Common Base BJT Amplifier Common Collector BJT Amplifier

Common Base BJT Amplifier Common Collector BJT Amplifier Common Collector (Emitter Follower) Configuration Common Base Configuration Small Signal Analysis Design Example Amplifier Input and Output Impedances 1

Basic Single BJT Amplifier Features CE Amplifier CC Amplifier CB Amplifier Voltage Gain (A V ) moderate (-R C /R E ) low (about 1) high Current Gain (A I ) moderate ( ) moderate ( 1) low (about 1) Input Resistance high high low Output Resistance high low high CE BJT amplifier => CS MOS amplifier CC BJT amplifier => CD MOS amplifier CB BJT amplifier => CG MOS amplifier 2

Common Collector ( Emitter Follower) Amplifier In the emitter follower, the output voltage is taken between emitter and ground. The voltage gain of this amplifier is nearly one the output follows the input - hence the name: emitter follower. 3

Emitter Follower Biasing i B Vb i E i C Then, choose/specified I E, and the rest of the design follows: R E = V E I E = V CC/2 0.7 I E Split bias voltage drops about equally across the transistor V CE (or V CB ) and V Re (or V B ). For simplicity,choose: V B = V CC 2 R 1 =R 2 For an assumed = 100: As with CE bias design, stable op. pt. => R B 1 R E, i.e. R B =R 1 R 2 = R 1 2 = 1 R E 10 10 R E R 1 =R 2 =20 R E 4

Typical Design Choose: I E =1mA V CC =12V And the rest of the design follows immediately: R E = V E I E = 12/2 0.7 10 3 =5.3k Use standard sizes! R E =5.1k R 1 =R 2 =100 k i B Vb i E i C 5

Equivalent Circuits R B =R 1 R 2 <=> Rb V CC /2 6

Multisim Bias Check I E V Rb =I B R B = 1 R B=0.495V i B <=> VRb - + Rb Identical results as expected! 7

Emitter Follower Small Signal Circuit Mid-band equivalent circuit: v sig ' = R B v R B R sig = 50 S 50.05 v v sig sig R TH =R S R B = 50 50.05 R S R S Rb Small signal mid-band circuit - where C B has negligible reactance (above ). Thevenin circuit consisting of R and R shows min S B effect of R B negligible, since it is much larger than R S. 8

Follower Small Signal Analysis - Voltage Gain Circuit analysis: v sig = R S r 1 R E i b i b Solving for i b i e i b = v sig R S r 1 R E v out = R E 1 v sig R S r 1 R E A V = v out v sig = R E v sig 1 R S r 1 R E 9

Small Signal Analysis Voltage Gain - cont. v out v sig = R E R S r 1 R E Since, typically: R S r 1 R E A V = v out R E =1 v sig R E Note: A V is non-inverting 10

Blocking Capacitor - C B - Selection + i b Use the base current expression: vbg Rb - v bg =r i b R E i E = r 1 i b i b = v bg r 1 R E R B =50 k R S R S =50 R in r bg = v bg i b =r 1 R E 1 R E =101 5.1 k=515k To obtain the base to ground resistance of the transistor: This transistor input resistance is in parallel with the 50 k R B, forming the total amplifier input resistance: R i n =R S R B r bg R B r bg = 515 50 k =45.6 k 515 50 11

C B Selection cont. Choose C B such that its reactance is 1/10 of R in at : min 1 = R i n C B 10 C B 10 min R i n C B 10 10 7 1.25 0.46 1.73 F Assume the lowest frequency is 20 Hz: min =2 20 125=1.25 10 2 =100 R i n 46 k Pick C B = 2 F (two 1 F caps in parallel), the nearest standard value in the RCA Lab. We could be (unnecessarily) more precise and include R s as part of the total resistance in the loop. It is very small compared to R in. 12

Final Design 2.0 uf 13

Multisim Simulation Results 20 Hz. Data 1 Khz. Data 14

Of What value is a Unity Gain Amplifier? i b i e To answer this question, we must examine the output impedance of the amplifier and its power gain. 15

Emitter Follower Output Resistance i b 0 i x R B =50 k R S Assume: v x R out I C =1 ma r = V T = V T =2500 I B I C =100 R S =50 i x = i b i b = 1 i b i b = i x 1 v x = i b Rs r = R S r 1 i x R out = v x = R S r i x 1 r 1 R out is the Thevenin resistance looking into the open-circuit output. R out 2550 100 =25.5 16

