BJT Load Line & Q point Analysis: Complete tutorial with solved examples
Learning Objectives:
- Why does the Q Point Vary?
- Analyse Q Point Variations With Different Circuit Parameters
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| Figure 1: circuit |
Variation In Current Gain β & Q Point:
Theoretically, β is constant but it is not practical. Now have a look at transistor Q point variations with beta β.
Consider the circuit in fig 1, β varies while other quantities remain the same.
Case 1: β = 50
Step 1: Find DC load line
KVL at the output circuit
-VCC + ICRC + VCE = 0
-10 + 1k*IC + VCE = 0 … eq 1.1
Substitute VCE = 0 in eq 1.1 for a point on y-axis
IC = 10mA
Substitute IC = 0 in eq 1.1 for a point on x-axis
VCE = 10V
Step 2: Find Q point
KVL at input circuit
-VCC + IBRB + VBE = 0
IC = β*IB
-10 + ICQ*140k/50 + 0.7 = 0
ICQ = 3.3mA
KVL at output circuit
-VCC + ICRC + VCEQ = 0
-10 + 3.3m*1k + VCEQ = 0
VCEQ = 6.7V
Step 3: Find AC load line
iC(sat) = ICQ + VCEQ/rC
rC = RC||RL
iC(sat) = 3.3m + 6.7/500
iC(sat) = 3.3m + 13.4m
iC(sat) = 16.7mA
vce(cut) = VCEQ + ICQ*rC
vce(cut) = 6.7 + 3.3m*500
vce(cut) = 8.35V
Case 2: β = 100
Step 1: Find DC load line
KVL at output circuit
-VCC + ICRC + VCE = 0
-10 + 1k*IC + VCE = 0 … eq 2.1
Substitute VCE = 0 in eq 2.1 for a point on y-axis
IC = 10mA
Substitute IC = 0 in eq 2.1 for a point on x-axis
VCE = 10V
Step 2: Find Q point
KVL at input circuit
-VCC + IBRB + VBE = 0
IC = β*IB
-10 + ICQ*140k/100 + 0.7 = 0
ICQ = 6.6mA
KVL at output circuit
-VCC + ICRC + VCEQ = 0
-10 + 6.6m*1k + VCEQ = 0
VCEQ = 3.4V
Step 3: Find AC load line
iC(sat) = ICQ + VCEQ/rC
rC = RC||RL
iC(sat) = 6.6m + 3.4/500
iC(sat) = 6.6m + 6.8m
iC(sat) = 13.4mA
vce(cut) = VCEQ + ICQ*rC
vce(cut) = 3.4 + 6.6m*500
vce(cut) = 6.7V
Case 3: β = 150
Step 1: Find DC load line
KVL at output circuit
-VCC + ICRC + VCE = 0
-10 + 1k*IC + VCE = 0 … eq 3.1
Substitute VCE = 0 in eq 3.1 for a point on y-axis
IC = 10mA
Substitute IC = 0 in eq 3.1 for a point on x-axis
VCE = 10V
Step 2: Find Q point
KVL at input circuit
-VCC + IBRB + VBE = 0
IC = β*IB
-10 + ICQ*140k/150 + 0.7 = 0
ICQ = 9.9mA
KVL at output circuit
-VCC + ICRC + VCEQ = 0
-10 + 9.9m*1k + VCEQ = 0
VCEQ = 0.1V
Step 3: Find AC load line
iC(sat) = ICQ + VCEQ/rC
rC = RC||RL
iC(sat) = 9.9m + 0.1/500
iC(sat) = 9.9m + 0.2m
iC(sat) = 10mA
vce(cut) = VCEQ + ICQ*rC
vce(cut) = 0.1 + 9.9m*500
vce(cut) = 5.1V
Results:
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| Figure 2: variations in current gain and Q point |
It is concluded that,
- Due to the change in β, there is no change in the DC load line
- Due to change in β, AC load line shifts
- Due to increase in β the Q point shifts towards the saturation region
Variation In RC & Q Point:
The magnitude of IC is not the function of the resistance RC. Change RC to any level and it does not affect IB or IC as long as the transistor is in the active region. RC determines the magnitude of VCE. Q point varies significantly with the variations in RC.
