Problem for Problems for Discrete Transistor Amplifiers ECE 65, - PowerPoint PPT Presentation

Problem for Problems for Discrete Transistor Amplifiers ECE 65, Winter2013, F. Najmabadi

Exercise 1: Find the amplifier parameters of the circuit below. (Si BJT with β = 200, V A = 150 V, ignore Early effect in bias calculations). When R sig & v sig are not given, it implies that v sig = v i and R sig = 0  This is a common collector amplifier (emitter follower) . o Input at the base, output at the emitter.  It has a emitter-degeneration bias with a voltage divider. F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (2/42)

Exercise 1 (cont’d): β = 200, V A = 150 V. Thevenin form of the Real circuit Bias circuit Voltage divider = = 22 k || 18 k 9 . 90 k R B 22 k = × = 9 4 . 95 V V + BB 22 k 18 k F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (3/42)

Exercise 1 (cont’d): β = 200, V A = 150 V. Thevenin form of the Bias Calculations Voltage divider = = 9 . 9 k 4 . 95 V R V B BB Assume BJT is in Active : = > ≥ 0 . 7 V, 0 and 0 . 7 V V I V BE C CE − = + + BE KVL : 4.95 I R V I R B B BE E E = × β + + + 3 3 4.95 9 . 9 10 /( 1 ) 0 . 7 10 I I E E = ≈ 4 . 05 mA I I E C = β = µ / 20 . 3 A I I × 3 - B C 4.05 10 I = = = C 156 mA/V g × m - 3 26 10 V − = + CE KVL : 9 V I R T CE E E 150 V = + × − × 3 3 ≈ = = 9 4 10 10 V r A 37 . 0 k CE o - 3 4 x 10 I = > = 5 V 0.7 V C V V 0 CE D β V = = = r T 1 . 28 k π I g B m F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (4/42)

Exercise 1 (cont’d): β = 200, V A = 150 V. = = = 156 mA/V, 38 . 8 k, 1 . 28 k g r r π m o Amplifier Parameters Emitter Follower ( || || ) v g r R R = o m o E L + 1 ( || || ) v g r R R i m o E L = = Ω ( || || ) 38 . 8 k || 1 k || 100 k 965 r R R o E L − = × × = 3 ( || || ) 156 10 965 151 g r R R m o E L 151 v = ≈ o 1 + Signal circuit 1 151 v i (IVC = 0) [ ] = + β || ( || || ) R R r r R R π i B o E L [ ] = + ≈ 9 . 9 k || 1 . 3 k 194 k 9 . 9 k R i + || r R R r π = = = Ω π B sig || || 6 . 4 R R R β β o E E v R v v = × = = o i o o 1 + v R R v v sig i sig i i F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (5/42)

Exercise 1 (cont’d): β = 200, V A = 150 V. Amplifier Cut-off frequency 2 caps: 2 poles 1 = f 1 π + p 2 ( ) R R C i sig c 1 1 = = 34 . 2 Hz f − π × + × × p 1 3 6 2 ( 9 . 9 10 0 ) 0 . 47 10 1 = f π + 2 p 2 ( ) R R C 2 o L c = 156 mA/V g 1 m = = = 3 . 39 Hz f = 9 . 9 k R 38 . 8 k r π + × × × − p 2 3 6 2 (6.4 100 10 ) 0 . 47 10 i o = Ω = 6 . 4 R 1 . 28 k r π o ≈ + = + = 34 . 2 3 . 4 37 . 6 Hz f f f 1 2 p p p F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (6/42)

Exercise 2: Find the amplifier parameters of the circuit below. (Si BJT with β = 200, V A = 150 V, ignore Early effect in bias calculations).  This is a common emitter amplifier with R E . o Input at the base, output at the collector.  It has a emitter-degeneration bias with a voltage divider. F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (7/42)

Exercise 2 (cont’d): β = 200, V A = 150 V Thevenin form of the Bias circuit Bias calculations Voltage divider Caps open = = 34 k || 5 . 9 k 5 . 0 k R B 5 . 9 k = × = 15 2 . 22 V V + BB 34 k 5 . 9 k Assume BJT is in Active : = > ≥ 0 . 7 V, 0 and 0 . 7 V V I V BE C CE − = + + BE KVL : 2.22 I R V I R B B BE E E = × β + + + 3 2.22 5 . 0 10 /( 1 ) 0 . 7 510 I I E E = ≈ 2 . 84 mA I I × E C 3 - 2.84 10 I = = = = β = µ C 10.9 mA/V g / 14 . 2 A I I × m - 3 B C 26 10 V T − = + + 150 V CE KVL : 15 I R V I R ≈ = = r A 52 . 8 k C C CE E E o × - 3 2 . 84 10 I − = + × × + 3 3 15 2 . 84 10 ( 10 510 ) V C CE β V = > = = = = 10 . 5 V 0.7 V V V r T 1 . 83 k π CE D 0 I g B m F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (8/42)

