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When a capacitor is charged in a first order RC circuit, it charges exponentially. I understand this behavior via equations. But can anyone explain the physical reason?

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    \$\begingroup\$ Simply notice that at the beginning when the voltage across the capacitor is 0V. All the input voltage will be present across the resistor. Thus, in the beginning, the charging current is the largest. But as the voltage across the capacitor increases the voltage drop across the resistor is reduced (VR = Vin - Vcap), thus decreases the charging current. So, the large the voltage across the capacitor is the smaller the charging current is. And smaller the charging current will be, the more time is needed to charge the capacitor. \$\endgroup\$ – G36 Dec 23 '20 at 16:30
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    \$\begingroup\$ For the physical reason, think about the discrete time approximation to this problem. For each time-step, the charge transferred is proportional to the resistor voltage. Iterating over the time steps we see that the charging is exponential (but probably not if the form \$e^{kt}\$, but of the form \$a^{KT}\$). Making the time steps smaller and smaller leads to the exponential. \$\endgroup\$ – AJN Dec 23 '20 at 17:01
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    \$\begingroup\$ The problem touches the question why we find very often an exponential function in nature. This e-function is very often the solution of diff. equations because the 1st derivation equals the origial function.: \$\endgroup\$ – LvW Dec 23 '20 at 17:25
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    \$\begingroup\$ I don't understand questions like this. You state that you understand this behavior via equations. So, what does it mean to ask "what is the physical reason?". The physical reason for what?. Physics consists essentially of finding mathematical models that adequately describe physical phenomena in the appropriate approximation. Are you asking why physical capacitors can be approximately described mathematically such that the (ideal) RC circuit has the characteristic exponential charge / discharge solutions? If not, please elaborate on what you're actually looking for in an answer. \$\endgroup\$ – Alfred Centauri Dec 24 '20 at 4:53
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schematic

simulate this circuit – Schematic created using CircuitLab

The current is determined by the voltage across the resistor, which is V1-Vc. As the capacitor charges, Vc increases while V1 stays the same, so the current decreases. The rate at which a capacitor charges is directly proportional to the current, so the rate at which it charges decreases proportional to its current state of charge--the classic differential equation for an exponential decay.

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    \$\begingroup\$ Yes, I understand that the Vc increases with time but why exponential? why not linear or some other nature? (I understand this mathematically but I want to know physical significance) \$\endgroup\$ – sumita sahu Dec 23 '20 at 16:28
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    \$\begingroup\$ I'm not sure how to explain that any more clearly than the mathematics already does. It's baked into how physics and mathematics work. \$\endgroup\$ – Hearth Dec 23 '20 at 16:30
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    \$\begingroup\$ Someone asked me this during an interview and I was blank, I searched a lot but can't find anything. \$\endgroup\$ – sumita sahu Dec 23 '20 at 16:39
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    \$\begingroup\$ And explaining the derivation from the differential equation didn't work? \$\endgroup\$ – Hearth Dec 23 '20 at 16:39
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    \$\begingroup\$ If someone asks for a physical reason with no equations, I question what kind of physics they've learned.... \$\endgroup\$ – Hearth Dec 23 '20 at 16:56
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When a series RC circuit is applied across a fixed DC voltage, the capacitor begins charging. It begins charging from 0 volts and, at that instant, the current that charges the capacitor is defined by the DC voltage and the value of the series resistor. That's simple ohm's law (if you are allowed to use that).

As the capacitor charges, the voltage across it rises from 0 volts and this means that the voltage across the resistor must reduce. Again, using ohm's law, if the resistor voltage reduces then, the charging current must also reduce. This is because R and C are in series.

So now, because the charging current has reduced, the rate at which the capacitor voltage charges also reduces. I don't know if you are allowed to use the charge formula in making an explanation but I guess, if you accept that current is the mechanism that forces a capacitor to charge up in voltage then, a reduction in charging current has to mean a slower rate in the rise of capacitor voltage.

Hence, the voltage rate of climb from 0 volts is starting to reduce as the capacitor charges. And, as the voltage climbs more there is even less voltage across the series resistor. In turn that means the charging current becomes even less and the rate of charge voltage across the capacitor slows down more.

More time passes; the rate at which voltage increases becomes less and the current into the capacitor is also less. Ultimately, as the capacitor voltage approaches the fixed DC voltage supply, the current through the resistor is getting very tiny indeed and so the rate of change of voltage of the capacitor is also very tiny.

