Chapter 8, Problem 8.9EP

(a)

To determine

The value of $I_{\text{Bias}}$ that induces quiescent collector current a $I_{CQ}$ of $1\text{mA}$ .

Answer to Problem 8.9EP

The value of the bias current is $0.6\times {10}^{-3}\text{A}$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

  

The conversion from $1\text{mA}$ into $\text{A}$ is given by,

  $1\text{mA}={\text{10}}^{-3}\text{A}$

The expression for the value of the base to emitter to voltage is given by,

  $V_{BE}=V_T\ln \left(\frac{I_{CQ}}{I_{SQ}}\right)$

Substitute $\text{0}\text{.026}\text{V}$ for $V_T$ , $2\times {10}^{-14}\text{A}$ for $I_{SQ}$ and $1\text{mA}$ for $I_{CQ}$ in the above equation.

  $\begin{array}{c}{V}_{BE}=\left(0.026\text{V}\right)\ln \left(\frac{1\times {10}^{-3}\text{A}}{2\times {10}^{-14}\text{A}}\right)\\ =0.6405\text{V}\end{array}$

The expression for the bias current is given by,

  $I_{\text{Bias}}=I_{SD}{e}^{\frac{V_{BE}}{V_T}}$

Substitute $0.026\text{V}$ for $V_T$ , $1.2\times {10}^{-14}\text{A}$ for $I_{SD}$ and $0.6405\text{V}$ for $V_{BE}$ in the above equation.

  $\begin{array}{c}{I}_{\text{Bias}}=\left(1.2\times {10}^{-14}\text{A}\right)e^{\frac{0.6405\text{V}}{0.026\text{V}}}\\ =0.6\times {10}^{-3}\text{A}\end{array}$

Conclusion:

Therefore, the value of the bias current is $0.6\times {10}^{-3}\text{A}$ .

(b)

To determine

The value of $i_{Cn}$ , $i_{Cp}$ , $v_{BEn}$ and $V_{E{B}_P}$ .

Answer to Problem 8.9EP

The value of current $i_{Cn}$ is $1.188\text{mA}$ , $I_D$ is $0.588\text{mA}$ , voltage $v_{BEn}$ is $0.645\text{V}$ , value of voltage is $V_{EBp}$ is $0.6349\text{V}$ and current $i_{Cp}$ is $0.8083\text{mA}$ .

Explanation of Solution

Calculation:

The value of peak voltage and output voltage are equal and the expression for the value of the load current is given by,

  $i_L=\frac{v_{O\left(\text{peak}\right)}}{R_L}$

Substitute $1\text{k}\Omega$ for $R_L$ and $1.2\text{V}$ for $v_{O\left(\text{peak}\right)}$ in the above equation.

  $\begin{array}{c}{i}_L=\frac{1.2\text{V}}{1\text{k}\Omega }\\ =1.2\text{mA}\end{array}$

The expression for the emitter current for3 the $Q_n$ transitory is equal to the load current and is given by,

  $i_{En}\approx i_L$

Substitute $1.2\text{mA}$ for $i_L$ in the above equation.

  $i_{En}\approx 1.2\text{mA}$

The expression for the base current of the $Q_n$ is given by,

  $i_{Bn}=\frac{i_{En}}{1+\beta }$

Substitute $1.2\text{mA}$ for $i_{En}$ and $100$ for $\beta$ in the above equation.

  $\begin{array}{c}{i}_{Bn}=\frac{1.2\text{mA}}{1+100}\\ =0.0118\text{mA}\end{array}$

The expression for the collector current of $Q_n$ is given by,

  $i_{Cn}=\beta i_{Bn}$

Substitute $0.0118\text{mA}$ for $i_{Bn}$ and $100$ for $\beta$ in the above equation.

  $\begin{array}{c}{i}_{Cn}=\left(100\right)\left(0.0118\text{mA}\right)\\ =1.18\text{mA}\end{array}$

The conversion from $\text{A}$ into $\text{mA}$ is given by,

  $1\text{A}={10}^3\text{mA}$

The conversion from $\text{0}\text{.6}\times {10}^{-3}\text{A}$ into $\text{mA}$ is given by,

  $\text{0}\text{.6}\times {10}^{-3}\text{A}=\text{0}\text{.6}\text{mA}$

The conversion from $\text{2}\times {10}^{-14}\text{A}$ into $\text{mA}$ is given by,

  $\text{2}\times {10}^{-14}\text{A}=\text{2}\times {10}^{-11}\text{mA}$

The expression to determine the value of the diode current is given by,

  $I_D=I_{\text{Bias}}-i_{Bn}$

Substitute $0.6\text{mA}$ for $I_{\text{Bias}}$ and $0.0118\text{mA}$ for $i_{Bn}$ in the above equation.

