Chapter 13, Problem 13.10P

a.

To determine

Currents $I_{ref},I_3,I_4\text{ and }{I}_5$

Answer to Problem 13.10P

Currents $I_{ref}=75\mu \text{A, }{I}_3=13\mu \text{A,}{I}_4=45\mu \text{A,}{I}_5=45\mu \text{A}$

Explanation of Solution

Given:

Circuit is given as;

  

  $\begin{array}{l}{V}^{+}=3\text{ V,}{V}^{-}=-3\text{ V,}\\ R_1=80\text{ k}\Omega \text{,}{R}_E=3.5\text{ k}\Omega \end{array}$

Current for transistors $Q_1,Q_2\text{ and }{Q}_3$ is $I_S=5\times {10}^{-15}\text{ A}$

For $Q_4,I_S=3\times {10}^{-15}\text{ A}$

For $Q_5,I_S={10}^{-15}\text{ A}$

Reference current is given by,

  $I_{ref}=\frac{V^{+}-V_{EB2}-V_{EB1}-V^{-}}{R_1}\mathrm{.....1}$

Base to emitter voltage for any transistor is given by,

  $V_{EB}=V_T\ln \left(\frac{I_S}{I_C}\right)$

Therefore base to emitter voltage for transistor $Q_2$ is , $V_{EB2}=V_T\ln \left(\frac{I_{S2}}{I_{C2}}\right)\mathrm{.......2}$

Now, base to emitter voltage for transistor $Q_1$ is , $V_{EB1}=V_T\ln \left(\frac{I_{S1}}{I_{C1}}\right)\mathrm{.......3}$

Now putting the value of equation 2 and equation 3 in equation 1.

  $I_{ref}=\frac{V^{+}-V_T\ln \left(\frac{I_{S2}}{I_{C2}}\right)-V_T\ln \left(\frac{I_{S1}}{I_{C1}}\right)-V^{-}}{R_1}.$ …..4

Using property of log equation 4 can be written as,

  $I_{ref}=\frac{V^{+}-V_T\ln \left(\frac{I_{S2}}{I_{C2}}\times \frac{I_{C1}}{I_{S1}}\right)-V^{-}}{R_1}.$

From the given data $I_{C1}=I_{C2}\text{ and }{I}_{S1}=I_{S2}$

Therefore, $\begin{array}{l}{I}_{ref}=\frac{V^{+}-V_T\ln \left(\frac{I_{S1}}{I_{C1}}\times \frac{I_{C1}}{I_{S1}}\right)-V^{-}}{R_1}.\\ =\frac{V^{+}-V_T\ln \left(1\right)-V^{-}}{R_1}.\\ =\frac{V^{+}-V^{-}}{R_1}.\end{array}$

Now putting the value of $V^{+},V^{-}\text{ and }{R}_1$

  $\begin{array}{l}{I}_{ref}=\frac{3-\left(-3\right)}{80\times {10}^3}\\ =75\times {10}^{-6}\text{ A}\\ \text{=75 }\mu \text{A}\end{array}$

Therefore reference current is equal to $\text{75 }\mu \text{A}$

Now, $I_3{R}_E=V_T\ln \left(\frac{I_{ref}}{I_3}\right)$

Now putting all values,

  $\begin{array}{l}{I}_3\times 3.5\times {10}^3=26\times {10}^{-3}\ln \left(\frac{75\times {10}^{-6}}{I_3}\right)\\ 1.346\times {10}^5{I}_3=\ln \left(\frac{75\times {10}^{-6}}{I_3}\right)\\ \left(\frac{75\times {10}^{-6}}{I_3}\right)=e^{1.346\times {10}^5{I}_3}\\ I_3=13\times {10}^{-6}\\ =13\mu \text{A}\end{array}$

Base to emitter voltage for transistor $Q_4$ is , $V_{EB4}=V_T\ln \left(\frac{I_{S4}}{I_4}\right)\mathrm{.......4}$

As $V_{EB4}=V_{EB2}$

Therefore from equation 2 and equation 4.