Multisim Verification of R out Rin Rout Thevenin equivalent for the short-circuited emitter follower. If was 200, as for most good NPN transistors, R out would be lower - close to 12. i sc =i x Av*vsig A V = 1 <=> + R in =R S r 1 R E R L 1 R L v oc =A V v sig - i sc =i x Multisim short circuit check ( = 100, v out = v sig ): i x = 1 i x v sig = R S i b r i b R out = A V v sig = R r S i x 1 R out = v oc i sc = A V v sig rms i sc rms = 1 0.0396 =25.25 17

Equivalent Circuits with Load R L i b i e RL Z in = v sig i e ' <=> Rin + - Rout Av*vsig + vload - RL Re R in =R S r 1 R E R L 1 R L R out = v sig rms i sc rms = 1 0.0396 =25.25 18

Emitter Follower Power Gain Consider the case where a R L = 50 load is connected through an infinite capacitor to the emitter of the follower we designed. Using its Thevenin equivalent: vsig isig Rin v load = R L A V v sig R L R out = 50 75 v sig= 2 3 v sig i load = A V v sig = v sig R out R L 75 p load =v load i load = 2 225 v 2 sig + - Rout 25 iload Av*vsig vth=g vsig C= + vload - R L 50 50 load is in parallel with 5.1k R E and dominates: i sig =i b = v sig R in A V 1 v sig 1 R E R L p sig =v sig i sig 1 5000 v 2 sig v sig 101 50 G pwr = p load = 2 5000 =44.4 1 p sig 225 19

The Common Base Amplifier Voltage Bias Design Current Bias Design 20

Common Base Configuration Both voltage and current biasing follow the same rules as those applied to the common emitter amplifier. As before, insert a blocking capacitor in the input signal path to avoid disturbing the dc bias. The common base amplifier uses a bypass capacitor or a direct connection from base to ground to hold the base at ground for the signal only! The common emitter amplifier (except for intentional R E feedback) holds the emitter at signal ground, while the common collector circuit does the same for the collector. 21

Voltage Bias Common Base Design 47k Ohm 4.7 k Ohm 5.1k Ohm 470 Ohm We keep the same bias that we established for the gain of 10 common emitter amplifier. All that we need to do is pick the capacitor values and calculate the circuit gain. 22

i' b Common Base Small Signal Analysis - C i b i' e i e 4.7 k Ohm i c I' c Determine C IN : (let C B = ) IN Find a equivalent impedance for the input circuit, R S, C IN, and R E2 : i sig 470 Ohm v Re2 = R E2 r e R E2 r e R S 1 j C IN v sig r e = r 1 Z in r e R E2 r e ideally v Re2 = v for R min E2 r e R sig S 1 R min C S R E2 r e 1 = R S r e 10 C IN min C IN 10 IN = min R S r e 23

Determine C IN cont. A suitable value for C IN for a 20 Hz lower frequency: 10 min C IN R S r e 1 C IN 2 min R S r e = 10 2 20 75 F or C IN = 10 125.6 75 1062 F! Not too practical! Must choose smaller value of C IN. 1. Choose: min C IN R S r e =1 2. Choose larger min 24

i b i' b R E2 >> R S i' e Small-signal Analysis - C i' c ib' i c B Determine C B : (let C IN = ) Note the reference current reversals (due to v sig polarity)! v sig =R S i ' r 1 e j C B v sig =R S i ' r e 1 j C B i ' b i e i e ' 1 Determine Z in Z in = v sig i e ' i e ' = 1 1 R S r 1 j C B v sig 25

Determine C B i' b i' c ib' i e ' = 1 1 R S r 1 j C B v sig i' e i e ' = v sig R E2 >> R S R S 1 1 r 1 j C B Z in Z in = v sig i e ' =R S 1 1 r 1 j C B ideally Z in R S r 1 1 C B 1 R S r for min 26

Determine - C B cont. i' b i' c ib' For min Z in R S r 1 1 C B 1 R S r R E2 >> R S i' e Choose: C B 10 min 1 R S r F i.e. 10 C B 2 20 100 1 50 2500 =10.5 F 27

Small-signal Analysis Voltage Gain R E2 >> R S ib' i e ' 1 R S r 1 v sig = 1 R S r e v sig ' v out =R C i c = R C i ' e = 1 R C R S r e v sig Assume: C B =C IN = A V = v out = R C = 100 v sig 1 R S r e 101 5100 50 25 67 28

Multisim Simulation 1062 uf 29

Multisim Frequency Response 20 Hz. response 1 KHZ. Response 30