Case 4: RC = 500
Step 1: Find DC load line
KVL at output circuit
-VCC + ICRC + VCE = 0
-10 + 500*IC + VCE = 0 … eq 4.1
Substitute VCE = 0 in eq 4.1 for a point on y-axis
IC = 20mA
Substitute IC = 0 in eq 4.1 for a point on x-axis
VCE = 10V
Step 2: Find Q point
KVL at input circuit
-VCC + IBRB + VBE = 0
IC = β*IB
-10 + ICQ*140k/100 + 0.7 = 0
ICQ = 6.6mA
KVL at output circuit
-VCC + ICRC + VCEQ = 0
-10 + 6.6m*500 + VCEQ = 0
VCEQ = 6.7V
Step 3: Find AC load line
iC(sat) = ICQ + VCEQ/rC
rC = RC||RL = 333.33 ohm
iC(sat) = 6.6m + 6.7/333.33
iC(sat) = 6.6m + 20m
iC(sat) = 27mA
vce(cut) = VCEQ + ICQ*rC
vce(cut) = 6.7 + 6.6m*333.33
vce(cut) = 8.86V
Case 5: RC = 1k
Step 1: Find DC load line
KVL at output circuit
-VCC + ICRC + VCE = 0
-10 + 1k*IC + VCE = 0 … eq 5.1
Substitute VCE = 0 in eq 5.1 for a point on y-axis
IC = 10mA
Substitute IC = 0 in eq 5.1 for a point on x-axis
VCE = 10V
Step 2: Find Q point
KVL at input circuit
-VCC + IBRB + VBE = 0
IC = β*IB
-10 + ICQ*140k/100 + 0.7 = 0
ICQ = 6.6mA
KVL at output circuit
-VCC + ICRC + VCEQ = 0
-10 + 6.6m*1k + VCEQ = 0
VCEQ = 3.4V
Step 3: Find AC load line
iC(sat) = ICQ + VCEQ/rC
rC = RC||RL
iC(sat) = 6.6m + 3.4/500
iC(sat) = 6.6m + 6.8m
iC(sat) = 13.4mA
vce(cut) = VCEQ + ICQ*rC
vce(cut) = 3.4 + 6.6m*500
vce(cut) = 6.7V
Case 6: RC = 1.5k
Step 1: Find DC load line
KVL at output circuit
-VCC + ICRC + VCE = 0
-10 + 1.5k*IC + VCE = 0 … eq 6.1
Substitute VCE = 0 in eq 6.1 for a point on y-axis
IC = 6.6mA
Substitute IC = 0 in eq 6.1 for a point on x-axis
VCE = 10V
Step 2: Find Q point
KVL at input circuit
-VCC + IBRB + VBE = 0
IC = β*IB
-10 + ICQ*140k/100 + 0.7 = 0
ICQ = 6.6mA
KVL at output circuit0?
-VCC + ICRC + VCEQ = 0
-10 + 6.6m*1.5k + VCEQ = 0
VCEQ = 0.1 V
Step 3: Find AC load line
iC(sat) = ICQ + VCEQ/rC
rC = RC||RL = 600
iC(sat) = 6.6m + 0.1/600
iC(sat) = 6.6m + 0.2m
iC(sat) = 6.7mA
vce(cut) = VCEQ + ICQ*rC
vce(cut) = 0.1 + 6.7m*600
vce(cut) = 3.5V
Results:
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| Figure 3: change in RC and point |
RIt is concluded that,
- Due to the change in RC, there is a change in the DC load line because it has a slope of 1/RC
- Due to change in RC, AC load line shifts
- Due to change in RC the Q point shifts towards the saturation region
Variation In RB & Q Point:
It is interesting to note that the base current is controlled by the level of RB and IB is related to IC. The base current is adjusted by RB.
Case 7: RB = 120k
Step 1: Find DC load line
KVL at output circuit
-VCC + ICRC + VCE = 0
-10 + 1k*IC + VCE = 0 … eq 7.1
Substitute VCE = 0 in eq 7.1 for a point on y-axis
IC = 10mA
Substitute IC = 0 in eq 7.1 for a point on x-axis
VCE = 10V
Step 2: Find Q point
KVL at input circuit
-VCC + IBRB + VBE = 0
IC = β*IB
-10 + IC*120k/100 + 0.7 = 0
ICQ = 7.75mA
KVL at output circuit
-VCC + ICRC + VCE = 0
VCEQ = 10 - 7.75
VCEQ = 2.25V
Step 3: Find AC load line
iC(sat) = ICQ + VCEQ/rC
rC = RC||RL
iC(sat) = 7.75m + 2.25/500
iC(sat) = 12.25mA
vce(cut) = VCEQ + ICQ*rC
vce(cut) = 2.25 + 7.75m*500
vce(cut) = 6.125V
Case 8: RB = 120k
Step 1: Find DC load line
KVL at output circuit
-VCC + ICRC + VCE = 0
-10 + 1k*IC + VCE = 0 … eq 8.1
Substitute VCE = 0 in eq 8.1 for a point on y-axis
IC = 10mA
Substitute IC = 0 in eq 8.1 for a point on x-axis
VCE = 10V
Step 2: Find Q point
KVL at input circuit
-VCC + IBRB + VBE = 0
IC = β*IB
-10 + ICQ*140k/100 + 0.7 = 0
ICQ = 6.6mA
KVL at output circuit
-VCC + ICRC + VCEQ = 0
-10 + 6.6m*1k + VCEQ = 0
VCEQ = 3.4V
Step 3: Find AC load line
iC(sat) = ICQ + VCEQ/rC
rC = RC||RL
iC(sat) = 6.6m + 3.4/500
iC(sat) = 6.6m + 6.8m
iC(sat) = 13.4mA
vce(cut) = VCEQ + ICQ*rC
vce(cut) = 3.4 + 6.6m*500
vce(cut) = 6.7V
