ECE3050 Homework Set 11. For V =18V, R1=39kΩ, R2=43kΩ,andR3=11kΩ,useOhm’sLaw,voltagedivision,and current division to solve for V1, V2, I1, I2,andI3.V1=1839 kΩ39 kΩ +43kΩk11 kΩ=14.7V V2=1843 kΩk11 kΩ39 kΩ +43kΩk11 kΩ=3.30 VI1=1839 kΩ +43kΩk11 kΩ= 376.8 µA I2=11 kΩ43 kΩ +11kΩI1=76.8 µAI3=43 kΩ43 kΩ +11kΩI1= 300 µA2. For I = 250 µA, R1= 100 kΩ, R2=68kΩ,andR3=82kΩ, use Ohm’s Law, voltage division,and current division to solve for I1, I2,andV3.I1= 250 µA68 kΩ +82kΩ100 kΩ +68kΩ +82kΩ= 150 µAI2= 250 µA100 kΩ100 kΩ +68kΩ +82kΩ= 100 µAV3= 100 µA × 82 kΩ =8.2V3. It is given that R1=1kΩ, Av=10−4, Ai=50,andR2=40kΩ.(a) With io=0, use superposition to write the equations for i1and vo(oc). Solve the equationsfor vo(oc)as a function of vs.i1=vs− AvvoR1vo(oc)= −Aii1R21vo(oc)=−AiR2R11 − AiAvR2R1vs= −2500vs(b) With vo=0, use superposition to write the equations for i1and io(sc). Solve the equationsfor io(sc)as a function of vs.io(sc)= −AivsR1= −0.05vs(c) Use the solutions for vo(oc)and io(sc)to show that rout=50kΩ .(d) Show that the Thévenin equivalen t circuit seen looking into the output is a voltage sourcevo(oc)= −2500vsin series with a resistance rout=50kΩ.(e) Show that the Norton equivalen t circuit seen looking into the output is a current sourceio(sc)= −0.05vsin parallel with a resistance rout=50kΩ.(f) If a load resistor RL=20kΩ is connected from the output node to ground, use both theThévenin and the Norton equivalent circuits to show that vo= −714.3vsfor both.4. It is given that R1=10Ω, Gm=0.5S, R2=5Ω, R3=50Ω,andAv=4.(a) With io=0, use superposition to solve for vo(oc)as a function of vs.vo(oc)=R3R2+ R3vs+ AvvsR2R2+ R3=R3+ AvR2R2+ R3vs=1.273vs(b) With vo=0, use superposition to solve for io(sc)as a function of vs.io(sc)=vsR2+AvvsR3= vsµ1R2+AvR3¶=0.28vs(c) Solve for rout.rout=vo(oc)io(sc)=4.545 Ω(d) Show that the Thévenin equivalen t circuit seen looking into the output is a voltage sourcevo(oc)=1.273vsin series with a resistance rout=4.545 Ω.(e) Show that the Norton equivalen t circuit seen looking into the output is a current sourceio(sc)=0.28vsin parallel with a resistance rout=4.545 Ω.(f) If a load resistor RL=5Ω is connected from the output node to ground, show vo=0.667vs.5. It is given that gm=0.025 S and R1=40kΩ.2(a) With the output terminals open-circuited, use superposition of vsand gmv to write equa-tions for vO(oc)and v. Solve the two equations for vO(oc)as a function of vS.vO(oc)= vS+ gmvR1v =0− gmvR1=⇒ v =0=⇒ vO(oc)= vs(b) For the output terminals short-circuited, use superposition of vSand gmv to write equa-tions for iO(sc)and v. Solve the two equations for iO(sc)as a function of vS.iO(sc)=vSR1+ gmvv= vS− 0=⇒ v = vS=⇒ iO(sc)= vSµ1R1+ gm¶=vS39.96(c) Solve for the Thévenin equivalent circuit looking in to the output terminals.vO(oc)= vSin series with rout=39.96 Ω(d) Solv e for the Norton equivalent circuit looking into the output terminals.iO(sc)=vS39.96in parallel with rout=39.96 Ω6. It is given that β =80, R1=75kΩ,andR2=39kΩ.