Eventually (and being practical) the rate at which voltage rises across the capacitor is seen to virtually stop and, the current into the capacitor is virtually zero. An "engineering" equilibrium is reached where the capacitor voltage is virtually the same value as the fixed DC voltage.

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Imagine a steel pressure vessel you are trying to charge with compressed air of constant pressure. This vessel will be your capacitor, the capacity -- amount of air mass it can store, being the capacitance. The compressor is the power source, outputting a constant air pressure -- the voltage.

There is a restriction valve on the pipeline between your compressor and the pressure vessel, which restricts the movement of air, thus becoming a resistor. The flow rate -- amount of air mass traveled through the pipeline per second is the current. Because of this restriction valve, the flow cannot be infinite.

As you charge the pressure vessel through the compressor and the restriction valve, the pressure in the vessel will gradually increase. Since the compressor only outputs a constant pressure, the pressure increase on the destination site causes the flow rate to decrease, reducing the speed at which the vessel is charged as it is being charged, until after an infinite amount of time (as in steady state), the compressor output pressure has equalized with the pressure of the vessel, and charging can no longer proceed.

The process of air mass increase slowing down is confirmed to be mathematically equivalent to the exponent representation.

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    \$\begingroup\$ Great Example John! \$\endgroup\$ – The Force Awakens Dec 23 '20 at 18:05
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You can think of the capacitor to be a voltage source.In the beginning when the capacitor is completely uncharged there isnt any voltage between the plates of the capacitor because no charge has come to sit on the plates and create a voltage difference. While the capacitor is being charged more and more charge sits on the plates and the result is a voltage differential. Now this opposes the voltage source which charged the capacitor and therefore less current must flow. This process will happen until the voltage of the capacitor becomes equal with the source which charged the capacitor.

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    \$\begingroup\$ No, the capacitor is not a voltage source. Saying such things will only cause more confusion later. \$\endgroup\$ – Elliot Alderson Dec 23 '20 at 18:48
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    \$\begingroup\$ @Elliot Alderson, For the purposes of this excellent intuitive explanation, a capacitor is a voltage source... like a battery.... but rechargeable battery. What is confusing is your formal and sterile thinking. It would be good to gain some practical experience to start feeling what is happening in circuits ... \$\endgroup\$ – Circuit fantasist Dec 24 '20 at 11:22
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You obviously see the circuit theory as a kind of symbol game which is disconnected from the physics. Actually you are right. Circuit theory doesn't care what voltage and current mean, they are only quantities which depend on time and the circuit. Voltage and current are physical in the sense they present the state of something which exists and which isn't only an imagined relation.

Electrodynamics based on Maxwell's field theory and some properties of materials is the physics behind the circuit theory. From there come such things as Ohm's law, Kirchoff's laws and equation I=C(dU/dt) for capacitors.

If it happens that you like to see a mechanical system which you understand intuitively and which is analoquous with the RC charging circuit think for example heating a mass. The voltage source is there some heating power, the resistor is the not perfectly heat conducting medium between the source and the mass to be heated and the capacitance is the heat capacity of the heated mass. It's temperature is the charged voltage.

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    \$\begingroup\$ I don't like how you state that quantities of a circuit are not physical. \$\endgroup\$ – The Force Awakens Dec 23 '20 at 18:04
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    \$\begingroup\$ @TheForceAwakens I don't see that anyone said they are not physical. The statement was that circuit theory doesn't care what they mean. In other words, circuit theory operates at a higher level of abstraction when compared to physics. Which is just fine. \$\endgroup\$ – Elliot Alderson Dec 23 '20 at 18:50
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    \$\begingroup\$ @ElliotAlderson Comments can be opinions. The Force Awakens writes about his feelings, he claims nothing about the rightness of my writings. \$\endgroup\$ – user287001 Dec 23 '20 at 19:52
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    \$\begingroup\$ @user287001 The Force Awakens made a statement "you state that quantities of a circuit are not physical". I did not see such a statement anywhere in your answer, so as far as I can see The Force Awakens made an objectively false statement, not just an opinion. \$\endgroup\$ – Elliot Alderson Dec 23 '20 at 20:30
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    \$\begingroup\$ Ok. Voltages and currents are physical in the sense they present the state of something existing which we do not consider to be only a relation. Voltages and currents can even be measured, so you are right. \$\endgroup\$ – user287001 Dec 23 '20 at 20:47
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Why not linear or some other nature?

