Chapter 5, Problem 5.9EP

The circuit elements in Figure 5.36(a) are $V^{+}=5\text{V}$ , $V_{BB}=-2\text{V}$ , $R_E=2\text{kΩ}$ , and $R_B=180\text{kΩ}$ . Assume $V_{EB}\text{(on)}=0.7\text{V}$ . Plot the Q-point on the load line for (a) $\beta =40$ , (b) $\beta =60$ , (c) $\beta =100$ , and (d) $\beta =150$ . (Ans. (a) $I_{CQ}=0.962\text{mA}$ , (b) $I_{CQ}=1.25\text{mA}$ , (c) $I_{CQ}=1.65\text{mA}$ , (d) $I_{CQ}=1.96\text{mA}$ )

To determine

To plot: The Q -point on the load line for the given circuit.

Answer to Problem 5.9EP

  

The Q-point is at $V_{ECQ}=3.028\text{ V}$ and $I_{CQ}=0.962\text{ mA}$ .

Explanation of Solution

Given Information:

  $V^{+}=5\text{ V,}{V}_{BB}=-2\text{ V, }{R}_E=2\text{ k}\Omega ,R_B=180\text{ k}\Omega \text{ , }\beta =\text{40 and }{V}_{EB(on)}=0.7\text{ V}$

Calculation:

Find the quiescent collector current and identify the load line equation. Then find the quiescent emitter to collector voltage. The figure shows the circuit

  

Applying Kirchhoff’s voltage law around E-B loop,

  $\begin{array}{l}{V}^{+}=I_{EQ}{R}_E\text{+ }{V}_{EB(on)}+I_{BQ}{R}_B+V_{BB}\\ V^{+}=\frac{1+\beta }{\beta }{I}_{CQ}{R}_E\text{+ }{V}_{EB(on)}+\frac{1}{\beta }{I}_{CQ}{R}_B+V_{BB}\end{array}$

The quiescent collector current is

  $I_{CQ}=\beta \left(\frac{V^{+}-V_{EB(on)}-V_{BB}}{R_B+\left(1+\beta \right)R_E}\right)$

  $I_{CQ}=40\left(\frac{5-0.7-\left(-2\right)}{180+\left(1+40\right)2}\right)\text{ mA}$

  $I_{CQ}=0.962\text{ mA}$

The load line is given by

  $\begin{array}{l}{V}_{EC}=V^{+}-I_E{R}_E\\ V_{EC}=V^{+}-\left(\frac{1+\beta }{\beta }\right)I_C{R}_E\\ V_{EC}=5-2.05I_C(1)\end{array}$

Then,

  $V_{ECQ}=5-2.05I_{CQ}=5-2.05\times 0.962\text{ V}$

  $V_{ECQ}=3.028\text{ V}$

Substituting $V_{EC}=5$ in equation (1)

  $I_C=0$

Substituting $V_{EC}=0$ in equation (1)

  $I_C=2.439\text{ mA}$

Then, draw the load line and mark the Q-point (red)on it as below.

  

To determine

To plot: The Q-point on the load line for the given circuit.

Answer to Problem 5.9EP

  

The Q -point is at $V_{ECQ}=2.46\text{ V}$ and $I_{CQ}=1.25\text{ mA}$ .

Explanation of Solution

Given Information:

  $V^{+}=5\text{ V,}{V}_{BB}=-2\text{ V, }{R}_E=2\text{ k}\Omega ,R_B=180\text{ k}\Omega \text{ , }\beta =60\text{ and }{V}_{EB(on)}=0.7\text{ V}$

Calculation:

Find the quiescent collector current and identify the load line equation. Then find the quiescent emitter to collector voltage. The figure shows the circuit

  

Applying Kirchhoff’s voltage law around E-B loop,

  $\begin{array}{l}{V}^{+}=I_{EQ}{R}_E\text{+ }{V}_{EB(on)}+I_{BQ}{R}_B+V_{BB}\\ V^{+}=\frac{1+\beta }{\beta }{I}_{CQ}{R}_E\text{+ }{V}_{EB(on)}+\frac{1}{\beta }{I}_{CQ}{R}_B+V_{BB}\end{array}$

The quiescent collector current is

  $I_{CQ}=\beta \left(\frac{V^{+}-V_{EB(on)}-V_{BB}}{R_B+\left(1+\beta \right)R_E}\right)$

  $I_{CQ}=60\left(\frac{5-0.7-\left(-2\right)}{180+\left(1+60\right)2}\right)\text{ mA}$

  $I_{CQ}=1.25\text{ mA}$

The load line is given by

  $\begin{array}{l}{V}_{EC}=V^{+}-I_E{R}_E\\ V_{EC}=V^{+}-\left(\frac{1+\beta }{\beta }\right)I_C{R}_E\end{array}$

  $V_{EC}=5-\frac{122}{60}{I}_C\to (1)$

Then,

  $V_{ECQ}=5-\frac{122}{60}{I}_{CQ}=5-\frac{122}{60}\times 1.25\text{ V}$

  $V_{ECQ}=2.46\text{ V}$

Substituting $V_{EC}=5$ in equation (1)

  $I_C=0$

Substituting $V_{EC}=0$ in equation (1)

  $I_C=2.459\text{ mA}$

Then, draw the load line and mark the Q-point (red)on it as below.

