homework Solved Questions - Microelectronics Technology | ECSE 2210, Assignments of Electrical and Electronics Engineering

Rensselaer Polytechnic Institute (RPI)Electrical and Electronics Engineering

Material Type: Assignment; Class: MICROELECTRONICS TECHNOLOGY; Subject: Electrical & Comp. Sys. Engr.; University: Rensselaer Polytechnic Institute; Term: Summer 2008;

Typology: Assignments

Pre 2010

Uploaded on 08/09/2009

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234 nF
2.34 S
ECSE-2210 Microelectronics Technology
Homework 6 – Solution
Reading list: Chapters 7, 8 and 14 (pages 301-318, 327-338 and 477 - 487).
1. An abrupt, one-sided p+-n junction has the following characteristics on the n-side.
N-side:
ND = 4 × 1016 cm-3
Dn = 25cm2/s; Dp = 10 cm2/s
τ
p =
τ
n = 10-7s
Area A = 1 cm2
Answer the following:
a. The diode is biased in the forward direction such that the forward voltage VA = 0.6 V.
Calculate the low-frequency diffusion capacitance, and the low frequency conductance of
the diode. Draw the equivalent circuit of the diode at low frequency.
First calculate the saturation current, I0, and hence I, using equation 6.29 and 6.30.
For a p+-n diode, we can neglect the saturation current caused by the p-side electrons
since the minority carrier concentration in p-side will be very small compared to the
minority carrier concentration in n-side.
Therefore, I0 = qA (Dp × pn)/Lp
I0 = 1.6 × 10-19 C × 1 cm2 × (10 cm2/s × 2500 cm-3)/(10-3 cm) = 4 × 10-12 A
[with Lp = (Dp ×
τ
p)1/2]
I = I0 exp(0.6/0.0256) = 0.06 A.
So, G0 = qI/kT = 2.34 S
Cd = qI
τ
p/kT = 234 nF.
pf2

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234 nF

2.34 S

ECSE-2210 Microelectronics Technology Homework 6 – Solution

Reading list: Chapters 7, 8 and 14 (pages 301-318, 327-338 and 477 - 487).

  1. An abrupt, one-sided p+-n junction has the following characteristics on the n-side. N-side:

N D = 4 × 10^16 cm- D n = 25cm^2 /s; D p = 10 cm^2 /s

τp = τn = 10-7^ s

Area A = 1 cm^2

Answer the following:

a. The diode is biased in the forward direction such that the forward voltage V A = 0.6 V. Calculate the low-frequency diffusion capacitance, and the low frequency conductance of the diode. Draw the equivalent circuit of the diode at low frequency.

First calculate the saturation current, I 0 , and hence I , using equation 6.29 and 6.30.

For a p+-n diode, we can neglect the saturation current caused by the p-side electrons since the minority carrier concentration in p-side will be very small compared to the minority carrier concentration in n-side. Therefore, I 0 = qA ( D p × p n )/ L p

I 0 = 1.6 × 10-19^ C × 1 cm^2 × (10 cm^2 /s × 2500 cm-3^ )/(10-3^ cm) = 4 × 10 -12^ A

[with L p = ( D p × τp ) 1/2^ ]

I = I 0 exp(0.6/0.0256) = 0.06 A.

So, G 0 = qI / kT = 2.34 S

C d = qI τp/ kT = 234 nF.

12.6 nF

b. The diode is biased in reverse such that the applied voltage | V A| = 20 V. Calculate the reverse bias capacitance (Hint: you can neglect V bi ). Draw the equivalent circuit, assuming an ideal diode. Explain briefly how the circuit will change if we start considering the non-ideal behavior of the diode.

First find the depletion layer width:

W = [(2 ε)/ qN D | V A|]1/2^ = 0.79 × 10 -4^ cm Æ C = ε/ W = 12.6 nF.

Reverse bias conductance is zero under reverse bias (for an ideal diode)

You have to add a parallel resistance across the capacitor above to account for the g-r current in the depletion layer (This accounts for the fact that the reverse bias current is voltage dependent). Also, you have to add a series resistance to account for the parasitic contact resistances, as well the resistance of the neutral regions.

2.) Refer to exercise 7.2. in the textbook.

R s R p

12.6 nF