Figure shows the voltage divider bias circuit. In this, biasing is provided by three resistors R1, R2 and RE.

**Voltage divider bias circuit:**

Figure
shows the voltage divider bias circuit. In this, biasing is provided by three
resistors R_{1}, R_{2} and R_{E.} The resistors R_{1}&
R_{2} act as a potential divider giving a fixed voltage to base. If
collector current increases due to change in temperature or change in β,
emitter current I_{E} also increases and voltage drop across R_{E}
increases thus reducing the voltage difference between base and emitter. Due to
reduction in base emitter voltage, base current and collector current reduces.
So we can say that negative feedback exists in emitter bias circuit. This
reduction in collector current compensates for the original change in I_{C}.

**Circuit analysis:**

**Base circuit:**

Fig. Base
circuit

Let us
consider the base circuit as shown in above figure. Voltage across R_{2}
is base voltage V_{B}. Applying voltage divider rule to find V_{B}

_{ }

**Collector circuit:**

Let us
consider the collector circuit as shown in above figure. Voltage across R_{E}
can be obtained as,

**Simplified circuit of voltage divider bias**

Fig.Thevenin’s
equivalent circuit for voltage divider bias

From
above figure, R_{1} and R_{2} are replaced by R_{B} and
V_{T}.

Where R_{B}
is the parallel combination of R=1 and R_{2}

V_{T}
is the thevenin’s voltage

**Problem 1: **

For the given circuit β=100 for silicon transistor. Calculate V_{CE} and I_{C}.

**Problem 2: **

For the
given figure find Q point with V_{CC} = 15V, V_{BE} = 0.7V and
β = 100.

**Stability factor for voltage divider bias: **

**Stability factor S:**

For
determining stability factor S for voltage divider bias, consider the
equivalent circuit. Thevenin’s voltage is given by,

R_{1},
R_{2} are replaced by R_{B} which is the parallel combination
of R_{1} and R_{2}.

Apply KVL
to base circuit,

Differentiating
with respect to I_{C} and considering V_{BE} to be independent
of I_{C,}

From
above equation, the following points are observed.

1. The
ratio R_{B} /R_{E} controls value of stability factor S. If R_{B}/R_{E}<<
1 then it is reduced to S = (1+β). 1/ (1+β) = 1

Practically
R_{B}/R_{E} not equal to zero. But to have better stability
factor S , we have to keep ratio R_{B}/R_{E} as small as
possible.

2. To keep R_{B}/R_{E}
small, it is necessary to keep R_{B} small. Due to small value of R_{1}
and R_{2}, potential divider circuit will draw more current from V_{CC}
reducing the life of the battery. Another important aspect is that reducing R_{B}
will reduce input impedance of the circuit, since R_{B} comes in
parallel with the input. This reduction of input impedance in amplifier circuit
is not desirable and hence R_{B} cannot be made very small.

3. Emitter
resistance R_{E} is another parameter, it is used to decrease the ratio
R_{B}/R_{E}. Drop across R_{C} will reduce. This shifts
the operating point Q which is not desirable and hence there is limit for
increasing R_{E}.

While
designing voltage divider bias circuit, the following conditions are to be
satisfied,

S – Small

RB - Reasonably small

RE - Not very large

4. If ratio
R_{B}/R_{E} is fixed, S
increases with β. So stability decreases with increasing β.

5. Stability
factor S is essentially independent of β for small value of S.

Substituting
the differentiation value of I_{B} /I_{C},

**Problem 1:**

For the
given circuit, V_{CC} = 20V, R_{C} = 2KΩ, β = 50, V_{BE}
= 0.2V, R_{1} = 100KΩ, R_{E} = 100Ω. Calculate I_{B}, V_{CE},
I_{C} and stability factor S.

Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail

Electronic Circuits : Biasing of Discrete BJT and MOSFET : Voltage divider bias circuit |

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