Load Line Analysis (BJT) With Solved Examples

Load line and Q point 

Outline:

• Determine biasing, Q point, and DC load line
• Compute Q point

BJT Load Line & Q point Analysis: Complete tutorial with solved examples

AC load line

Variation of Q point and voltage swing limitations

Variation of Q point with different circuit parameters

I talked about the different modes of operation. To operate a transistor in a particular mode of opration, you need proper biasing. In simple terms, biasing means apply potential or voltage to the input and output terminal of the transistor to get the desired mode of operation. 

In BJT analysis, a load line is a line drawn over collector curves to show every possible operating point of a transistor. Or on a load line, there are valid values of IC and VCE.

I am going to consider a base bias amplifier circuit. Although this biasing technique has many flaws it is a simple circuit and a beginner can easily understand this. 

Figure: Base biased amplifier circuit

The purpose of this section is to plot the DC load line:

The DC load line contains possible DC values of IC and VCE for a given amplifier.

To draw a line you need two points (according to mathematics). The load line is drawn between these two extremes is the saturation point and the cut-off point. We need to find out at what point collector current is maximum (that is the saturation point). And the second point is the maximum possible collector to emitter voltage (that is the cut off point).

Method:

You can locate saturation and cut off point with the help of circuit analysis methods that is KVL and KCL.

Look at the circuit in figure 1. The DC equivalent of this circuit is given below. 

Figure: DC equivalent circuit

Look at the simple circuit below. Apply KVL to the output side.

-VCC + ICRC + VCE  = 0

ICRC = VCC - VCE

IC = (VCC - VCE )/RC 

IC = VCC/RC - VCE/RC .. equation 1

Equation 1 is the equation for the load line. We have to draw it over output characteristics curves. 

Look at equation 1, and compare it with the straight-line equation.

y = mx + c

Here, c = VCC/RC 

And  m = slope = -1/RC 

For y-intercept, x = 0 (that is VCE = 0)

IC = VCC/RC … Point 1

For x-intercept, y = 0 (that is IC = 0)

VCE = VCC … Point 2

Figure: load line and Q point general formula

Look at point 1, it is the extreme of saturation region while point 2 is the extreme of the cut off region.

I have plotted the line by joining Point 1 and Point 2 over the characteristic curves. This is the load line. The line intersects the curve at the Q point. It means the Q point is determined by IC and VCE. 

Point 1 intersects the curve at the saturation region of collector curves. It tells you the maximum collector current for the circuit.  While point 2 intersects the curve at the cut off region of collector curves. This point tells you the maximum collector-emitter voltage for the circuit. 

Plotting The Q Point:

You have successfully drawn a load line. Now the second step is to locate a Q point. 

Consider DC equivalent circuit in fig 2,

Apply KVL to the input side

-VCC + IBRB + VBE = 0

IB = (VCC - VBE)/RB

ICQ = βIB

Apply KVL to the output side

-VCC + ICRC + VCE = 0

VCEQ = VCC - ICQRC

Steps involved in load line and Q point analysis:

• Draw DC equivalent circuit
• In DC analysis we assume all the capacitors are open circuit
• Apply KVL to the input side and find out ICQ
• Apply KVL to the output side and find out VCEQ
• Find out maximum collector current IC(sat). You need to consider VCE = 0
• Find out a maximum collector to emitter voltage VCE(cut off). You need to consider IC = 0
• Draw the load line
• Locate the Q point with the help of calculated values of ICQ and VCEQ

Example#1:

Figure: Base biased amplifier

Apply KVL to the input side:

-VCC + IBRB + VBE = 0 

IB = (VCC - VBE)/RB

IB = (10-0.7)/300k

IB = 31 μA

IC = βIB

Consider β = 100

ICQ= 3.1 mA

Apply KVL to the input side:

-VCC + ICRC + VCE = 0… equation 1

VCE = VCC - ICRC

VCE = 10 - 3.1 m*2k

VCEQ = 3.8 V 

Equation 1 is the load line equation

-VCC + ICRC + VCE = 0

For maximum collector current IC(sat), substitute VCE = 0

IC(sat) = VCC/RC

IC(sat) = 10/2k

IC(sat) = 5 mA

For maximum collector to emitter voltage VCE(cut off), substitute IC = 0

  VCE = VCC

VCE(cut off) = 10 V

Figure: Load line and Q point 

Conclusion:

It is all about DC load line analysis. There is an AC load line as well. It is an important topic in designing a BJT based circuit. The two load lines must consider while designing an amplifier circuit.