## What Are Methods Of Transistor Biasing?

In the transistor amplifier circuits drawn so far biasing was done with the aid of a battery VBB which was separate from the battery VCC used in the output circuit. However, in the interest of simplicity and economy, it is desirable that transistor circuit should have a single source of supply—the one in the output circuit (i.e. VCC). The following are the most commonly used methods of obtaining transistor biasing from one source of supply (i.e. VCC ) :

(i) Base resistor method
(ii) Emitter bias method
(iii) Biasing with collector-feedback resistor
(iv) Voltage-divider bias

In all these methods, the same basic principle is employed i.e. required value of base current (and hence IC) is obtained from VCC in the zero signal conditions. The value of collector load RC is selected keeping in view that VCE should not fall below 0.5 V for germanium transistors and 1 V for silicon transistors.

For example, if β = 100 and the zero signal collector current IC is to be set at 1mA, then IB is made equal to IC /β = 1/100 = 10 µA. Thus, the biasing network should be so designed that a base current of 10 µA flows in the zero signal conditions.

## Base Resistor Biasing

In this method, a high resistance RB (several hundred kΩ) is connected between the base and +ve end of supply for npn transistor (See Fig. 9.6) and between base and negative end of supply for pnp transistor. Here, the required zero signal base current is provided by VCC and it flows through RB. It is because now base is positive w.r.t. emitter i.e. base-emitter junction is forward biased. The required value of zero signal base current IB (and hence IC= βIB) can be made to flow by selecting the proper value of base resistor RB.

## Circuit AnalysisOf Base Resistor Biasing

It is required to find the value of RB so that required collector current flows in the zero signal conditions. Let IC be the required zero signal collector current.

As VCC and IB are known and VBE can be seen from the transistor manual, therefore, value of RB can be readily found from exp. (I). Since VBE is generally quite small as compared to VCC, the former can be neglected with little error. It then follows from exp. (i) that :

RB= Vcc/Ib

It may be noted that VCC is a fixed known quantity and IB is chosen at some suitable value. Hence, RB can always be found directly, and for this reason, this method is sometimes called fixed-bias method.

### Stability factor

As shown in Art. 9.6,

In fixed-bias method of biasing, IB is independent of IC so that dIB/dIC = 0. Putting the value of dIB/ dIC = 0 in the above expression, we have,

Stability factor, S = β + 1

Thus the stability factor in a fixed bias is (β + 1). This means that IC changes (β + 1) times as much as any change in ICO. For instance, if β = 100, then S = 101 which means that IC increases 101 times faster than ICO. Due to the large value of S in a fixed bias, it has poor thermal stability.

1. This biasing circuit is very simple as only one resistance RB is required.
2. Biasing conditions can easily be set and the calculations are simple.
3. There is no loading of the source by the biasing circuit since no resistor is employed across
4. base-emitter junction.

### Disadvantages Base Resistor Biasing :

(i) This method provides poor stabilisation. It is because there is no means to stop a selfincrease in collector current due to temperature rise and individual variations. For example, if β increases due to transistor replacement, then IC also increases by the same factor as IB is constant.

(ii) The stability factor is very high. Therefore, there are strong chances of thermal runaway. Due to these disadvantages, this method of biasing is rarely employed.

Example 9.3. Fig. 9.7 (i) shows biasing with base resistor method.

(i) Determine the collector current IC and collector-emitter voltage VCE. Neglect small base-emitter voltage. Given that β = 50.
(ii) If RB in this circuit is changed to 50 kΩ, find the new operating point

In the circuit shown in Fig. 9.7 (i), biasing is provided by a battery VBB (= 2V) in the base circuit which is separate from the battery VCC (= 9V) used in the output circuit. The same circuit is shown in a simplified way in Fig. 9.7 (ii). Here, we need show only the supply voltages, + 2V and + 9V. It may be noted that negative terminals of the power supplies are grounded to get a complete path of current.

(i) Referring to Fig.9.7 (ii) and applying Kirchhoff ’s voltage law to the circuit ABEN, we get,

Example 9.4. Fig. 9.8 (i) shows that a silicon transistor with β = 100 is biased by base resistor method. Draw the d.c. load line and determine the operating point. What is the stability factor ?
Solution.

