Question

V.+w Operation in the triode reglon Condition v. e Wov 20 Vos uov os os-V (2) p V, so onl+Pala Characteristics Same relationships as for NMOS trasistos tCharacteristics: a CuGs- V,) ®os- } ip.C Replace .and NA with p,,and Nprespectively. V.V V, and yare negative. 2 wov ps For vos 2( -V) e Conditions for operation in the triode region ip lvi Q1. (10 points) For the following configuration of the given figure below, with the following parameters: VDD= +10 V, Vss = -10 V Rsig 1.2kO, RI 100k2 ,RG= 40k, RD=3.3kO KN (kn'W/L)= 0.4 mA/V2, gm 5mS ,VIN= IV 1) Draw the ac small signal equivalent network, 2) Determine Rin 3) Determine Rou, 4) Find Ay(VoVsie) without the effect of R 5) Sketch the output waveform if Vsig 10 mV sine wave. Lo+ 2. (10 points) or the given FET network shown in the given figure etermine the following if kn'(W/L)=4 mA/V2 and Vt 2V: 1) VGs 2) Vs 3) VD 4) VDs 5) Verify the operation mode for transistor if it is working in saturation or triode and why? VDD 15 V 4mA NE=2V ヒn 0.5 mA Rp 10k o VD V +75V o -o Vs g=+7s Rs 10 k 7 MQ rerdrtre voltage D the ariode region: Operation Conditions Overdrive voltage: tor 20 nw e-V. Q4. (10 the circuit in the below figure (Find the values of RD, s operates 1V from the edge of saturation: VDs 4mA/V2. Make sure the values of Ri and R2 in the range of Mega Ohm. R2), so that the transistor 3V and k'pW/L= VGS-V-1 +15 V VDs=Vas-Ne-1 R Rs VD= o Vs VGO ID=2mA o Vp=SN R2 Rp KPWAMA Ove tar Operation in the triode region: Conditions: V Overdrive voltage Vow ee-V. Q3. (10 points) the small AC equivalent circuit for the network shown below. b) Find c) Determine ro- d) Determine Gv Vo/ Vsig- e) Determine the output voltage (Vo) for the network shown below if 35mV, ID= 1mA, Vas=2V, kn' (W/L) = 0.4 mA/V2, V 0.75V and VA1 put with Vsig f) Sketch the waveform of the output voltage (Vo). a sine wave G=O 4.7 kn 47 MQ o V 100 kn 0.1 F 10kn 0.01 F 10 F 10 Mn2kN

Answer 1

quívalent ciucuut ac 12K EEsu Poat Tlk Rot33k Reut 3.1945 Kn 4ok1.2K Now, m3.3k -m3-3k = -5m (s-3k) - 16.5 Vv - toYn -lam 165m 65 m

t15V 2 lok -Vp 4Mушh MA Re lok FET in Sataialion o.5m 2.5V GS Gsa5 7.5-V 5 -) Vc 15 o5m(1ok 10V 4) p-Vs a.5- o-5 VosVas- FET n Satuaation

3 Ca) ac equavalent civcuit o1 h equivalmt cicit 40-75V 1.25 cok. - 100k +(un) Nou - oq88 (rtm) (tcok{ly7k/o) - 4.895 Yv (e) 35mv 3 U^ = -17.33 mv +359 -35 +131 m -1m

41 3v V -3V m2 by chms law m ven thst - 4- - V-3)- VGt 3V 5 VGt Rs4k 15 8 V 3v XI5V FR2Prf,R = 10M