  $\begin{array}{c}{I}_D=0.6\text{mA}-0.0118\text{mA}\\ =0.588\text{mA}\end{array}$

The value of the collector current for $Q_n$ is given by,

  $i_{Cn}=I_D$

Substitute $0.588\text{mA}$ for

  $I_D$ in the above equation.

  $i_{Cn}=0.588\text{mA}$

The expression for the base emitter voltage of

  $Q_n$ is given by,

  $v_{BEn}=V_T\ln \left(\frac{i_{Cn}}{I_{SQ}}\right)$

Substitute

  $\text{2}\times {10}^{-11}\text{mA}$ for

  $I_{SQ}$ , $1.188\text{mA}$ for $i_{Cn}$ and $0.026\text{V}$ for $V_T$ in the above equation.

  $\begin{array}{c}{v}_{BEn}=0.026\text{V}\ln \left(\frac{1.188\text{mA}}{\text{2}\times {10}^{-11}\text{mA}}\right)\\ =0.645\text{V}\end{array}$

The expression for the value of $V_{BB}$ is given by,

  $V_{BB}=2V_T\ln \left(\frac{I_D}{I_{SD}}\right)$

Substitute $0.026\text{V}$ for $V_T$ , $1.2\times {10}^{-11}\text{mA}$ for $I_{SD}$ and $0.588\text{mA}$ for $I_D$ in the above equation.

  $\begin{array}{c}{V}_{BB}=2\left(26\text{mV}\right)\ln \left(\frac{0.588\text{mA}}{1.2\times {10}^{-11}\text{mA}}\right)\\ =1.2799\text{V}\end{array}$

The expression for the base emitter voltage of $Q_p$ is given by,

  $V_{EBp}=V_{BB}-v_{BEn}$

Substitute $1.2799\text{V}$ for $V_{BB}$ and $0.645\text{V}$ for $v_{BEn}$ in the above equation.

  $\begin{array}{c}{V}_{EBp}=1.2799\text{V}-0.645\text{V}\\ =\text{0}\text{.6349}\text{V}\end{array}$

The expression for the collector current $Q_p$ is given by,

  $i_{Cp}=I_{SQ}{e}^{\frac{v_{EBp}}{V_T}}$

Substitute $2\times {10}^{-11}\text{mA}$ for $I_{SQ}$ , $0.026\text{V}$ for $V_T$ and $0.6349\text{V}$ for $v_{EBp}$ in the above equation.

  $\begin{array}{c}{i}_{Cp}=\left(2\times {10}^{-11}\text{mA}\right)e^{\frac{0.6349\text{V}}{0.026\text{V}}}\\ =0.8083\text{mA}\end{array}$

Conclusion:

Therefore, the value of current $i_{Cn}$ is $1.188\text{mA}$ , $I_D$ is $0.588\text{mA}$ , voltage $v_{BEn}$ is $0.645\text{V}$ , value of voltage is $V_{EBp}$ is $0.6349\text{V}$ and current $i_{Cp}$ is $0.8083\text{mA}$ .

(c)

To determine

The value of $i_{Cn}$ , $i_{Cp}$ , $v_{BEn}$ and $V_{E{B}_P}$ .

Answer to Problem 8.9EP

The value of current $i_{Cn}$ is $2.97\text{mA}$ , $I_D$ is $0.5703\text{mA}$ , voltage $v_{BEn}$ is $0.668\text{V}$ , value of voltage $V_{EBp}$ is $0.6104\text{V}$ and current $i_{Cp}$ is $0.3139\text{mA}$ .

Explanation of Solution

Calculation:

The value of peak voltage and output voltage are equal and the expression for the value of the load current is given by,

  $i_L=\frac{v_{O\left(\text{peak}\right)}}{R_L}$

Substitute $1\text{k}\Omega$ for $R_L$ and $3\text{V}$ for $v_{O\left(\text{peak}\right)}$ in the above equation.

  $\begin{array}{c}{i}_L=\frac{3\text{V}}{1\text{k}\Omega }\\ =3\text{mA}\end{array}$

The expression for the emitter current for3 the $Q_n$ transitory is equal to the load current and is given by,

  $i_{En}\approx i_L$

Substitute $3\text{mA}$ for $i_L$ in the above equation.

