Question

Sketch circuit diagram of common source FET Amplifier. State working principle of it.

Answer 1

Working: - 

* When small a.c. signal is applied to the gate, it produces variation in the gate to source voltage. This produces variation in the drain current. As the gate to source voltage increases, the drain current also increases. As the result of this voltage drop across R_D also increases. This causes the drain voltage to decreases. 

* As the input voltage rises, gate to source voltage becomes less negative, it will increase the channel width and increase the level of drain current I_D. 

* As the input voltage falls, it will decrease the channel width and decrease the level of drain current I_D. * Thus I_D varies sinusoidally above its Q point value. 

* The drain to source voltage VDS is given by 

* V_DS = V_DD – I_DRD 

* Therefore as I_D increases the voltage drop IDRD will also increase and voltage VDS will decrease. 

* If ΔI_D is large for a small value of ΔV_GS; the ΔV_DS will also be large and we get amplification. Thus the AC output voltage V_DS is 180º out of phase with AC input voltage.