Chapter 15, Problem 15.49P

Consider the Schmitt trigger in Figure 15.30(a). (a) Derive the expressionfor the switching point and crossover voltages as given in Equations (15.76)and (15.77). (b) Let $V_H=+12\text{V}$ and $V_L=-12\text{V}$ . The minimum resistance is to be $4\text{k}\Omega$ . Determine $R_1,R_2$ , and $V_{\text{REF}}$ such that the crossovervoltages are $V_{TH}=-1.5\text{V}$ and $V_{TL}=-2\text{V}$ . (c) What are the currents in the resistors when (i) ${\upsilon }_O=V_H$ , and (ii) ${\upsilon }_O=V_L$ ?

To determine

To derive: the expression for the switching point and crossover voltages

Answer to Problem 15.49P

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

The upper crossover voltage of Schmitt trigger is

  $V_{TH}=V_S+\left(\frac{R_1}{R_1+R_2}\right)V_H$

The lower crossover voltage of Schmitt trigger is

  $V_{TL}=V_S+\left(\frac{R_1}{R_1+R_2}\right)V_L$

Explanation of Solution

Given:

Consider the Schmitt trigger as shown below.

  

Calculation:

In an ideal op-amp, the inverting and non-inverting terminal currents are zero. And the inverting and non-inverting node voltages are equal. Given circuit can be represented as

  

Applying Kirchhoff’s current law at inverting node:

  $\begin{array}{l}\frac{v^{+}-v_1}{R_1\left|\right|R_2}+0=0\\ \frac{v^{+}-v_1}{R_1\left|\right|R_2}=0\\ v^{+}-v_1=0\\ v^{+}=v_1\end{array}$

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

  $\begin{array}{l}\frac{v^{+}-V_{REF}}{R_1}+\frac{v^{+}-v_0}{R_2}=0\\ \frac{R_2\left(v^{+}-V_{REF}\right)+R_1\left(v^{+}-v_0\right)}{R_1{R}_2}=0\\ R_2\left(v^{+}-V_{REF}\right)+R_1\left(v^{+}-v_0\right)=0\\ R_2{v}^{+}-R_2{V}_{REF}+R_1{v}^{+}-R_1{v}_0=0\\ \left(R_1+R_2\right)v^{+}-R_2{V}_{REF}-R_1{v}_0=0\\ \left(R_1+R_2\right)v^{+}=R_2{V}_{REF}+R_1{v}_0\\ v^{+}=\left(\frac{R_2}{R_1+R_2}\right)V_{REF}+\left(\frac{R_1}{R_1+R_2}\right)v_0\\ V_1=\left(\frac{R_2}{R_1+R_2}\right)V_{REF}+\left(\frac{R_1}{R_1+R_2}\right)v_0(\sin ce,v^{+}=v_1)\\ V_1=\left(\frac{R_2}{R_1+R_2}\right)V_{REF}+\left(\frac{R_1}{R_1+R_2}\right)v_0\mathrm{.......}(1)\end{array}$

Assuming $V_H$ and $V_L$ are symmetrical about zero (That is $v_0=0$ ), The switching voltage $V_S$ becomes:

  $V_S=\left(\frac{R_2}{R_1+R_2}\right)V_{REF}$ (From equation-(1))

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

When $v_0=V_H$ and $v_1=V_{TH}$ in equation-(1),

The upper crossover voltage off Schmitt trigger is

  $\begin{array}{l}{V}_{TH}=\left(\frac{R_2}{R_1+R_2}\right)V_{REF}+\left(\frac{R_1}{R_1+R_2}\right)V_H\\ V_{TH}=V_S+\left(\frac{R_1}{R_1+R_2}\right)V_H(Since,V_S=\left(\frac{R_2}{R_1+R_2}\right)V_{REF})\end{array}$

Therefore, the upper crossover voltage of Schmitt trigger is

  $V_{TH}=V_S+\left(\frac{R_1}{R_1+R_2}\right)V_H$

When $v_0=V_L$ and $v_1=V_{TL}$ in equation-(1),

The lower cross over voltage of Schmitt trigger is

  $\begin{array}{l}{V}_{TL}=\left(\frac{R_2}{R_1+R_2}\right)V_{REF}+\left(\frac{R_1}{R_1+R_2}\right)V_L\\ V_{TL}=V_S+\left(\frac{R_1}{R_1+R_2}\right)V_L(Since,V_S=\left(\frac{R_2}{R_1+R_2}\right)V_{REF})\end{array}$

Therefore, the lower crossover voltage of Schmitt trigger is

  $V_{TL}=V_S+\left(\frac{R_1}{R_1+R_2}\right)V_L$

Conclusion:

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

The upper crossover voltage of Schmitt trigger is

  $V_{TH}=V_S+\left(\frac{R_1}{R_1+R_2}\right)V_H$

The lower crossover voltage of Schmitt trigger is

  $V_{TL}=V_S+\left(\frac{R_1}{R_1+R_2}\right)V_L$

To determine

The values of $R_1$ , $R_2$ , and $V_{REF}$

Answer to Problem 15.49P

The resistor values are $R_1=4k\Omega$ and $R_2=188k\Omega$ .

