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DC Analysis of
Four-Resistor
Biasing Circuit
Assoc Prof Chang Chip Hong email: echchang@ntu.edu.sg
EE2002 Analog Electronics
EE2002 Analog Electronics
Lesson Objectives
DC Analysis of Four-Resistor Biasing Circuit 2
At the end of this lesson, you should be able to:
• Draw DC equivalent circuits for four-resistor biasing BJT and MOSFET
amplifiers
• Calculate the Q-points from the DC equivalent circuits of BJT and MOSFET
amplifiers by using appropriate circuit analysis techniques
EE2002 Analog Electronics
DC Analysis
DC and AC Analyses
DC Analysis of Four-Resistor Biasing Circuit 3
• Obtain DC equivalent circuit by replacing all capacitors by open circuits,
inductors by short circuit, AC voltage sources by ground connections, and
AC current sources by open circuits.
• Find Q-point from DC equivalent circuit by using appropriate circuit analysis
techniques, such as Thevenin equivalent circuit, KVL, and KCL.
EE2002 Analog Electronics
• Obtain AC equivalent circuit by replacing all capacitors by short circuits,
inductors by open circuits, DC voltage sources by ground connections, and
DC current sources by open circuits.
• Replace transistor by its small signal model.
• Use small signal AC equivalent to analyse AC characteristics of amplifier.
DC and AC Analyses
DC Analysis of Four-Resistor Biasing Circuit 4
AC Analysis
Combine end results of DC and AC analyses to yield total voltages and currents
in the network.
EE2002 Analog Electronics
All capacitors in original amplifier circuits are replaced by open circuits,
disconnecting vi, RI, and RL from circuit.
DC Equivalent Circuit for BJT Amplifier
DC Analysis of Four-Resistor Biasing Circuit 5
RB2
Q
Vcc
RB1 RC
RE
RI
RB2vi
Q
Rin
Vcc
RB1
RC
vo
Rout
RE
RL
C1  
C2  
C3  
+


1
ωC
CX  ω 0 rad/s (DC), CX   (open circuit)
EE2002 Analog Electronics
DC Analysis Example: Four-Resistor
BJT Biasing Circuit
DC Analysis of Four-Resistor Biasing Circuit
RB1
100 kW
+15 V
Q
RB2
50 kW
RC
5 kW
RE
3 kW
b100
RB1
100 kW
RB2
50 kW
+15 V +15 V
Q
RC
5 kW
RE
3 kW
B
Req
Veq
B+

33.3 kW
5 V
KVL 1: Veq = IBReq + VBE + IERE
5 = 33.3IB + 0.7 + 101×IB×3
IB = 0.0128 mA
IC = bIB = 1.28 mA, IE = (b+1) IB =1.29 mA
2
1 2
50
15 5 V
100 50
B
eq CC
B B
R
V V
R R
 
  
+ 
  
+
1 2
100 50
33.3
eq B BR R R
k


 W
+15 V
Q
RC
5 kW
RE
3 kW
2
1
KVL 2: VCE = 15 – ICRC – IERE = 15 – 1.28×5 – 1.29×3 = 4.73 V
6
EE2002 Analog Electronics DC Analysis of Four-Resistor Biasing Circuit
VB = VBE + IERE = 0.7 + 3.87 = 4.57 V
VC = VCC  ICRC = 15 – 1.28 × 5 = 8.6 V
VBC = VB – VC = 4.57 – 8.6 =  4.03 V
7
BCJ is reverse biased, Q is indeed in active mode as had been assumed.
RB1
100 kW
+15 V
Q
RB2
50 kW
RC
5 kW
RE
3 kW
b100
RB1
100 kW
RB2
50 kW
+15 V +15 V
Q
RC
5 kW
RE
3 kW
Req
Veq
B+

33.3 kW
5 V
+15 V
Q
RC
5 kW
RE
3 kW
3.87 V
8.6 V
1.28 mA
1.29 mA
0.0128 mA
0.103 mA
4.57 V
0.09 mA
0.0128 mA 4.57 V
DC Analysis Example: Four-Resistor
BJT Biasing Circuit
EE2002 Analog Electronics
Req
Veq
G
+

DC Analysis of Four-Resistor Biasing Circuit
RG1
1.5 MW
+10 V
M
RG2
1 MW
RD
75 kW
RS
39 kW
RG1
1.5 MW
RG2
1 MW
+10 V
G
600 kW
4 V
1 2
1.5 1 0.6 M
600 k
eq G GR R R
  W
 W
2
1
+10 V
M
RD
75 kW
RS
39 kW
+10 V
M
RD
75 kW
RS
39 kW
1
10
1 1.5
4 V
eqV
 
  
+ 

KVL 1: Since IG = 0, Veq = VGS + IDRS
4 = VGS + 0.5 × 25m(VGS – 1)2 ×39 k
VGS
2 + 0.05 VGS – 7.21 = 0
VGS = 2.71 or 2.66 V
Since VGS = 2.71 < VTN = 1, VGS = 2.66 V.
KVL 2: VDS = 10 – IDRD – IDRS
VDS = 10 – 0.0344×(75 +39) = 6.08 V
Since VDS > VGS – VTN = 1.66,
M is in saturation region.
Kn =
25 mA/V2
VTN = 1 V
ID = (4  2.66)/39k = 34.4 mA
8
DC Analysis Example: Four-Resistor
MOSFET Biasing Circuit

