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The biasing in transistor circuits is done by using two DC sources V_{BB} and V_{CC}. It is economical to minimize the DC source to one supply instead of two which also makes the circuit simple.

The commonly used methods of transistor biasing are

- Base Resistor method
- Collector to Base bias
- Biasing with Collector feedback resistor
- Voltage-divider bias

All of these methods have the same basic principle of obtaining the required value of I_{B} and I_{C} from V_{CC} in the zero signal conditions.

In this method, a resistor R_{B} of high resistance is connected in base, as the name implies. The required zero signal base current is provided by V_{CC} which flows through R_{B}. The base emitter junction is forward biased, as base is positive with respect to emitter.

The required value of zero signal base current and hence the collector current (as I_{C} = βI_{B}) can be made to flow by selecting the proper value of base resistor RB. Hence the value of R_{B} is to be known. The figure below shows how a base resistor method of biasing circuit looks like.

Let I_{C} be the required zero signal collector current. Therefore,

$$I_B = \frac{I_C}{\beta}$$

Considering the closed circuit from V_{CC}, base, emitter and ground, while applying the Kirchhoff’s voltage law, we get,

$$V_{CC} = I_B R_B + V_{BE}$$

Or

$$I_B R_B = V_{CC} - V_{BE}$$

Therefore

$$R_B = \frac{V_{CC} - V_{BE}}{I_B}$$

Since V_{BE} is generally quite small as compared to V_{CC}, the former can be neglected with little error. Then,

$$R_B = \frac{V_{CC}}{I_B}$$

We know that V_{CC} is a fixed known quantity and I_{B} is chosen at some suitable value. As R_{B} can be found directly, this method is called as **fixed bias method**.

Stability factor

$$S = \frac{\beta + 1}{1 - \beta \left ( \frac{d I_B}{d I_C} \right )}$$

In fixed-bias method of biasing, I_{B} is independent of I_{C} so that,

$$\frac{d I_B}{d I_C} = 0$$

Substituting the above value in the previous equation,

Stability factor, $S = \beta + 1$

Thus the stability factor in a fixed bias is (β+1) which means that I_{C} changes (β+1) times as much as any change in I_{CO}.

- The circuit is simple.
- Only one resistor R
_{E}is required. - Biasing conditions are set easily.
- No loading effect as no resistor is present at base-emitter junction.

The stabilization is poor as heat development can’t be stopped.

The stability factor is very high. So, there are strong chances of thermal run away.

Hence, this method is rarely employed.

The collector to base bias circuit is same as base bias circuit except that the base resistor R_{B} is returned to collector, rather than to V_{CC} supply as shown in the figure below.

This circuit helps in improving the stability considerably. If the value of I_{C} increases, the voltage across R_{L} increases and hence the V_{CE} also increases. This in turn reduces the base current I_{B}. This action somewhat compensates the original increase.

The required value of R_{B} needed to give the zero signal collector current I_{C} can be calculated as follows.

Voltage drop across R_{L} will be

$$R_L = (I_C + I_B)R_L \cong I_C R_L$$

From the figure,

$$I_C R_L + I_B R_B + V_{BE} = V_{CC}$$

Or

$$I_B R_B = V_{CC} - V_{BE} - I_C R_L$$

Therefore

$$R_B = \frac{V_{CC} - V_{BE} - I_C R_L}{I_B}$$

Or

$$R_B = \frac{(V_{CC} - V_{BE} - I_C R_L)\beta}{I_C}$$

Applying KVL we have

$$(I_B + I_C)R_L + I_B R_B + V_{BE} = V_{CC}$$

Or

$$I_B(R_L + R_B) + I_C R_L + V_{BE} = V_{CC}$$

Therefore

$$I_B = \frac{V_{CC} - V_{BE} - I_C R_L}{R_L + R_B}$$

Since V_{BE} is almost independent of collector current, we get

$$\frac{d I_B}{d I_C} = - \frac{R_L}{R_L + R_B}$$

We know that

$$S = \frac{1 + \beta}{1 - \beta (d I_B / d I_C)}$$

Therefore

$$S = \frac{1 + \beta}{1 + \beta \left ( \frac{R_L}{R_L + R_B} \right )}$$

This value is smaller than (1+β) which is obtained for fixed bias circuit. Thus there is an improvement in the stability.

This circuit provides a negative feedback which reduces the gain of the amplifier. So the increased stability of the collector to base bias circuit is obtained at the cost of AC voltage gain.

In this method, the base resistor R_{B} has its one end connected to base and the other to the collector as its name implies. In this circuit, the zero signal base current is determined by V_{CB} but not by V_{CC}.

