Chapter 5, Problem 5.32P

The current gain of the transistor in the circuit shown in Figure P5.32 is $\beta =150$ . Determine $I_C,I_E$ , and $V_C$ . for (a) $V_B=0.2\text{V}$ , (b) $V_B=0.9\text{V}$ , (c) $V_B=1.5\text{V}$ , and (d) $V_B=2.2\text{V}$ .

Figure P5.32

To determine

The values of $I_C,I_E,V_C$ for the given circuit.

Answer to Problem 5.32P

  $\begin{array}{l}{V}_C=6\text{V}\\ I_C=0\\ I_E=0\end{array}$

Explanation of Solution

Given Information:

  

  $\begin{array}{l}\beta =150\\ V_B=0.2\text{V}\end{array}$

Calculation:

Assuming the NPN transistor operates in cutoff region.

The value of $V_{BE}$ is:

  $\begin{array}{l}{V}_{BE}=V_B-V_E\\ =0.2-0\\ =0.2\text{V}\\ V_{BE}<0.7\text{V}\end{array}$

Hence, the assumption is correct, and transistor operates in cutoff region.

  $\begin{array}{l}{I}_C=0\\ I_E=0\end{array}$

The value of collector voltage is:

  $\begin{array}{l}{V}_C=6-I_C{R}_C\\ =6-0\times R_C\\ =6\text{V}\end{array}$

To determine

The values of $I_C,I_E,V_C$ for the given circuit.

Answer to Problem 5.32P

  $\begin{array}{l}{I}_C=0.199\text{mA}\\ I_E=0.2\text{mA}\\ V_C=4\text{V}\end{array}$

Explanation of Solution

Given Information:

  

  $\begin{array}{l}\beta =150\\ V_B=0.9\text{V}\end{array}$

Calculation:

The value of emitter current is:

Applying Kirchhoff’s voltage law in base-emitter loop:

  $\begin{array}{l}-V_B+V_{BE}+I_E{R}_E=0\\ -0.9+0.7+I_E\left(\text{1}\text{k}\right)=0\\ I_E=0.2\text{mA}\end{array}$

The value of collector current is:

  $\begin{array}{l}{I}_C=\left(\frac{\beta }{1+\beta }\right)I_E\\ I_C=\left(\frac{150}{151}\right)\times 0.2\text{mA}\\ I_C=0.199\text{mA}\end{array}$

The value of collector voltage is:

  $\begin{array}{l}{V}_C=6-I_C{R}_C\\ =6-\left(0.199\right)\left(\text{10}\text{k}\right)\\ =4\text{V}\end{array}$

To determine

The values of $I_C,I_E,V_C$ for the given circuit.

Answer to Problem 5.32P

  $\begin{array}{l}{I}_C=0.5\text{mA}\\ I_E=0.8\text{mA}\\ V_C=1\text{V}\end{array}$

Explanation of Solution

Given Information:

  

  $\begin{array}{l}\beta =150\\ V_B=1.5\text{V}\end{array}$

Calculation:

The value of emitter current is:

  $\begin{array}{l}-V_B+V_{BE}+I_E{R}_E=0\\ -1.5+0.7+I_E\left(\text{1}\text{k}\right)=0\\ I_E=\frac{0.8}{\text{1}\text{k}}\\ I_E=0.8\text{mA}\end{array}$

The value of collector current is:

  $\begin{array}{l}{I}_C=\left(\frac{\beta }{1+\beta }\right)0.8\text{mA}\\ I_C=\left(\frac{150}{151}\right)\times 0.8\text{mA}\\ I_C=0.795\text{mA}\end{array}$

The value of collector voltage is:

  $\begin{array}{l}{V}_C=6-I_C{R}_C\\ =6-\left(0.795\text{mA}\right)\left(\text{10}\text{k}\right)\\ =-1.95\text{V}\end{array}$

The value of $V_{CB}$ is:

  $\begin{array}{l}{V}_{CB}=V_C-V_B\\ =-1.95-1.5\\ =-3.45\text{V}\end{array}$

The value of $V_{CB}$ is negative, so it is forward biased (NPN transistor). When collector base junction is forward biased then transistor operates in saturation region.

