Charging of capacitor in RC circuit

Question

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?

Answer 1

^{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.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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.

Answer 6

EDIT

I have added some additional pictures and explanations to my answer hoping to get the rating it deserves. I have used more water analogies to illustrate the circuit evolution and the problem solved by the active version. The circuit diagrams are conceptual - they are "geometrically drawn" and visualized by voltage bars and current loops.

Perfect C integrator

If we supply a capacitor by an "ideal" current source (pump), there is no problem - the voltage (water level) linearly increases.

Imperfect RC integrator at ideal load conditions

But we want to drive the integrator, as usual, by a voltage source. So, we connect a resistor R to convert the input voltage into current. And indeed, at the first moment everything is fine - the output voltage begins to change linearly.

Imperfect RC integrator at non-ideal load conditions

But after a while a problem appears - the voltage VC across the capacitor begins not linearly changing because it is subtracted from the input voltage and the current decreases - I = (VIN - VC)/R.

So this was my answer "why not linear"... but I will continue to show how we can made it linear. In fact, this is the goal in most cases of practice; exponential relation, with few exceptions, is undesirable.

Active RC integrator

Obviously, we have to compensate somehow the "undesired" voltage drop. For this purpose, we can 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.

The hydraulic analogy on the right is represented by a little unusual communicating vessels and corresponds to the op-amp circuit below. Here are 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".

Op-amp inverting integrator

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".

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).

Conclusion

This is a great idea that we can see everywhere. We can figuratively call it "removing a disturbance by an anti-disturbance".