Question

A MOSFET is wired as a common-source amplifier as shown below. The input voltage v_IN is the total of the source for biasing the circuit at its operating point (v_BIAS), and a small signal ac source providing the signal that we want to amplify (v_in). The total output voltage is v_O.

ملتی BLAS

a) Assume V_DD = 5?, V_IN = 2?, and ? = 4?Ω in the circuit and the MOSFET parameters are K = 0.5??/?², V_TH 1?, and ? = 0.05V^-1. First verify that you have all the parameters you need. Also, rewrite any parameters if they are given to you in a different form than you need.

b) Now, assuming a region of operation of the MOSFET, solve for V_O (the output bias point) and check if the assumption checks ignore λ for this part.

c) Draw a small signal equivalent circuit. Label all the elements (including one for the input signal source) and all the nodes (calling out the MOS D, G, and S). If any sources are zeroed out, indicate this on the equivalent circuit. Also write symbolic expressions for every small signal element in terms of the large signal bias point (no numbers). Ignore λ when calculating g_m.

d) Write a symbolic expression for the small signal voltage at the output, v_out. Then solve numerically for small signal output gain, v_out/v_in.

e) Now assuming R is extremely large – write a symbolic expression for v_out/v_in. Is it a function of the DC input voltage ?

f) Now assuming R is extremely small – rewrite a symbolic expression for v_out/v_in. Is this voltage a function of the DC input voltage, V_IN ? Is it a function of the DC output voltage, V_OUT ? Solve numerically for the small signal output gain, v_out/v_in.

Answer 1

Vds DC Analysi " T UDD=50 3 R = 416 VOS Gau I + vgs = av Ids=ke (Vgs- utro] | Ids = 0.25mA | Vos = luos = 5-ux 4 volts Vo = 4 vots Region of operation ģ (saturation region VDS=v0 =uu Vou = vgs to Elu

2 finding Ac parameters om ro) 19m knCwga utan = 0.5[2-1] 19m = o.sms] ro= ! tIds 0.05* 1 = 4x10% x 103 rds =80= 80kh small signal equivalut (Ac) cercent Ig=0 ERO ILY vod Birds (oof Rao @gmxgs | Boke | 4162 =R vout - IRO I=-gm vgs rotRO robo - current division] I- Im vout = -gm[Ro. ro] an Ro. o .vgs Rotro vout = -9m [Rollso] vgs

but vgs = vin voot =-9m (Roll ro] vin Ivout = Avangm[Rollro) vin Voltage gain expressia Au = -0.5*( 41180 Au =-0.5*3.809 Au=-1.90 ulu (=vel symbol phase shift b/w Yp af op ie. common source amplits

but vgs = vin voot =-9m (Roll ro] vin Ivout = Avangm[Rollro) vin Voltage gain expressia Au = -0.5*( 41180 Au =-0.5*3.809 Au=-1.90 ulu (=vel symbol phase shift b/w Yp af op ie. common source amplits

@ if R = 4 very large ellrds=rds Vout= -9 mrds ugs AU= vout a-gmids vin Au = -406

if then Ŕ is very large, the gain at MOSFET is increased from 1.9 ulu to dovo ® lf R is veey small Rllods = R Nout=-gm R. vgs Vout = -0.5x4 vin -20/0 -vout Uin