MOSFET Amplifiers
MOSFETs
- 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
- 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
- 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!