Chapter 15, Problem 15.50P

a.

To determine

The expression for the switching point and the crossover voltages.

Answer to Problem 15.50P

The switching voltage $V_s$ is $V_s=\left(1+\frac{R_1}{R_2}\right)V_{REF}$ .

The lower crossover voltage of Schmitt trigger is

  $V_{TL}=V_S-\left(\frac{R_1}{R_2}\right)V_H$

The upper crossover voltage of Schmitt trigger is

  $V_{TH}=V_S-\left(\frac{R_1}{R_2}\right)V_L$

Explanation of Solution

Given:

The circuit is given as:

  

For the ideal operational amplifier, the currents in the inverting and the non-inverting terminal will be zero. By the virtual ground concept the inverting and the non-inverting node voltages will be equal.

Redrawing the circuit:

  

Applying nodal analysis at inverting node:

  $\begin{array}{l}\frac{V_{REF}-v_I}{R_1\parallel R_2}+0=0\\ \frac{V_{REF}-v_I}{R_1\parallel R_2}=0\\ V_{REF}-v_I=0\\ V_{REF}=v_I\end{array}$

Applying nodal analysis at the non-inverting node:

$\begin{array}{l}\frac{v^{+}-v_I}{R_1}+\frac{v^{+}-v_0}{R_2}=0\\ \frac{R_2\left(v^{+}-v_I\right)+R_1\left(v^{+}-v_0\right)}{R_1{R}_2}=0\\ R_2\left(v^{+}-v_I\right)+R_1\left(v^{+}-v_0\right)=0\\ R_2{v}^{+}-R_2{v}_I+R_1{v}^{+}-R_1{v}_0=0\end{array}$

$\begin{array}{l}\left(R_1+R_2\right)v^{+}-R_2{v}_I-R_1{v}_0=0\\ \left(R_1+R_2\right)v^{+}=R_2{v}_l-R_1{v}_0\\ R_2{v}_l=\left(R_1+R_2\right)v^{+}-R_1{v}_0\\ v_I=\left(\frac{R_1+R_2}{R_2}\right)v^{+}-\left(\frac{R_1}{R_2}\right)v_0\\ v_l=\left(1+\frac{R_1}{R_2}\right)v^{+}-\left(\frac{R_1}{R_2}\right)v_0\end{array}$

$\begin{array}{l}{v}_I=\left(1+\frac{R_1}{R_2}\right)v_{REF}-\left(\frac{R_1}{R_2}\right)v_0\\ v_I=\left(1+\frac{R_1}{R_2}\right)v_{REF}-\left(\frac{R_1}{R_2}\right)v_0\mathrm{............}\left(1\right)\end{array}$

Considering the $V_H$ and $V_L$ symmetrical about the zero ( $v_0=0$ ). The switching voltage $V_s$ is given as:

  $V_s=\left(1+\frac{R_1}{R_2}\right)V_{REF}$

Hence, the switching voltage $V_s$ is $V_s=\left(1+\frac{R_1}{R_2}\right)V_{REF}$ .

$V_{TH}$ will develop when the $v_0=V_L$ and $v_I=V_{TH}$ in equation 1:

Evaluating the upper crossover voltage of Schmitt trigger:

  $\begin{array}{l}{V}_{TH}=\left(1+\frac{R_1}{R_2}\right)v_{REF}-\left(\frac{R_1}{R_2}\right)V_L\\ V_{TH}=V_S-\left(\frac{R_1}{R_2}\right)V_L\left(\text{Since,}{V}_s=\left(1+\frac{R_1}{R_2}\right)V_{REF}\right)\end{array}$

Hence, the upper crossover voltage of Schmitt trigger is

  $V_{TH}=V_S-\left(\frac{R_1}{R_2}\right)V_L$

$V_{TL}$ will take place when $v_0=V_H$ and $v_I=V_{TL}$ In equation 1.

Evaluating the lower crossover voltage of Schmitt trigger:

  $\begin{array}{l}{V}_{TL}=\left(1+\frac{R_1}{R_2}\right)v_{REF}-\left(\frac{R_1}{R_2}\right)V_H\\ V_{TL}=V_S-\left(\frac{R_1}{R_2}\right)V_H\left(\because V_s=\left(1+\frac{R_1}{R_2}\right)V_{REF}\right)\end{array}$

Hence, the lower crossover voltage of Schmitt trigger is

  $V_{TL}=V_S-\left(\frac{R_1}{R_2}\right)V_H$

b.

To determine

The R₁ and V_REF for the given condition.

Answer to Problem 15.50P

The reference voltage is $V_{REF}=-1.44V$ .

  $R_1=0.833\text{kΩ}$

Explanation of Solution

Given:

The circuit is given as:

  

$V_H=+12\text{V}$ and $V_L=-12\text{V}$

  $R_2=20\text{kΩ}$

The crossover voltages are $V_{TH}=-1\text{V}\text{and}{V}_{TL}=-2\text{V}$

Substituting the $R_2=20\text{kΩ}$ , $V_{TH}=-1\text{V}\text{and}{V}_L=-12\text{V}$ in the upper crossover voltage equation:

  $\begin{array}{l}-1=V_s-\left(\frac{R_1}{20\times {10}^3}\right)\left(-12\right)\\ -1=V_s+12\left(\frac{R_1}{20\times {10}^3}\right)\\ V_S=-1-12\left(\frac{R_1}{20\times {10}^3}\right)\end{array}$

Hence, $V_S=-1-12\left(\frac{R_1}{20\times {10}^3}\right)$ .

Substitute, $R_2=20k\Omega$ , $V_{TL}=-2Vand{V}_H=+12V$ In the lower crossover voltages equation:

  $-2=V_s-\left(\frac{R_1}{20\times {10}^3}\right)\left(12\right)$

Substitute $V_s$ in the above equation

  $\begin{array}{l}-2=\left(-1-12\left(\frac{R_1}{20\times {10}^3}\right)\right)-\left(\frac{R_1}{20\times {10}^3}\right)\left(12\right)\\ -2=-1-12\left(\frac{R_1}{20\times {10}^3}\right)-\left(\frac{R_1}{20\times {10}^3}\right)\left(12\right)\\ -2=-1-24\left(\frac{R_1}{20\times {10}^3}\right)\\ R_1=\frac{20\times {10}^3}{24}\\ R_1=0.833\times {10}^3\end{array}$

Hence, $R_1=0.833\text{kΩ}$

Substituting, $V_{TH}=-1\text{V}\text{and}{V}_H=+12\text{V}$ , $R_1=0.833\text{kΩ}$ and $R_2=20\text{kΩ}$ in switching voltage equation,

  $\begin{array}{l}-1=V_s-\left(\frac{0.833\times {10}^3}{20\times {10}^3}\right)\left(-12\right)\\ -1=V_s+0.4998\\ V_s=-1.4998\text{V}\end{array}$

Therefore, the reference voltage is $V_s=-1.4998\text{V}$ .

Substituting, $V_s=-1.4998\text{V}$ , $R_1=0.833\text{kΩ}$ and $R_2=20\text{kΩ}$ in switching the voltage equation.

  $\begin{array}{l}-1.4998=\left(1+\frac{0.833\times {10}^3}{20\times {10}^3}\right)V_{REF}\\ -1.4998=\left(1+0.04165\right)V_{REF}\\ -1.4998=\left(1.04165\right)V_{REF}\\ V_{REF}=-1.439\\ V_{REF}=-1.4\text{V}\end{array}$

Hence, the reference voltage is $V_{REF}=-1.44V$ .