ECE 123
Spring 2025
Solved Practice Problems Set #5
DC Analysis of BJT Circuits – part 1
Version 1.0
Posted 2/23/2025
Practice Problem #1
In the following circuit, assume that  is 80 when the transistor in in forward active mode
and that the base emitter junction has a voltage drop of 0.7 V, i.e., VBE = 0.7 V, when it is
forward biased. Find the emitter, collector and base voltages and currents.
VE
VB
VC
IE
IB
IC
Solution Practice Problem #1
In the following circuit, assume that  is 80 when the transistor in in forward active mode
and that the base emitter junction has a voltage drop of 0.7 V, i.e., VBE = 0.7 V, when it is
forward biased. Find the emitter, collector and base voltages and currents.
VE
0.8 V
VB
1.5 V
VC
1.62 V
IE = VE / 4.7k = 0.17 mA
IE
0.17 mA
IB = IE / (+) =  
IB
2.1 A
IC = [   (+)  IE =  m
IC
0.168mA
Assuming Forward Active:
VB = 1.5 V as connected directly to
a DC supply of +1.5 V.
VE = VB – 0.7 V = 0.8 V
VC = 3 V – IC 8.2k = 1.62 V
Confirming Forward Active:
VC-VE = 1.62V - 0.8V = 0.82V > 0.3V
Practice Problem #2
For the circuit shown below assume that  is 100 when the device is in forward active mode; VCE
= 0.2 V when the device is in saturation; and that VBE = 0.7 V when the base-emitter diode is
forward biased. Find the corresponding voltages and currents at the base, emitter, and collector
when: a) RB = 100 k, and b) RB = 10 k.
a)
b)
VB
IB
VB
IB
VE
IE
VE
IE
VC
IC
VC
IC
Solution Practice Problem #2
For the circuit shown below assume that  is 100 when the device is in forward active mode; VCE
= 0.2 V when the device is in saturation; and that VBE = 0.7 V when the base-emitter diode is
forward biased. Find the corresponding voltages and currents at the base, emitter, and collector
when: a) RB = 100 k, and b) RB = 10 k.
Following similar approach as part (a)
it can be shown that BJT is not in Active.
Assuming Saturation:
+5V = IBRB + 0.7V + IE 1k (1)
+5V = IC1k +VCE +IE 1k = IC1k +0.2V+IE 1k (2)
IE = IC + IB (3)
Assuming Forward Active:
+5V = IBRB + 0.7V + IE 1k =
= IBRB + 0.7V + 101 IB 1k
Solving for IB we get:
IB = (5V - 0.7V) / (100k +101k) =
= 0.0214 mA
VB=5 V - IB 100k = 2.86 V
Then:
IC = 100 IB =2.14 mA and
VC = 5V – IC 1 k = 2.86 V
and
IE = 101 IB =2.16 mA and
VE = IE 1 k = 2.16 V
Three equations: (1), (2), (3)
Three unknowns: IE , IC , IB
Solving for the three currents we get:
IB = 0.18 mA,
IC = 2.31 mA
IE = 2.49 mA
Then: VB= 5 V - IB 10k = 3.2 V
VC = 5 V – IC 1 k = 2.69 V
VE = IE 1 k = 2.49 V
Confirming Forward Active:
VC-VE=2.9V-2.12V=0.78V > 0.3V
a)
Confirming Saturation:
b)
VC-VE = 2.69V - 2.49V = 0.2V < 0.3V
VB
2.86 V
IB
21.4 A
VB
3.2 V
IB
0.18 mA
VE
2.16 V
IE
2.16 mA
VE
2.49 V
IE
2.49 mA
VC
2.86 V
IC
2.14 mA
VC
2.69 V
IC
2.31 mA
Practice Problem #3
The transistor amplifier in the circuit shown below is biased with a current source I where I = 0.2 mA. This BJT
transistor has a beta of  = 100 and a voltage drop VBE = 0.7 V when the base-emitter diode is forward biased.
Calculate all the dc bias currents and voltages and fill the Tables below.
Hint: Recall that when doing the DC analysis all capacitors act as open circuits.
IC
VC
IB
VB
VE
IE
VB
IB
VE
IE
VC
IC
Solution Practice Problem #3
The transistor amplifier in the circuit shown below is biased with a current source I where I = 0.2 mA. This BJT
transistor has a beta of  = 100 and a voltage drop VBE = 0.7 V when the base-emitter diode is forward biased.
Calculate all the dc bias currents and voltages and fill the Tables below.
Hint: Recall that when doing the DC analysis all capacitors act as open circuits.
VB = 0 V as base is connected directly to ground.
IC
Assume Forward Active:
VC
VE = VB – 0.7 V = 0 − 0.7 V = − 0.7 V
IB
IE = 0.2 mA as emitter current is in series with the
current set of the 0.2 mA current source and thus
the two currents must be equal.
