cyeokpeng Posted November 15, 2005 Posted November 15, 2005 Hi, I know how to bias a BJT transistor to operate in the active region, so that the amplifier circuit can operate as a small-signal amplifier. Two methods (i) Use biasing resistors to form a self-biasing circuit. (ii) Use diode biasing, coupled with biasing resistors. Qn1 However, how do we bias the transistor circuit to act as small signal amplifier, if MOSFETs are used instead? Design of dc biasing circuit for MOSFET as an amplifier??? I read from my textbook that we need to bias the MOSFET in the saturation region, which is quite different from the BJT amplifier dc biasing design. The relation of MOSFET in saturation region is Id = K(Vgs - Vt)^2 where Id = dc drain current K = a property of MOSFET (What exactly is this?) Vgs = dc gate-source voltage Vt = threshold voltage of MOSFET Qn2 Is there any difference in dc biasing design of n-type enhancement MOSFETs with n-type depletion type MOSFETs? What is the difference? Qn3 After the MOSFET biasing point is fixed (saturated), we input an small ac signal in the input port, to get an amplified output signal in the output port, with the midband gain depending on the amplifier design. How is the small signal equivalent circuit model different for enhancement and depletion type MOSFETs? My answer is, they are essentially the same, but with one slight difference: vgs for enhancement type is positive, while vgs for depletion type is negative, am I correct? Qn4 In the small signal model of MOSFET, the ac drain current id = transconductance * vgs ---Linear, voltage-controlled current source Why is it this formula, and NOT Id = K(Vgs - Vt)^2? transconductance = 2K(Vgs - Vt) --Enhancement and Depletion type MOSFET
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