It is clear that V_{CB} forward biases the base-emitter junction and hence base current I_{B} flows through R_{B}. This causes the zero signal collector current to flow in the circuit. The below figure shows the biasing with collector feedback resistor circuit.

The required value of R_{B} needed to give the zero signal current I_{C} can be determined as follows.

$$V_{CC} = I_C R_C + I_B R_B + V_{BE}$$

Or

$$R_B = \frac{V_{CC} - V_{BE} - I_C R_C}{I_B}$$

$$= \frac{V_{CC} - V_{BE} - \beta I_B R_C}{I_B}$$

Since $I_C = \beta I_B$

Alternatively,

$$V_{CE} = V_{BE} + V_{CB}$$

Or

$$V_{CB} = V_{CE} - V_{BE}$$

Since

$$R_B = \frac{V_{CB}}{I_B} = \frac{V_{CE} - V_{BE}}{I_B}$$

Where

$$I_B = \frac{I_C}{\beta}$$

Mathematically,

Stability factor, $S < (\beta + 1)$

Therefore, this method provides better thermal stability than the fixed bias.

The Q-point values for the circuit are shown as

$$I_C = \frac{V_{CC} - V_{BE}}{R_B/ \beta + R_C}$$

$$V_{CE} = V_{CC} - I_C R_C$$

- The circuit is simple as it needs only one resistor.
- This circuit provides some stabilization, for lesser changes.

- The circuit doesn’t provide good stabilization.
- The circuit provides negative feedback.

Among all the methods of providing biasing and stabilization, the **voltage divider bias method** is the most prominent one. Here, two resistors R_{1} and R_{2} are employed, which are connected to V_{CC} and provide biasing. The resistor R_{E} employed in the emitter provides stabilization.

The name voltage divider comes from the voltage divider formed by R_{1} and R_{2}. The voltage drop across R_{2} forward biases the base-emitter junction. This causes the base current and hence collector current flow in the zero signal conditions. The figure below shows the circuit of voltage divider bias method.

Suppose that the current flowing through resistance R_{1} is I_{1}. As base current I_{B} is very small, therefore, it can be assumed with reasonable accuracy that current flowing through R_{2} is also I_{1}.

Now let us try to derive the expressions for collector current and collector voltage.

From the circuit, it is evident that,

$$I_1 = \frac{V_{CC}}{R_1 + R_2}$$

Therefore, the voltage across resistance R_{2} is

$$V_2 = \left ( \frac{V_{CC}}{R_1 + R_2}\right ) R_2$$

Applying Kirchhoff’s voltage law to the base circuit,

$$V_2 = V_{BE} + V_E$$

$$V_2 = V_{BE} + I_E R_E$$

$$I_E = \frac{V_2 - V_{BE}}{R_E}$$

Since I_{E} ≈ I_{C},

$$I_C = \frac{V_2 - V_{BE}}{R_E}$$

From the above expression, it is evident that I_{C} doesn’t depend upon β. V_{BE} is very small that I_{C} doesn’t get affected by V_{BE} at all. Thus I_{C} in this circuit is almost independent of transistor parameters and hence good stabilization is achieved.

Applying Kirchhoff’s voltage law to the collector side,

$$V_{CC} = I_C R_C + V_{CE} + I_E R_E$$

Since I_{E} ≅ I_{C}

$$= I_C R_C + V_{CE} + I_C R_E$$

$$= I_C(R_C + R_E) + V_{CE}$$

Therefore,

$$V_{CE} = V_{CC} - I_C(R_C + R_E)$$

R_{E} provides excellent stabilization in this circuit.

$$V_2 = V_{BE} + I_C R_E$$

Suppose there is a rise in temperature, then the collector current I_{C} decreases, which causes the voltage drop across R_{E} to increase. As the voltage drop across R_{2} is V_{2}, which is independent of I_{C}, the value of V_{BE} decreases. The reduced value of I_{B} tends to restore I_{C} to the original value.

The equation for **Stability factor** of this circuit is obtained as

Stability Factor = $S = \frac{(\beta + 1) (R_0 + R_3)}{R_0 + R_E + \beta R_E}$

$$= (\beta + 1) \times \frac{1 + \frac{R_0}{R_E}}{\beta + 1 + \frac{R_0}{R_E}}$$

Where

$$R_0 = \frac{R_1 R_2}{R_1 + R_2}$$

If the ratio R_{0}/R_{E} is very small, then R0/RE can be neglected as compared to 1 and the stability factor becomes

Stability Factor = $S = (\beta + 1) \times \frac{1}{\beta + 1} = 1$

This is the smallest possible value of S and leads to the maximum possible thermal stability.

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