  

To determine

To plot: The Q -point on the load line for the given circuit.

Answer to Problem 5.9EP

  

The Q-point is at $V_{ECQ}=1.67\text{ V}$ and $I_{CQ}=1.65\text{ mA}$ .

Explanation of Solution

Given Information:

  $V^{+}=5\text{ V,}{V}_{BB}=-2\text{ V, }{R}_E=2\text{ k}\Omega ,R_B=180\text{ k}\Omega \text{ , }\beta =\text{100 and }{V}_{EB(on)}=0.7\text{ V}$

Calculation:

Find the quiescent collector current and identify the load line equation. Then find the quiescent emitter to collector voltage. The figure shows the circuit

  

Applying Kirchhoff’s voltage law around E-B loop,

  $\begin{array}{l}{V}^{+}=I_{EQ}{R}_E\text{+ }{V}_{EB(on)}+I_{BQ}{R}_B+V_{BB}\\ V^{+}=\frac{1+\beta }{\beta }{I}_{CQ}{R}_E\text{+ }{V}_{EB(on)}+\frac{1}{\beta }{I}_{CQ}{R}_B+V_{BB}\end{array}$

The quiescent collector current is

  $I_{CQ}=\beta \left(\frac{V^{+}-V_{EB(on)}-V_{BB}}{R_B+\left(1+\beta \right)R_E}\right)$

  $I_{CQ}=100\left(\frac{5-0.7-\left(-2\right)}{180+\left(1+100\right)2}\right)\text{ mA}$

  $I_{CQ}=1.65\text{ mA}$

The load line is given by

  $V_{EC}=V^{+}-I_E{R}_E$

  $V_{EC}=V^{+}-\left(\frac{1+\beta }{\beta }\right)I_C{R}_E$

  $V_{EC}=5-2.02I_C\to (1)$

Then,

  $V_{ECQ}=5-2.02I_{CQ}=5-2.02\times 1.65\text{ V}$

  $V_{ECQ}=1.67\text{ V}$

Substituting $V_{EC}=5$ in equation (1)

  $I_C=0$

Substituting $V_{EC}=0$ in equation (1) we get

  $I_C=2.4752\text{ mA}$

Then we can draw the load line and mark the Q-point (red)on it as below.

  

To determine

To plot: TheQ -point on the load line for the given circuit.

Answer to Problem 5.9EP

  

The Q -point is at $V_{ECQ}=1.05\text{ V}$ and $I_{CQ}=1.96\text{ mA}$ .

Explanation of Solution

Given Information:

  $V^{+}=5\text{ V,}{V}_{BB}=-2\text{ V, }{R}_E=2\text{ k}\Omega ,R_B=180\text{ k}\Omega \text{ , }\beta =150\text{ and }{V}_{EB(on)}=0.7\text{ V}$

Calculation:

Find the quiescent collector current and identify the load line equation. Then find the quiescent emitter to collector voltage. The figure shows the circuit

  

Applying Kirchhoff’s voltage law around E-B loop,

  $\begin{array}{l}{V}^{+}=I_{EQ}{R}_E\text{+ }{V}_{EB(on)}+I_{BQ}{R}_B+V_{BB}\\ V^{+}=\frac{1+\beta }{\beta }{I}_{CQ}{R}_E\text{+ }{V}_{EB(on)}+\frac{1}{\beta }{I}_{CQ}{R}_B+V_{BB}\end{array}$

The quiescent collector current is

  $I_{CQ}=\beta \left(\frac{V^{+}-V_{EB(on)}-V_{BB}}{R_B+\left(1+\beta \right)R_E}\right)$

  $I_{CQ}=150\left(\frac{5-0.7-\left(-2\right)}{180+\left(1+150\right)2}\right)\text{ mA}$

  $I_{CQ}=1.96\text{ mA}$

The load line is given by

  $\begin{array}{l}{V}_{EC}=V^{+}-I_E{R}_E\\ V_{EC}=V^{+}-\left(\frac{1+\beta }{\beta }\right)I_C{R}_E\end{array}$

  $V_{EC}=5-\frac{302}{150}{I}_C\to (1)$

Then,

  $V_{ECQ}=5-\frac{302}{150}{I}_{CQ}=5-\frac{302}{150}\times 1.96\text{ V}$

  $V_{ECQ}=1.05\text{ V}$

Substituting $V_{EC}=5$ in equation (1)

  $I_C=0$

Substituting $V_{EC}=0$ in equation (1)

  $I_C=2.4834\text{ mA}$

Then we can draw the load line and mark the Q -point (red)on it as below.