Mumbai University > Computer Engineering > Sem 3 > Electronic Circuits and Communication Fundamentals

Marks: 5 Marks

Year: Dec 2015

The transfer characteristics are generated as figure using the procedure. Each operating point is then identified and a tangent is drawn ateach point to best reflect the slope of the transfer curve in the region. An appropriate increament is then choosen for $V_{GS}$ to reflect a variation to either side of each Q-point. Equation is then applied to determine $g_m$.

$$1. \ \ g_m=\dfrac{\triangle I_D}{\triangle V_{GS}}\cong \dfrac{2.1 mA}{0.6V}=3.25mS \\2. \ \ g_m=\dfrac{\triangle I_D}{\triangle V_{GS}}\cong \dfrac{1.8mA}{0.7V} \cong2.57mS \\ 3. \ \ g_m=\dfrac{\triangle I_D}{\triangle V_{GS}}= \dfrac{1.5mA}{1.0V}=1.5mS$$

The decrease in $g_m$ as $V_{GS}$ approaches $V_P$.

In transfer characteristics we have seen that the relation between the drain current $I_D$ and gate to source voltage $V_{GS}$ is non-linear. The relationship is defined by Schockley’s equation.

$$I_D=I_{DSS}\bigg(1-\dfrac{V_{GS}}{V_P}\bigg)^2$$

The squared term of the equation will result in non-linear relationship between $I_D$ and $V_{GS}$, producing a curve that grows exponentially with decreasing magnitudes of $V_{GS}$.

In equation above $I_D$ represents saturation drain current. $I_{DSS}$ is the value of $I_D$ when $V_{GS}$=0, and $V_P$ is the pinch-off voltage. Differentiating this expression we get.

$$\dfrac {\partial I_D}{\partial V_{GS}}=I_{DSS}\times 2\bigg(1-\dfrac{V_{GS}}{V_P}\bigg)\bigg(-\dfrac{1}{V_P}\bigg) \\ g_m=\dfrac{-2 I_{DSS}}{V_P} \bigg(1- \dfrac{V_{GS}}{V_P} \bigg) \ \ since \ \ \dfrac{\partial I_D}{\partial V_{GS}}|V_{DS \ \ constant}=g_m$$

From expression we also have

$$\bigg(1-\dfrac{v_{GS}}{V_P}\bigg)=\sqrt{\dfrac{I_D}{I_{DSS}}}$$

Substituting this value in equation above we have

$$g_m=\dfrac{-2 I_{DSS}}{V_P}\sqrt{\dfrac{I_D}{I_{DSS}}} \\ =\dfrac{-2\sqrt{I_DI_{DSS}}}{V_P}$$