Chapter 15, Problem 15.16EP

(a)

To determine

Bias currents $I_{C3}\text{ and }{I}_{C4}$ and the temperature-compensated portion of the reference voltage $V_{B7}$

Answer to Problem 15.16EP

The value of current $I_{C3}$ is 0.482mA.

The value of current $I_{C4}$ is 0.213mA

The reference voltage of temperature compensated portion is 3.08 V

Explanation of Solution

Given:

Circuit diagram for voltage regulator is shown below.

  

The voltage of Zener diode $V_z=5.6\text{ V}$

Transistor parameters of $V_{BE(npn)}=V_{EB(pnp)}=0.6\text{V}$

Also the resistor in the emitter of $Q_4$ be $R_4=100\Omega$

The expression for bias current $\left(I_{C3}\right)$ neglecting base currents is,

  $I_{C3}=\frac{V_Z-3V_{BE}(npn)}{R_1+R_2+R_3}$

Substituting 5.6 V for $V_Z$ , 0.6V for $V_{BE\left(npn\right)}$ , $3.9\text{k}\Omega \text{ for }{R}_1,3.4\text{k}\Omega \text{ for }{R}_2\text{ and }576\Omega \text{ for }{R}_3.$

  $\begin{array}{l}{I}_{C3}=\frac{5.6\text{V}-3(0.6\text{V})}{3.9\text{k}\Omega +3.4\text{k}\Omega +576\Omega }\hfill \\ =\frac{5.6\text{V}-1.8\text{V}}{3.9\text{k}\Omega +3.4\text{k}\Omega +576\Omega }\hfill \\ =\frac{3.8\text{V}}{3900\Omega +3400\Omega +576\Omega }\hfill \\ \hfill \\ I_{C3}=\frac{3.8\text{V}}{7876\Omega }\hfill \\ =0.482\times {10}^{-3}\text{A}\hfill \\ =0.482\text{mA}\hfill \end{array}$

The value of current $I_{C3}$ is 0.482mA.

The expression for Emitter-base voltage of transistor $Q_3\left(V_{EB3}\right)$

  $V_{EB3}=I_{C4}{R}_4+V_{EB4}\mathrm{........}\left(1\right)$

Also,

  $V_{EB4}$ is emitter-base voltage of transistor $Q_4$ ,

  $I_{Z2}$ is current in Zener diode $D_2$ , and$R_4$ is resistance.

The expression for emitter-base voltage of transistor $Q_4$ .

  $V_{EB4}=V_T\ln \left(\frac{I_{C4}}{I_S}\right)$

Also,

  $V_T$ is thermal voltage, which is 0.026 V, and$I_S$ is reverse-saturation current.

The expression for emitter-base voltage of transistor $Q_5$ .

  $V_{EB3}=V_T\ln \left(\frac{I_{C3}}{I_S}\right)$

Substituting values of $V_T\ln \left(\frac{I_{C3}}{I_S}\right)$ for $V_{EB3}$ and $V_T\ln \left(\frac{I_{C4}}{I_S}\right)$ for $V_{EB4}$ in equation (1) and re-arrange expression to find the resistance $I_{C4}$ .

  $\begin{array}{l}{V}_T\ln \left(\frac{I_{C3}}{I_S}\right)=I_{C4}{R}_4+V_T\ln \left(\frac{I_{C4}}{I_S}\right)\hfill \\ I_{C4}{R}_4=V_T\ln \left(\frac{I_{C3}}{I_S}\right)–V_T\ln \left(\frac{I_{C4}}{I_S}\right)\hfill \\ I_{C4}{R}_4=V_T\left[\ln \left(\frac{I_{C3}}{I_S}\right)–\ln \left(\frac{I_{C4}}{I_S}\right)\right]\hfill \\ I_{C4}{R}_4=V_T\ln \left(\frac{I_{C3}}{I_S}\times \frac{I_S}{I_{C4}}\right) \hfill \\ I_{C4}{R}_4=V_T\ln \left(\frac{I_{C3}}{I_{C4}}\right)\hfill \\ \hfill \end{array}$

Substituting value of $100\Omega \text{ for }{R}_4,0.026\text{V for }{V}_T,\text{ and }0.482\text{ mA for }{I}_{C3}$ .

  $\begin{array}{l}{I}_{C4}\left(100\Omega \right)=\left(0.026\text{ V}\right)\ln \left(\frac{0.482\text{mA}}{I_{C4}}\right)\hfill \\ I_{C4}\left(0.1\times {10}^3\Omega \right)=\left(0.026\text{V}\right)\ln \left(\frac{0.482\text{mA}}{I_{C4}}\right)\hfill \\ I_{C4}=\left(\frac{0.026\text{V}}{0.1\times {10}^3\Omega }\right)\ln \left(\frac{0.482\text{mA}}{I_{C4}}\right)\hfill \\ I_{C4}=\left(0.26\times {10}^{-3}\right)\ln \left(\frac{0.482\text{mA}}{I_{C4}}\right)\text{A}\mathrm{.......}\left(2\right)\hfill \end{array}$

Let’s consider the current values 0.205 mA, 0.213 mA, and 0.220 mA for $I_{C4}$ to implement trial and error method.