Case 9: RB = 180k
Step 1: Find DC load line
KVL at output circuit
-VCC + ICRC + VCE = 0
-10 + 1k*IC + VCE = 0 … eq 9.1
Substitute VCE = 0 in eq 9.1 for a point on y-axis
IC = 10mA
Substitute IC = 0 in eq 9.1 for a point on x-axis
VCE = 10V
Step 2: Find Q point
KVL at input circuit
-VCC + IBRB + VBE = 0
IC = β*IB
-10 + ICQ*180k/β + 0.7 = 0
ICQ = 5.2mA
KVL at output circuit
-VCC + ICRC + VCE = 0
-10 + 5.2m*1k + VCEQ = 0
VCEQ = 4.8V
Step 3: Find AC load line
iC(sat) = ICQ + VCEQ/rC
rC = RC||RL
iC(sat) = 5.2m + 4.8/500
iC(sat) = 14.8mA
vce(cut) = VCEQ + ICQ*rC
vce(cut) = 4.8 + 5.2m*500
vce(cut) = 7.4V
Results:
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| Figure 4: change in RB and Q point |
It is concluded that,
- Due to the change in RB, there is no change in the DC load line
- Due to change in RB AC load line shifts
- Due to the change in RC the Q point shifts towards the cut off region
Variation In VCC & Q Point:
Case 10: VCC = 10V
Step 1: Find DC load line
KVL at output circuit
-VCC + ICRC + VCE = 0
-10 + 1k*IC + VCE = 0 … eq 10.1
Substitute VCE = 0 in eq 10.1 for a point on y-axis
IC = 10mA
Substitute IC = 0 in eq 10.1 for a point on x-axis
VCE = 10V
Step 2: Find Q point
KVL at input circuit
-VCC + IBRB + VBE = 0
IC = β*IB
-10 + ICQ*140k/100 + 0.7 = 0
ICQ = 6.6mA
KVL at output circuit
-VCC + ICRC + VCEQ = 0
-10 + 6.6m*1k + VCEQ = 0
VCEQ = 3.4V
Step 3: Find AC load line
iC(sat) = ICQ + VCEQ/rC
rC = RC||RL
iC(sat) = 6.6m + 3.4/500
iC(sat) = 6.6m + 6.8m
iC(sat) = 13.4mA
vce(cut) = VCEQ + ICQ*rC
vce(cut) = 3.4 + 6.6m*500
vce(cut) = 6.7V
Case 11: VCC = 12V
Step 1: Find DC load line
KVL at output circuit
-VCC + ICRC + VCE = 0
-12 + 1k*IC + VCE = 0 … eq 11.1
Substitute VCE = 0 in eq 11.1 for a point on y-axis
IC = 12mA
Substitute IC = 0 in eq 11.1 for a point on x-axis
VCE = 12V
Step 2: Find Q point
KVL at input circuit
-VCC + IBRB + VBE = 0
IC = β*IB
-12 + ICQ*140k/100 + 0.7 = 0
ICQ = 8mA
KVL at output circuit
-VCC + ICRC + VCEQ = 0
-12 + 8m*1k + VCEQ = 0
VCEQ = 4V
Step 3: Find AC load line
iC(sat) = ICQ + VCEQ/rC
rC = RC||RL
iC(sat) = 8m + 4/500
iC(sat) = 8m + 8m
iC(sat) = 16mA
vce(cut) = VCEQ + ICQ*rC
vce(cut) = 4 + 8m*500
vce(cut) = 8V
Case 12: VCC = 15V
Step 1: Find DC load line
KVL at output circuit
-VCC + ICRC + VCE = 0
-10 + 1k*IC + VCE = 0 … eq 10.1
Substitute VCE = 0 in eq 10.1 for a point on y-axis
IC = 15mA
Substitute IC = 0 in eq 10.1 for a point on x-axis
VCE = 15V
Step 2: Find Q point
KVL at input circuit
-VCC + IBRB + VBE = 0
IC = β*IB
-15 + ICQ*140k/100 + 0.7 = 0
ICQ = 10.2mA
KVL at output circuit
-VCC + ICRC + VCEQ = 0
-15 + 10.2m*1k + VCEQ = 0
VCEQ = 4.8V
Step 3: Find AC load line
iC(sat) = ICQ + VCEQ/rC
rC = RC||RL
iC(sat) = 10.2m + 4.8/500
iC(sat) = 10.2m + 9.6m
iC(sat) = 20.5mA
vce(cut) = VCEQ + ICQ*rC
vce(cut) = 4.8 + 10.2m*500
vce(cut) = 9.9V
Results:
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| Figure 5: change in VCC and Q point |
It is concluded that,
- Due to a change in VCC, there is a change in the DC load line.
- Due to change in VCC, AC load line shifts as well
- Due to change in VCC the Q point shifts as well but the mode of operation doesn't change by changing the VCC value. You can observe from the graph, the amplifier remains in the active region for all three values of VCC
Conclusion:
This is an easy topic. It only requires circuit solving skills. The topic helps in designing a BJT circuit either it is an amplifier or a switch. Now, you can choose the Q point wisely. I have written a series of articles on Q point and load line with my immense efforts. If you like my posts please leave sincere comments.
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