Chapter 15, Problem 15.68P

(a)

To determine

To find: The quiescent collector currents in transistors.

Answer to Problem 15.68P

The quiescent collector currents in transistors are $I_{C1}=I_{C2}=\text{0}\text{.019mA}$

  $I_{C3}=I_{C4}=I_{C5}=I_{C6}=0.398\text{mA}$

Explanation of Solution

Given:

Given LM380 power amplifier circuit as

  

  $\begin{array}{l}{V}^{+}=22V\\ {\beta }_n=100\\ {\beta }_p=20\end{array}$

Calculation:

Assuming matched input transistor and neglecting the base currents. For zero input voltages, the currents in $Q_3$ and $Q_4$ are equal then

  $\begin{array}{l}{I}_{C3}=\frac{V^{+}-3V_{EB}}{R_{1A}+R_{1A}}\\ =\frac{22-3\left(0.7\right)}{\left(25+25\right)\times {10}^3}\\ =\frac{19.9}{50\times {10}^3}\\ =0.398mA\\ I_{C4}=\frac{V_O-2V_{EB}}{R_2}\end{array}$

Since $I_{C3}=I_{C4},$ the quiescent output voltages are

  $\begin{array}{l}{V}_o=\frac{1}{2}{V}^{+}+\frac{1}{2}{V}_{EB}\\ \frac{1}{2}\left(22\right)+\frac{1}{2}(0.7)\\ =11+0.35\\ =11.35V\end{array}$

Hence, the collector current is

  $\begin{array}{l}{I}_{C4}=\frac{11.35-2\left(0.7\right)}{25\times {10}^3}\\ =\frac{9.95}{25\times {10}^3}\\ =0.398mA\end{array}$

Since, the collector currents of $Q_3,Q_4,Q_5\text{ and }{Q}_6$ are equal,

  $I_{C3}=I_{C4}=I_{C5}=I_{C6}=0.398\text{mA}$

  $\begin{array}{l}{I}_{C1}=I_{C2}\\ =\frac{I_{C3}}{{\beta }_P+1}\\ =\frac{0.398\times {10}^{-3}}{20+1}\\ =\text{0}\text{.019mA}\end{array}$

Therefore, $I_{C1}=I_{C2}=\text{0}\text{.019mA}$

Conclusion:

Therefore, the quiescent collector currents in transistors are $I_{C1}=I_{C2}=\text{0}\text{.019mA}$

  $I_{C3}=I_{C4}=I_{C5}=I_{C6}=0.398\text{mA}$

(b)

To determine

To find: The quiescent currents inD1, D2, Q7, Q8 and Q9.

Answer to Problem 15.68P

The quiescent currents are

  $I_{C9}=3.74\text{mA}$ $I_{C7}=4.16\text{mA}$ $I_{C8}=37.4\mu \text{A}$ $I_{D1}=I_{D2}=0.398\text{mA}$

Explanation of Solution

Given: $I_s={10}^{-13}A$

Diodes D1 and D2 and transistors Q7, Q8, Q9 are all matched.

Calculation:

The currents in $D_1{\text{ and D}}_2$ are equal to the emitter current of $Q_3\text{ or }{Q}_4$ as ${\beta }_n$ is higher than ${\beta }_p$

Hence, the emitter currents of transistor $Q_3\text{ or }{Q}_4$ is

  $\begin{array}{l}{I}_{E3}=\left(\frac{{\beta }_p}{1+{\beta }_p}\right)I_{C3}\\ =\left(\frac{20}{20+1}\right)\left(0.398\times {10}^{-3}\right)\\ =0.398\text{mA}\end{array}$

Therefore, the diode currents are

  $I_{D1}=I_{D2}=0.398\text{mA}$

The voltage across base of $Q_7$ and base of $Q_8$ is

  $\begin{array}{l}{V}_{BB}=V_{BE7}+V_{BE7}\\ =2V_D\\ =2V_T\ln \left(\frac{I_D}{I_S}\right)\\ =2\left(0.026\right)\ln \left(\frac{0.418\times {10}^{-3}}{{10}^{-3}}\right)\\ =2\left(0.026\right)\left(22.154\right)\\ =1.152\text{V}\end{array}$

From the circuit given

  $\begin{array}{l}{I}_{B9}=I_{C8}\\ =\left(\frac{{\beta }_p}{{\beta }_p+1}\right)I_{E8}\\ =\left(\frac{20}{20+1}\right)I_{E8}\\ =\left(0.952\right)I_{E8}\\ I_{E8}=1.05I_{B9}\\ =\left(1.05\right)\left(\frac{I_{C9}}{{\beta }_n}\right)\\ I_{C9}=\left(\frac{100}{1.05}\right)I_{E8}\mathrm{.......}(1)\end{array}$

Thus, the collector current through the transistor $Q_7$ is

  $I_{C7}=I_{E4}+I_{C9}+I_{E8}$

Now, substitute all values in the above expression we have

  $\begin{array}{l}{I}_{C7}=0.398\text{mA}+\left(\frac{100}{1.05}\right)I_{E8}+\left(\frac{1+{\beta }_p}{{\beta }_p}\right)I_{C8}\\ =0.398\text{mA}+\left(95.24\right)I_{E8}+\left(\frac{20+1}{20}\right)I_{C8}\\ =0.398\text{mA}+\left(95.24\right)\left(\frac{20+1}{20}\right)I_{C8}+\left(1.05\right)I_{C8}\\ =0.398\text{mA}+\left(95.24+1\right)\left(1.05\right)I_{C8}\\ I_{C7}=0.398\text{mA}+101I_{C8}\mathrm{......}\left(2\right)\end{array}$