VCC = 6 V, RB = 530 kΩ, RC = 2 kΩ
D.C. load line. Referring to Fig. 9.8 (i), VCE = VCC − IC RC
When IC = 0, VCE = VCC = 6 V. This locates the first point B (OB = 6V) of the load line on collector-emitter voltage axis as shown in Fig. 9.8 (ii).
When V CE = 0, IC= VCC/RC= 6V/2 kΩ = 3 mA.

This locates the second point A (OA = 3mA) of the load line on the colleector current axis. By joining points A and B, d.c. load line AB is constructed [See Fig. 9.8 (ii)].

Operating point Q. As it is a silicon transistor, therefore, VBE = 0.7V. Referring to Fig. 9.8 (I), it is clear that :

Example 9.5. (i) A germanium transistor is to be operated at zero signal IC = 1mA. If the collector supply VCC = 12V, what is the value of RB in the base resistor method ? Take β = 100.

(ii) If another transistor of the same batch with β = 50 is used, what will be the new value of zero signal IC for the same RB ?

Comments. It is clear from the above example that with the change in transistor parameter β, the zero signal collector current has changed from 1mA to 0.5mA. Therefore, base resistor method cannot provide stabilisation.

Example 9.6. Calculate the values of three currents in the circuit shown in Fig. 9.9.

Solution. Applying Kirchhoff ‘s voltage law to the base side and taking resistances in kΩ and currents in mA, we have,

Example 9.7. Design base resistor bias circuit for a CE amplifier such that operating point is VCE = 8V and IC = 2 mA. You are supplied with a fixed 15V d.c. supply and a silicon transistor with β = 100. Take base-emitter voltage VBE = 0.6V. Calculate also the value of load resistance that would be employed.

Example 9.8. A *base bias circuit in Fig.9.11 is subjected to an increase in temperature from 25°C to 75°C. If β = 100 at 25°C and 150 at 75°C, determine the percentage change in Q-point values (VCE and IC) over this temperature range. Neglect any
change in VBE and the effects of any leakage current.

Comments. It is clear from the above example that Q-point is extremely dependent on β in a base bias circuit. Therefore, base bias circuit is very unstable. Consequently, this method is normally not used if linear operation is required. However, it can be used for switching operation.

Example 9.9. In base bias method, how Q-point is affected by changes in VBE and ICBO.
Solution. In addition to being affected by change in β, the Q-point is also affected by changes in VBE and ICBO in the base bias method.

(i) Effect of VBE. The base-emitter-voltage VBE decreases with the increase in temperature (and vice-versa). The expression for IB in base bias method is given by ;

It is clear that decrease in VBE increases IB. This will shift the Q-point (IC= βIB and VCE = VCC – ICRC). The effect of change in VBE is negligible if VCC >> VBE (VCC at least 10 times greater than VBE).

(ii) Effect of ICBO. The reverse leakage current ICBO has the effect of decreasing the net base current and thus increasing the base voltage. It is because the flow of ICBO creates a voltage drop across RB that adds to the base voltage as shown in Fig. 9.12. Therefore, change in ICBO shifts the Q-point of the base bias circuit. However, in modern transistors, ICBO is usually less than 100nA and its effect on the bias is negligible if VBB >> ICBO RB.

Example 9.10. Fig. 9.13 (i) shows the base resistor transistor circuit. The device (i.e. transistor) has the characteristics shown in Fig. 9.13 (ii).

Determine VCC, RC and RB

Example 9.11. What fault is indicated in (i) Fig. 9.14 (i) and (ii) Fig. 9.14 (ii)?
Solution.

(i) The obvious fault in Fig. 9.14 (i) is that the base is internally open. It is because 3V at the base and 9V at the collector mean that transistor is in cut-off state.

(ii) The obvious fault in Fig. 9.14 (ii) is that collector is internally open. The voltage at the base is correct. The voltage of 9V appears at the collector because the ‘open’ prevents collector current.

Common Emitter | Common Collector Connection base resistor method

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