Chapter 15, Problem 15.47P

The saturated output voltages are $\pm V_P$ for the Schmitt trigger in Figure P15.47. (a) Derive the expressions for the crossover voltages $V_{TH}$ and $V_{TL}$ (b) if $V_P=12\text{V}$ , $V_{\text{REF}}=-10\text{V}$ , and $R_3=10\text{k}\Omega$ , find $R_1$ and $R_2$ such that the switching point is $V_S=-5\text{V}$ and the hysteresis width is 0.2 V. (c) Sketch the voltage transfer characteristics.

Figure P15.47

To determine

To find: The expression for given crossover voltages.

Answer to Problem 15.47P

The upper crossover voltage of Schmitt trigger is $V_{TH}=\frac{\frac{V_{\text{REF}}}{R_3}+\frac{V_P}{R_2}}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}}$

The lower crossover voltage of Schmitt trigger is $V_{TL}=\frac{\frac{V_{\text{REF}}}{R_3}-\frac{V_P}{R_2}}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}}$

Explanation of Solution

Given:

The given circuit is shown below.

  

Calculation:

  

From the above circuit

The inverting node is, $v^{+}=v_i$

Applying Kirchhoff's current law at non-inverting node:

  $\frac{v^{+}}{R_1}+\frac{v^{+}-v_o}{R_2}+\frac{v^{+}-V_{\text{REF}}}{R_3}=0$

  $\frac{v^{+}}{R_1}+\frac{v^{+}}{R_2}-\frac{v_o}{R_2}+\frac{v^{+}}{R_3}-\frac{V_{\text{REF}}}{R_3}=0$

  $v^{+}\left[\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\right]-\frac{v_o}{R_2}-\frac{V_{\text{REF}}}{R_3}=0$

  $v^{+}\left[\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\right]=\frac{V_{\text{REF}}}{R_3}+\frac{v_o}{R_2}$

  $v^{*}=\frac{\frac{V_{\text{REF}}}{R_3}+\frac{v_o}{R_2}}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}}$

Substitute $v^{+}=v,$ in the above equation,

  $v_l=\frac{\frac{V_{\text{RFF}}}{R_3}+\frac{v_o}{R_2}}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}}\dots \dots$

When $v_o=V_P$ and $v_I=V_{TH}$ in equation-(1),

The upper crossover voltage of Schmitt trigger is $V_{TH}=\frac{\frac{V_{\text{REF}}}{R_3}+\frac{V_P}{R_2}}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}}$

When $v_o=-V_p$ and $v_t=V_{TL}$ in equation-(1)

The lower crossover voltage of Schmitt trigger is $V_{TL}=\frac{\frac{V_{\text{REF}}}{R_3}-\frac{V_P}{R_2}}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}}$

Conclusion:

The upper crossover voltage of Schmitt trigger is $V_{TH}=\frac{\frac{V_{\text{REF}}}{R_3}+\frac{V_P}{R_2}}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}}$

The lower crossover voltage of Schmitt trigger is $V_{TL}=\frac{\frac{V_{\text{REF}}}{R_3}-\frac{V_P}{R_2}}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}}$

To determine

To find: The values of $R_1\text{ and }{R}_2$

Answer to Problem 15.47P

The required values are $R_1=10.17\text{k}\Omega$ and $R_2=600\text{k}\Omega$

Explanation of Solution

Given:

Saturated output voltage is, $V_P=12\text{V}$

Reference voltage is, $V_{\text{REF}}=-10\text{V}$

One resistor value is, $R_3=10\text{k}\Omega$

Switching point is, $V_s=-5\text{V}$

Hysteresis width is $V_{TH}-V_{TL}=0.2\text{V}$

Calculation:

Substitute $V_p,V_{\text{REF}}$ and $R_3$ in upper crossover voltage,

  $V_{TH}=\frac{\frac{-10}{10\times {10}^3}+\frac{12}{R_2}}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{10\times {10}^3}}$

Hence, $V_{TH}=\frac{\frac{-10}{10\times {10}^3}+\frac{12}{R_2}}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{10\times {10}^3}}$

Substitute $V_p,V_{\text{REF}}$ and $R_3$ in lower crossover voltage,

  $V_{TL}=\frac{\frac{-10}{10\times {10}^3}-\frac{12}{R_2}}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{10\times {10}^3}}$

Substitute $V_{TH}$ and $V_{TL}$ in equation $V_{TH}-V_{TL}=\frac{\frac{-10}{10\times {10}^3}+\frac{12}{R_2}}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{10\times {10}^3}}-\frac{\frac{-10}{10\times {10}^3}-\frac{12}{R_2}}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{10\times {10}^3}}$

  $\frac{\frac{-10}{10\times {10}^3}+\frac{12}{R_2}}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{10\times {10}^3}}-\frac{\frac{-10}{10\times {10}^3}-\frac{12}{R_2}}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{10\times {10}^3}}=0.2$

