CE - CC Amplifier E xam pleFor the circuits in the figure, it is given that V+=10V, V−= −10 V, Rs=5kΩ, R1= 100 kΩ,R2= 120 kΩ, RE1=2kΩ, R3=51Ω, RC=2.4kΩ, RE2=2kΩ, RL=1kΩ, VBE=0.65 V,VT=0.025 V, α =0.99, β =99, rx=20Ω,andr0=50kΩ. The capacitors are ac short circuitsanddcopencircuits.DC SolutionThe dc solution for Q1isthesameasfortheCEamplifier and is repeated. To solve for IE1,replace the capacitors with open circuits. Look out the base and form a Thévenin equivalent circuit.We haveVBB1=V+R2+ V−R1R1+ R2=10120 kΩ100 kΩ +120kΩ− 10100 kΩ100 kΩ + 120 kΩ=1011RBB1= R1kR2= 100 kΩk120 kΩ =54.55 kΩVEE1= V−= −10REE1= RE=2kΩThe emitter current in Q1is given byIE1=VBB1− VBE1− VEE1RBB1/ (1 + β)+REE1=10/11 − 0.65 − (−10)54.55 kΩ/ (1 + 99) + 2 kΩ=4.031 mAThe ac emitter intrinsic resistance of Q1isre1=VTIE1=25 mV4.031 mA=6.202 ΩLook out of the base and emitter of Q2and form Thévenin equivalent circuits. We haveVBB2= V+− αIE1RC1=10− 0.99 × 4.031 mA × 2.4kΩ =0.4223 VRBB2=2.4kΩVEE2= V−= −10 VREE2= RE2=2kΩThe emitter current in Q2is given byIE2=VBB2− VBE2− VEE2RBB2/ (1 + β)+REE2=0.4223 − 0.65 − (−10)2.4kΩ/ (1 + 99) + 2 kΩ=4.828 mA1The ac emitter intrinsic resistance of Q2isre2=VTIE2=25 mV4.828 mA=5.178 ΩAC Solution - Method 1Zero the dc supplies and short the capacitors. Look out the base of Q1and make a Théveninequivalent circuit. We havevtb1= vsR1kR2Rs+ R1kR2= vs100 kΩk120 kΩ5kΩ + 100 kΩk120 kΩ=vs1.092=0.9160vsRtb1= RskR1kR2=5kΩk100 kΩk120 kΩ =4.580 kΩThe Thévenin equivalent circuit looking into the i0e1branc h is vtb1in series with r0e1,wherer0e1=Rtb1+ rx11+β1+ re1=4.580 kΩ +201+99+6.202 = 52.20 ΩThe resistance looking out of the emitter of Q1isRte1= REkR3=2kΩk51 = 49.73 ΩThe resistance looking into the collector of Q1isric1=r01+ r0e1kRte11 − α1Rte1/ (r0e1+ Rte1)=50 kΩ +52.20k49.731 − 0.99 × 49.73/ (49.73 + 52.20)=97.76 kΩThe short circuit collector output current from Q1isic1(sc)= Gmb1vtb1=α1r0e1+ Rte1r01− Rte1/β1r01+ r0e1kRte1vtb1=0.9952.20 + 49.7350 kΩ − 49.73/9950 kΩ +52.20k49.73vtb1=vtb1103.0=vs112.4Look out of the base of Q2and make a Thévenin equivalent circuit. We ha vevtb2= −ic(sc)RCkric1=−vs112.4× 2.4kΩk97.76 kΩ = −20.84vsRtb2= RCkric1=2.4kΩk97.76 kΩ =2.342 kΩReplace Q2with its simplified T model. Looking into the r0e2branch, we see vtb2in series with r0e2given byr0e2=Rtb2+ rx21+β2+ re2=2.342 kΩ +201+99+5.178 = 28.80The resistance seen looking out of the emitter of Q2isRte2= RE2kRL= 666.7 ΩBy voltage division, vois giv en b yvo= vtb2r02kRte2r0e2+ r02kRte2= −20.84vs50 kΩk666.728.80 + 50 kΩk666.7= −19.97vsThus the voltage gain isvovs= −19.972The output resistance isrout= RE2kr02kr0e2=2kΩk50 kΩk28.80 = 28.38 ΩTo solve for the input resistance, we need rib1. To calculate this, we need Rtc1, which requires usto know rib2. For the latter, we haverib2= rx2+(1+β2) re2+ Rte2(1 + β2) r02+ Rtc2r02+ Rte2+ Rtc2= 20 + (1 + 99) 5.178 + 666.7(1 + 99) 50 kΩ50 kΩ + 666.7=66.33 kΩThus the resistance seen looking out of the collector of Q1isRtc1= RCkrib2=2.4kΩk66.33 kΩ =2.316 kΩThe resistance looking into the base of Q1isrib1= rx1+(1+β1) re1+ Rte1(1 + β1) r01+ Rtc1r01+ Rte1+ Rtc1=5.391 kΩThe input resistance isrin= R1kR2krib1= 100 kΩk120 kΩk5.613 kΩ =5.089 kΩIf Q2and RE2are omitted from the circuit and the left node of C2is connected to the collectorof Q1, we have a common-emitter amplifier. In this case, the output voltage isvo= −ic1(sc)RCkric1kRL=−vs112.4× 2.4kΩk97.76 kΩk1kΩ = −6.235voThus the voltage gain drops tovovs= −6.235This is lower than with the CC stage by a factor of 3.25 or b y 10.2 dB. This illustrates how a stagethat has a gain less than unity can increase the gain of a circuit when it is used to drive the loadresistor.AC Solution - Method 2For this solution, we use the r0approximations for Q1. That is, we neglect the current throughr01in calculating ic1(sc)but not in calculating ric1. The short circuit collector output current of Q1isic1(sc)= Gm1vtb1=αr0e1+ Rte1vtb1=0.99vtb152.20 + 49.73=vtb1103.0=vs1111.3Look out of the base of Q2and make a Thévenin equivalent circuit. We ha vevtb2= −ic1(sc)RCkric1=−vs111.3× 2.4kΩk97.76 kΩ = −21.05vsRtb2= RCkric1=2.4kΩk97.76 kΩ =2.342 kΩReplace Q2with its simplified T model. Looking into the r0e2branch, we see vtb2in series with r0e2given byr0e2=Rtb2+ rx21+β2+ re2=2.342 kΩ +201+99+5.178 = 28.803By voltage division, vois giv en b yvo= vtb2r02kRte2r0e2+ r02kRte2= −21.05vs50 kΩk666.728.80 + 50 kΩk666.7= −20.17vsThus the voltage gain isvovs= −20.17This differs from the answer by Method 1 by 0.99%.The solutions for routand rinare the same as for Method