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, 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
F. Najmabadi, ECE65, Winter 2012
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
Exercise 2: Compute transistor parameters (Si BJT with β = 100).
EB - KVL : 12 = vEB + 40 × 103 iB + 8 → 4 = vEB + 40 × 103 iB
EC - KVL : 12 = vEC + 103 iC
Assume Cut - off : iB = 0 and vEB < VD 0 = 0.7 V
EB - KVL : 4 = 40 × 103 × 0 + vEB → vEB = 4 V
vEB = 4 V > VD 0 = 0.7 V → Assumption incorrect
EB ON : vEB = VD 0 = 0.7 V and iB ≥ 0
EB - KVL : 4 = 40 × 103 × iB + 0.7 → iB = 82.5 µ A > 0
Assume Active : iC = β iB and vEC ≥ VD 0 = 0.7 V
iC = β iB = 100 × 82.5 × 10 −6 = 8.25 mA
EC - KVL : 12 = vEC + 103 × 8.25 × 10 −3 → vEC = 3.75 V
vEC = 3.75 V > VD 0 = 0.7 V → Assumption correct
F. Najmabadi, ECE65, Winter 2012
PNP Transistor!
Exercise 3: Compute transistor parameters (Si BJT with β = 100).
BE - KVL : 4 = 40 × 103 iB + vBE + 103 iE
CE - KVL : 12 = 103 iC + vCE + 103 iE
Assume Cut - off : iB = 0, iC = 0 and vBE < VD 0 = 0.7 V
iE = iB + iC = 0
BE - KVL : 4 = 40 × 103 × 0 + vBE + 103 × 0 → vBE = 4 V
vBE = 4 V > VD 0 = 0.7 V → Assumption incorrect
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
F. Najmabadi, ECE65, Winter 2012
Exercise 3 (cont’d): Compute transistor parameters (Si BJT with β = 100).
BE - KVL : 4 = 40 × 103 iB + vBE + 103 iE
CE - KVL : 12 = 103 iC + vCE + 103 iE
Assume Active : iC = β iB and vCE ≥ VD 0 = 0.7 V
BE ON : vBE = VD 0 = 0.7 V and iB ≥ 0
iE = iB + iC = ( β + 1) iB = 101iB
BE - KVL : 4 = 40 × 103 iB + vBE + 103 × 101 iB
4 = (40 +101) × 103 iB + 0.7 → iB = 23.4 µA
iC = β iB = 100 × 23.4 × 10 −6 = 2.34 mA
iE = iB + iC = 2.36 mA
CE - KVL : 12 = 103 × 2.34 × 10 −3 + vCE + 103 × 2.36 × 10 −3 → vCE = 7.3 V
vCE = 7.3 V > VD 0 = 0.7 V → Assumption correct
It is a very good approximation to set iE ≈ iC in the active mode!
F. Najmabadi, ECE65, Winter 2012
Exercise 4: Compute transistor parameters (Si BJT with β = 100).
EB - KVL : 10 = 103 iE + vEB
EC - KVL : 10 = 103 iC + vEC + 103 iE − 10
Since a 10-V supply is in the EB circuit, EB junction is probably ON
Assume Active : iC = β iB and vEC ≥ VD 0 = 0.7 V
EB ON : vEB = VD 0 = 0.7 V and iB ≥ 0
10 − vEB
= 4.65 mA > 0
3
2 × 10
i
iB = E = 46.0 µA
β +1
iC = β iB = 4.60 mA
iE =
(EB ON justified!)
PNP Transistor!
EC - KVL : 20 = 2 × 103 × 4.65 × 10 −3 + vEC + 103 × 4.60 × 10 −3 → vEC = 6.10 V
vEC = 6.10 V > VD 0 = 0.7 V → Assumption correct
F. Najmabadi, ECE65, Winter 2012
Exercise 5: Find iC2 (Si BJTs with β 1 = 100 and β 2 = 50 ).
Darlington Pair:
If Q1 is ON: iE1 > 0 → iB2 > 0 → Q2 is ON!
If Q1 is OFF: iE1 = 0 → iB2 = 0 → Q2 is OFF!
If both in active:
iE1 = ( β1 + 1) iB1
iB 2 = iE1 = ( β1 + 1) iB1
iC 2 = β 2 iB 2 = β 2 ( β1 + 1) iB1 ≈ β1β 2 iB1
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
Darlington Pair
iE 1 = iB 2
Exercise 5 (cont’d): Find iC2 (Si BJTs with β 1 = 100 and β 2 = 50 ).
BE1 + BE2 - KVL : 3 = 470 × 103 iB1 + vBE1 + vBE 2
CE1 - KVL :
10 = 4.7 × 103 iC1 + vCE1 + vBE 2
CE2 - KVL :
10 = 470 iC 2 + vCE 2
Darlington Pair :
iE1 = iB 2
Since a 3-V supply is in the BE1+BE2 circuit,
both BJTs are probably ON.
BEs ON : vBE1 = vBE 2 = VD 0 = 0.7 V and iB1 ≥ 0 & iB 2 ≥ 0
BE1 + BE2 - KVL : 3 = 470 × 103 × iB1 + 0.7 + 0.7 → iB1 = 3.40 µ A > 0
Darlington Pair with Q1 ON → Q2 is ON
Assume Q1 Active : iC1 = β 1iB1 and vCE1 ≥ VD 0 = 0.7 V
iC1 = β 1iB1 = 100 × 3.40 × 10 −6 = 0.340 mA
CE1 - KVL : 10 = 4.7 × 103 iC1 + vCE1 + vBE 2 → vCE1 = 7.70 V
vCE1 = 7.70 V > VD 0 = 0.7 V → Assumption correct
F. Najmabadi, ECE65, Winter 2012
Exercise 5 (cont’d): Find iC2 (Si BJTs with β 1 = 100 and β 2 = 50 ).
BE1 + BE2 - KVL : 3 = 470 × 103 iB1 + vBE1 + vBE 2
CE1 - KVL :
10 = 4.7 × 103 iC1 + vCE1 + vBE 2
CE2 - KVL :
10 = 470 iC 2 + vCE 2
Darlington Pair :
iE1 = iB 2
From previous slide:
vBE1 = vBE 2 = 0.7 V
iB1 = 3.40 µ A
iC1 = 0.340 mA (Q1 active)
vCE1 = 7.70 V
iB 2 = iE1 = ( β 1+ 1)iB1 = 0.343 mA
Assume Q2 Active : iC 2 = β 2iB 2 and vCE 2 ≥ VD 0 = 0.7 V
iC 2 = β 2iB 2 = 50 × 0.343 × 10 −3 = 17.2 mA
CE2 - KVL : 10 = 470 iC 2 + vCE 2 → vCE 2 = 1.94 V
vCE 2 = 1.94 V > VD 0 = 0.7 V → Assumption correct
F. Najmabadi, ECE65, Winter 2012