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

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

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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 = 100, min = 50 ). PNP Transistor i C = 1 mA > 0 : BJT is NOT in cut-off i

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SLIDE 1

Problems for BJT Section

Lecture notes: Sec. 3

  • F. Najmabadi, ECE65, Winter 2012
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SLIDE 2
  • F. Najmabadi, ECE65, Winter 2012

Exercise 1: Find state of transistor and its currents/voltages. (Si BJT with β = 100, βmin = 50).

PNP Transistor iC = 1 mA > 0 : BJT is NOT in cut-off iE = 1.2 mA iB = iE − iC = 0.2 mA iC / iB = 1/0.2 = 5 < βmin BJT is in saturation: vCE = Vsat = 0.2 V vBE = VD0 = 0.7 V iB = 10 µA > 0: BJT is NOT in cut-off vEC = 5 V > VD0 = 0.7 V BJT is in active mode: iC = β iB = 1 mA vEB = VD0 = 0.7 V

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SLIDE 3
  • F. Najmabadi, ECE65, Winter 2012

Exercise 2: Compute transistor parameters (Si BJT with β = 100).

C EC B EB B EB

i v i v i v 10 12 : KVL

  • EC

10 40 4 8 10 40 12 : KVL

  • EB

3 3 3

+ = × + = → + × + = incorrect Assumption V 7 . V 4 V 4 10 40 4 : KVL

  • EB

V 7 . and :

  • ff
  • Cut

Assume

3

→ = > = = → + × × = = < =

D EB EB EB D EB B

V v v v V v i A 5 . 82 7 . 10 40 4 : KVL

  • EB

and V 7 . : ON EB

3

> = → + × × = ≥ = = µ

B B B D EB

i i i V v correct Assumption V 7 . V 75 . 3 V 75 . 3 10 25 . 8 10 12 : KVL

  • EC

mA 25 . 8 10 5 . 82 100 V 7 . and : Active Assume

3 3 6

→ = > = = → × × + = = × × = = = ≥ =

− − D EC EC EC B C D EC B C

V v v v i i V v i i β β

PNP Transistor!

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SLIDE 4
  • F. Najmabadi, ECE65, Winter 2012

Exercise 3: Compute transistor parameters (Si BJT with β = 100).

E CE C E BE B

i v i i v i 10 10 12 : KVL

  • CE

10 10 40 4 : KVL

  • BE

3 3 3 3

+ + = + + × = incorrect Assumption V 7 . V 4 V 4 10 10 40 4 : KVL

  • BE

V 7 . and , :

  • ff
  • Cut

Assume

3 3

→ = > = = → × + + × × = = + = = < = =

D BE BE BE C B E D BE C B

V v v v i i i V v i i

Because BE-KVL depends on iE (there is a resistor in the emitter circuit), iB would depend on the state of transistor (active or saturation)e

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SLIDE 5
  • F. Najmabadi, ECE65, Winter 2012

Exercise 3 (cont’d): Compute transistor parameters (Si BJT with β = 100).

E CE C E BE B

i v i i v i 10 10 12 : KVL

  • CE

10 10 40 4 : KVL

  • BE

3 3 3 3

+ + = + + × = mA 36 . 2 mA 34 . 2 10 4 . 23 100 A 4 . 23 7 . 10 ) 101 40 ( 4 101 10 10 40 4 : KVL

  • BE

101 ) 1 ( and V 7 . : ON BE V 7 . and : Active Assume

6 3 3 3

= + = = × × = = = → + × + = × + + × = = + = + = ≥ = = = ≥ =

− C B E B C B B B BE B B B C B E B D BE D CE B C

i i i i i i i i v i i i i i i i V v V v i i β µ β β correct Assumption V 7 . V 3 . 7 V 3 . 7 10 36 . 2 10 10 34 . 2 10 12 : KVL

  • CE

3 3 3 3

→ = > = = → × × + + × × =

− − D CE CE CE

V v v v

It is a very good approximation to set iE ≈ iC in the active mode!

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SLIDE 6
  • F. Najmabadi, ECE65, Winter 2012

Exercise 4: Compute transistor parameters (Si BJT with β = 100).

