0
32kviews
Derive an expression for trans-conductance for JFET.
1 Answer
1
3.6kviews

The transconductance gm, is the change in the drain current for given change in gate to source voltage with the drain to source voltage constant as shown in Fig.

Looking at Fig. we can say that it is the slope of the transfer characteristic. Since the slope varies, gm also varies. gm has a greater value near the top of the curve than it does near the bottom. The transconductance gm is defined as

$$g_m=\dfrac{\triangle I_D}{\triangle V_{GS}}\bigg| V_{DS \ \ constant}$$

The transconductance gm is also called mutual conductance. The practical unit for gm is mS (millisiemen) or mA/V. For given gm, we can calculate an approximate value for gm at any point on the transfer characteristic curve using the following equation

$$g_m=g_{mo}\bigg[1-\dfrac{V_{GS}}{V_{GS(off)}}\bigg]$$

where gmo is the value of gm for VGS = 0, and is given by,

$$g_{mo}=\dfrac{-2 I_{DSS}}{V_P}$$

This can be proved as given below. We know that,

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

enter image description here

Differentiating this equation w.r.t $V_{GS}$ is given as

$$g_m=\dfrac{\triangle I_D}{\triangle V_{GS}}=\dfrac{-2 I_{DSS}}{V_P} \bigg[1-\dfrac{V_{GS}}{V_P}\bigg] \\ g_{mo}\bigg[1-\dfrac{V_{GS}}{V_P}\bigg] \ \ \ \ \ \ where \ \ g_{mo}\cong \dfrac{-2 I_{DSS}}{V_P}$$

Please log in to add an answer.