ECE3050 — Assignment 141. The figure sho w s a CS amplifier with a current-mirror active load.(a) If the r0approximations are used, show that the small-signal short-circuit output currentis give n byio(sc)= −vi1+χ11r0s1+ RS=gm11+gm1(1 + χ1) RSvi(b) Show that the small-signal output resistance is given byrout= rid1kr02=·r01µ1+RSr0s1¶+ RS¸kr02(c) Show that the small-signal open-circuit voltage is given byvo(oc)= io(sc)rout= −gm11+gm1(1 + χ1) RS·r01µ1+RSr0s1¶+ RS¸kr02× vi2. For the CS amplifier of Problem 1, each MOSF ET has the parameters K0=0.002 A/ V2,VTO=1.4V, λ =0.02 V−1, γ =1.5V1/2,andϕ =0.6V. It is given that V+=10V,V−= −10 V, Iref=1mA, RS= 200 Ω.(a) Use the equation ID= K0(1 + λVSD)(VSG− VTO)2to show that VSG3=2.09 V.Notethat VSD3= VSG3.(b) Use the equation ID= K0(1 + λVSD)(VSG− VTO)2to show that ID2=1.15 mA.Notethat VSG2= VSG3.(c) Show that VDS1=9.77 V and VBS1= −0.23 V. Note that the curren t through RSisequal to ID2.(d) Use the equation χ =0.5γ/√ϕ − VBSto show that χ1=0.823.(e) Use the equation K = K0(1 + λVDS) to show that K1=4.78 × 10−3A/ V2.(f) Use the equations gm=2√KIDand r0=(VDS+1/λ) /IDto show that gm1=3.32 ×10−3S and r01=51.9kΩ.(g) Use the equation r0=(VSD+1/λ) /IDto show that r02=52.1kΩ.1(h) Show that the small-signal short-circuit output current isio(sc)= −vi1+χ11r0s1+ RS=1.50 × 10−3viAssume r01= ∞ and use the simplified T model.(i) Show that the small-signal output resistance and open-circuit output voltage are rout=rid1kr02=35.8kΩ and vo(oc)= io(sc)× rout= −53.8vi.(j) If a load resistor RL=10kΩ is connected from output to ground, show that the outputvoltage changes by the factor RL/ (rout+ RL)=0.218 (or by −13.2dB) and the newv oltage gain is vo/vi= −11.7.(k) If RS=0, show that vo(oc)increases to the value vo(oc)= −86.28vi. Show that the gainincreases by 4.10 dB.3. The figure shows a CG amplifier with a current-mirror active load. The v oltage VGis a dcbias voltage. Each MOSFET has the parameters gm=2.5 × 10−3S, r0=40kΩ,andχ =0.5.It is given that RS= 200 Ω. It can be assumed that the dc value of the output voltage iszero.(a) Solve for the Norton short-circuit output current. Assume r01= ∞ and use the simplifiedTmodel. Answer: io(sc)= vi/ (r0s1+ RS)=2.14 × 10−3vi.(b) Solve for the Thévenin equivalent circuit seen looking into the vonode, i.e. solve for routand vo(oc).Answers:rout= rid1kr02=25.5kΩ and vo(oc)= io(sc)× rout=54.6vi.(c) By what factor does vochange if a load resistor RL=10kΩ is connected from output toground? What is the new voltage gain? Answer: RL/ (rout+ RL)=0.282 or b y −11 dBand vo/vi=15.4.(d) Show that the input resistance is rin= r0s1= 267 Ω.4. The figure shows a CD amplifier with a curren t-mirror active load. The voltage VGis a dcbias voltage. Each MOSFET has the parameters gm=2.5 × 10−3S, r0=40kΩ,andχ =0.5.2(a) Solve for the Norton short-circuit output current. Use the simplified T model. Answ er:io(sc)= vi/rs1= gm1vi=2.5 × 10−3vi. Note that the body effect cancels when vo=0.(b) Solve for the Thévenin equivalent circuit seen looking into the vonode, i.e. solve for routand vo(oc).Answers:rout= r0s1kr01kr02= 263 Ω and vo(oc)= io(sc)× rout=0.658vi.(c) By what factor does vochange if a load resistor RL=10kΩ is connected from outputto ground? What is the new voltage gain? Answers: RL/ (rout+ RL)=0.974 or by−0.226 dB and vo/vi=0.641.5. The figure shows a cascode amplifier. M1is operated as a CS amplifier with a small-signalvoltage vsand a dc bias voltage VB1applied to its gate. M2is operated as a CG amplifierwith a dc bias voltage VB2applied to its gate. M3and M4form a current mirror with aninput dc current IREF. For each MOSFET, it is given that gm=0.005 S, gmb=0.0025 S,and r0=50kΩ.(a) To simplify the solution for io(sc),assumer01= r02= ∞. Show that io(sc)/vs= i0d2=i0d1=0.005 S.(b) With r01= r02=50kΩ, show that the output resistance isrout= r03k£r02¡1+r01/r0s2¢+ r02¤=49.87 kΩ(c) Show that vo(oc)/vs= −£io(sc)/vs¤× routkr03=