As shown before, the DC operating point of a transistor amplification circuit needs to be set up properly (in the middle of the linear region) to avoid signal distortion. We now consider how the operating point is determined by the biasing circuit, in terms of , , and .

**Fixed Biasing**

By properly setting the voltage (not too low) and (not too large), the voltage can be approximated as a constant value of , as shown in the input characteristic plot:

Then the base current can be estimated to be:

(29) |

(30) |

(31) |

**Example 1**

In the fixed biasing transistor circuit shown above, , , , find so that the DC operating point is in the middle of the linear region of the output characteristic plot, i.e., . We assume (may not be valid if is too large) and get

(32) |

i.e. | (33) |

i.e. | (34) |

(35) |

**Example 2:**

In the same circuit above, , , . Find the operating point for .

The load line is determined by these two points:

- Open-circuit voltage:
- Short-circuit current:

(36) |

(37) |

To minimize distortion, the DC operating point needs to be in the middle of the load line at . But in this case, we see that

- , too close to cutoff region.
- , in the middle of linear region as desired.
- , too close to the saturation region.

The DC operating point of this fixed biasing circuit is not completely determined by the parameters of the circuit such as the resistors, as it is also directly affected by factors such as value and temperature. This situation can be improved by introducing negative feedback into the circuit.

**Self-Biasing**

To correct the problem above, the self-biasing circuit shown below can be used to decrease the effect of changing by negative feed back due to the introduction of .

Qualitatively, an increased (caused by reasons such as increased due to temperature change) will cause the following to happen:

(38) |

Quantitatively, we can further carry out analysis of the circuit:

- If the resistances of and are small so that the
current through is much larger than the base current
i.e.,
, then the basis voltage can be
approximated to be (voltage divider):
(39) (40) (41) - If the condition
is not satisfied, the method
above is no longer valid. In this case, we can use Thevenin's theorem
to replace the base circuit by an open circuit voltage
(already found above), in series with the internal resistance :
(42) Applying KVL to the base loop we get

(43) and (44) (45) For this approximation above to be valid, we desire to have smaller so that is less affected by , and large for stronger negative feedback. However, as the voltage gain of the circuit will be reduced due to the negative feedback, cannot be too large.

**Example 3:**

In the circuit of self-biasing, , , , , , Assume . The load line is determined by this equation:

(46) |

- and (short-circuit current)
- and (open-circuit voltage)

To minimize distortion, the desired operating point should be in the middle of the load line at and .

- Based on the voltage divider approximation, we get the DC
operating point independent of :

(47)

- Based on the Thevenin theorem, we get more accurate results:

(48)

The DC operating points corresponding to the three values can be found to be:(49)

We see that in all three cases, , , i.e., the DC operating point is always close to the middle of the load line.

**Example 4**

In a self-biasing transistor circuit, , , , , find so that the DC operating point is in the middle of the linear region of the output characteristic plot.

We first convert the base circuit into its Thevenin's equivalent voltage source composed of

(50) |

(51) |

(52) |

**Example 5**

The circuit below shows yet another way to introduce feedback to stablize the DC operating point.

- The resistor connecting the collector to the base forms
a feedback from the output to as well as providing
the forward baising needed for the base-emitter PN junction:
(53) - The DC operating point can be found by applying KVL:
(54) (55) (56) - Given
, , and a desired , find
and so that the DC operating point is in the middle of the
linear region.
We want the DC operating point to be at and , and get

(57) (58) (59) - Following the equations given above, we can find the DC
operating pointt for each of the three values, which
are all approximately in the middle of the linear region:
(60)