MOSFET Amplifiers

Posted on Jun 11, 2022

MOSFETs

MOSFET

  • MOSFET’s drain, gate and source analogous to BJT collector, source and emitter
    MOSFET saturation mode analogous to BJT forward active mode
  • k_n = transconduction constant, given to us (defined in terms of semiconductor geometry, etc), V_TN = threshold voltage
  • MOSFET is in saturation if the following holds: $$V_{DS}>V{DS(Sat)}(=V_{GS}-V_{TN})$$ $$V_{GS}>V_{TN}$$ In this case: $$i_{D}=k_n(V_{GS}-V_{TN})^2$$
  • MOSFET is in linear region if the following holds: $$V_{DS}<V{DS(Sat)}(=V_{GS}-V_{TN})$$

DC analysis

  • Similar circuit to BJTs for biasing

Voltage divider (Four Resistor) bias circuit (MOSFET)

  • Consider: $$R_{Th}=R_1||R_2, V_{Th}=V_{DD}\frac{R_2}{R_1+R_2}$$ If split power: $$V_{Th}=(V_{DD}-V_{EE})\frac{R_2}{R_1+R_2}$$
  • In MOSFETs: $$I_G=0\therefore I_D=I_S$$
  • Unlike BJTs, V_GS isn’t constant: $$i_D=k_n(V_{GS}-V_{TN})^2\therefore V_{GS}=\sqrt{i_D/k_n}+V_{TN}$$
  • When analysing using Thévenin equivalent circuit of above, we can solve for (root) I_D: $$I_DR_S+\frac{1}{\sqrt{k_n}}\sqrt{I_D}+V_{TN}-V_{Th}=0$$ Quadratic yields 2 values for I_D - need to test which one is correct by applying the criteria for saturation as listed above and checking consistency

AC analysis

Small signal model for a MOSFET.

(Including bias resistors)

  • Transconductance of the MOSFET: $$g_m=\frac{2I_{DQ}}{V_{GS}-V_{TN}}$$
  • Input impedence @ gate $$R_{ig}=\infty$$ Impedence @ ri $$R_i=R_{ig}||R_1||R_2=\infty||R_{Th}=R_{Th}$$ Output impedence $$R_o=R_D$$ (In a source follower): $$R_o=\frac{1}{g_m}$$
  • Voltage in $$V_{in}=V_{gs}+g_mV_{gs}R_S$$ Voltage out $$V_o=-g_mV_{gs}R_D$$ Therefore: Voltage gain: $$A_v=\frac{V_o}{V_{in}}=\frac{-g_mV_{gs}R_D}{V_{gs}+g_mV_{gs}R_S}=\frac{-g_mR_D}{1+g_mR_S}$$ Assuming g_mR_S » 1: $$A_v=\frac{-R_D}{R_S}$$ …just like a BJT!