Chapter 17, Problem 17.2TYU

To determine

To design: The ECL circuit.

To find: All resistor values and the value of $V_R$ .

Answer to Problem 17.2TYU

The redesign of the ECL circuit is shown in Figure 2 and the value of resistors are $R_{C1}=320\Omega$ , $R_E=1.52\text{kΩ}$ , $R_{C2}=358\Omega$ , $R_3=1.8\text{kΩ}$ and $R_4=1.8\text{kΩ}$ , the value of voltage is $V_R=-1.1\text{V}$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

  

The redesign circuit is shown below.

The required diagram is shown in Figure 2

  

The expression for $v_{O1}$ is given by,

  $v_{O1}=v_{\text{NOR}}+V_{BE}\left(\text{on}\right)$

Substitute $-1.5\text{V}$ for $v_{\text{NOR}}$ and $0.7\text{V}$ for $V_{BE}\left(\text{on}\right)$ in the above equation.

  $\begin{array}{c}{v}_{O1}=v_{\text{NOR}}+V_{BE}\left(\text{on}\right)\\ =-1.5+0.7\\ =-0.8\text{V}\end{array}$

If $\text{logic}0$ value at the $v_{\text{NOR}}$ terminal is $-1.5\text{V}$ , then the transistors $Q_1\text{and}{Q}_2$ are in on states and the transistor $Q_R$ is in off state. So, that the value of $i_{CXY}$ is given by,

  $i_{CXY}=i_E\left(\max \right)$

The expression for $R_{C1}$ is given by,

  $R_{C1}=\frac{-v_{O1}}{i_{CXY}}$

Substitute $i_E\left(\max \right)$ for $i_{CXY}$ in the above equation.

  $R_{C1}=\frac{-v_{O1}}{i_E\left(\max \right)}$

Substitute $-0.8\text{V}$ for $v_{O1}$ and $2.5\text{mA}$ for $i_E\left(\max \right)$ in the above equation.

  $\begin{array}{c}{R}_{C1}=\frac{-\left(-0.8\right)\text{V}}{2.5\text{mA}}\\ =\frac{0.8\text{V}}{\left(2.5\text{mA}\right)\times \frac{\left({10}^{-3}\text{A}\right)}{\left(1\text{mA}\right)}}\\ =320\Omega \end{array}$

If $\text{logic}0$ value at the $v_{\text{NOR}}$ terminal is $-1.5\text{V}$ , then the transistors $Q_1\text{and}{Q}_2$ are in on states and the transistor $Q_R$ is in off state. So, that the value of $v_X\text{and}{v}_Y$ are equal and it is equal to the $-0.7\text{V}$ .

The expression for $v_E$ is given by,

  $v_E=v_X-V_{BE}\left(\text{on}\right)$

Substitute $-0.7\text{V}$ for $v_X$ and $0.7\text{V}$ for $V_{BE}\left(\text{on}\right)$ in the above equation.

  $\begin{array}{c}{v}_E=\left(-0.7\right)-0.7\\ =-1.4\text{V}\end{array}$

The expression for $R_E$ is given by,

  $R_E=\frac{v_E-V^{-}}{i_E\left(\max \right)}$

Substitute $-1.4\text{V}$ for $v_E$ , $-5.2\text{V}$ for ${\text{V}}^{-}$ and $2.5\text{mA}$ for $i_E\left(\max \right)$ in the above equation.

  $\begin{array}{c}{R}_E=\frac{\left[-1.4-\left(-5.2\right)\right]\text{V}}{2.5\text{mA}}\\ \text{=}1.52\text{kΩ}\end{array}$

Input voltages $v_X\text{and}{v}_Y$ are greater than $V_R$ in $\text{logic}1$ state and they are less than $V_R$ in $\text{logic}0$ state. If $V_R$ is set to the midpoint between the $\text{logic}\text{1}\text{and}\text{logic}\text{0}$ states then the expression for the $V_R$ is given by,

  $V_R=\frac{V_R\left(\text{logic}0\right)+V_R\left(\text{logic}1\right)}{2}$

Substitute $-1.5\text{V}$ for $V_R\left(\text{logic}1\right)$ and $-0.7\text{V}$ for $V_R\left(\text{logic}0\right)$ in the above equation.

