Showing posts with label load line analysis. Show all posts
Showing posts with label load line analysis. Show all posts

Variation Of Q Point With Different Circuit Parameters (BJT)

Change in RB and Q point | Change in RC and Q point | Change in supply voltage and Q point | Change in current gain and Q point
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

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:

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:
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:
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:
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.


Popular Posts