Chapter 5, Problem 5.79P

For each transistor in the circuit in Figure P5.79, $\beta =120$ and the B−E turn on voltage is 0.7 V. Determine the quiescent base, collector, and emitter currents in Q and $Q_2$ . Also determine $V_{CEQ1}$ and $V_{CEQ2}$ .

Figure P5.79

To determine

The quiescent base collector, and emitter current foe each transistor in given circuit and also $V_{CEQ1}$ and $V_{CE{Q}_2}$ in $Q_1$ and $Q_2$ .

Answer to Problem 5.79P

  $\begin{array}{l}{\text{I}}_{\text{BQ1}}\text{=0}\text{.0144}\text{mA}\\ {\text{I}}_{\text{BQ2}}=0.0232\text{mA}\\ {\text{I}}_{\text{CQ1}}\text{=1}\text{.73}\text{mA}\\ {\text{I}}_{\text{CQ2}}\text{=2}\text{.785}\text{mA}\\ {\text{I}}_{\text{EQ1}}\text{=1}\text{.75}\text{mA}\\ {\text{I}}_{\text{EQ2}}\text{=2}\text{.808 mA}\\ {\text{V}}_{\text{CEQ1}}\text{=2}\text{.99}\text{V}\\ {\text{V}}_{\text{CEQ2}}\text{=5}\text{.96}\text{V}\end{array}$

Explanation of Solution

Given:

  $\begin{array}{l}\text{β=120}\\ {\text{V}}_{\text{BE}}\text{=0}\text{.7}\text{V}\end{array}$

The given circuit is:

  

Drawing the Thevenin equivalent circuit according to the given fig.

  

Now finding the Thevenin resistance ${\text{R}}_{\text{TH}}$ .

  ${\text{R}}_{\text{TH}}=R_1\parallel R_2$

Putting the values for ${\text{R}}_{\text{1}}$ and ${\text{R}}_{\text{2}}$ .

  $\begin{array}{l}{R}_{TH}=100\times {10}^3\parallel 40\times {10}^3\\ =\frac{100\times {10}^3\times 40\times {10}^3}{100\times {10}^3+40\times {10}^3}\\ =\text{28}\text{.6}\text{kΩ}\end{array}$

Now evaluating Thevenin voltage ${\text{V}}_{\text{TH}}$ :

  $V_{TH}=\left(\frac{R_2}{R_1+R_2}\right)\left(10\right)$

Putting the value of ${\text{R}}_{\text{1}}$ and ${\text{R}}_{\text{2}}$ .

  $\begin{array}{l}{V}_{TH}=\left(\frac{40\times {10}^3}{100\times {10}^3+40\times {10}^3}\right)\left(10\right)\\ =\left(\frac{40}{140}\right)\left(10\right)\\ =2.8\text{6}\text{V}\end{array}$

Now evaluating ${\text{I}}_{\text{BQ1}}$ .

  $I_{BQ1}=\frac{V_{TH}-V_{BE}\left(\text{on}\right)}{R_{TH}+\left(1+\beta \right)R_{E1}}$

Putting all values:

  $\begin{array}{l}{I}_{BQ1}=\frac{2.86-0.7}{28.6\times {10}^3+\left(1+120\right){10}^3}\\ =\frac{2.16}{149.6}\times {10}^{-3}\\ =0.0144\text{mA}\end{array}$

Now evaluating ${\text{I}}_{\text{CQ1}}$ :

  $I_{CQ1}=\beta I_{BQ1}$

  $\begin{array}{l}{I}_{CQ1}=120\times 0.0144\times {10}^{-3}\\ =1.73\text{mA}\end{array}$

Now evaluating ${\text{I}}_{\text{EQ1}}$ ;

  ${\text{I}}_{EQ1}=I_{BQ1}+I_{CQ1}$

Putting all values;

  $\begin{array}{l}{\text{I}}_{EQ1}=0.0144\times {10}^{-3}+1.73\times {10}^{-3}\\ =1.75\text{mA}\end{array}$

