Applied Electronics and Instrumentation 3(2+1)

LESSON 17. Application of operated amplifier-Summing-Integrating-Differentiating amplifier

^{Fig.17.1 shows an inverting amplifier, used to sum two input voltages.  The input voltages v₁ and v₂ are applied through the resistors R₁ and R₂ to the summing junction (P) and R_f is the feedback resistor.  At the point P,}

                        ^{i₁ + i₂ = i_f}

^{Since the voltage at the point P is ideally 0,}

            ^{(V₁ / R₁) + (V₂/R₂) = V_out / R_f}

            ^{Hence the output voltage,}

                        ^{V_out = (R_f / R₁) V₁  +  (R_f / R₂)V₂}

^{If R₁ = R₂ = R_f = R_v then V_out = - (V₁ + V₂)}

^{Hence the output voltage is equal to the sum of the input voltages and the circuit acts as a summing amplifier.  The negative sign indicates than OP-AMP is used in the inverting mode.}

^{Difference amplifier}

The difference amplifier is shown in Fig.17.2.  The output voltage can be obtained by using superposition  principle.  To find the output voltage V₀₁ due to V₁ alone, assume that V₂ is shorted to ground.  Then

            ^{V+  =  [R₂ / (R₁ + R₂)] V₁}

^{And V₀₁ = [(R₃ + R₄) / R₃] V+  =  [(R₃ + R₄) / R₃]  [R₂ /( R₁ + R₂)] V₁}

^{Now assuming that V₁ is shorted to ground, the output voltage V₀₂ due to V₂ alone is given by}

                       ^{V₀₂ =  (R₄/R₃)  V₂}

^{Therefore, with both inputs present, the output is}

            ^{V₀ = V₀₁ + V₀₂}

            ^{= [(R₃ + R₄) / R₃]  [R₂ / (R₁ + R₂)] V₁ – (R₄/R₃)V₂}

            ^{If R₁ = R₂ = R₃ = R₄ = R}

            ^{Then V_o = V₁ – V₂}

            ^{If all the external resistors are equal, the voltage difference amplifier functions as a voltage subtractor.}

^{Integrating amplifier}

^{If a capacitor is inserted in the feedback path, we got an output, which is the time, integral of the input signal.  Figure 17.3 (a) shows the circuit and Fig.17.3 (b) shows the equivalent circuit (with virtual ground).}

 Thus, the output voltage is proportional to the integral of input voltage.

 

If the input voltage is constant, i.e., V_in = V_s = V, the output will be –Vt/RC which is a ramp.  Thus, the circuit of Fig 17.3 (a) can be used as a ramp generator.

^{Differentiating amplifier}

^{The insertion of a capacitor in the input path and resistor in the feed back path (Fig.17.4 (a)) gives a differentiating amplifier.   Figure17.4 (b) shows the equivalent current.}

 

 Where, V_in = V_s is the input voltage.