# Op-amp Integrator Network-Operational Amplifiers Types Tutorials Series

http://www.ingenuitydias.com/2014/06/op-amp-integrator-network-operational.html

1. Shown below is an op-amp integrator network

–The output will be equal to the integral of the input, as long as the op-amp remains in its linear region

–Due to the virtual short property of the op-amp input, we can write i1 = vIN/R1

–This current i1 starts charging the capacitor C according to the relation i1 = C(dvC/dt).

2. Since v- remains at GND, the output drops below GND as C charges and the time derivative of vOUT becomes the negative of the time derivative of vC

–since vC = 0 - vOUT

3. Combining the above equations, we obtain

–dvOUT/dt = -i1/C = -vIN/R1C

4. Solving for vOUT(t) and assuming C is initially uncharged, we obtain

–vOUT(t) = (-1/R1C) ¤ vIN dt where the integral is from 0 to t

*Given an input signal of 4V square wave for 10 ms duration, what is the integrator output versus time for the integrator circuit?

–The current into the capacitor during the square wave is constant at 4V/5Kohm = 0.8 mA

–Using the integral expression from the previous chart, the capacitor voltage will increase linearly in time (1/R1C) 4t = 0.8t V/ms during the square wave duration

–The output will therefore reduce linearly in time by – 0.8t V/ms during the pulse duration, falling from 0 to –8 volts, as shown in the figure at left

–Since at 10 ms the output will be –8 V > VNEG, the op-amp will not saturate during the 10 ms input pulse

*Consider a case with an infinitely long 4V pulse

–The capacitor will continue to charge linearly in time, but will eventually reach 10V which will force vOUT to –10V (= VNEG) and saturate the op-amp (at 12.5 ms)

–After this time, the op-amp will no longer be able to maintain v- at 0 volts

–Since vOUT is clamped at –10V, the capacitor will continue to charge exponentially with time constant R1C until v- = +4V

*During this time the capacitor voltage will be given by

vC(t) = 10 + 4[1 – exp(t1 – t)/R1C] where t1 = 12.5 ms

*At t = t1 , vC = 10 V and at t = infinity, vC = 14 V

–The resulting capacitor and output waveforms are shown above.

–The output will be equal to the integral of the input, as long as the op-amp remains in its linear region

–Due to the virtual short property of the op-amp input, we can write i1 = vIN/R1

–This current i1 starts charging the capacitor C according to the relation i1 = C(dvC/dt).

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–since vC = 0 - vOUT

3. Combining the above equations, we obtain

–dvOUT/dt = -i1/C = -vIN/R1C

4. Solving for vOUT(t) and assuming C is initially uncharged, we obtain

–vOUT(t) = (-1/R1C) ¤ vIN dt where the integral is from 0 to t

__Example__

*Given an input signal of 4V square wave for 10 ms duration, what is the integrator output versus time for the integrator circuit?

–The current into the capacitor during the square wave is constant at 4V/5Kohm = 0.8 mA

–Using the integral expression from the previous chart, the capacitor voltage will increase linearly in time (1/R1C) 4t = 0.8t V/ms during the square wave duration

–The output will therefore reduce linearly in time by – 0.8t V/ms during the pulse duration, falling from 0 to –8 volts, as shown in the figure at left

–Since at 10 ms the output will be –8 V > VNEG, the op-amp will not saturate during the 10 ms input pulse

__Op-amp Integrator Example - Long Pulse__

*Consider a case with an infinitely long 4V pulse

–The capacitor will continue to charge linearly in time, but will eventually reach 10V which will force vOUT to –10V (= VNEG) and saturate the op-amp (at 12.5 ms)

–After this time, the op-amp will no longer be able to maintain v- at 0 volts

–Since vOUT is clamped at –10V, the capacitor will continue to charge exponentially with time constant R1C until v- = +4V

*During this time the capacitor voltage will be given by

vC(t) = 10 + 4[1 – exp(t1 – t)/R1C] where t1 = 12.5 ms

*At t = t1 , vC = 10 V and at t = infinity, vC = 14 V

–The resulting capacitor and output waveforms are shown above.