I continue to develop the powerful “voltage compensation” idea by asking more and more questions about its ubiquitous implementations. So far, we have considered how to virtually decrease the resistance (almost) up to zero thus making an “ideal” ammeter and a current-to-voltage converter (transimpedance amplifier):
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?
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?
Now it is interesting to see if we can apply this trick to the time-dependent capacitor. Let’s try.
People get an impression of the capacity by observing the voltage across the capacitor when charging it - the larger the capacity, the lower the voltage (at the same moment of the time). So, if we add a voltage that is a part of the voltage drop across the capacitor, we will create the illusion that the capacity has increased... and if the additional voltage is equal to the voltage drop, we will create an even stronger illusion that the capacity has become infinite! People like illusions, so let's give them that pleasure:-)
So, let’s add a humble variable voltage source in series with the capacitor and adjust its voltage equal to the voltage drop across the capacitor. As a result, the voltage drop as though disappears. What a magic - the people continue charging the capacitor but its voltage stays zero as though the capacitance has become infinite! They may even try to use this huge capacitor as a source of energy... but is this possible?
Finally, to obtain a real circuit, we just replace the “helping” voltage source with a properly supplied op-amp and make it “observe” the difference between the voltage drop across the capacitor and its output voltage... and change the latter so that to keep the virtual ground. Practically, this means to connect the capacitor between the op-amp output and its inverting input and to inject the input current into the node at the inverting input. As a result, its output voltage will always be equal to the voltage drop and the voltage across the whole network (seen by the input source) will be almost zero. Thus we have used a real capacitor to create a “virtual capacitor” with multiplied capacitance (the convential name of this circuit is “current integrator”).
http://www.circuit-fantasia.com/tutorial/intro/question4.swf
In this circuit, like in the op-amp ammeter and transimpedance amplifier, the op-amp output voltage (energy) is a “mirror” copy of the voltage drop (loss) across the capacitor and the current is the same. So, we have two elements with equal voltages and currents – the first is a real capacitor consuming a voltage while the second is a voltage source producing the same voltage. Then can we say the op-amp is a "negative capacitor" and the whole circuit is an "infinite capacitor"? Or, even more figuratively, the current integrator is a "neutralized positive capacitor"? If so, it would be a good prelude to the subject of negative capacitor...
You probably have already guessed that this is a story about the ubiquitous Miller effect; see also
http://en.wikipedia.org/wiki/Talk:Miller_effect#How_to_modify_impedance
What do you think about it? Have you ever seen such an explanation before? Is it reliable? Is it useful for understanding compared with the conventional Miller effect explanation? I expect your opinion. Cyril
And to arouse interest in the subject, I would ask a little "provocative" question at the very beginning:
By adding a voltage that is less and equal to the voltage drop, we have obtained a virtual capacitor with enlarged and infinite capacitance. But what will happen if we add a voltage that is higher than the voltage drop across the capacitor?
Regards, Cyril
When C --> infinity, you can model it as an ideal voltage source! I think I was the first to suggest this back in 1996. See page 10, here:
http://www.eleceng.adelaide.edu.au/Personal/dabbott/publications/EDU_abbott1996.pdf
Hi Derek,
I like "... in the limit as C ---> oo, the capacitor becomes simultaneously an infinite store of charge and a perfect short..." Do you remember the question below?
https://www.researchgate.net/post/Is_a_charged_capacitor_a_voltage_source_Is_a_charged_inductor_a_current_source_Is_there_any_difference_between_them_and_true_electrical_sources#share
So it seems we have two kinds of voltage sources in this world - "converters" and "stores". It remains only to decide what batteries are because they both store chemical energy and convert it to electrical energy...
Regards, Cyril
But still Derek, can we use this "supercapacitor" to solve the world's energy problems? Most likely, oil kings will not let it :-)
Yes we can have infinite capacitance....we just hvve to replace the conducting plates of caps with one such plates available metal in nature in which electron takes years to come back to original shell once being excited with constant field
Yes exactly....as it will have trapped carriers on one side & area with lesser charge on another side..one convention goes on to say that polarity in capacitor is subjective to material used
Shaikh, it sounds wonderful but I would like to discuss here the possible circuit implementations of the great Miller idea. The question is not about physical effects; it is about circuit solutions...
According to the "Miller idea" a capacitance in the feedback path of an inverting amplifier with gain -G is multplied by (1+|G|). This is a finite value - thus the capacitance cannot be infinite.
if we talk of infinite capacitance this could simply be synthesize as on caps is series with DC source in circuit...but this infinite case is dependent on factors governing the displacement of electrons on the capacitance plate..so infinite sort of case could be realized for longer time by changing the material used...but every infinity caused by ckt condition will fail to exist as an independent capacitance model
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