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

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:

$g_m=\sqrt{2{\mu }_n{C}_{ox}\left(\frac{W}{L}\right)I_D}$

$=\frac{1}{300\mathrm{\Omega }}$

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

$r_0=\frac{1}{\lambda I_D}=100k$

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

= 2.19 k

For NMOS:

$gm=\sqrt{2\left(u_n{C}_{ox}\right)\left(\frac{W}{\mathrm{L}}\right)I_d}$

$=\sqrt{2\times 267\times 10\times 100}$

= 0.73 mA/V

$r_0=\frac{V_A^{\prime }L}{I_D}=\frac{5\times 0.4}{0.1}=20k\mathrm{\Omega }$

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

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

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

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

$g_{m1}=\sqrt{2k_n^{\prime }{\left(\frac{\omega }{L}\right)}_1{I}_{ref}}$

$=\sqrt{2\times 200\times 10\times 100}=0.63mA/V$

$r_{01}=\frac{V_{AN}}{I_{D1}}=\frac{20V}{0.1mA}=100k\mathrm{\Omega }$

$r_{02}=\frac{V_{AP}}{I_{D2}}=\frac{10V}{0.1mA}=100k\mathrm{\Omega }$

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

$\begin{array}{l}{V}_{DS}=\frac{1}{k\left(5-0.5\right)}\\ =\frac{1}{\frac{4.5}{750}}\\ =\frac{750}{4.5}=166.67\mathrm{\Omega }\end{array}$