The reference voltage is $V_{REF}=-1.787V$

Explanation of Solution

Given:

The crossover voltages are $V_{TH}=-1.5V$ and $V_{TL}=-2V$ .

The minimum resistance is to be $4k\Omega$ .

  

Calculation:

Let $V_H=+12V$ and $V_L=-12V$

The minimum resistance is to be $4k\Omega$

The crossover voltages are $V_{TH}=-1.5V$ and $V_{TL}=-2V$ .

Substitute $V_{TH}=-1.5V$ and $V_H=+12V$ in the upper crossover voltage equation

  $\begin{array}{l}-1.5=V_S+\left(\frac{R_1}{R_1+R_2}\right)(12)\\ V_S=-1.5-\left(\frac{R_1}{R_1+R_2}\right)(12)\end{array}$

Substitute $V_{TL}=-2V$ and $V_L=-12V$ in the lower crossover voltage equation

  $-2=-1.5-\left(\frac{R_1}{R_1+R_2}\right)(-12)$

Substitute $V_S$ in the above equation

  $\begin{array}{l}-2=-1.5-\left(\frac{R_1}{R_1+R_2}\right)(12)+\left(\frac{R_1}{R_1+R_2}\right)(-12)\\ -2=-1.5-\left(\frac{R_1}{R_1+R_2}\right)(24)\\ -0.5=-\left(\frac{R_1}{R_1+R_2}\right)(24)\\ R_1+R_2=48R_1\end{array}$

Choose $R_1$ is the minimum resistance (That is, $R_1=4k\Omega$ ) . Then

  $\begin{array}{l}4\times {10}^3+R_2=48(4\times {10}^3)\\ 4\times {10}^3+R_2=192\times {10}^3\\ R_2=188\times {10}^3\end{array}$

Therefore, the resistor values are $R_1=4k\Omega$ and $R_2=188k\Omega$ .

Substitute $V_{TH}=-1.5V$ , $V_H=+12V$ , $R_1=4k\Omega$ and $R_2=188k\Omega$ in the upper crossover voltage equation,

  $\begin{array}{l}-1.5=V_s+\left(\frac{4\times {10}^3}{4\times {10}^3+188\times {10}^3}\right)\left(12\right)\\ -1.5=V_s+\left(\frac{4\times {10}^3}{192\times {10}^3}\right)\left(12\right)\\ -1.5=V_s+0.25\\ V_s=-1.75\end{array}$

Substitute $V_s=-1.75V$ , $R_1=4k\Omega$ and $R_2=188k\Omega$ in switching voltage equation,

  $\begin{array}{l}-1.75=\left(\frac{188\times {10}^3}{4\times {10}^3+188\times {10}^3}\right)V_{REF}\\ -1.75=\left(\frac{188\times {10}^3}{192\times {10}^3}\right)V_{REF}\\ -1.75=\left(0.9792\right)V_{REF}\\ V_{REF}=-\frac{1.75}{0.9792}\\ V_{REF}=-1.787\text{V}\end{array}$

Conclusion:

Therefore, the resistor values are $R_1=4k\Omega$ and $R_2=188k\Omega$ and the reference voltage is $V_{REF}=-1.787\text{V}$

To determine

To find: the currents in the resistors when $(i)v_0=V_H$ and $(ii)v_0=V_L$

Answer to Problem 15.49P

When $v_0=V_H$ , $V^{+}=V_{TH}$ , then the current is $\left|i\right|=71.8\mu A$

When $v_0=V_L$ , $V^{+}=V_{TL}$ , then the current is $\left|i\right|=71.8\mu A$ .

Explanation of Solution

Given:

Consider the Schmitt trigger as shown below.

  

Calculation:

The current in the resistor is given by

  $i=\frac{V_0-V^{+}}{R_2}$

( i ) When $v_0=V_H$ , $V^{+}=V_{TH}$ ,

The current in the resistor is

  $\begin{array}{l}i=\frac{V_H-V_{TH}}{R_2}\\ =\frac{12-(-1.5)}{188\times {10}^3}\\ =\frac{13.5}{188\times {10}^3}\\ i=71.8\times {10}^6\end{array}$

Therefore, the current is $\left|i\right|=71.8\mu A$

( ii ) When $v_0=V_L$ , $V^{+}=V_{TL}$ ,

The current in the resistor is

  $\begin{array}{l}i=\frac{V_L-V_{TL}}{R_2}\\ =\frac{12-(-2)}{188\times {10}^3}\\ =\frac{-10}{188\times {10}^3}\\ i=-53.19\times {10}^6\end{array}$

Therefore, the current is $\left|i\right|=71.8\mu A$ .

Conclusion:

Therefore,

When $v_0=V_H$ , $V^{+}=V_{TH}$ , then the current is $\left|i\right|=71.8\mu A$

When $v_0=V_L$ , $V^{+}=V_{TL}$ , then the current is $\left|i\right|=71.8\mu A$ .