Electronic Devices Test 5

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Electronic Devices Test 5

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Question 1
1 / -0

The ideal current-voltage parameters of a MOSFET with a very small threshold voltage operating in non-saturation (triode region) is scaled by a constant k such that W, L, V_GS, V_DS, C_ox changes to kW, kL, kV_GS, kV_DS, \(\frac{{{C_{ox}}}}{k}\), respectively.
The drain current I_D changes to (approximately):

A
k² I_D

B
kI_D

C
\(\frac{{{I_D}}}{{{k^2}}}\)

D
\(\frac{{{I_D}}}{k}\)

Solution

Concept:
The current equation for a MOSFET operating in the triode region is given by:
\({I_D} = \frac{{W\;{\mu _n}{C_{ox}}}}{{2L}}\left[ {2\;\left( {{V_{GS}} - {V_T}} \right){V_{DS}} - V_{DS}^2} \right]\)
V_GS = Gate to source voltage
V_DS = Drain to source voltage
V_T = Threshold voltage
Application:
After scaling the parameters by the given scaling factor k, the current equation becomes:
\(I_D' = \frac{1}{2}{\mu _n}\left( {\frac{{{C_{ox}}}}{k}} \right)\left( {\frac{{kW}}{{kL}}} \right)\left[ {2\left( {k{V_{GS}} - {V_T}} \right)\left( {k{V_{DS}}} \right) - {{\left( {k{V_{DS}}} \right)}^2}} \right]\)
Given V_T < < 1, it can be neglected, i.e.
\(I_D' = \frac{1}{2}{\mu _n}\left( {\frac{{{C_{ox}}}}{k}} \right)\left( {\frac{{kW}}{{kL}}} \right)\left[ {2\left( {k{V_{GS}}} \right)\left( {k{V_{DS}}} \right) - {{\left( {k{V_{DS}}} \right)}^2}} \right]\)
\(I_D' = k\left[ {\frac{{W{\mu _n}{C_{0x}}}}{{2L}}\left( {2\left( {{V_{GS}} - {V_T}} \right){V_{DS}} - V_{DS}^2} \right)} \right]\)
\(I_D' \cong k{I_D}\)
Question 2
1 / -0

Ignoring channel length modulation, the transconductance of a MOSFET in saturation mode does not depend on:

A
Channel length

B
Mobility of carrier

C
Drain to source voltage

D
Gate to source voltage

Solution

Concept:
The transconductance of a MOSFET is defined as the change in drain current(I_D) with respect to the corresponding change in gate voltage (V_GS), i.e.
\({g_m} = \frac{{\partial {I_D}}}{{\partial {V_{GS}}}}\)
For a MOSFET in saturation, the current is given by:
\({I_{D\left( {sat} \right)}} = \frac{{W{μ _x}{C_{ox}}}}{{2L}}{\left( {{V_{GS}} - {V_{th}}} \right)^2}\)
W = Width of the Gate
C_ox = Oxide Capacitance
μ = Mobility of the carrier
L = Channel Length
V_th = Threshold voltage
Application:
The transconductance in saturation is given by differentiating the above w.r.t. V_GS, i.e.
\({g_{m\left( {sat.} \right)}} = \frac{{\partial {I_{D\left( {sat.} \right)}}}}{{\partial {V_{GS}}}} = \frac{{W{μ _x}{C_{ox}}}}{L}{\left( {{V_{GS}} - {V_{th}}} \right)^\;}\)
We observe that the g_m(sat.) of a MOSFET is independent of V_DS and depends on W, μ_n, C_ox, L, V_GS, and V_th.
Question 3
1 / -0

Consider a MOS capacitor with a p-type silicon substrate doped with N_a = 10¹⁶ cm^-3, a silicon dioxide insulator with a thickness of t_ox = 20 nm and an n⁺ polysilicon gate. Assume Q’_ss = 5 × 10¹⁰ electronic charges per cm² are the trapped oxide charge and ϕ_ms = - 1.1 V, the flat-band voltage for the MOS capacitor is approximately:
(Take ϵ_ox ≈ 4 × 10^-13F/cm )

A
-1.14 V

B
-2.9 V

C
1.15 V

D
2.9 V

Solution

Concept:
Flat band voltage is defined as the voltage to be applied to align the Fermi level of the metal oxide semiconductor. Mathematically it is defined as:
\( \Rightarrow {V_{FB}} = {\phi _{ms}} - \frac{{{{Q'}_{ss}}}}{{{C_{on}}}}\;\) (For Polysilicon Gate)
Where ϕ_ms = Work function difference between the metal and the semiconductor.
Q’_ss = Trapped oxide charge (in C)
C_ox = Capacitance of the oxide.
Calculation:
Given, t_ox = 20 nm = 20 × 10^-7 cm
\(C_{ox} = \frac{\epsilon_{ox}}{t_{ox}} = \frac{4\times10^{-13}}{20 \times 10^{-7}} = 2 \times 10^{-7} \ F/cm^2\)
The equivalent oxide charge density is:
Q’_ss = (5 × 10¹⁰)(1.6 × 10^-19) C/cm²
= 8 × 10^-9 C/cm²
The Flat Band voltage is, therefore:
\(\Rightarrow V_{FB} = \phi_{ms} - \frac{Q'_{ss}}{C_{ox}} =- 1.1 - \frac{(8 \times 10^{-9})}{2 \times 10^{-7}} =- 1.14 \ V\)