Question

5. Cui For the circuit at left, Vo = 2.4V, Von = 0.4V, VTP = -0.4V are given. i) Assuming that Vout = OV at the beginning and a step voltage rising from zero volt to Vpp applied to input (Vin), find and explain the voltage at Vout ii) As the continuation of the above process, assume that the voltage at input dropped back to o volt, find and explain the final voltage at Vout V our C

Answer 1

Ang! NOD Giveni sol ITO NDD=2.40 cs) I van = 0.40 CGO Cha = -0.4V cs) ayo Yout Cl2 in N for nogós bransistor ON?- Ves for pros transistor on :- VSG ? to VTpl Innros transistor - CP) vorano = Voo=QAN Jeon cov NGERA Ža, 2014) i Nords transistor is on and it is shooted, then von OV. die when Ninis changed ch to Noda- usão - Vin= NDD 2.40 "GCP) z Now "D(n) "GCP) OV NSGTCP) -0.4 = 2.4 > CAMP ". Pros transistor is on and it is shorted, then "D (PS = Yout = Vin = 2.40

2 therefore, Yout = 2.40 The change stored in 'Coy stored in Cas is 2.40. voltage when Vin dropped back to 'o' volta "Scp) Vch = 2.41 Cador IS Yout at NDCP)= VIR SOV Near Chz COV "GCP) NSGCP) Ov=in 245 Noter For MOSFET = 2.4 Z CINTA pros transistor is On and it is shelted then Yout Vincov source and drain terminals are not fixed. For pros the terminal which is having highest voltage acts as drat source terminal. Yout=on