  $\begin{array}{l}{V}_T\ln \left(\frac{I_{S4}}{I_4}\right)=V_T\ln \left(\frac{I_{S2}}{I_{C2}}\right)\\ \frac{I_{S4}}{I_4}=\frac{I_{S2}}{I_{C2}}\\ \text{As }{I}_{C2}=I_{ref}\\ \frac{I_{S4}}{I_4}=\frac{I_{S2}}{I_{ref}}\end{array}$

Now putting all values,

  $\begin{array}{l}\frac{3\times {10}^{-15}}{I_4}=\frac{75\times {10}^{-6}}{5\times {10}^{-15}}\\ I_4=45\times {10}^{-6}\text{ A}\\ =45\mu \text{A}\end{array}$

Base to emitter voltage for transistor $Q_5$ is , $V_{EB5}=V_T\ln \left(\frac{I_{S5}}{I_5}\right)\mathrm{.......5}$

As $V_{EB5}=V_{EB2}$

Therefore from equation 2 and equation 4.

  $\begin{array}{l}{V}_T\ln \left(\frac{I_{S5}}{I_5}\right)=V_T\ln \left(\frac{I_{S2}}{I_{C2}}\right)\\ \frac{I_{S5}}{I_5}=\frac{I_{S2}}{I_{C2}}\\ \text{As }{I}_{C2}=I_{ref}\\ \frac{I_{S5}}{I_5}=\frac{I_{S2}}{I_{ref}}\end{array}$

Now putting all values,

  $\begin{array}{l}\frac{{10}^{-15}}{I_5}=\frac{75\times {10}^{-6}}{5\times {10}^{-15}}\\ I_4=15\times {10}^{-6}\text{ A}\\ \text{=15 }\mu \text{A}\end{array}$

b.

To determine

Currents $I_4\text{ and }{I}_5$

Answer to Problem 13.10P

Currents $I_4=120\mu \text{A,}{I}_5=30\mu \text{A}$

Explanation of Solution

Given:

Circuit is given as;

  

  $\begin{array}{l}{V}^{+}=3\text{ V,}{V}^{-}=-3\text{ V,}\\ R_1=80\text{ k}\Omega \text{,}{R}_E=3.5\text{ k}\Omega \end{array}$

Current for transistors $Q_1,Q_2\text{ and }{Q}_3$ is $I_S=5\times {10}^{-15}\text{ A}$

For $Q_4,I_S=8\times {10}^{-15}\text{ A}$

For $Q_5,I_S=2\times {10}^{-15}\text{ A}$

Base to emitter voltage for transistor $Q_4$ is , $V_{EB4}=V_T\ln \left(\frac{I_{S4}}{I_4}\right)\mathrm{.......4}$

As $V_{EB4}=V_{EB2}$

Therefore from equation 2 and equation 4.

  $\begin{array}{l}{V}_T\ln \left(\frac{I_{S4}}{I_4}\right)=V_T\ln \left(\frac{I_{S2}}{I_{C2}}\right)\\ \frac{I_{S4}}{I_4}=\frac{I_{S2}}{I_{C2}}\\ \text{As }{I}_{C2}=I_{ref}\\ \frac{I_{S4}}{I_4}=\frac{I_{S2}}{I_{ref}}\end{array}$

Now putting all values,

  $\begin{array}{l}\frac{8\times {10}^{-15}}{I_4}=\frac{75\times {10}^{-6}}{5\times {10}^{-15}}\\ I_4=120\times {10}^{-6}\text{ A}\\ =120\mu \text{A}\end{array}$

Base to emitter voltage for transistor $Q_5$ is , $V_{EB5}=V_T\ln \left(\frac{I_{S5}}{I_5}\right)\mathrm{.......5}$

As $V_{EB5}=V_{EB2}$

Therefore from equation 2 and equation 4.

  $\begin{array}{l}{V}_T\ln \left(\frac{I_{S5}}{I_5}\right)=V_T\ln \left(\frac{I_{S2}}{I_{C2}}\right)\\ \frac{I_{S5}}{I_5}=\frac{I_{S2}}{I_{C2}}\\ \text{As }{I}_{C2}=I_{ref}\\ \frac{I_{S5}}{I_5}=\frac{I_{S2}}{I_{ref}}\end{array}$

Now putting all values,

  $\begin{array}{l}\frac{2\times {10}^{-15}}{I_5}=\frac{75\times {10}^{-6}}{5\times {10}^{-15}}\\ I_4=30\times {10}^{-6}\text{ A}\\ \text{=30 }\mu \text{A}\end{array}$