The base emitter voltages across the transistor $Q_7\text{ and }{Q}_8$ are

  $\begin{array}{l}{V}_{BE7}=V_T\ln \left(\frac{I_{C7}}{I_S}\right)\\ V_{BE8}=V_T\ln \left(\frac{I_{C8}}{I_S}\right)\end{array}$

  $\begin{array}{l}{V}_{BE7}+V_{BE8}=1.152\\ 1.152=V_T\left[\ln \left(\frac{I_{C7}}{I_S}\right)+\ln \left(\frac{I_{C8}}{I_S}\right)\right]\\ 1.152=V_T\left[\ln \left(\frac{0.398\text{mA}+101I_{C8}}{I_S}\right)+\ln \left(\frac{I_{C8}}{I_S}\right)\right]\end{array}$

Since,

  $\begin{array}{l}\ln \left(a\right)+\ln \left(b\right)=\ln \left(ab\right)\\ 1.152=0.026\ln \left[\frac{I_{C8}\left(0.398\text{mA}+101I_{C8}\right)}{{\left({10}^{-13}\right)}^2}\right]\\ 44.31=\ln \left(\frac{I_{C8}0.398\text{mA}+101I_{C8}}{{10}^{-26}}\right)\end{array}$

Taking exponential on both sides,

  $\begin{array}{l}{10}^{26}\left(e^{44.31}\right)=I_{C8}0.398\times {10}^{-3}+101I_{C8}{}^2\\ 0=101I_{C8}{}^2+I_{C8}0.398\times {10}^{-3}-1.752\times {10}^{-7}\end{array}$

As the above equation is in the form of polynomial equation, hence the roots are

  $\begin{array}{l}{I}_8=\frac{-\left(3.79\times {10}^{-3}\right)\pm \sqrt{{\left(0.418\times {10}^{-3}\right)}^2-4\left(101\right)\left(-1.752\times {10}^{-7}\right)}}{2\left(101\right)}\\ =\frac{-\left(3.79\times {10}^{-3}\right)\pm \sqrt{7.0956\times {10}^{-5}}}{202}\\ =\frac{-3.79\times {10}^{-3}+8.424\times 0^{-3}}{202},\frac{-3.79\times {10}^{-3}-8.424\times 0^{-3}}{202}\\ =37.4\times {10}^{-5},-4.377\times {10}^{-5}\end{array}$

Thus, the required collector current at transistor $Q_8$ as

  ${\text{I}}_{\text{C8}}\text{=37}\text{.4μA}$

Now, substitute the value of $I_{C8}$ in equation (2),

  $\begin{array}{l}{I}_{C7}=0.398\times {10}^{-3}+101\left(0.0374\right)\\ =4.16\times {10}^{-3}\end{array}$

Thus, the required collector current at transistor $Q_7$ as

  $I_{C7}=4.16\text{mA}$

Now, substitute the value of $I_{C8}$ in equation (1),

  $\begin{array}{l}{I}_{C9}=\left(\frac{100}{1.05}\right)I_{E8}\\ =\left(\frac{100}{1.05}\right)\left(\frac{{\beta }_n+1}{{\beta }_n}\right)I_{C8}\\ =\left(\frac{100}{1.05}\right)\left(\frac{20+1}{20}\right)\left(\text{37}\text{.4}\times {10}^{-6}\right)\\ =\left(\frac{100}{1.05}\right)\left(1.05\right)\left(\text{37}\text{.4}\times {10}^{-6}\right)\\ =\left(100\right)\left(\text{37}\text{.4}\times {10}^{-6}\right)\end{array}$

Thus, the required collector current at transistor $Q_9$ as

  $I_{C9}=3.74\text{mA}$

Conclusion:

Thus, the quiescent currents are

  $I_{C9}=3.74\text{mA}$ $I_{C7}=4.16\text{mA}$ $I_{C8}=37.4\mu \text{A}$ $I_{D1}=I_{D2}=0.398\text{mA}$

(c)

To determine

To calculate: The quiescent power dissipated in the amplifier.

Answer to Problem 15.68P

The required power is $P=115.54\text{mW}$

Explanation of Solution

Given LM380 power amplifier circuit as

  

  $\begin{array}{l}{V}^{+}=22V\\ {\beta }_n=100\\ {\beta }_p=20\end{array}$

Calculation:

For no load,

The power dissipated in the amplifier is

  $\begin{array}{l}P=\left(I_{D1}+I_{D2}+I_{C7}\right)V^{+}\\ =\left(0.418+0.418+4.416\right)\times {10}^{-3}\left(22\right)\\ =\left(5.252\right)\left(22\right)\times {10}^{-3}\\ =115.54\times {10}^{-3}\end{array}$

Conclusion:

Therefore, the required power is $P=115.54\text{mW}$