Transconductance MCQ Quiz in मल्याळम - Objective Question with Answer for Transconductance - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

• The drain current (I_D) of a MOSFET in the saturation region is given by the equation:
  
  I_D = (1/2) × μ_n × C_ox × (W/L) × (V_GS - V_TH)²,
  
  where:
  • μ_n: Electron mobility
  • C_ox: Gate-oxide capacitance per unit area
  • W: Width of the MOSFET channel
  • L: Length of the MOSFET channel
  • V_GS: Gate-to-source voltage
  • V_TH: Threshold voltage
• All other parameters (μ_n, C_ox, and L) are constant in this problem.
• Therefore, the drain current is directly proportional to both the width of the MOSFET (W) and the square of the overdrive voltage (V_GS - V_TH).

Solution:

Let the parameters of the second MOSFET (MOSFET 2) be:

• Width: W
• Overdrive voltage: (V_GS - V_TH)
• Drain current: I_D2

For the first MOSFET (MOSFET 1), the parameters are:

• Width: 2W (double the width of MOSFET 2)
• Overdrive voltage: 2(V_GS - V_TH) (double the overdrive voltage of MOSFET 2)
• Drain current: I_D1

The drain current for MOSFET 2 is:

I_D2 = (1/2) × μ_n × C_ox × (W/L) × (V_GS - V_TH)²

The drain current for MOSFET 1 is:

I_D1 = (1/2) × μ_n × C_ox × (2W/L) × [2(V_GS - V_TH)]²

Simplifying I_D1:

I_D1 = (1/2) × μ_n × C_ox × (2W/L) × 4(V_GS - V_TH)²

I_D1 = 4 × (1/2) × μ_n × C_ox × (W/L) × (V_GS - V_TH)²

I_D1 = 8 × I_D2

Conclusion:

The ratio of the drain currents of the two MOSFETs is:

I_D1 : I_D2 = 8 : 1

The correct answer is Option 3.

Concept:

Voltage gain is given by:

A_v = - g_mR_D

\({{g}_{m}}=\sqrt{2{{\mu }_{n}}{{C}_{ox}}\left( \frac{W}{L} \right){{I}_{D}}}\)

Calculation:

\({{g}_{m}}=\sqrt{2{{\mu }_{n}}{{C}_{ox}}\left( \frac{W}{L} \right){{I}_{D}}}\)

\(=\frac{1}{300\text{ }\!\!\Omega\!\!\text{ }}\)

A_V = - g_mR_D

\(=-\frac{1}{300}\times 1000\)

= -3.33

The small signal model with a test voltage V_x is shown.

The output resistance is given by \({R_0} = \frac{{{V_x}}}{{{I_x}}}\)

From the circuit

V_gs = V_x

Applying KCL

\(- {I_x} + {g_m}{V_x} + \frac{{{V_x}}}{{{r_0}}} = 0\)

\(\frac{{{V_x}}}{{{I_x}}} = {R_0} = \frac{{{r_0}}}{{1 + {r_0}gm}} = {r_0}/\frac{1}{{{g_m}}}\)

Calculation of g_m

\({I_D} = \frac{1}{2}{\mu _n}{C_{ox}}\left( {\frac{W}{L}} \right){\left( {{V_{gs}} - {V_t}} \right)^2}\)

\({I_D} = k{\left( {{V_{gs}} - {V_t}} \right)^2}\)

\(\sqrt {\frac{{{I_D}}}{k}} = \left( {{V_{gs}} - {V_t}} \right)\)

\(\frac{{d{I_D}}}{{d{V_{GS}}}} = 2k\left( {{V_{gs}} - {V_t}} \right)\)

\(\frac{{d{I_D}}}{{d{V_{GS}}}} = 2k\sqrt {\frac{{{I_D}}}{k}} = 2\sqrt {k{I_D}} \)

\({g_m} = 2\sqrt {{k_n}{I_D}}\)

= 0.447 mA/V

\({r_0} = \frac{1}{{\lambda {I_D}}} = 100\;k\)

\({R_0} = \frac{1}{{{g_m}}}\left| {\left| {{r_0} = 2.24k} \right|} \right|100k\)

= 2.19 k

For NMOS:

\(gm = \sqrt {2\left( {{u_n}{C_{ox}}} \right){{\left( {\frac{W}{{\rm{L}}}} \right)}{I_d}}} \)

\( = \sqrt {2 \times 267 \times 10 \times 100} \;\)

= 0.73 mA/V

\({r_0} = \frac{{V_A'L}}{{{I_D}}} = \frac{{5 \times 0.4}}{{0.1}} = 20\;k{\rm{\Omega }}\)

A₀ = gm r₀ = 0.73 × 20

= 14.6 V/V

The circuit shown is CMOS circuit implementation of the common-source amplifier.

Here load = Q₂ transistor and output is taken across Q₁

\({A_v} = \frac{{{v_0}}}{{{v_i}}} = {A_{{v_0}}}\left( {\frac{{{R_L}}}{{{R_L} + {R_0}}}} \right)\)

\( = - \left( {{g_{m1}}{r_{01}}} \right)\left( {\frac{{{r_{02}}}}{{{r_{02}} + {r_{01}}}}} \right)\)

\( = - {g_{m1}}({r_{01}}|{\rm{|}}{r_{02}}{\rm{)}}\)      ......(1)

\({g_{m1}} = \sqrt {2k_n'{{\left( {\frac{\omega }{L}} \right)}_1}{I_{ref}}} \)

\( = \sqrt {2 \times 200 \times 10 \times 100} = 0.63\;mA/V\)

\({r_{01}} = \frac{{{V_{AN}}}}{{{I_{D1}}}} = \frac{{20\;V}}{{0.1\;mA}} = 100\;k{\rm{\Omega }}\)

\({r_{02}} = \frac{{{V_{AP}}}}{{{I_{D2}}}} = \frac{{10\;V}}{{0.1\;mA}} = 100\;k{\rm{\Omega }}\)

A_v = -g_m1 (r₀₁||r₀₂)

= - 0.63 (mA/V) × (200||100) kΩ

= -42 V/V

In linear region MOSFET drain current

\({I_D} = {\mu _n}{C_o} \times \frac{W}{L}\left[ {\left( {{V_{GS}} - {V_T}} \right){V_{DS}} - \frac{1}{2}V_{DS}^2} \right]\)

In linear region with very small V_DS the equation can be written as

\(\begin{array}{l} {I_D} \cong {\mu _n}{C_o} \times \frac{W}{L}\left( {{V_{GS}} - {V_T}} \right){V_{DS}}\\ {V_{DS}} = \frac{{{V_{DS}}}}{{{I_D}}} = \frac{1}{{{\mu _n}{C_o} \times \frac{W}{L}\left( {{V_{GS}} - {V_T}} \right)}} \end{array}\)

V_DS = 500 Ω for V_GS = 2V

\(\begin{array}{l} \Rightarrow 500 = \frac{1}{{k\left( {2 - 0.5} \right)}}\\ k = \frac{1}{{500 \times 1.5}}\\ = \frac{1}{{750}} \end{array}\)

When V_GS = 5V

\(\begin{array}{l} {V_{DS}} = \frac{1}{{k\left( {5 - 0.5} \right)}}\\ = \frac{1}{{\frac{{4.5}}{{750}}}}\\ = \frac{{750}}{{4.5}} = 166.67{\rm\:{\Omega }} \end{array}\)