Chapter 5, Problem 5.82P

Consider the circuit shown in Figure P5.82. The current gain for the npn transistor is ${\beta }_n=120$ and for the pnp transistor is ${\beta }_p=80$ . Determine $I_{B1},I_{C1},I_{B2},I_{C2},V_{CE1}$ , and $V_{EC2}$ .

Figure P5.82

To determine

The values of $I_{B1},I_{C1},I_{B2},V_{CE1}and{V}_{EC2}$ for the gain in current for the given transistor.

Answer to Problem 5.82P

  $\begin{array}{l}{I}_{B1}=8.56\mu A\\ I_{C1}=1.0272mA\\ I_{B2}=0.134mA\\ V_{CE1}=5.142V\\ V_{EC2}=5.77V\end{array}$

Explanation of Solution

Given:

  $\beta n=120$ for the transistor npn.

  $\beta p=80$ for the transistor pnp

The circuit is given as:

  

Resistance $R_{TH}$ for given fig.,

  $\begin{array}{l}{R}_{TH}=R_1\parallel R_2\\ =\frac{R_1{R}_2}{R_1+R_2}\end{array}$

Substituting the value,

  $R_1=80\times {10}^{3}and{R}_2=40\times {10}^3$ according to fig.

  $\begin{array}{l}{R}_{TH}=\frac{80\times {10}^3\times 40\times {10}^3}{80\times {10}^3+40\times {10}^3}\\ =\frac{3200\times {10}^3}{120}\\ =26.66k\Omega \end{array}$

Now calculating the voltage $V_{TH.}$

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

Substituting the known values.

  $\begin{array}{l}{V}_{TH}=\left(\frac{40\times {10}^3}{80\times {10}^3+40\times {10}^3}\right)\times 9\\ =3V\end{array}$

Now the equivalent circuit is drawn as:

  

For given figure applying Kirchoff’s voltage law for base emitter loop of the circuit for transistor $Q_1$ .

  $\begin{array}{l}-V_{TH}+I_{B1}{R}_{TH}+V_{BE\left(on\right)}+I_{E1}{R}_{E1}=0\\ -V_{TH}+I_{B1}{R}_{TH}+V_{BE\left(on\right)}+\left(1+{\beta }_n\right)I_{B1}{R}_{E1}=0\\ I_{B1}=\frac{V_{TH}-V_{BE\left(on\right)}}{R_{TH}+\left(1+{\beta }_n\right)R_{E1}}\end{array}$

Substituting all values:

  $\begin{array}{l}{I}_{B1}=\frac{3-0.7}{26.66\times {10}^3+\left(1+120\right)\times 2\times {10}^3}\\ =\frac{2.3}{268.66}\times {10}^{-3}\\ =8.56\mu A\end{array}$

Now calculating the collector current $I_{C1}$ .

  $I_{C1}={\beta }_n{I}_{B1}$

Putting the values:

  $\begin{array}{l}{I}_{C1}={\beta }_n{I}_{B1}\\ I_{C1}=120\times 8.56\times {10}^{-6}\\ =1.0272mA\end{array}$

Now calculating the value of emitter current $I_{E1}$ .

  $I_{E1}=\left(1+{\beta }_n\right)I_{B1}$

Again substituting all values:

  $\begin{array}{l}{I}_{E1}=\left(1+120\right)\times 8.56\times {10}^{-6}\\ =81\times 0.134\times {10}^{-3}\\ =1.036mA\end{array}$

According to Kirchoff’s law of current at the collector node of transistor $Q_1$ in respective figure.

  $\begin{array}{l}{I}_{R1}+I_{B2}=I_{C1}\\ \frac{9-V_{C1}}{R_{C1}}+I_{B2}=I_{C\mathrm{1.............}\left(1\right)}\end{array}$

Now applying the Kirchoff’s voltage law in emitter- base loop for transistor $Q_2$ .

  $\begin{array}{l}{V}^{+}=I_{E2}(100)+V_{EB}\left(on\right)+V_{C1}\\ V^{+}=\left(1+{\beta }_p\right)I_{B2}\left(100\right)+V_{EB}\left(on\right)+V_{C1}\\ I_{B2}=\frac{V^{+}-V_{EB}\left(on\right)-V_{C1}}{\left(1+{\beta }_p\right)\times 100}\mathrm{........}\left(2\right)\end{array}$

Substituting the value of $I_{B2}$ in equation $\left(1\right)$

  $\frac{9-V_{C1}}{R_{C1}}+\frac{V^{+}-V_{EB}\left(on\right)-V_{C1}}{\left(1+{\beta }_p\right)\times 100}=I_{C1}$

Now putting all values:

  $\begin{array}{l}\frac{9-V_{C1}}{2\times {10}^3}+\frac{9-0.7-V_{C1}}{\left(1+80\right)\times 100}=1.0272\times {10}^{-3}\\ \frac{9-V_{C1}}{2}+\frac{8.3-V_{C1}}{8.1}=1.0272\\ V_{C1}\left(\frac{1}{2}+\frac{1}{8.1}\right)=\frac{9}{2}+\frac{8.3}{8.1}-1.0272\\ V_{C1}=7.214V\end{array}$

According to equation $\left(2\right)$

  $I_{B2}=\frac{V^{+}-V_{EB}\left(on\right)-V_{C1}}{\left(1+{\beta }_p\right)\times 100}$

Putting all values:

  $\begin{array}{l}{I}_{B2}=\frac{9-0.7-7.214}{\left(1+80\right)\times 100}\\ =0.134mA\end{array}$

Now calculating $I_{C2}$

  $I_{C2}={\beta }_p{I}_{B2}$

Substituting all values:

  $\begin{array}{l}{I}_{C2}=80\times 0.134\times {10}^{-3}\\ =10.73mA\end{array}$

Now calculating $I_{E2}$ .

  $I_{E2}=\left(1+{\beta }_p\right)I_{B2}$

Putting all values:

  $\begin{array}{l}{I}_{E2}=\left(1+80\right)\times 0.134\times {10}^{-3}\\ =81\times 0.134\times {10}^{-3}\\ =10.86mA\end{array}$

Calculating $V_{E1}$ now,

  $V_{E1}=I_{E1}{R}_{E1}$

Putting all known values:

  $\begin{array}{l}{V}_{E1}=1.036\times {10}^{-3}\times 2\times {10}^{-3}\\ =2.072V\end{array}$

Now calculating $V_{CE1}$ .

  $V_{CE1}=V_{C1}-V_{E1}$

Putting the values:

  $\begin{array}{l}{V}_{CE1}=7.214-2.072\\ =5.142V\end{array}$

Now calculating $V_{EC2}$ .

According to Kirchoff’s Law:

  $\begin{array}{l}{V}_{EC2}=V^{+}-I_{C2}{R}_{C2}-I_{E2}{R}_{E2}\\ \end{array}$

Again substituting all values:

  $\begin{array}{l}{V}_{EC2}=9-\left(10.73\times {10}^{-3}\right)\left(200\right)-\left(10.86\times {10}^{-3}\right)\left(100\right)\\ =5.77V\\ \end{array}$