  $\begin{array}{l}\begin{array}{l}0.205\text{ mA}=\left(0.26\times {10}^{-3}\right)\ln \left(\frac{0.482\text{mA}}{I_{C4}}\right)\text{A}\hfill \\ =\left(0.26\times {10}^{-3}\right)\ln (2.35)\text{A}\hfill \\ =\left(0.26\right)\left(0.8544\right)\text{mA}\hfill \end{array} \\ \ne 0.222\text{mA}\end{array}$

Substituting values 0.213 mA for $I_{C4}$ in equation(2) and check for the result on both sides by the use of trial and error method.

  $\begin{array}{l}0.213\text{mA}=\left(0.26\times {10}^{-3}\right)\ln \left(\frac{0.482\text{mA}}{0.213\text{mA}}\right)\text{A}\hfill \\ =\left(0.26\times {10}^{-3}\right)\ln \left(2.2629\right)\text{A}\hfill \\ =\left(0.26\right)\left(0.82\right)\text{mA}\hfill \\ =0.213\text{mA}\hfill \end{array}$

Substituting again 0.220mA for $I_{C4}$ in equation (2) and check for the result on both sides by the use of trail and error method.

  $\begin{array}{l}\begin{array}{l}0.220mA=\left(0.26\times {10}^{-3}\right)\ln \left(\frac{0.482\text{mA}}{0.220\text{mA}}\right)\text{A}\hfill \\ =\left(0.26\times {10}^{-3}\right)\ln \left(2.1909\right)\text{A}\hfill \\ =\left(0.26\right)\left(0.7843\right)\text{mA}\hfill \end{array} \\ \ne 0.204\text{ mA}\end{array}$

Hence, the current value 0.213mA is satisfying the equation (2)

After trial and error method it is found that the value of current $I_{C4}$ is 0.213mA.

The reference voltage of temperature compensated portion ( $V_{B7}$ ) can be expressed by.

  $V_{B7}=2V_{BE\left(npn\right)}+I_{C3}{R}_1$

Substitute 0.6 V for $V_{BE\left(npn\right)}$ ,0.482mA for $I_{C3}$ , and $3.9\text{k}\Omega$ for $R_1$ .

  $\begin{array}{l}{V}_{B7}=2\left(0.6\text{V}\right)+\left[\left(0.482\text{mA}\right)\left(3.9\text{k}\Omega \right)\right]\hfill \\ =1.2\text{V}+\left[\left(0.482\times {10}^{-3}\text{A}\right)\left(3.9\times {10}^3\Omega \right)\right]\hfill \\ =1.2\text{V }+1.879\text{ V}\hfill \\ =3.079\text{ V}\hfill \\ V_{B7}\approx 3.08\text{ V}\hfill \end{array}$

Hence, the reference voltage of temperature compensated portion is 3.08 V.

(b)

To determine

Resistance $R_{12}$ for given voltage.

Answer to Problem 15.16EP

The value of resistance $R_{12}$ is

  $1.39\text{k}\Omega$ .

Explanation of Solution

Given:

The voltage of Zener diode $V_z=5.6\text{ V}$

Transistor parameters of $V_{BE(npn)}=V_{EB(pnp)}=0.6\text{V}$

Also the resistor in the emitter of $Q_4$ be $R_4=100\Omega$

Given voltage $V_O=5\text{V}$

The expression for reference voltage ( $V_{B7}$ ) of temperature-compensated portion which is input of base of $Q_7$

  $V_{B7}=\left(\frac{R_{13}}{R_{12}+R_{13}}\right)V_O$

Here,

  $R_{12}$ and $R_{13}$ are resistances, and$V_O$ is output voltage.

Let’s re-arrange above expression to find the resistance $R_{12}$ .

  $\begin{array}{l}\frac{V_{B7}}{V_O}=\frac{R_{13}}{R_{12}+R_{13}}\\ R_{12}+R_{13}=R_{13}\left(\frac{V_O}{V_{B7}}\right)\\ R_{12}=R_{13}\left(\frac{V_O}{V_{B7}}\right)–R_{13}\end{array}$

Substituting $2.23\text{k}\Omega \text{ for }{R}_{13},5\text{ V for }{V}_O,\text{ and }3.08\text{ V for }{V}_{B7}$ .

  $\begin{array}{l}\begin{array}{l}{R}_{12}=\left[\left(2.23\text{k}\Omega \right)\left(\frac{5\text{V}}{3.08\text{V}}\right)\right]–2.23\text{k}\Omega \hfill \\ =\left[\left(2.23\text{k}\Omega \right)\left(1.623\right)\right]–2.23\text{k}\Omega \hfill \\ =3.62\text{k}\Omega –2.23\text{k}\Omega \hfill \end{array} \\ =1.39\text{k}\Omega \end{array}$

The value of resistance $R_{12}$ is

  $1.39\text{k}\Omega$ .