Chapter 15, Problem 15.46P

Consider the Schmitt trigger in Figure P15.46. Assume the saturated outputvoltages are $\pm V_P$ . (a) Derive the expression for the crossover voltages $V_{TH}$ and $V_{TL}$ . (b) Let $R_A=10\text{k}\Omega$ , $R_B=20\text{k}\Omega$ , $R_1=5\text{k}\Omega$ , $R_2=20\text{k}\Omega$ , $V_P=10\text{V}$ , and $V_{\text{REF}}=2\text{V}$ . (a) Find $V_{TH}$ and $V_{TL}$ . (b) Sketch the voltagetransfer characteristics.

Figure P15.46

To determine

To derive: The expression for the crossover voltage $V_{TH}$ and $V_{TL}$

Answer to Problem 15.46P

The expression for the crossover voltage $V_{TH}$ and $V_{TL}$

  $V_{TH}=\frac{R_1}{R_B}\left(\frac{R_A+R_B}{R_1+R_2}\right)\left(V_P\right)-\left(\frac{R_A}{R_B}\right)V_{REF}$

  $V_{TL}=\frac{R_1}{R_B}\left(\frac{R_A+R_B}{R_1+R_2}\right)\left(-V_P\right)-\left(\frac{R_A}{R_B}\right)V_{REF}$

Explanation of Solution

Given:

The upper saturated output voltage, $V_H=V_{P}V$

The lowest saturated output voltage, $V_L=-V_{P}V$

  

Calculation:

In an ideal op-amp, the inverting and non-inverting terminal currents are zero. Due to the virtual ground concept, the inverting and non-inverting node voltages are equal.

  $v^{-}=v^{+}$

Given circuit can be represented as

  

Apply Kirchhoff’s current law at inverting node.

  $\begin{array}{l}\frac{v^{+}-v_t}{R_A}+\frac{v^{+}-V_{REF}}{R_B}+0=0\\ \frac{R_B\left(v^{+}-v_t\right)+R_A\left(v^{+}-V_{REF}\right)}{R_A{R}_B}=0\\ R_B\left(v^{+}-v_t\right)+R_A\left(v^{+}-V_{REF}\right)=0\\ R_B{v}^{+}-R_B{v}_t+R_A{v}^{+}-R_A{V}_{REF}=0\\ \left(R_A+R_B\right)v^{+}-R_B{v}_t-R_A{V}_{REF}=0\\ \left(R_A+R_B\right)v^{+}=R_A{V}_{REF}+R_B{v}_t\\ v^{+}=\left(\frac{R_A}{R_A+R_B}\right)V_{REF}+\left(\frac{R_B}{R_A+R_B}\right)v_t.\end{array}$

Hence, we get $v^{+}=\left(\frac{R_A}{R_A+R_B}\right)V_{REF}+\left(\frac{R_B}{R_A+R_B}\right)v_t.$

Apply Kirchhoff’s current law at non-inverting node.

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

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

  $\begin{array}{l}\left(\frac{R_1+R_2}{R_1}\right)\left[\frac{R_A}{R_A+R_B}{V}_{REF}+\left(\frac{R_B}{R_A+R_B}\right)v_t\right]-v_0=0\\ \left(\frac{R_A}{R_1}\right)V_{REF}+\left(\frac{R_B}{R_1}\right)v_t-\left(\frac{R_A+R_B}{R_1+R_2}\right)v_o=0\\ \left(\frac{R_B}{R_1}\right)v_I=\left(\frac{R_A+R_B}{R_1+R_2}\right)v_o-\left(\frac{R_A}{R_1}\right)V_{REF}\\ v_I=\frac{R_1}{R_B}\left[\left(\frac{R_A+R_B}{R_1+R_2}\right)v_o-\left(\frac{R_A}{R_1}\right)V_{REF}\right]\\ v_I=\frac{R_1}{R_B}\left(\frac{R_A+R_B}{R_1+R_2}\right)v_o-\left(\frac{R_A}{R_1}\right)V_{REF}\mathrm{...}\left(1\right)\end{array}$

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

The upper crossover voltage of Schmitt trigger $V_{TH}=\frac{R_1}{R_B}\left(\frac{R_A+R_B}{R_1+R_2}\right)\left(V_P\right)-\left(\frac{R_A}{R_B}\right)V_{REF}$ .