10 10 10 10 : KVL

  • EC

10 10 : KVL

  • EB

3 3 3

− + + = + =

E EC C EB E

i v i v i PNP Transistor! mA 60 . 4 A . 46 1 ) justified! ON (EB mA 65 . 4 10 2 10 and V 7 . : ON EB V 7 . and : Active Assume

3

= = = + = > = × − = ≥ = = = ≥ =

B C E B EB E B D EB D EC B C

i i i i v i i V v V v i i β µ β β correct Assumption V 7 . V 10 . 6 V 10 . 6 10 60 . 4 10 10 65 . 4 10 2 20 : KVL

  • EC

3 3 3 3

→ = > = = → × × + + × × × =

− − D EC EC EC

V v v v

Since a 10-V supply is in the EB circuit, EB junction is probably ON

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SLIDE 7

If both in active: Q1 & Q2 act as a super-high-β BJT

  • F. Najmabadi, ECE65, Winter 2012

Exercise 5: Find iC2 (Si BJTs with β 1 = 100 and β 2 = 50 ).

Darlington Pair

1 2 1 1 1 2 2 2 2 1 1 1 2 1 1 1

1) ( 1) ( 1) (

B B B C B E B B E

i i i i i i i i i β β β β β β β ≈ + = = + = = + = If Q1 is ON: iE1 > 0 → iB2 > 0 → Q2 is ON! If Q1 is OFF: iE1 = 0 → iB2 = 0 → Q2 is OFF!

2 1 B E

i i =

Darlington Pair: Note: It is possible that one BJT be in active and one in saturation

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SLIDE 8
  • F. Najmabadi, ECE65, Winter 2012

Exercise 5 (cont’d): Find iC2 (Si BJTs with β 1 = 100 and β 2 = 50 ).

Since a 3-V supply is in the BE1+BE2 circuit, both BJTs are probably ON.

2 1 2 2 2 1 1 3 2 1 1 3

: Pair Darlington 470 10 : KVL

  • CE2

10 4.7 10 : KVL

  • CE1

10 470 3 : KVL

  • BE2

BE1

B E CE C BE CE C BE BE B

i i v i v v i v v i = + = + + × = + + × = + ON is Q2 ON Q1 Pair with Darlington A 40 . 3 7 . 7 . 10 470 3 : KVL

  • BE2

BE1 & and V 7 . : ON BEs

1 1 3 2 1 2 1

→ > = → + + × × = + ≥ ≥ = = = µ

B B B B D BE BE

i i i i V v v correct Assumption V 7 . V 70 . 7 V 70 . 7 10 4.7 10 : KVL

  • CE1

mA 340 . 10 40 . 3 100 V 7 . and : Active Q1 Assume

1 1 2 1 1 3 6 1 1 1 1 1 1 1

→ = > = = → + + × = = × × = = = ≥ =

− D CE CE BE CE C B C D CE B C

V v v v v i i i V v i i β β

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SLIDE 9
  • F. Najmabadi, ECE65, Winter 2012

Exercise 5 (cont’d): Find iC2 (Si BJTs with β 1 = 100 and β 2 = 50 ).

2 1 2 2 2 1 1 3 2 1 1 3

: Pair Darlington 470 10 : KVL

  • CE2

10 4.7 10 : KVL

  • CE1

10 470 3 : KVL

  • BE2

BE1

B E CE C BE CE C BE BE B

i i v i v v i v v i = + = + + × = + + × = + V 70 . 7 active) (Q1 mA 340 . A 40 . 3 V 7 .

1 1 1 2 1

= = = = =

CE C B BE BE

v i i v v µ mA 343 . ) 1 (

1 1 1 2

= + = =

B E B

i i i β

From previous slide:

correct Assumption V 7 . V 94 . 1 V 94 . 1 470 10 : KVL

  • CE2

mA 2 . 17 10 343 . 50 V 7 . and : Active Q2 Assume

2 2 2 2 3 2 2 2 2 2 2 2

→ = > = = → + = = × × = = = ≥ =

− D CE CE CE C B C D CE B C

V v v v i i i V v i i β β