Now applying Kirchhoff’s current law at ${\text{Q}}_{\text{2}}$ ;

  $\begin{array}{l}\frac{10-V_{BQ2}}{R_{C1}}=I_{CQ1}+I_{BQ2}\\ \frac{10-V_{BQ2}}{R_{C1}}=I_{CQ1}+\frac{I_{EQ2}}{\left(1+\beta \right)}\mathrm{...........}\left(1\right)\end{array}$

Again applying Kirchhoff’s voltage law at ${\text{Q}}_{\text{2}}$ ;

  $I_{EQ2}=\frac{V_{BQ2}-V_{BE}\left(\text{on}\right)-\left(-10\right)}{R_{E2}}\mathrm{.............}\left(2\right)$

Now substituting the value of ${\text{I}}_{\text{EQ2}}$ in equation $\left(1\right)$

  $\begin{array}{l}\frac{10-V_{BQ2}}{R_{C1}}=I_{CQ1}+\frac{V_{BQ2}-V_{BE}\left(\text{on}\right)-\left(-10\right)}{\left(1+\beta \right)R_{E2}}\\ \frac{10-V_{BQ2}}{R_{C1}}=I_{CQ1}+\frac{V_{BQ2}-V_{BE}\left(\text{on}\right)+10}{\left(1+\beta \right)R_{E2}}\end{array}$

Putting all values now;

  $\begin{array}{l}\frac{10-V_{BQ2}}{3\times {10}^3}=1.73\times {10}^{-3}+\frac{V_{BQ2}-0.7+10}{\left(121\right)\times 5\times {10}^3}\\ \frac{10-V_{BQ2}}{3}=1.73+\frac{V_{BQ2}+9.3}{121\times 5}\\ V_{BQ2}\left(\frac{1}{5\times 121}+\frac{1}{3}\right)=\frac{10}{3}-1.73-\frac{9.3}{121\times 5}\\ \end{array}$

After further simplification;

  $\begin{array}{l}{V}_{BQ2}\left(0.335\right)=1.588\\ V_{BQ2}=4.74\text{V}\end{array}$

Now from equation $\left(2\right)$ ;

  $\begin{array}{l}{I}_{EQ2}=\frac{V_{BQ2}-V_{BE}\left(\text{on}\right)-\left(-10\right)}{R_{E2}}\\ I_{EQ2}=\frac{4.74-0.7+10}{5\times {10}^3}\\ =2.808\text{mA}\end{array}$

Now calculating ${\text{I}}_{\text{BQ2}}$ ;

  $\begin{array}{l}{\text{I}}_{BQ2}=\frac{I_{EQ2}}{\left(1+\beta \right)}\\ {\text{I}}_{BQ2}=\frac{2.808\times {10}^{-3}}{\left(1+120\right)}\\ =0.0232\text{mA}\\ \end{array}$

Now evaluating ${\text{I}}_{\text{CQ2}}$ ;

  $\begin{array}{l}{I}_{CQ2}=\beta I_{BQ2}\\ I_{CQ2}=120\times 0.0232\times {10}^{-3}\\ =2.785\text{mA}\end{array}$

Now evaluating ${\text{V}}_{\text{CEQ1}}$ ;

Applying Kirchhoff’s Voltage law at ${\text{Q}}_{\text{1}}$ ;

  $\begin{array}{l}{V}_{CEQ1}=V_{CQ1}-I_{EQ1}{R}_{E1}\\ =V_{BQ2}-I_{EQ1}{R}_{E1}\\ =4.74-1.75\times {10}^{-3}\times {10}^3\\ =2.99\text{V}\end{array}$

Now evaluating ${\text{V}}_{\text{CEQ2}}$ ;

Applying Kirchhoff’s Voltage law at ${\text{Q}}_{\text{2}}$ ;

  $\begin{array}{l}{V}_{CEQ2}=10-\left[V_{BQ2}-V_{BE}\left(\text{on}\right)\right]\\ V_{CEQ2}=10-\left[4.74-0.7\right]\\ =5.96\text{V}\end{array}$