Exercise 2 (cont’d): β = 200, V A = 150 V Amplifier Parameters CE amplifier with R E ( || ) v g R R = − o m C L + + + 1 [( || ) / ]( 1 / ) v g R R R r R r π i m E C L o E = = || 1 k || 100 k 0.990k R R C L v = − o 1 . 64 v i   β R = + Signal circuit E ||   R R r π + + i B  1 [( || ) / ]( 1 / )  R R r R r (IVC = 0) π C L o E = = 5 . 0 k || 104k 4 . 8 k R i     β R   ≈ +   E || 1 R R r   + + o C o   || r R R R     π E B sig ≈ = 1 . 0 k R R o C = 10.9 mA/V, g m 4 , 800 v R v = × = × − = − = = o i o ( 1 . 64 ) 1 . 59 52 . 8 k, 1 . 83 k r r + + π 4 , 800 100 o v R R v sig i sig i F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (9/42)

Exercise 2 (cont’d): β = 200, V A = 150 V Amplifier Cut-off frequency (2 caps: 2 poles) 1 = f π + p 1 2 ( ) R R C i sig c 1 1 = = 6 . 91 Hz f − π × + × × p 1 3 6 2 ( 4 . 8 10 100 ) 4 . 7 10 1 = = 10.9 mA/V g f π + m 2 p 2 ( ) R R C = = 2 o L c r 52 . 8 k r 1 . 83 k π o 1 = = = = 15 . 8 Hz f 4 . 8 k 1 . 0 k R R π + × × × − 2 p 3 3 9 i o 2 ( 10 100 10 ) 100 10 ≈ + = + = 6 . 9 15 . 8 22 . 7 Hz f f f p p 1 p 2 F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (10/42)

Exercise 3: Find the amplifier parameters of the circuit below. (Si BJT with β = 200, V A = 150 V, ignore Early effect in bias calculations).  This is a PNP common emitter amplifier (no R E ). o Input at the base, output at the collector ( R E is shorted out for signal because of the by-pass capacitor)  It has a emitter-degeneration bias with two voltage sources ( R B = ∞ ) F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (11/42)

Exercise 3 (cont’d). β = 200, V A = 150 V. Assume BJT is in Active : = > ≥ 0 . 7 V, 0 and 0 . 7 V V I V EB C EC − = × + + 3 BE KVL : 3 2 . 3 10 100 I V I E EB B = ≈ 1 . 0 mA I I E C = β = µ / 5 . 0 A I I B C − = × + + × − 3 3 CE KVL : 3 2 . 3 10 2 . 3 10 3 I V I E EC C Bias circuit = − × − × 3 3 6 4 . 6 10 10 V (Signal = 0, Caps open ) EC = > = 1 . 4 V 0.7 V V V 0 EC D -3 10 I = = = C 38.5 mA/V g × m - 3 26 10 V T 150 V ≈ = = r A 150 k o - 3 10 I C β V = = = r T 5 . 26 k π I g B m F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (12/42)

Exercise 3 (cont’d). β = 200, V A = 150 V. Amplifier Parameters 1) CE amplifier with no R E 2) No voltage divider biasing: no R B R B is open circuit, R B = ∞ v = − o ( || || ) g r R R m o C L v i v = − × − = − 3 o 38.5 10 ( 150 k || 2 . 3 k || 100 k ) 85 . 3 v Signal circuit i (IVC = 0) = = = || 5 . 26 k R R r r π π i B = || R R r o C o = = 150 k || 2 . 3 k 2 . 27 k R o v R v v = × ≈ = − o i o o 85 . 3 + v R R v v sig i sig i i = = = 38.5 mA/V, 150 k, 5 . 26 k g r r << π m o R R sig i F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (13/42)

Exercise 3 (cont’d). β = 200, V A = 150 V. Amplifier Cut-off frequency Signal circuit 1) 2 caps: 2 poles (IVC = 0) 2) no coupling capacitors at the input 1 = = 0 f π + p 1 2 ( ) R R C i sig c 1 1 = f 2 π + p 2 ( ) R R C 2 o L c 1 = = 15 . 8 Hz f − 2 π × + × × p 3 5 9 2 (2.3 10 10 ) 100 10 1 = f 3 π + β p 2 { || [ 1 / ( || ) / ]} R g R R C E m B sig E = 38.5 mA/V g = 5 . 26 k R 1 m = = i 129 Hz f = r 150 k = − 3 π × × × p 3 6 2 . 27 k 2 (2.3 10 || 26 . 5 ) 47 10 R o o = r 5 . 26 k π ≈ + + = + + = 0 15 . 8 129 145 Hz f f f f 1 2 3 p p p p F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (14/42)

Exercise 4: Find the amplifier parameters of this circuit. ( µ n C ox = 100 µ A/V 2 , ( W/L ) = 6/0.1, V t = 0.5 V, λ = 0.1 V -1 . Assume capacitors are large and ignore channel width modulation in biasing. )  This is a MOS common source amplifier (no R S ). o Input at the gate, output at the drain ( R S is shorted out for the signal because of the by-pass capacitor)  It has a source-degeneration bias with a voltage divider. F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (15/42)