What is Load Line Analysis | Voltage Gain

What Is Load Line?

The output characteristics are determined experimentally and indicate the relation between VCE and IC. However, the same information can be obtained in a much simpler way by representing the mathematical relation between IC and VCE graphically. As discussed before, the relationship between VCE and IC is linear so that it can be represented by a straight line on the output characteristics. This is known as a load line. The points lying on the load line give the possible values of VCE and IC in the output circuit. As in a transistor circuit both d.c. and a.c. conditions exist, therefore, there are two types of load lines, namely; d.c. load line and a.c. load line. The former determines the locus of IC and VCE in the zero signal conditions and the latter shows these values when the signal is applied.

(i) d.c. load line.

It is the line on the output characteristics of a transistor circuit which gives the values of IC and VCE corresponding to zero signal or d.c. conditions. Consider the transistor amplifier shown in Fig. 10.12. In the absence of signal, d.c. conditions prevail in the circuit as shown in Fig. 10.13 (i). Referring to this circuit and applying Kirchhoff’s voltage law

d.c. load line.

As for a given circuit, VCC and (RC + RE) are constant, therefore, it is a first degree *equation and can be represented by a straight line on the output characteristics. This is known as d.c. load line and determines the loci of VCE and IC points in the zero signal conditions. The d.c. load line can be readily plotted by locating two end points of the straight line.

This is known as d.c. load line

The value of VCE will be maximum when IC = 0. Therefore, by putting IC = 0 in exp. (i), we get,

Max. VCE = VCC

This locates the first point B (OB = VCC) of the d.c. load line.

The value of IC will be maximum when VCE = 0.

 first point B (OB = VCC) of the d.c. load line.

This locates the second point A (OA = VCC/RC + RE) of the d.c. load line. By joining points A and B, d.c. load line AB is constructed [See Fig. 10.13 (ii)]. Alternatively. The two endpoints of the d.c. load line can also be determined in another way.

With the construction of d.c. load line on the output characteristics, we get the complete information about the output circuit of transistor amplifier in the zero signal conditions. All the points showing zero signal IC and VCE will obviously lie on the d.c. load line. At the same time IC and VCE conditions in the circuit are also represented by the output characteristics. Therefore, actual operating conditions in the circuit will be represented by the point where d.c. load line intersects the base current curve under study. Thus, referring to Fig. 10.13 (ii), if IB = 5 µA is set by the biasing circuit, then Q (i.e. intersection of 5 µA curve and load line) is the operating point.

(ii) a.c. load line.

This is the line on the output characteristics of a transistor circuit which gives the values of iC and vCE when signal is applied. Referring back to the transistor amplifier shown in Fig. 10.12, its a.c. equivalent circuit as far as output circuit is concerned is as shown in Fig. 10.14 (i). To add a.c. load line to the output characteristics, we again require two endpoints–one maximum collector-emitter voltage point and the other maximum collector current point. Under the application of a.c. signal, these values are (refer to example 10.4) :

Max. collector-emitter voltage = VCE + IC RAC. This locates the point C of the a.c. loadline on the collector-emitter voltage axis.

To draw d.c. load line, we require two endpoints viz

This locates the point D of a.c. loadline on the collector-current axis. By joining points C and D,
the a.c.load line CD is constructed [See Fig. 10.14 (ii)].


Example 10.5. For the transistor amplifier shown in Fig. 10.15, R1 = 10 kΩ, R2 = 5 kΩ, RC =
1 kΩ, RE = 2 kΩ and RL = 1 kΩ.
(i) Draw d.c. loadline (ii) Determine the operating point (iii) Draw a.c. load line.
Assume VBE = 0.7 V.


Solution. (i) d.c. load line :
To draw d.c. load line, we require two endpoints viz maximum VCE point and maximum IC point.

To draw d.c. load line, we require two endpoints viz maximum VCE point and maximum IC point

(ii) Operating point Q.

The voltage across R2 (= 5 kΩ ) is *5 V i.e. V2 = 5 V.

Operating point Q.

∴ Operating point Q is 8.55 V, 2.15 mA. This is shown on the d.c. load line.


(iii) d.c. load line. To draw a.c. load line, we require two endpoints viz. maximum collector emitter voltage point and maximum collector current point when signal is applied.

This locates the point D (OD = 19.25mA) on the i C axis. By joining points C and D, a.c. loadline CD is constructed [See Fig. 10.16 (ii)].

Example 10.6. In the transistor amplifier shown in Fig. 10.15, RC = 10 kΩ, RL = 30 kΩ and VCC = 20V. The values R1 and R2 are such so as to fix the operating point at 10V, 1mA. Draw the d.c. and a.c. load lines. Assume RE is negligible.

Solution. d.c. load line. For drawing d.c. load line, two endpoints viz. maximum VCE point and maximum IC point are needed. Maximum VCE = 20 V. This locates the point B (OB = 20V) of the d.c. loadline on the VCE axis.

 By joining points A and B, the d.c. load line AB is constructed (See Fig. 10.17). a.c. load line.