My answer will be a little unexpected for you because I will answer not "why not linear" but I will show how it can be made linear. In fact, this is the goal in most cases of practice; exponential relation, with few exceptions, is undesirable.

The trick is extremely simple if only you can guess. The voltage VC across the capacitor does not linearly change because it is subtracted from the input voltage (it is a loss) and the current decreases - I = (VIN - VC)/R. So we have to compensate this voltage drop.

For this purpose, we connect a variable voltage source in series to the capacitor and with the same polarity as the input voltage source (travelling the loop)... and adjust its voltage equal to the voltage drop across the capacitor. As a result, the voltage drop will be removed and the current will be as in the beginning - I = (VIN - VC +VC)/R = VIN/R.

If we feel bored doing this tedious job, we assign it to an op-amp. This is the idea of the op-amp inverting integrator - Fig. 1.

Op-amp RC integrator

Fig. 1. Op-amp RC integrator (a geometrical interpretation visualized by voltage bars and current loops).

I created this Corel Draw picture in the 90's (the element designations do not correspond to the generally accepted ones). Here are some explanations for the inscriptions inside the figure. The circled op-amp (including the bipolar power supply) is a "helping voltage source"; UOUT (VOUT) is a "copy" and UC (VC) is the "original" voltage. So, the op-amp "copies" the capacitor voltage and adds it in series to the input voltage EIN (VIN).

And here is the hydraulic analogy (a little unusual communicating vessels) of the inverting integrator - Fig. 2. This conceptual picture corresponds to the op-amp circuit above.

Hydraulic analogy of the inverting integrator

Fig. 2. Hydraulic analogy of the inverting integrator.

Some explanations about the text inside the figure: The little man on the left is a "helper" and the capacitor on the right is a "thief":) So the "thief" steals voltage... but the "helper" restores it and adds it to the input voltage. The left vessel is a constant pressure source. The right vessel is a "bottomless vessel" - when its water level tries to increase, the little man oh the right lowers the vessel thus keeping up a "hydraulic virtual ground".

EDIT: This is a great idea (removing disturbance by anti-disturbance) that we can see everywhere... even in SE EE. There are contributors here for whom the powerful explanations of others are disturbance for his ego; so they try to compensate it with an anti-disturbance (-4)...

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    \$\begingroup\$ I'm not the downvoter but you really need to stop taking it as a personal attack when people downvote you. I would guess the downvoter did so because this doesn't actually answer the question that was asked, instead going off on a tangent to explain an interesting concept instead. And yes, linearization via feedback is an interesting topic, but it's not an answer to the question that was asked here. \$\endgroup\$ – Hearth Dec 24 '20 at 17:20
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    \$\begingroup\$ So you declare that you have no intention whatsoever of answering the actual question, and you are then surprised when someone says "This answer is not useful"? You could have written a detailed and "powerful" explanation of quantum theory but it would still be useless to the OP. And if I were you I would not say anything about anyone else's ego. \$\endgroup\$ – Elliot Alderson Dec 24 '20 at 17:47
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    \$\begingroup\$ @Hearth, Thanks for the interpretation. There is no problem; we all know each other very well and understand what we are talking about. A little humor is never superfluous. I just saw the downvoter's reaction and made the connection with the great principle. BTW it can be implemented without negative feedback, e.g., by a "negative capacitor". Really, this is not the exact answer but it is closely related to it; this is the answer to the next question that logically follows, "How do we make the capacitor charge linearly?" In your place, I would first admire the idea and then criticize... \$\endgroup\$ – Circuit fantasist Dec 24 '20 at 18:51
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    \$\begingroup\$ @Elliot Alderson, My answer is closely related to this question; so it is useful. If you take the trouble to follow the link above, you will see a 5-step scenario; the third step is dedicated to this question. Showing how something nonlinear can become linear is an indirect (and more original) way to explain what causes this nonlinearity over time... \$\endgroup\$ – Circuit fantasist Dec 24 '20 at 19:11
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    \$\begingroup\$ @tlfong01, I think it has become clear that this cannot be done through intuition but through mathematics. Unfortunately, intuition only tells us that the current is progressively decreasing but it can't tell us exactly how ... But nevertheless, I keep thinking about this phenomenon of "communicating vessels". But now the question arose how to make a -1 ohm resistor. Interesting topics have no end... \$\endgroup\$ – Circuit fantasist Dec 25 '20 at 7:58

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