By adding a “helping” voltage to the voltage drop across a resistor (real ammeter with internal resistance), we have virtually zeroed its resistance thus building the perfect op-amp ammeter
https://www.researchgate.net/post/How_do_we_improve_the_real_ammeter_How_do_we_create_an_almost_ideal_ammeter_What_does_the_op-amp_really_do_in_the_circuit_of_an_op-amp_ammeter?;
then by using the “helping” voltage as an output, we have built the ubiquitous transimpedance amplifier
https://www.researchgate.net/post/Is_there_any_connection_between_the_humble_resistor_and_the_transimpedance_amplifier_What_does_the_op-amp_really_do_in_this_electronic_circuit?
Doing the same trick with a capacitor, we have increased its capacitance up to infinity and created the current integrator
https://www.researchgate.net/post/Can_we_virtually_increase_the_capacitance_up_to_infinity_How_do_we_make_such_virtual_capacitors_Have_they_the_same_properties_as_real_capacitors?
In electronics we prefer to deal with voltages; so it would be useful to make a voltage-to-voltage integrator. For this purpose, we have only to connect a voltage-to-current converter before the current integrator and to use the “helping” voltage as an output. I have done it step-by-step in the stories below:
http://www.circuit-fantasia.com/circuit_stories/building_circuits/integrator/op-amp_rc_integrator.htm (Circuit stories on the whiteboard)
http://www.circuit-fantasia.com/collections/tools/build_integrator.html (Flash movie)
http://en.wikibooks.org/wiki/Circuit_Idea/How_to_Make_a_Perfect_RC-integrator (Circuit Idea wikibook)
and I have developed this idea in the Wikipedia disputation below
http://en.wikibooks.org/wiki/Talk:Circuit_Idea/How_to_Make_a_Perfect_RC-integrator#About_the_op-amp_inverting_integrator
I will illustrate this building approach with a picture scenario immediately below the question and will wait for your opinions and questions.
1. C INTEGRATOR. The bare capacitor is the simplest integrator. It is perfect, if only we drive it by a constant current.
2. RC integrator (at the beginning). Most frequently, we need to drive the capacitor by a voltage. For this purpose, we connect before a resistor R acting as a voltage-to-current converter. In the beginning, the capacitor is discharged... there is no voltage drop across it... there is no problem...
3. RC integrator (after a while). The capacitor is continuously charging and a disturbing voltage drop appears across it...
4. ACTIVE RC INTEGRATOR. To compensate the undesired voltage drop, we connect an additional "helping" voltage source and take its voltage as an output voltage...
5. OP-AMP RC INTEGRATOR. Finally, we replace the "helping" voltage source with a properly supplied op-amp...
Cyril, I am sure I have somewhere mentioned already another simple view of the active RC integrator:
It is simply an application of the well-known MILLER effect (enlargement of the effective capacitance if seen from the input).
Example: Instead of using a capacitance of 1E-3 F we can use 1E-6 F in the feedback path of an amplifier having a gain of -999.
Welcome, Lutz! I am glad to see you again because there are no signs of life in RG discussions without you. Really, I expected more enthusiastic comments of my insights, but I'm happy with the little:-)
Best regards, Cyril
Thank you Cyril!
I forgot to add that the output of this RC integrator (if seen as Miller capacitance) is - like for each passive RC circuit - BETWEEN R and the grounded C. As this node is identical to the inverting opamp input this tiny voltage is amplified by a gain of -999 and appears at a low-resistive opamp output.
Maybe, it's time to discuss the great Miller effect and the theorem devoted to him?
About your last remark... We have at least three outputs of this circuit: the voltage drop across the capacitor ("floating", unbuffered, unsuitable for use); the op-amp output voltage (a "mirror copy" of the voltage drop across the capacitor, grounded, buffered, inverted, excellent for use); and finally, your output (with an almost zero voltage, grounded, usually useless)...
So, we ask yourself, "Where do we take the output from?" The same question arises when we reinvent the transimpedance amplifier (see the attached picture below).
I think, it was discussed here already recently. Did it not?
However, my description in short is as follows:
The current through a specific part X is caused by two voltage sources (here: input and inverted output signal). However, we treat this current as a result of the input source only. Thus, the part X appears with a reduced impedance (as seen from the input only).
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