For the circuit shown in Figure P5.52, let
Figure P5.52
a.
The ICQ and VCEQ and then sketching the load line and plotting the Q-point.
Answer to Problem 5.52P
The ICQ and VCEQ :
ICQ =0.9128mA.
VCEQ =14.87V.
Explanation of Solution
Given:
The value of the attenuation factor,
The circuit diagram is shown below:
The Thevenin resistance is evaluated as:
Applying the voltage division rule to evaluate the Thevenin voltage:
Redrawing the given circuit:
Applying the Kirchhoff’s voltage law in the base-emitter loop:
By the expression of the common emitter current gain:
The quiescent collector emitter voltage is given as:
Hence, the load line equation is given as:
The coordinates of the end points of the load line.
Since, VCE =0-V.
Hence,
Now, substituting IC =0A, then:
Hence, the coordinates of the two extremities of the load line is:
Sketching the load line graph:
b.
The range of the ICQ and VCEQ and plotting the various Q-point on the load line.
Explanation of Solution
Given:
The value of the attenuation factor,
The circuit diagram is shown below:
The resistor R1 and R2 vary by the
Now, evaluating the range of the resistor:
Now, evaluating the range of the R2:
Now taking
as R1 and R2 respectively:
Evaluating the Thevenin resistance using the equation 1:
Applying the voltage division rule to evaluate the Thevenin voltage using the equation 2:
By the use of the equation 3:
Now, from the equation 4:
Now taking
as R1 and R2 respectively:
Evaluating the Thevenin resistance using the equation 1:
Applying the voltage division rule to evaluate the Thevenin voltage using the equation 2:
By the use of the equation 3:
Now, from the equation 4:
Now taking
as R1 and R2 respectively:
Evaluating the Thevenin resistance using the equation 1:
Applying the voltage division rule to evaluate the Thevenin voltage using the equation 2:
By the use of the equation 3:
Now, from the equation 4:
Hence, the plot for the various Q points on the load line is:
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Chapter 5 Solutions
Microelectronics: Circuit Analysis and Design
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