Consider an Op-AMP circuit shown in the figure The input is symm

SOLUTION

Input:

(i) with positive slope:

(ii) with negative slope:

Now,

Suppose output V₀ = +15 V

\({V_ + } = \left( {\frac{5}{{15}}} \right) \times 15 = 5\;volt\) 

V_in > 5 V output switches to -15 V

V_in < 5 V output remains at +15V

V₀ = -15 V

\({V_ + } = \left( {\frac{5}{{15}}} \right) \times - 15 = - 5\;V\) 

V_in > -5 V output remains at – 15 V

V_in < -5 V output switches to +15V

The transfer characteristics is as shown:

Hysteresis width = 10

Input with positive slope:

The input goes from -10 V to +10V,

∴ As seen from the Transfer characteristics.

The output will be +15V from t = 0 to the time t₁, when input reaches 5 V.

After that, the output switches to -15 V and remains at -15 V, till the input is again – 10 V.

Then the cycle repeats.

To find t₁

\(\frac{{10 + 10}}{{2T}} = \frac{{{v_{in}} + 10}}{{{t_1}}}\) 

\({v_{in}} = \frac{{20}}{{2T}}{t_1} - 10\) 

\(5 = \frac{{20}}{{2T}}{t_1} - 10\) 

\({t_1} = \frac{{3T}}{2}\) 

The average value can be calculated as follows:

 \( = \frac{{\frac{{15 \times 3T}}{2} - \frac{{15 \times T}}{2}}}{{2T}}\) 

= 7.5

Input with a negative slope:

The input goes from +10V to -10V,

∴ As seen from the Transfer characteristics.

The output will be -15 V from t = 0, fill the time t₂, the input is -5.

After t₂, the output switches to +15 V and remains at +15 V, fill the input is again + 10V.

Then the cycle repeats.

To find t₂

\(\frac{{10 + 10}}{{0 - 2T}} = \frac{{{v_{in}} + 10}}{{t - 2T}}\) 

20t – 40T = -2v_in T – 20T

-10T + 10t = -v_in T

\(\frac{{ - 10t}}{T} + 10 = {v_{in}}\) 

\( - 15 = \frac{{ - 10{t_2}}}{T}\) 

\({t_2} = \frac{{3T}}{2}\) 

The average value can be calculated as follows:

\(= \frac{{\frac{{ - 15 \times 3T}}{2} + \frac{{15\;T}}{2}}}{{2T}}\) 

= -7.5 V