  $i_{En}\approx 3\text{mA}$

The expression for the base current of the $Q_n$ is given by,

  $i_{Bn}=\frac{i_{En}}{1+\beta }$

Substitute $1.2\text{mA}$ for $i_{En}$ and $100$ for $\beta$ in the above equation.

  $\begin{array}{c}{i}_{Bn}=\frac{3\text{mA}}{1+100}\\ =0.0297\text{mA}\end{array}$

The expression for the collector current of $Q_n$ is given by,

  $i_{Cn}=\beta i_{Bn}$

Substitute $0.0297\text{mA}$ for $i_{Bn}$ and $100$ for $\beta$ in the above equation.

  $\begin{array}{c}{i}_{Cn}=\left(100\right)\left(0.0297\text{mA}\right)\\ =2.97\text{mA}\end{array}$

The conversion from $\text{A}$ into $\text{mA}$ is given by,

  $1\text{A}={10}^3\text{mA}$

The conversion from $\text{2}\times {10}^{-14}\text{A}$ into $\text{mA}$ is given by,

  $\text{2}\times {10}^{-14}\text{A}=\text{2}\times {10}^{-11}\text{mA}$

The expression to determine the value of the diode current is given by,

  $I_D=I_{\text{Bias}}-i_{Bn}$

Substitute $0.6\text{mA}$ for $I_{\text{Bias}}$ and $2.97\text{mA}$ for $i_{Bn}$ in the above equation.

  $\begin{array}{c}{I}_D=0.6\text{mA}-2.97\text{mA}\\ =0.5703\text{mA}\end{array}$

The value of the collector current for $Q_n$ is given by,

  $i_{Cn}=I_D$

Substitute $0.5703\text{mA}$ for $I_D$ in the above equation.

  $i_{Cn}=0.5703\text{mA}$

The expression for the base emitter voltage of $Q_n$ is given by,

  $v_{BEn}=V_T\ln \left(\frac{i_{Cn}}{I_{SQ}}\right)$

Substitute $\text{2}\times {10}^{-11}\text{mA}$ for $I_{SQ}$ , $0.5703\text{mA}$ for $i_{Cn}$ and $0.026\text{V}$ for $V_T$ in the above equation.

  $\begin{array}{c}{v}_{BEn}=0.026\text{V}\ln \left(\frac{0.5703\text{mA}}{\text{2}\times {10}^{-11}\text{mA}}\right)\\ =0.668\text{V}\end{array}$

The expression for the value of $V_{BB}$ is given by,

  $V_{BB}=2V_T\ln \left(\frac{I_D}{I_{SD}}\right)$

Substitute $0.026\text{V}$ for $V_T$ , $1.2\times {10}^{-11}\text{mA}$ for $I_{SD}$ and $0.5703\text{mA}$ for $I_D$ in the above equation.

  $\begin{array}{c}{V}_{BB}=2\left(26\text{mV}\right)\ln \left(\frac{0.5703\text{mA}}{1.2\times {10}^{-11}\text{mA}}\right)\\ =1.2784\text{V}\end{array}$

The expression for the base emitter voltage of $Q_p$ is given by,

  $V_{EBp}=V_{BB}-v_{BEn}$

Substitute $1.2784\text{V}$ for $V_{BB}$ and $0.668\text{V}$ for $v_{BEn}$ in the above equation.

  $\begin{array}{c}{V}_{EBp}=1.2784\text{V}-0.668\text{V}\\ =\text{0}\text{.6104}\text{V}\end{array}$

The expression for the collector current $Q_p$ is given by,

  $i_{Cp}=I_{SQ}{e}^{\frac{v_{EBp}}{V_T}}$

Substitute $2\times {10}^{-11}\text{mA}$ for $I_{SQ}$ , $0.026\text{V}$ for $V_T$ and $\text{0}\text{.6104}\text{V}$ for $v_{EBp}$ in the above equation.

  $\begin{array}{c}{i}_{Cp}=\left(2\times {10}^{-11}\text{mA}\right)e^{\frac{0.6104\text{V}}{0.026\text{V}}}\\ =0.3139\text{mA}\end{array}$

Conclusion:

Therefore, the value of current $i_{Cn}$ is $2.97\text{mA}$ , $I_D$ is $0.5703\text{mA}$ , voltage $v_{BEn}$ is $0.668\text{V}$ , value of voltage $V_{EBp}$ is $0.6104\text{V}$ and current $i_{Cp}$ is $0.3139\text{mA}$ .