This locates the point A (OA = 2 mA) on the IC axis. By joining points A and B, the d.c. load line AB is constructed (See Fig. 10.17). a.c. load line. To draw a.c. load line, we require two end points viz maximum collector-emitter voltage point and maximum collector current point when signal is applied.

This locates the point A (OA = 2 mA) on the IC axis. By joining points A and B, the d.c. load line AB is constructed (See Fig. 10.17). a.c. load line.

This locates the point C (OC = 2.33 mA) on the iC axis. By joining points C and D, a.c. loadline CD is constructed (See Fig. 10.17).

This locates the point C (OC = 2.33 mA) on the iC axis. By joining points C and D, a.c. loadline CD is constructed

Comments. The reader may see that the operating point lies on both a.c. and d.c. load lines. It is not surprising because signal is a.c. and it becomes zero after every half-cycle. When the signal is zero, we have the exact d.c. conditions. Therefore, key point to keep in mind is that the point of intersection of d.c. and a.c. load lines is the operating point Q.

Example 10.7. In a transistor amplifier, the operating point Q is fixed at 8V, 1mA. When a.c. signal is applied, the collector current and collector-emitter voltage change about this point. During the positive peak of signal, iC = 1.5 mA and vCE = 7 V and during negative peak, iC = 0.5mA and vCE = 9V. Show this phenomenon with the help of a.c. load line.

In a transistor amplifier, the operating point Q is fixed at 8V, 1mA. When a.c. signal is applied, the collector current and collector-emitter voltage

Solution. Fig. 10.18 shows the whole process. When no signal is applied, vCE = 8 V and i C = 1mA. This is represented by the operating point Q on the a.c. load line. During the positive halfcycle of a.c. signal, i C swings from 1 mA to 1.5 mA and vCE swings from 8 V to 7 V. This is represented by point A on the a.c. load line. During the negative half-cycle of the signal, i C swings from 1 mA to 0.5 mA and vCE swings from 8 V to 9 V. This is represented by the point B on the a.c. load line. The following points may be noted :

(i) When a.c. signal is applied, the collector current and collector-emitter voltage variations take place about the operating point Q.

(ii) When a.c. signal is applied, operating point moves along the a.c. load line. In other words, at any instant of a.c. signal, the co-ordinates of collector current and collector-emitter voltage are on the a.c. load line.

What is voltage gain?

The basic function of an amplifier is to raise the strength of an a.c. input signal. The voltage gain of the amplifier is the ratio of a.c. output voltage to the a.c. input signal voltage. Therefore, in order to find the voltage gain, we should consider only the a.c. currents and voltages in the circuit. For this purpose, we should look at the a.c. equivalent circuit of transistor amplifier. For facility of reference,the a.c. equivalent circuit of transistor amplifier is redrawn in Fig. 10.19.

What is voltage gain?

It is clear that as far as a.c. signal is concerned, load RC appears in parallel with RL. Therefore, effective load for a.c. is given by :

 In the circuit shown in Fig. 10.20, find the voltage gain. Given that β = 60 and input resistance

Example 10.8. In the circuit shown in Fig. 10.20, find the voltage gain. Given that β = 60 and input resistance Rin = 1 kΩ.

 In the circuit shown in Fig. 10.20, find the voltage gain. Given that β = 60 and input resistance Rin = 1 kΩ.

Example 10.9. In the circuit shown in Fig. 10.20, if RC = 10 kΩ, RL = 10 kΩ, Rin = 2.5 kΩ, β = 100, find the output voltage for an input voltage of 1mV r.m.s.

In a transistor amplifier, when the signal changes by 0.02V, the base current changes by 10 µA

Example 10.10. In a transistor amplifier, when the signal changes by 0.02V, the base current changes by 10 µA and collector current by 1mA. If collector load RC = 5 kΩ and RL = 10 kΩ, find: (i) current gain (ii) input impedance (iii) a.c. load (iv) voltage gain (v) power gain.

In a transistor amplifier, when the signal changes by 0.02V, the base current changes by 10 µA and collector
In a transistor amplifier, when the signal changes by 0.02V, the base current changes by 10 µA and collector current by 1mA

Example 10.11. In Fig. 10.21, the transistor has β = 50. Find the output voltage if input resistance Rin = 0.5 kΩ.

In Fig. 10.21, the transistor has β = 50. Find the output voltage if input resistance Rin = 0.5 kΩ.

Example 10.12. Fig. 10.22 shows a transistor circuit. The manufacturer of the circuit shows that collector potential is to be + 6V. The voltage measured at point B by a technician is found to be + 4V. Is the circuit operating properly ?

Solution. The voltage at point B is equal to the voltage across R1. Now total voltage VT across the series combination of R1 and R2 is 6 V. Therefore, using voltage divider method, we have,

The voltage at point B is equal to the voltage across R1. Now total voltage VT across the series combination of R1 and R2 is 6 V

The circuit is not operating properly. It is because the voltage at point B should be 2 V instead of 4V.

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Reference: Principles Of Electronics By V K Mehta And Rohit Mehta

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