VB
VE
IE
IB = IE / (+) =  
IC =  IB =  m
VC = 3 V – IC 10k = 3 V – 1.98 V = 1.02 V
VB
0V
IB
1.98 A
VE
- 0.7 V
IE
0.2 mA
VC − VE = 1.02V − (−0.7V) = 1.72V > 0.3V
VC
1.02 V
IC
0.198 mA
Since VCE is greater than 0.3 V then BJT is in
Forward Active!
Confirm Forward Active:
Practice Problem #4
V1
V2
V5
V6
V7
Practice Problem #4
Practice Problem #5
A BJT amplifier circuit is shown below, the power supply VCC is 15 V, R1 =27 k, R2 = 15 k, RE =
3.9 k and RC = 3.9 k. The transistor has a beta of  = 100 and a voltage drop VBE =0.7 V when
the base-emitter diode is forward biased. Calculate all dc currents and voltages and fill the
Tables below. Hint: Recall that when doing the DC analysis all capacitors act as open circuits.
VT
VCC
RT
VB
IB
VE
IE
VC
IC
Solution Practice Problem #5
A BJT amplifier circuit is shown below, the power supply VCC is 15 V, R1 =27 k, R2 = 15 k, RE =
3.9 k and RC = 3.9 k. The transistor has a beta of  = 100 and a voltage drop VBE =0.7 V when
the base-emitter diode is forward biased. Calculate all dc currents and voltages and fill the
Tables below. Hint: Recall that when doing the DC analysis all capacitors act as open circuits.
VT
VCC
One can replace the R1 and R2 bias resistors with a
Thevenin equivalent circuit as shown, where:
RT = R1//R2 = 9.64 k and
RT
VT = VCC [R2/(R1+R2)]= 5.36 V
Assuming Forward Active:
+5.36 V = IBRT + 0.7V + IE 3.9k =
= IBRT + 0.7V + 101 IB 3.9k
Solving for IB = 0.0115 mA
IE = IB (+) =  mA
IC =  IB =  mA
VE = IE 3.9 k = 4.53 V
VB
5.23 V
IB
0.0115 mA
VB = VE + 0.7 V = 5.23 V
VE
4.53 V
IE
1.161 mA
VC = VCC − IC 3.9 k = 10.51 V
VC
10.51 V
IC
1.15 mA
Confirming Forward Active:
VC − VE = 10.51 V − 4.53 V = 5.98 V > 0.3V
Practice Problem #6
The transistor in the circuit below has a  of 100 and and a base emitter voltage drop of 0.7 V when the base
emitter diode is forward biased. Determine the VCE voltage. Consider that when a diode is forward biased the
voltage drop across it is 0.7 V.
HINT: Determine VC and VE separately and then calculate the difference VCE = VC - VE.
VC
VE
VCE
IC
VB
IB
VC
VE
IE
Solution Practice Problem #6
The transistor in the circuit below has a  of 100 and and a base emitter voltage drop of 0.7 V when the base
emitter diode is forward biased. Determine the VCE voltage. Consider that when a diode is forward biased the
voltage drop across it is 0.7 V.
HINT: Determine VC and VE separately and then calculate the difference VCE = VC - VE.
VB
6.914 V
VB = 3 x 0.7 V = 2.1 V
VE
1.4 V
Assume Forward Active:
VCE
5.514 V
VE = VB - 0.7 V = 1.4 V
IC
IB
VC
IE = (VE - 0) / 2 k = 1.4 V / 2k = 0.7 mA
VC
IB = IE / ( +1) = IE / (100+1) = 0.00693 mA
IC =  IB = 100 IB = 0.693 mA
VE
IE
VC = 9 V - IC x 2k - 0.7 V = 6.914 V
VCE = VC - VE = 6.914V – 1.4 V = 5.514V
Confirm that BJT is in Forward Active:
Since VCE is greater than 0.3 V then BJT is operating
indeed in Forward Active!
Practice Problem #7
The BJT transistor in the circuit shown below has a  = 100 and a base emitter voltage drop of 0.7 V
when the base emitter diode is forward biased. Each of the two diodes, D1 and D2, can be modelled
with a constant voltage drop of 0.7 V when forward biased and as an open circuit when reversed
biased. Determine the value of the currents ID1 and ID2, that flow through these two diodes.
ID1
10 V
ID1
D1
1 k
ID2
100 k
D2
ID2
10 k
Solution Practice Problem #7
The BJT transistor in the circuit shown below has a  = 100 and a base emitter voltage drop of 0.7 V
when the base emitter diode is forward biased. Each of the two diodes, D1 and D2, can be modelled
with a constant voltage drop of 0.7 V when forward biased and as an open circuit when reversed
biased. Determine the value of the currents ID1 and ID2, that flow through these two diodes.
