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

Marks: 10 Marks

Year: Dec 2015

Characteristics of JFET

In JFET characteristics we can find three regions,

1) The linear or the ohmic region: Here the drain to source voltage is small and drain current in nearly proportional to the drain to source voltage. When a positive drain to source voltage is applied, this voltage increases from zero to a small value, the depletion region width remain very small and under this condition the semi-conductor bar behaves just like a resistor. So, drain current increases almost linearly with drain to source voltage.

2) The saturation of the active region: Here the drain current is almost constant and it is not dependent on the drain to source voltage actually. When the drain to source voltage continuous to increase the channel resistance increases and at some point, the depletion regions meet near the drain to pinch off the channel. Beyond that pinch off voltage, the drain, current attains saturation.

3) The breakdown voltage: Here the drain current increases rapidly with a small increase of the drain to source voltage. Actually for large value of drain to source voltage, a breakdown of the gate junction takes place which results a sharp increase of the drain current.

Transfer Characteristics:

The graphical characteristics plot of the saturation drain current against the gate to source voltage is known as the transfer characteristics of JFET. It can be obtained from static characteristics very easily. The transfer characteristics of an n- channel is shown below

Proof

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

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

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

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

Hence , $\boxed{g_m=\dfrac{2}{|V_P|}\sqrt{I_{DSS}I_{DS}}}$