Assignment – 2
Chapter 7
1. Doping concentrations of a Silicon pn junction are 1016 and 1017 per cm3 on p and n side
respectively.
a) Determine the zero bias junction voltage, depletion layer width, maximum electric
field, span of depletion region on both sides.
b) The changes in depletion layer width, junction voltage and maximum electric field
when a 5V reverse bias voltage is applied across the junction. Also calculate the
capacitance per cm2 in reverse bias
2. Design a pn junction having the given specifications scenarios
a) Zero bias
i)
ii)
iii)
0.7 Volts built in potential and depletion width of 0.9 µm
Hint – Might have to solve a quadratic equation
Maximum electric field of -13kV/cm and 0.2µm span of depletion layer into
the p region.
***Bonus: 0.7 Volts built in potential and 0.5µm span of depletion layer into
the n region.
b) Reverse Bias
i)
ii)
At a junction voltage of 6 Volts, 2 nF/cm2 junction capacitance per unit area
and depletion layer span of 0.3 µm into the p region.
***Bonus: At a reverse bias voltage of 5 Volts, Maximum electric field of 14kV/cm and 1.2 µm depletion layer.
Chapter 8
1. Doping concentrations of a Silicon pn junction are 1016 and 1017 per cm3 on p and n side
respectively. A 0.3 Volts forward bias is applied across it.
a) Determine
i) Thermal equilibrium minority carrier concentrations
ii) Excess minority carrier concentrations at edges of depletion layer
iii) Electron and hole diffusion current densities at edges of depletion layer
iv) The total current density and the ideal reverse saturation current density
v) Electric Field at ohmic contact edges of both sides
b) Repeat iii, iv and v for a short diode with bulk region widths 1/10th of the
corresponding diffusion lengths
c) Minority carrier concentrations at edge of depletion layer at a reverse bias voltage of
6 Volts
2. Design a Silicon pn junction that should have diffusion current densities of 15 A/cm2 and
10 A/cm2 at edges of depletion layer on p and n sides respectively at a 0.5 Volts applied
forward bias voltage.
Hint – Example 8.3
Chapter 10
A p-channel MOSFET with a p+ polysilicon gate has a substrate doping concentration of
1016/cm3 and effective oxide charge of 1010/cm2 at oxide-semiconductor interface. Given oxide
material is SiO2 with thickness of 125 β«.
a) Determine
i)
The metal semiconductor work function difference from graph
ii)
The flat-band voltage
iii)
Maximum width of depletion layer in the substrate
iv)
Maximum depletion layer charge in the substrate and its type
v)
Threshold voltage
vi)
Mode of the MOSFET (Enhancement/ Depletion)
b) ***Bonus - Roughly draw the energy band diagram of the MOS structure in this
MOSFET at zero applied gate voltage.
Formula
Chapter – 7
πππ + ππ
→ π
ππ£πππ π π΅πππ
πππ → ππππ π΅πππ
Junction voltage, ππ½ = {
}
πππ − ππ → πΉπππ€πππ π΅πππ
ππ
= πππππππ πππ£πππ π π£πππ‘πππ
ππ = πππππππ ππππ€πππ π£πππ‘πππ
π π
built in potential, πππ = ππ‘ ln ( ππ2 π ) , ππ‘ = π‘βπππππ π£πππ‘πππ = 0.0259 ππππ‘π ππ‘ π = 300πΎ
π
π πππ ππ ππππππ‘πππ πππ¦ππ πππ‘π π − ππππππ, π₯π =
ππ
π
ππ + ππ
π
π
πππππππππ¦, π₯π = π +π
π, where the depletion layer width is W.
π
ππ½ =
π
π ππ ππ
π2
2ππ ππ + ππ
π
In a p+n junction, π ≈ π₯π and ππ½ = 2π ππ π₯π2
π
π
In an n p junction, π ≈ π₯π and ππ½ = 2π ππ π₯π2
+
π
2π
maximum electric field, πΈπππ₯ = ππ½ =
πππ π₯π
2ππ
=
πππ π₯π
2ππ
π
Junction capacitance per unit area under reverse bias, πΆ ′ = ππ
Chapter 8
Normal
Short Diode
ππ2
ππ2
π‘βπππππ πππ’πππππππ’π πππππππ‘π¦ πππππππ‘πππ‘ππππ : ππ0 =
πππ ππ0 =
ππ
ππ
Minority concentrations at edges of depletion region under a forward bias voltage ππ :
ππ
ππ
ππ (π₯π ) = ππ0 exp ( ) πππ ππ (−π₯π ) = ππ0 exp ( )
ππ‘
ππ‘
For a reverse bias voltage, use ππ = −ππ
Excess minority concentrations at edges of depletion region under a forward bias voltage ππ :
πΏππ (π₯π ) = ππ (π₯π ) − ππ0 πππ πΏππ (−π₯π ) = ππ (−π₯π ) − ππ0
Minority diffusion lengths:
πΏπ = √π·π ππ0 πππ πΏπ = √π·π ππ0
πππππππ‘π¦ πππππ’π πππ ππ’πππππ‘ ππππ ππ‘πππ :
π½π =
ππ πππ π‘πππ ππ πΏπ πππ ππ πππ π‘πππ ππ πΏπ
ππ·π πΏππ (−π₯π )
πΏπ
πππ π½π =
Same formula but use
ππ·π πΏππ (π₯π )
πΏπ
Total forward current density, π½ = π½π + π½π
Ideal Reverse Saturation Current Density,
π½π =
Electric Field in the bulk:
πΈπ =
π½
πππ ππ
πππ πΈπ =
π½
πππ ππ
π½
π
exp (ππ ) − 1
π‘
Electric Field in the bulk:
πΈπ =
π½π
π½π
πππ πΈπ =
πππ ππ
πππ ππ
0