ID1
10 V
D1
100 k
VB
IB
I1
1 k
ID1
0.086 mA
ID2
0.0636 mA
VE = 0 V as base is connected directly to ground.
VC
IC
D2
ID2
10 k
VE
Assume Forward Active:
VB = VE + 0.7 V = 0 + 0.7 V = 0.7 V
ID1 = ( 10 V – 0.7 V – VB ) / 100 k = 8.6 V / 100 k = 0.086 mA
IB = ID1 = 0.086 mA = 86 A
IC =  IB =  mA
From KCL we have: I1 – IC – ID2 = 0
Since: I1 = (10V – VC) / 1k, ID2 = (VC – 0.7V) / 10k and IC = 8.6 mA, we have:
[(10V – VC) / 1k] − 8.6 mA − (VC – 0.7V) / 10k ] = 0
Solving for VC we get: VC = 1.336 V and thus ID2 = 0.0636 mA
Confirm that BJT is in Forward Active:
VCE = VC − VE = 1.336 V − 0 V = 1.336 V > 0.3V
Since VCE is greater than 0.3 V then BJT is indeed in Forward Active!
Practice Problem #8
In the following circuit, assume that  is 80 when the BJT transistor in in forward active mode
and that the emitter base junction has a voltage drop of 0.7 V, i.e., VEB = 0.7 V, when it is
forward biased. Find the emitter, collector and base voltages and currents.
VE
VB
VC
IE
IB
IC
Solution Practice Problem #8
In the following circuit, assume that  is 80 when the BJT transistor in in forward active mode
and that the emitter base junction has a voltage drop of 0.7 V, i.e., VEB = 0.7 V, when it is
forward biased. Find the emitter, collector and base voltages and currents.
VE
-2.0 V
VB
-2.7 V
VC
-4.07 V
IE = (12 - VE) / 5.6 k = 2.5 mA
IE
2.5 mA
IB = IE / (+) =  m
IB
0.031 mA
IC = [   (+)  IE =  m
IC
2.47 mA
Assuming Forward Active:
VB = −2.7 V as connected directly
to a DC supply of -2.7 V.
VE = VB + 0.7 V = -2.0 V
VC = −10 + IC 2.4 k = −4.07 V
Confirming Forward Active:
VE-VC =-2V – (-4.07)V=2.07V > 0.3V
Practice Problem #9
Solution Practice Problem #9
Practice Problem #10
Solution Practice Problem #10
Practice Problem #11
For the emitter follower in the circuit shown below, the signal source is directly coupled
to the transistor base. If the dc component of the sig is zero, find the dc emitter and
collector currents. Assume  = 100.
IE
IB
IC
Solution Practice Problem #11
For the emitter follower in the circuit shown below, the signal source is directly coupled
to the transistor base. If the dc component of the sig is zero, find the dc emitter and
collector currents. Assume  = 100.
At DC capacitor is open circuit and considering the DC
component of the signal source is zero the circuit can be
simplified as shown below.
IE
Assuming forward active mode of operation then we can
write an equation starting from the +3 V to the GRD
potential on the base side as:
IB
IC
Confirm that this pnp BJT is indeed in forward active mode of operation. First we need to calculate VB
VB = 0 + IB 100k = (IC /100) 100k =0.53 V and VE = 3V - IE 3.3k = 1.23 V
Then at the BC junction we have:
VC - VB= -5 -0.53V =-5.53V < 0 thus R.B. while at the BE junction: VE - VB= 1.23 - 0.53V = 0.7 V thus F.B.
Moreover: VCE = VE - VC= 1.23V – (-5V) = 6.23V > 0.3 V
Practice Problem #12
In the circuit shown below, the transistor has a  = 200. What are the dc voltages at the
base, collector and emitter terminals?
VE
VB
VC
VE
Solution Practice Problem #12
In the circuit shown below, the transistor has a  = 200. What are the dc voltages at the
base, collector and emitter terminals?
VE
VB
Both capacitors at DC
conditions act as open
circuit. So for the DC
analysis the circuit can
be simplified to the
one shown on the
right.
VE
Recall that since:
VC
 = 
VB
VC
then
 =  /( +1) = 0.995
IE = 10 mA same as the 10 mA current source. Assuming forward active mode of operation then:
IC =  IE = 9.95 mA and VC = 0 +100 IC = 0.995 V Moreover:
IB = IC /  = 0.05 mA, and thus VB = 1.5V + 10k IB = 2 V then VE = VB + 0.7V = 2.7 V
Confirm that this pnp BJT is indeed in forward active mode of operation:
BC junction: VC – VB = 0.995 – 2.7V = – 1.7V < 0 thus R.B.
Moreover: VCE = VE – VC = 2.7V – 0.995V = 1.7 V > 0.3 V