Problems for BJT Section Lecture notes: Sec. 3 F. Najmabadi, ECE65, - PowerPoint PPT Presentation

Problems for BJT Section Lecture notes: Sec. 3 F. Najmabadi, ECE65, Winter 2012

Exercise 1: Find state of transistor and its currents/voltages. (Si BJT with β = 100, β min = 50 ). PNP Transistor i C = 1 mA > 0 : BJT is NOT in cut-off i B = 10 µ A > 0 : BJT is NOT in cut-off i E = 1.2 mA v EC = 5 V > V D 0 = 0.7 V i B = i E − i C = 0.2 mA BJT is in active mode: i C / i B = 1/0.2 = 5 < β min i C = β i B = 1 mA v EB = V D 0 = 0.7 V BJT is in saturation: v CE = V sat = 0.2 V v BE = V D 0 = 0.7 V F. Najmabadi, ECE65, Winter 2012

Exercise 2: Compute transistor parameters (Si BJT with β = 100 ). = + × + → = + × 3 3 EB - KVL : 12 40 10 8 4 40 10 v i v i EB B EB B = + 3 EC - KVL : 12 v 10 i EC C = < = Assume Cut - off : i 0 and v V 0 . 7 V 0 B EB D = × × + → = 3 EB - KVL : 4 40 10 0 v v 4 V EB EB = > = → v 4 V V 0 . 7 V Assumption incorrect EB D 0 = = ≥ EB ON : 0 . 7 V and 0 v V i PNP Transistor! EB D 0 B = × × + → = µ > 3 EB - KVL : 4 40 10 i 0 . 7 i 82 . 5 A 0 B B = β ≥ = Assume Active : and 0 . 7 V i i v V C B EC D 0 = β = × × − = 6 100 82 . 5 10 8 . 25 mA i i C B = + × × − → = 3 3 EC - KVL : 12 v 10 8 . 25 10 v 3 . 75 V EC EC = > = → 3 . 75 V 0 . 7 V Assumption correct v V EC D 0 F. Najmabadi, ECE65, Winter 2012

Exercise 3: Compute transistor parameters (Si BJT with β = 100 ). = × + + 3 3 BE - KVL : 4 40 10 10 i v i B BE E = + + 3 3 CE - KVL : 12 10 i v 10 i C CE E = = < = Assume Cut - off : 0 , 0 and 0 . 7 V i i v V B C BE D 0 = + = i i i 0 E B C = × × + + × → = 3 3 BE - KVL : 4 40 10 0 v 10 0 v 4 V BE BE = > = → v 4 V V 0 . 7 V Assumption incorrect BE D 0 Because BE-KVL depends on i E (there is a resistor in the emitter circuit), i B would depend on the state of transistor (active or saturation)e F. Najmabadi, ECE65, Winter 2012

Exercise 3 (cont’d): Compute transistor parameters (Si BJT with β = 100 ). = × + + 3 3 BE - KVL : 4 40 10 10 i v i B BE E = + + 3 3 CE - KVL : 12 10 i v 10 i C CE E = β ≥ = Assume Active : i i and v V 0 . 7 V C B CE D 0 = = ≥ BE ON : 0 . 7 V and 0 v V i BE D 0 B = + = β + = i i i ( 1 ) i 101 i E B C B B = × + + × 3 3 BE - KVL : 4 40 10 i v 10 101 i B BE B = + × + → = µ 3 4 ( 40 101 ) 10 0 . 7 23 . 4 A i i B B = β = × × − = 6 100 23 . 4 10 2 . 34 mA i i C B = + = i i i 2 . 36 mA E B C − − = × × + + × × → = 3 3 3 3 CE - KVL : 12 10 2 . 34 10 10 2 . 36 10 7 . 3 V v v CE CE = > = → v 7 . 3 V V 0 . 7 V Assumption correct CE D 0 It is a very good approximation to set i E ≈ i C in the active mode! F. Najmabadi, ECE65, Winter 2012

