Question

the differential andlifier. Assume the gate al consider the differential amplifie of Q2 is grounded: V = 0 and VG1 - vid Find value of Vid reeded to ste through Q2 to make 12 = . value of Vid reeded to steen the entire tail current x2 to make 62= 1 mA, Round to ² decimal. 7 +2.5V IMA V = – 0.8V kp - 16 MAN ² Vol 11 Q4V 62 B2 4kr 24kt 32 -2.5 V =b] If the output voltage is taken as Vo=VD2-B1 and input as Vid=VE 1 - VG2 Find the magnitude of small signal diff voltage gain Ad at the center of the transfer characteristics (for Vid=0). Also, 2-0. Express answer in VV, type the value of 1A.

Answer 1

solution Given differential Amplifier with [VGI=vid) VCA (9) +2.50 Vtp = -0.8V 4) |๓ ก Kp = 16mA/2 vai ke J ID, Izzy B e e 3 4ku. & 4K2 sv -2.5 0. -2.50 W - Now here, we want entire current to pan througer 82, ID2 = Imali so, da is in Saturation & s, is in off Condit Now, ID a= IMA = KP (Usapz + Vep) ² Imá = 1600 A (Vis apete VED)'. - (vsagt vtp.) ² = usarzt (p = 0.3534 Usap2 = 0.3536, - vtp = 0.3536-(-0.8). Usap2= 10154 N / Vaspe=-1.1544) 2 Vid= UGI- Vaz = Vasi - Vasz, - kastt letsit & Now, VG2- Us2 = - 10154 + Us 2=71.154v [192=10154 v = vsil For &, to be off, USGpl + Utp co. Vsi-VGI + vtpco

7 1.154 - vid + (-0.8) Co. (vid > 0.354 v So, s vid is greater than 0. 35 volt, I will be off & 2 will conduct YMA. 6 Now, Vo = VD2-VD), vid - Vai-VG2) Given, (vidzou, A=0 Here, we will apply, vid to a so doing Ac- Analysis. - As noitia (i= gomvid I (ide=-gomvid ) Vid=1 8i 8 - vid to 22 vid to apply, e will safely write , we can 0 - t = No = VD2 - Voi= lipz-201) (AKP). = -gmyil - gmvid aku. - gm vid (4kv) Vo = - gm (Aku). vid Now, given (vid=ov - middle of transfer-Characteristian 30 both 8,8 82 are in saturation hence, ID = ID2 = OosmĄ. d iz sin IND So, gm> √ 2 kp ID. = 2)(1681043) 4 kr Cosx103) 30, AD = Vo a [gm = (41 103) A/V) 15-(4410-3)(4x103) TA - 16. (AD) = 16 VITOLY