Question

Suppose valt) = 0.25 sin 2000xV in the MOSFET amplifier in Fig. Write expressions for vGO). VE/!), and VDs/t). (b) What is the maximum value of In that corresponds to the active region of operation? 40 mA 9.10 V 6.7V! Ver 5V Rou 3.0 mA 4.8V Ves - 40 V 9 2.7 ince 20mA 4.00 V 1.0 MA 31.50 V M Var 35V Q-point VV Vo = 25 V Vos = 20 2V 4V 6V 8V 12V Drain-source voltage 35V Vas 3.00 VF OA OV 10V

Please solve step by step otherwise down thumb

Answer 1

for Vasct) - 3.5 + 0.5 Sin (wt) Given Data: 10V Ro 3.3k ID Vos 19 3.5V VGA Vos 6.7V M 4.8V 9.7V + 4 35 the output Vosct) is Vasct) = 4.8-9 sin(wt)

from given chariocleistic We find Q-poist QC VOS, ID) - (vos= 4.8V, Ip - 4.515 Topp Small signal gains of giron Amplifier is Av] = IVasim 2 Tugsler 11 Au = 4 | Avi = 9m. RD gm. 3.3k : 4 Im = 3.36 Gm 1.212ms from above data we can derive all required result.

given that: Vgs (t) = 0.45 sin(Qooort) a) Expression for Vasco) Vgs (t) = Ves + Vgs Ct) VESE) - 3.5 +0.25 Sin (20001Tt) Expression for vasct) since it is an inverting amplifier Vds - - Avl.vgs Ct) Vas Ct) = -4 (0-25 in (2000 11t) Ves(t) = - Sin (20001Tt) Vos (t) Vosa + Vas (ty VDS(t) = 4008 sin (20001t)

tets draw graphs of VGs(t) and Vos (t) Avascts 3.75 3.5V mm t 3.25 VOSC) 5.8 4.88 un 3.8 b) ID is maximum coher Vas is minimum Vos 2 VOD- Ip .RO ID z VDO - VOS RD Tocmax) VOD - Voscmio) Rp : Voscming 3.8V Ipcmax) 2 10-3.8 3.3k ID (max) = 1.8787 MA