Chapter 5, Problem D5.31P

(a) The bias voltage in the circuit in Figure P5.3 1 is changed to $V_{CC}=9\text{V}$ . The transistor current gain is $\beta =80$ . Design the circuit such that $I_{CQ}\text{=0}\text{.25mA}$ and $V_{CEQ}=4.5\text{V}$ . (b) If the transistor is replaced by a new one with $\beta =120$ , find the new values of $I_{CQ}$ and $V_{CEQ}$ . (c) Sketch the load line and Q−point for both parts (a) and (b).

Figure P5.31

To determine

The design parameters of the circuit.

Answer to Problem D5.31P

The required values are

  $I_{BQ}=3.125\mu A$

  $R_C=18k\Omega$

  $R_B=2.656M\Omega$

  $I_{EQ}=0.2531mA$

Explanation of Solution

Given:

The bias voltage is $V_{CC}=9V$ .

The transistor current gain is $\beta =80$

  $I_{CQ}=0.25mA$ and $V_{CEQ}=4.5V$

  

Calculation:

Considering the BJT (Bipolar Junction Transistor) as single node, then by Kirchhoff’s current law the emitter current is:

  $I_E=I_B+I_C(1)$

We know that in common emitter mode, the current gain is related as

  $\begin{array}{l}{I}_{CQ}=\beta I_{BQ}(2)\\ I_{BQ}=\frac{I_{CQ}}{\beta }\\ I_{BQ}=\frac{0.25\times {10}^{-3}}{80}\\ {\text{I}}_{\text{BQ}}\text{=3}\text{.125μA}\end{array}$

Applying Kirchhoff’s voltage law to the input circuit

  $\begin{array}{l}-I_{BQ}{R}_B-V_{BE(on)}+V_{CC}=0(3)\\ R_B=\frac{V_{CC}-V_{BE(on)}}{I_{BQ}}\\ R_B=\frac{9-0.7}{3.125\times {10}^{-6}}\\ R_B=\frac{8.3}{3.125}\times {10}^{-6}\\ {\text{R}}_{\text{B}}\text{=2}\text{.656MΩ}\text{(4)}\end{array}$

From equation (1), quiescent values as

  $\begin{array}{l}{I}_{EQ}=I_{BQ}+I_{CQ}\\ I_{EQ}=\left(3.125\times {10}^{-3}+0.25\right)\times {10}^{-3}\\ \therefore I_{EQ}=0.2531\text{mA}\end{array}$

Applying Kirchhoff’s voltage law at the output for the given circuit as

  $\begin{array}{l}-I_{CQ}{R}_C-V_{CEQ(on)}+V_{CC}=0(5)\\ R_C=\frac{V_{CC}-V_{CEQ}}{I_{CQ}}\\ R_C=\frac{9-4.5}{0.25\times {10}^{-3}}\\ R_C=\frac{4.5}{0.25\times {10}^{-3}}\\ R_C\text{=18kΩ}\text{(6)}\end{array}$

Conclusion:

Therefore, the required values are

  $I_{BQ}=3.125\mu A$

  $R_C=18k\Omega$

  $R_B=2.656M\Omega$

  $I_{EQ}=0.2531mA$

To determine

To find: the new values of $I_{CQ}$ and $V_{CEQ}$ .

Answer to Problem D5.31P

The required values are

  ${\text{V}}_{\text{CEQ}}\text{=2}\text{.25V}$

  ${\text{I}}_{\text{CQ}}\text{=0}\text{.375mA}$

Explanation of Solution

Given:

  $\beta =120$

Calculation:

From equation (4) and (6),

  $R_B=2.656M\Omega$ , $R_C=18M\Omega$ and $\beta =120$

From equation (3),

  $\begin{array}{l}-I_{BQ}{R}_B-V_{BE(on)}+V_{CC}=0\\ I_{BQ}=\frac{V_{CC}-V_{BE(on)}}{R_B}\\ I_{BQ}=\frac{9-0.7}{2.656\times {10}^6}\\ I_{BQ}=\frac{8.3}{2.656}\times {10}^6\\ \therefore I_{BQ}=3.125\text{μA}\end{array}$

From equation (2),

  $\begin{array}{l}{I}_{CQ}=\beta I_{BQ}\\ I_{CQ}=(120)(3.125\times {10}^{-6})\\ \therefore I_{CQ}=0.375\text{mA}\end{array}$

From equation (5),

  $\begin{array}{l}{V}_{CEQ}=V_{CC}-I_{CQ}{R}_C\\ V_{CEQ}=9-(0.375\times {10}^{-3})(18\times {10}^3)\\ V_{CEQ}=9-6.75\\ \therefore V_{CEQ}=2.25\text{V}\end{array}$

Conclusion:

Therefore, the required values are

  $V_{CEQ}=2.25V$

  $I_{CQ}=0.375mA$

To determine

To sketch: the load the line and $Q$ -point for both parts (a) and (b).

Answer to Problem D5.31P

When $\beta =80$

  

When $\beta =120$

  

Explanation of Solution

Given:

The bias voltage is $V_{CC}=9V$ .

The transistor current gain is $\beta =80$

  $I_{CQ}=0.25mA$ and $V_{CEQ}=4.5V$

  

Calculation:

When $\beta =80$

Thus, the load line equation is

From equation (5) can be written as

  $\begin{array}{l}-I_C{R}_C-V_{CE}+V_{CC}=0\\ V_{CE}=V_{CC}-I_C{R}_C\\ V_{CE}=9-18\times {10}^3{l}_C\end{array}$

Now, put $I_C=0A$ in above equation then

  $\begin{array}{l}{V}_{CE}=9-18\times {10}^3(0)\\ \therefore V_{CE}=9\text{V}\end{array}$

Now, put $V_{CE}=0V$ in above equation then

  $\begin{array}{l}0=9-18\times {10}^3{l}_C\\ 18\times {10}^3{l}_C=9\\ \therefore l_C=0.5\text{mA}\end{array}$

Thus, the co-ordinates of $Q$ -points are

  $V_{CEQ}=4.5\text{V}$ and $I_{CQ}=0.25\text{mA}$

Thus, the load line and $Q$ -Point are plotted as

  

When $\beta =120$

Thus, the load line equation is

From equation (5) can be written as

  $\begin{array}{l}-I_C{R}_C-V_{CE}+V_{CC}=0\\ V_{CE}=V_{CC}-I_C{R}_C\\ V_{CE}=9-18\times {10}^3{l}_C\end{array}$

Now, put $I_C=0A$ in above equation then

  $\begin{array}{l}{V}_{CE}=9-18\times {10}^3(0)\\ \therefore V_{CE}=9\text{V}\end{array}$

Now, put $V_{CE}=0V$ in above equation then

  $\begin{array}{l}0=9-18\times {10}^3{l}_C\\ 18\times {10}^3{l}_C=9\\ \therefore l_C=0.5\text{mA}\end{array}$

Thus, the co-ordinate of $Q$ -points are

  $V_{CEQ}=2.25V$ and $I_{CQ}=0.375mA$

Thus, the load and $Q$ -Point are plotted as