(a) With the output terminals open-circuited, use superposition of vSand βi to write equa-tions for vO(oc)and i. Solve the two equations for vO(oc)as a function of vSby eliminatingi.vO(oc)=0− βiR2i =−vSR1=⇒ vO(oc)=βR2R1vS=41.6vSV(b) With the output terminals short-circuited, use superposition of vSand βi to write equa-tions for iO(sc)and i. Solve the two equations for iO(sc)as a function of vSby eliminatingi.iO(sc)=0− βi i =−vSR1=⇒ iO(oc)=βR1vS=1.067vSmA(c) Solve for the Thévenin equivalent circuit looking into the output terminals. [vO(oc)=41.6vSin series with 39 kΩ](d) Solve for the Norton equivalen t circuit looking into the output terminals. [iO(sc)=1.067×10−3vSin parallel with 39 kΩ]37. It is given that gm=0.002 S, R1= 100 kΩ,andR2=1MΩ.(a) With the output terminals open-circuited, solve for v as a function of iSand vO(oc)as afunction of v. Solve the t wo equations for vO(oc)as a function of isby eliminating v.vO(oc)= −gmvR2v = iSR1=⇒ vO(oc)= −gmiSR2R1= −2 × 108iS(b) With the output terminals short-circuited, solve for v as a function of iSand iO(sc)as afunction of v. Solve the t wo equations for iO(sc)as a function of iSby eliminating v.iO(sc)= −gmvv= iSR1=⇒ iO(sc)= −gmiSR1= −200iS(c) Show that the Thév enin equivalent circuit looking into the output terminals is a voltagesource vO(oc)= −2 × 108iSin series with a resistance rout=1MΩ](d) Show that the Norton equivalent circuit looking into the output terminals i s a currentsource iO(sc)= −200iSin parallel with a resistance rout=1MΩ.8. It is giv en that R1=3kΩ, R2=2kΩ,andgm=0.1. This problem illustrates two solutionsfor the input resistance to the circuit. In one solution, the source is a voltage source. In theother it is a current source.(a) Use superposition of vSand gmv1to solve for iS.iS=vSR1+ R2− gmv1R2R1+ R2(b) Use superposition of vSand gmv1to solve for v1.v1= vSR1R1+ R2− gmv1R1kR2=⇒ v1= vSR1R1+ R2×11+gmR1kR2(c) Solve the two equations for iSas a function of vSby eliminating v1.iS=vSR1+ R2µ1 −gmR2R1+ R2R11+gmR1kR2¶=vSR1+ R2µ1 −gmR1kR21+gmR1kR2¶=vSR1+ R211+gmR1kR2=vS605 kΩ(d) Solve for the input resistance to the circuit.rin=vSiS= 605 kΩ4(e) Replace vSwith an independent current source iS. Repeat the problem to solve for rin.Which solution is simpler?vS= iS(R1+ R2)+gmv1R2v1= iSR1=⇒ vS= iS(R1+ R2+ gmR1R2)rin=vSiS= R1+ R2+ gmR1R2=605kΩ9. The figure sho w s an amplifier equivalent circuit. It is given that Ai=0.99, RS=1kΩ,R1=25Ω, R2= 100 Ω,andR3=30kΩ.(a) With RL= ∞, use superposition of vsand Aiiαto show that vo(oc)and iαare given byvo(oc)= vsR2RS+ R1+ R2− AiiαµR3+R1RS+ R1+ R2R2¶=vs11.25− 29702.2iαiα=vsRS+ R1+ R2+ AiiαRS+ R2RS+ R1+ R2=vs1125+iα1.03306(b) Solv e the equations to show that vo(oc)= −824.97vs.(c) With RL=0, use superposition of vsand Aiiαto show that io(sc)and iαare given byio(sc)=vsRS+ R1+ R2kR3R2R2+ R3− Aiiαµ1 −RSRS+ R1+ R2kR3R2R2+ R3¶=vs338525−iα1.01309iα=vsRS+ R1+ R2kR3+ AiiαRSRS+ R1+ R2kR3=vs1124.67−iα1.13603(d) Solv e the equations to show that io(sc)= −vs/136.486.(e) Solve for rout= vo(oc)/io(sc).[rout= 112.597 kΩ](f) With RL=10kΩ,solveforvo.[vo= −67.2913vs]510. For the circuit shown, it is given that is=4mA, R1=2kΩ, R2=1kΩ, R3=9kΩ,andgm=0.005 S.(a) Use superposition to solve for