Exercise 4: Compute transistor parameters (Si BJT with β = 100 ). = + 3 EB - KVL : 10 10 i v E EB = + + − 3 3 EC - KVL : 10 10 i v 10 i 10 C EC E Since a 10-V supply is in the EB circuit, EB junction is probably ON = β ≥ = Assume Active : i i and v V 0 . 7 V C B EC D 0 = = ≥ EB ON : v V 0 . 7 V and i 0 0 EB D B − 10 v = = > EB i 4 . 65 mA 0 (EB ON justified! ) × E 3 2 10 i = = µ E 46 . 0 A i β + PNP Transistor! B 1 = β = i i 4 . 60 mA C B = × × × − + + × × − → = 3 3 3 3 EC - KVL : 20 2 10 4 . 65 10 v 10 4 . 60 10 v 6 . 10 V EC EC = > = → v 6 . 10 V V 0 . 7 V Assumption correct EC D 0 F. Najmabadi, ECE65, Winter 2012

Exercise 5: Find i C 2 (Si BJTs with β 1 = 100 and β 2 = 50 ). Darlington Pair: If Q1 is ON: i E1 > 0 → i B2 > 0 → Q2 is ON! If Q1 is OFF: i E1 = 0 → i B2 = 0 → Q2 is OFF! If both in active: = β + ( 1) i i E 1 1 B 1 Darlington Pair = = β + ( 1) i i i B 2 E 1 1 B 1 = = β = β β + ≈ β β i i ( 1) i i i 1 i C 2 2 B 2 2 1 B 1 1 2 B 1 E B 2 Q1 & Q2 act as a super-high- β BJT Note: It is possible that one BJT be in active and one in saturation F. Najmabadi, ECE65, Winter 2012

Exercise 5 (cont’d): Find i C 2 (Si BJTs with β 1 = 100 and β 2 = 50 ). + = × + + 3 BE1 BE2 - KVL : 3 470 10 i v v B 1 BE 1 BE 2 = × + + 3 CE1 - KVL : 10 4.7 10 i v v C 1 CE 1 BE 2 = + CE2 - KVL : 10 470 i v 2 2 C CE = Darlington Pair : i i E 1 B 2 Since a 3-V supply is in the BE1+BE2 circuit, both BJTs are probably ON. = = = ≥ ≥ BEs ON : 0 . 7 V and 0 & 0 v v V i i BE 1 BE 2 D 0 B 1 B 2 + = × × + + → = µ > 3 BE1 BE2 - KVL : 3 470 10 i 0 . 7 0 . 7 i 3 . 40 A 0 B 1 B 1 → Darlington Pair with Q1 ON Q2 is ON = β ≥ = Assume Q1 Active : and 0 . 7 V i i v V C 1 1 B 1 CE 1 D 0 = β = × × − = 6 i i 100 3 . 40 10 0 . 340 mA C 1 1 B 1 = × + + → = 3 CE1 - KVL : 10 4.7 10 i v v v 7 . 70 V 1 1 2 1 C CE BE CE = > = → 7 . 70 V 0 . 7 V Assumption correct v V CE 1 D 0 F. Najmabadi, ECE65, Winter 2012

Exercise 5 (cont’d): Find i C 2 (Si BJTs with β 1 = 100 and β 2 = 50 ). + = × + + 3 BE1 BE2 - KVL : 3 470 10 i v v B 1 BE 1 BE 2 = × + + 3 CE1 - KVL : 10 4.7 10 i v v C 1 CE 1 BE 2 = + CE2 - KVL : 10 470 i v 2 2 C CE = Darlington Pair : i i E 1 B 2 = = From previous slide: 0 . 7 V v v BE 1 BE 2 = µ i 3 . 40 A B 1 = i 0 . 340 mA (Q1 active) C 1 = 7 . 70 V v CE 1 = = β + = i i ( 1 ) i 0 . 343 mA 2 1 1 1 B E B = β ≥ = Assume Q2 Active : and 0 . 7 V i i v V C 2 2 B 2 CE 2 D 0 = β = × × − = 3 i i 50 0 . 343 10 17 . 2 mA C 2 2 B 2 = + → = CE2 - KVL : 10 470 1 . 94 V i v v C 2 CE 2 CE 2 = > = → v 1 . 94 V V 0 . 7 V Assumption correct CE 2 D 0 F. Najmabadi, ECE65, Winter 2012