  $\begin{array}{c}{V}_R=\frac{V_R\left(\text{logic}0\right)+V_R\left(\text{logic}1\right)}{2}\\ =\frac{-0.7+\left(-1.5\right)}{2}\\ =-1.1\text{V}\end{array}$

If $\text{logic}\text{0}$ value at the $v_{\text{OR}}$ terminal is $-1.5\text{V}$ then the transistors $Q_1\text{and}{Q}_2$ are in off states and the transistor $Q_R$ is in on state, then $v_E$ is given by,

  $v_E=V_R-V_{BE}\left(\text{on}\right)$

Substitute $-1.1\text{V}$ for $V_R$ and $\text{0}\text{.7}\text{V}$ for $V_{BE}\left(\text{on}\right)$ in the above equation.

  $\begin{array}{c}{v}_E=-1.1-0.7\\ =-1.8\text{V}\end{array}$

The expression for $i_E$ is given by,

  $i_E=\frac{v_E-V^{-}}{R_E}$

Substitute $-1.8\text{V}$ for $v_E$ , $1.52\text{kΩ}$ for $R_E$ and $-5.2\text{V}$ for $V^{-}$ in the above equation.

  $\begin{array}{c}{i}_E=\frac{\left[-1.8-\left(-5.2\right)\right]\text{V}}{1.52\text{kΩ}}\\ i_E=\frac{\left[-1.8-\left(-5.2\right)\right]\text{V}}{1.52\times {10}^{-3}\Omega }\\ =2.236\times {10}^{-3}\text{A}\\ =2.236\text{mA}\end{array}$

The value of $i_{CXY}$ is given by,

  $i_{CXY}=i_E$

The expression for $R_{C2}$ is given by,

  $R_{C2}=\frac{-v_{O2}}{i_{CXY}}$

Substitute $-0.8\text{V}$ for $v_{O2}$ , and $2.236\text{mA}$ for $i_E$ in the above equation.

  $\begin{array}{c}{R}_{C2}=\frac{-v_{O2}}{i_{CXY}}\\ =\frac{-\left(-0.8\right)\text{V}}{2.236\text{mA}}\\ =\frac{-\left(-0.8\right)\text{V}}{\left(2.236\text{mA}\right)\times \frac{\left({10}^{-3}\text{A}\right)}{\left(1\text{mA}\right)}}\\ =358\Omega \end{array}$

The expression for $R_3$ is given by,

  $R_3=\frac{v_{\text{OR}}\left(\text{logic}1\right)-V^{-}}{i_3\left(\max \right)}$

Substitute $-0.7\text{V}$ for $v_{\text{OR}}\left(\text{logic}1\right)$ , $-5.2\text{V}$ for $V^{-}$ and $2.5\text{mA}$ for $i_3\left(\max \right)$ in the above equation.

  $\begin{array}{c}{R}_3=\frac{v_{\text{OR}}\left(\text{logic}1\right)-V^{-}}{i_3\left(\max \right)}\\ =\frac{\left[-0.7-\left(-5.2\right)\right]\text{V}}{2.5\text{mA}}\\ =1.8\text{kΩ}\end{array}$

The expression for $R_4$ is given by,

  $R_4=\frac{v_{\text{NOR}}\left(\text{logic}1\right)-V^{-}}{i_4\left(\max \right)}$

Substitute $-0.7\text{V}$ for $v_{\text{NOR}}\left(\text{logic}1\right)$ , $-5.2\text{V}$ for $V^{-}$ and $2.5\text{mA}$ for $i_4\left(\max \right)$ in the above equation.

  $\begin{array}{c}{R}_4=\frac{v_{\text{NOR}}\left(\text{logic}1\right)-V^{-}}{i_4\left(\max \right)}\\ =\frac{\left[-0.7-\left(-5.2\right)\right]\text{V}}{2.5\text{mA}}\\ =1.8\text{kΩ}\end{array}$

Conclusion:

Therefore, the redesign of the ECL circuit is shown in Figure 2 and the value of resistors are $R_{C1}=320\Omega$ , $R_E=1.52\text{kΩ}$ , $R_{C2}=358\Omega$ , $R_3=1.8\text{kΩ}$ and $R_4=1.8\text{kΩ}$ , the value of voltage is $V_R=-1.1\text{V}$ .