  $\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{10\times {10}^3}}\left(\frac{-10}{10\times {10}^3}+\frac{12}{R_2}-\left(\frac{-10}{10\times {10}^3}-\frac{12}{R_2}\right)\right)=0.2$

  $\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{10\times {10}^3}}\left(\frac{-10}{10\times {10}^3}+\frac{12}{R_2}+\frac{10}{10\times {10}^3}+\frac{12}{R_2}\right)=0.2$

  $\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{10\times {10}^3}}\left(\frac{24}{R_2}\right)=0.2$

  $\frac{24}{R_2}=0.2\left(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{10\times {10}^3}\right)$

  $\frac{24}{0.2}=R_2\left(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{10\times {10}^3}\right)$

  $120=R_2\left(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{10\times {10}^3}\right)\mathrm{.......}\left(3\right)$

Assuming $V_H$ and $V_L$ are symmetrical about zero. The switching voltage becomes,

  $\begin{array}{l}{V}_s=\frac{\frac{V_{\text{REF}}}{R_3}}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}}\hfill \\ -5=\frac{\frac{1}{10\times {10}^3}}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{10\times {10}^3}}\hfill \\ 5=\frac{10}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{10\times {10}^3}}\hfill \\ \frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{10\times {10}^3}=\frac{\frac{10}{10\times {10}^3}}{5}\hfill \\ \frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{10\times {10}^3}=\frac{1}{5\times {10}^3}\hfill \end{array}$

Hence, we get $\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{10\times {10}^3}=\frac{1}{5\times {10}^3}$

Substitute $\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{10\times {10}^3}=\frac{1}{5\times {10}^3}$ in equation- (3)

  $\begin{array}{l}120=R_2\left(\frac{1}{5\times {10}^3}\right)\hfill \\ R_2=120\left(5\times {10}^3\right)\hfill \\ R_2=600\times {10}^3\hfill \end{array}$

Therefore, value of $R_2$ is $R_2=600\text{k}\Omega$

  $\text{ We have, }\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{10\times {10}^3}=\frac{1}{5\times {10}^3}$

Substitute $R_2=600\text{k}\Omega$ in the above equation

  $\begin{array}{l}\frac{1}{R_1}+\frac{1}{600\times {10}^3}+\frac{1}{10\times {10}^3}=\frac{1}{5\times {10}^3}\hfill \\ \frac{1}{R_1}=\frac{1}{5\times {10}^3}-\frac{1}{600\times {10}^3}-\frac{1}{10\times {10}^3}\hfill \\ \frac{1}{R_1}=\frac{120-1-60}{600\times {10}^3}\hfill \end{array}$

Therefore, the value of $R_1$ is $R_1=10.17\text{k}\Omega$

Conclusion:

Therefore, the required values are $R_1=10.17\text{k}\Omega$ and $R_2=600\text{k}\Omega$

To determine

To sketch: The voltage transfer characteristics.

Answer to Problem 15.47P

The voltage transfer characteristics are shown in Figure 1.

Explanation of Solution

Given:

  

Calculation:

The upper crossover voltage of Schmitt trigger is

  $\begin{array}{l}{V}_{TH}=\frac{\frac{V_{\text{REF}}}{R_3}+\frac{V_P}{R_2}}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}}\\ =\frac{\frac{-10}{10\times {10}^3}+\frac{12}{600\times {10}^3}}{10.17\times {10}^3+\frac{1}{600\times {10}^3}+\frac{1}{10\times {10}^3}}\\ =\frac{\frac{-10}{10}+\frac{12}{600}}{\frac{1}{10.17}++\frac{1}{600}+\frac{1}{10}}\end{array}$

  $\begin{array}{l}=\frac{-1+0.02}{0.09833+0.00167+0.1}\hfill \\ =\frac{-0.98}{0.2}\hfill \\ V_{TH}=-4.9\hfill \end{array}$

Therefore, the upper crossover voltage of Schmitt trigger is $V_{TH}=-4.9\text{V}$

The lower crossover voltage of Schmitt trigger is

  $\begin{array}{l}{V}_{TL}=\frac{\frac{V_{\text{REF}}}{R_3}-\frac{V_P}{R_2}}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}}\\ =\frac{\frac{-10}{10\times {10}^3}-\frac{12}{600\times {10}^3}}{\frac{1}{10.17\times {10}^3}+\frac{1}{600\times {10}^3}+\frac{1}{10\times {10}^3}}\\ =\frac{\frac{-10}{10}-\frac{12}{600}}{\frac{1}{10.17}+\frac{1}{600}+\frac{1}{10}}\\ =\frac{1}{0.09833+0.00167+0.1}\\ =\frac{-1.02}{0.2}\\ V_{TL}=-5.1\end{array}$

Therefore, the lower crossover voltage of Schmitt trigger is $V_{TL}=-5.1\text{V}$

  

Figure 1

Conclusion:

Therefore, the voltage transfer characteristics are shown in Figure 1.