Hence, the transistor operates in saturation region.

  $V_{CE}\left(\text{sat}\right)=0.2\text{V}$

The emitter current is:

  $\begin{array}{l}-V_B+V_{BE}+I_E{R}_E=0\\ -1.5+0.7+I_E\left(\text{1}\text{k}\right)=0\\ I_E=\frac{0.8}{\text{1}\text{k}}\\ I_E=0.8\text{mA}\end{array}$

The collector current is:

Applying Kirchhoff’s voltage law in collector-emitter loop:

  $\begin{array}{l}-6+I_C{R}_C+V_{CE}+I_E{R}_E=0\\ I_C\left(\text{10}\text{k}\right)+0.2+\left(\text{0}\text{.8}\text{m}\right)\left(\text{1}\text{k}\right)=6\\ I_C=\frac{6-1}{10}\\ I_C=0.5\text{mA}\end{array}$

The value of collector voltage:

  $\begin{array}{l}{V}_C=6-I_C{R}_C\\ =6-0.5\left(\text{m}\right)\times 10\left(\text{k}\right)\\ =6-5\\ =1\text{V}\end{array}$

To determine

The values of $I_C,I_E,V_C$ for the given circuit.

Answer to Problem 5.32P

  $\begin{array}{l}{I}_C=0.43\text{mA}\\ I_E=1.5\text{mA}\\ V_C=1.7\text{V}\end{array}$

Explanation of Solution

Given Information:

  

  $\begin{array}{l}\beta =150\\ V_B=2.2\text{V}\end{array}$

Calculation:

The value of emitter current is:

  $\begin{array}{l}-V_B+V_{BE}+I_E{R}_E=0\\ -2.2+0.7+I_E\left(\text{1}\text{k}\right)=0\\ I_E=\frac{1.5}{\text{1}\text{k}}\\ I_E=1.5\text{mA}\end{array}$

The value of collector current is:

  $\begin{array}{l}{I}_C=\left(\frac{\beta }{1+\beta }\right)1.5\text{mA}\\ I_C=\left(\frac{150}{151}\right)\times 1.5\text{mA}\\ I_C=1.49\text{mA}\end{array}$

The value of collector voltage is:

  $\begin{array}{l}{V}_C=6-I_C{R}_C\\ =6-\left(1.49\text{mA}\right)\left(\text{10}\text{k}\right)\\ =-8.9\text{V}\end{array}$

The value of $V_{CB}$ is:

  $\begin{array}{l}{V}_{CB}=V_C-V_B\\ =-8.9-2.2\\ =-11.1\text{V}\end{array}$

The value of $V_{CB}$ is negative, so it is forward biased (NPN transistor). When collector base junction is forward biased then transistor operates in saturation region.

Hence, the transistor operates in saturation region.

  $V_{CE}\left(\text{sat}\right)=0.2\text{V}$

The emitter current is:

  $\begin{array}{l}-V_B+V_{BE}+I_E{R}_E=0\\ -2.2+0.7+I_E\left(\text{1}\text{k}\right)=0\\ I_E=\frac{1.5}{\text{1}\text{k}}\\ I_E=1.5\text{mA}\end{array}$

The collector current is:

Applying Kirchhoff’s voltage law in collector-emitter loop:

  $\begin{array}{l}-6+I_C{R}_C+V_{CE}+I_E{R}_E=0\\ I_C\left(\text{10}\text{k}\right)+0.2+\left(1.5\text{m}\right)\left(\text{1}\text{k}\right)=6\\ I_C=\frac{6-1.7}{10}\\ I_C=0.43\text{mA}\end{array}$

The value of collector voltage:

  $\begin{array}{l}{V}_C=6-I_C{R}_C\\ =6-0.43\left(\text{m}\right)\times 10\left(\text{k}\right)\\ =6-4.3\\ =1.7\text{V}\end{array}$