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Dc analysis of four resistor biasing circuit

  • 1. DC Analysis of Four-Resistor Biasing Circuit Assoc Prof Chang Chip Hong email: echchang@ntu.edu.sg EE2002 Analog Electronics
  • 2. EE2002 Analog Electronics Lesson Objectives DC Analysis of Four-Resistor Biasing Circuit 2 At the end of this lesson, you should be able to: • Draw DC equivalent circuits for four-resistor biasing BJT and MOSFET amplifiers • Calculate the Q-points from the DC equivalent circuits of BJT and MOSFET amplifiers by using appropriate circuit analysis techniques
  • 3. EE2002 Analog Electronics DC Analysis DC and AC Analyses DC Analysis of Four-Resistor Biasing Circuit 3 • Obtain DC equivalent circuit by replacing all capacitors by open circuits, inductors by short circuit, AC voltage sources by ground connections, and AC current sources by open circuits. • Find Q-point from DC equivalent circuit by using appropriate circuit analysis techniques, such as Thevenin equivalent circuit, KVL, and KCL.
  • 4. EE2002 Analog Electronics • Obtain AC equivalent circuit by replacing all capacitors by short circuits, inductors by open circuits, DC voltage sources by ground connections, and DC current sources by open circuits. • Replace transistor by its small signal model. • Use small signal AC equivalent to analyse AC characteristics of amplifier. DC and AC Analyses DC Analysis of Four-Resistor Biasing Circuit 4 AC Analysis Combine end results of DC and AC analyses to yield total voltages and currents in the network.
  • 5. EE2002 Analog Electronics All capacitors in original amplifier circuits are replaced by open circuits, disconnecting vi, RI, and RL from circuit. DC Equivalent Circuit for BJT Amplifier DC Analysis of Four-Resistor Biasing Circuit 5 RB2 Q Vcc RB1 RC RE RI RB2vi Q Rin Vcc RB1 RC vo Rout RE RL C1   C2   C3   +   1 ωC CX  ω 0 rad/s (DC), CX   (open circuit)
  • 6. EE2002 Analog Electronics DC Analysis Example: Four-Resistor BJT Biasing Circuit DC Analysis of Four-Resistor Biasing Circuit RB1 100 kW +15 V Q RB2 50 kW RC 5 kW RE 3 kW b100 RB1 100 kW RB2 50 kW +15 V +15 V Q RC 5 kW RE 3 kW B Req Veq B+  33.3 kW 5 V KVL 1: Veq = IBReq + VBE + IERE 5 = 33.3IB + 0.7 + 101×IB×3 IB = 0.0128 mA IC = bIB = 1.28 mA, IE = (b+1) IB =1.29 mA 2 1 2 50 15 5 V 100 50 B eq CC B B R V V R R      +     + 1 2 100 50 33.3 eq B BR R R k    W +15 V Q RC 5 kW RE 3 kW 2 1 KVL 2: VCE = 15 – ICRC – IERE = 15 – 1.28×5 – 1.29×3 = 4.73 V 6
  • 7. EE2002 Analog Electronics DC Analysis of Four-Resistor Biasing Circuit VB = VBE + IERE = 0.7 + 3.87 = 4.57 V VC = VCC  ICRC = 15 – 1.28 × 5 = 8.6 V VBC = VB – VC = 4.57 – 8.6 =  4.03 V 7 BCJ is reverse biased, Q is indeed in active mode as had been assumed. RB1 100 kW +15 V Q RB2 50 kW RC 5 kW RE 3 kW b100 RB1 100 kW RB2 50 kW +15 V +15 V Q RC 5 kW RE 3 kW Req Veq B+  33.3 kW 5 V +15 V Q RC 5 kW RE 3 kW 3.87 V 8.6 V 1.28 mA 1.29 mA 0.0128 mA 0.103 mA 4.57 V 0.09 mA 0.0128 mA 4.57 V DC Analysis Example: Four-Resistor BJT Biasing Circuit
  • 8. EE2002 Analog Electronics Req Veq G +  DC Analysis of Four-Resistor Biasing Circuit RG1 1.5 MW +10 V M RG2 1 MW RD 75 kW RS 39 kW RG1 1.5 MW RG2 1 MW +10 V G 600 kW 4 V 1 2 1.5 1 0.6 M 600 k eq G GR R R   W  W 2 1 +10 V M RD 75 kW RS 39 kW +10 V M RD 75 kW RS 39 kW 1 10 1 1.5 4 V eqV      +   KVL 1: Since IG = 0, Veq = VGS + IDRS 4 = VGS + 0.5 × 25m(VGS – 1)2 ×39 k VGS 2 + 0.05 VGS – 7.21 = 0 VGS = 2.71 or 2.66 V Since VGS = 2.71 < VTN = 1, VGS = 2.66 V. KVL 2: VDS = 10 – IDRD – IDRS VDS = 10 – 0.0344×(75 +39) = 6.08 V Since VDS > VGS – VTN = 1.66, M is in saturation region. Kn = 25 mA/V2 VTN = 1 V ID = (4  2.66)/39k = 34.4 mA 8 DC Analysis Example: Four-Resistor MOSFET Biasing Circuit