Similarly, when $v_o=-V_P$ and $v_I=V_{TL}$ in equation-(1),

The lower crossover voltage of Schmitt trigger $V_{TL}=\frac{R_1}{R_B}\left(\frac{R_A+R_B}{R_1+R_2}\right)\left(-V_P\right)-\left(\frac{R_A}{R_B}\right)V_{REF}$ .

Conclusion:

Therefore, the expression for the crossover voltage $V_{TH}$ and $V_{TL}$

  $V_{TH}=\frac{R_1}{R_B}\left(\frac{R_A+R_B}{R_1+R_2}\right)\left(V_P\right)-\left(\frac{R_A}{R_B}\right)V_{REF}$

  $V_{TL}=\frac{R_1}{R_B}\left(\frac{R_A+R_B}{R_1+R_2}\right)\left(-V_P\right)-\left(\frac{R_A}{R_B}\right)V_{REF}$

To determine

To find: the values of $V_{TH}$ and $V_{TL}$ .

To sketch: The voltage transfer characteristics.

Answer to Problem 15.46P

The required values are

  $V_{TH}=2V$

  $V_{TL}=-4V$

The voltage transfer characteristics are shown in Figure 1

Explanation of Solution

Given:

  $\begin{array}{l}{R}_A=10k\Omega \\ R_B=20k\Omega \\ R_1=5k\Omega \\ R_2=20k\Omega \end{array}$

The circuit voltages are:

  $V_P=10V$ And

  $V_{REF}=2V$

Calculation:

The upper crossover voltage is

  $\begin{array}{l}{V}_{TH}=\left(\frac{R_1}{R_B}\right)\left(\frac{R_A+R_B}{R_1+R_2}\right)\left(V_P\right)-\left(\frac{R_A}{R_B}\right)V_{REF}\\ =\left(\frac{5\times {10}^3}{20\times {10}^3}\right)\left(\frac{10\times {10}^3+20\times {10}^3}{5\times {10}^3+20\times {10}^3}\right)\left(10\right)-\left(\frac{10\times {10}^3}{20\times {10}^3}\right)\left(2\right)\\ =\left(\frac{5}{20}\right)\left(\frac{30\times {10}^3}{25\times {10}^3}\right)\left(10\right)-\left(\frac{10}{20}\right)\left(2\right)\\ =3-1\\ V_{TH}=2V\end{array}$

Therefore, the upper crossover voltage is $V_{TH}=2V$ .

The lower crossover voltage is given by,

  $\begin{array}{l}{V}_{TL}=\left(\frac{R_1}{R_B}\right)\left(\frac{R_A+R_B}{R_1+R_2}\right)\left(-V_P\right)-\left(\frac{R_A}{R_B}\right)V_{REF}\\ =\left(\frac{5\times {10}^3}{20\times {10}^3}\right)\left(\frac{10\times {10}^3+20\times {10}^3}{5\times {10}^3+20\times {10}^3}\right)\left(-10\right)-\left(\frac{10\times {10}^3}{20\times {10}^3}\right)\left(2\right)\\ =\left(\frac{5}{20}\right)\left(\frac{30\times {10}^3}{25\times {10}^3}\right)\left(-10\right)-\left(\frac{10}{20}\right)\left(2\right)\\ =\left(\frac{1}{4}\right)\left(\frac{6}{5}\right)\left(-10\right)-\left(\frac{1}{2}\right)\left(2\right)\\ =-3-1\\ V_{TL}=-4V\end{array}$

Therefore, the lower crossover voltage is $V_{TL}=-4V$ .

The given Schmitt trigger is positive feedback comparator.

Here, the transfer characteristic is a rectangle. Hence, it is called hysteresis which is a dead band. It is so called as the output is not changing

The output remains in the state indefinitely until input voltage crosses the any of the threshold levels.

The voltage transfer characteristics of the given Schmitt trigger is,

  

Figure 1

Conclusion:

Therefore, the required values are

  $V_{TH}=2V$

  $V_{TL}=-4V$

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