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

Concept:

n-channel MOSFET biased in saturation region will result in a current given by:

\(\Rightarrow {I_D} = \frac{{W\mu {C_{ox}}}}{{2L}}{\left( {{V_{GS}} - {V_{th}}} \right)^2}\)

And the transconductance is defined as \({g_m} = \frac{{\partial {I_D}}}{{\partial {V_{GS}}}}\)

\(\Rightarrow {g_m} = \frac{{W\mu {C_{ox}}}}{{2L}}\left( {{V_{GS}} - {V_{th}}} \right)\)

Calculation:

Given, two transistors T₁ and T₂ which are identical in all respects except for width and gate overdrive voltage (V_GS - V_th).

For T₁ MOSFET,

g_m1 ∝ W₁ (V_GS - V_th)₁  and  I_D1 ∝ W₁ (V_GS - V_th)²₁

Similarly for T₂, g_m2 ∝ W₂ (V_GS - V_th)₂  and  I_D2 ∝ W₂ (V_GS - V_th)²₂

\(\Rightarrow \frac{{{I_{D1}}}}{{{I_{D2}}}} = \frac{{{W_1}\left( {{V_{GS}} - {V_{th}}} \right)_1^2}}{{{W_2}\left( {{V_{GS}} - {V_{th}}} \right)_2^2}}\;\;\;\; \ldots \left( 1 \right)\)

Given W₂ = 2W₁,

And (V_GS - V­_th)₂ = 2 (V_GS - V_th)₁

Putting these in Equation (1), we get,

\(\Rightarrow \frac{{{I_{D1}}}}{{{I_{D2}}}} = \frac{{{W_1}\left( {{V_{GS}} - {V_{th}}} \right)_1^2}}{{2{W_1}{{\left( 2 \right)}^2}\left( {{V_{GS}} - {V_{th}}} \right)_1^2}}\)

\(\Rightarrow \frac{{{I_{D1}}}}{{{I_{D2}}}} = \frac{{{W_1}\left( {{V_{GS}} - {V_{th}}} \right)_1^2}}{{8{W_1}\left( {{V_{GS}} - {V_{th}}} \right)_1^2}} \Rightarrow {I_{D2}} = 8{I_{D1}}\)

Similarly,

\(\frac{{{g_{m1}}}}{{{g_{m2}}}} = \frac{{{W_1}{{\left( {{V_{GS}} - {V_{th}}} \right)}_1}}}{{{W_2}{{\left( {{V_{GS}} - {V_{th}}} \right)}_2}}}\)

\(\frac{{{g_{m1}}}}{{{g_{m2}}}} = \frac{{{W_1}{{\left( {{V_{GS}} - {V_{th}}} \right)}_1}}}{{2{W_1}\left( 2 \right){{\left( {{V_{GS}} - {V_{th}}} \right)}_1}}}\)

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

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