Funny this is my third question in a row about RC circuits, but I wanted to figure out some stuff I'd been wondering for a while. What I was confused about the power while charging. Like say that you have a battery that is able to produce 1 amp at 1 volt of electricity, and you have it connected to a 1 volt, 2 farad capacitor (for simplicity). The energy it can store is 1/2CV^2 which is equal to 1 joule. What confuses me is if the time constant is low, like 0.01 seconds, it would suggest that the capacitor would be able to become 62.3 percent charged in that little time, but if the power source can generally only supply 1 watt of DC power, isn't that a limiting factor which would mean it would charge in 0.623 seconds. In other words, is the power supply a limitation in this situation or could it still be able to charge in that amount of time?

  • \$\begingroup\$ rapidtables.com/convert/electric/Watt_to_Joule.html \$\endgroup\$ – jsotola May 24 '18 at 23:54
  • \$\begingroup\$ You've limited the current to 1A, but you've also said that the time constant is 0.01 seconds, meaning that the resistance is 0.005ohms. Since you've limited the current, when the capacitor is initially connected, the output voltage of the supply will be 0.005ohms*1A=0.005V, or 5mV, or 5mW. As the capacitor charges up, this voltage will increase, but the current will stay the same. Due to the extremely low resistance, this will act like a constant current charger, and will take ~2 seconds to charge to 1V \$\endgroup\$ – BeB00 May 24 '18 at 23:57
  • \$\begingroup\$ So what you're also saying is that the charging time of a capacitor can be limited by what the power supply can supply. \$\endgroup\$ – Tom May 25 '18 at 0:07
  • \$\begingroup\$ @Tom Yes, if you limit the supply, it can limit the time \$\endgroup\$ – BeB00 May 25 '18 at 0:16

Before spending time on capacitors, it might be good to understand a highly simplified, but still useful, model of a voltage supply. That model is an ideal voltage source (which doesn't exist in our universe) together with a series resistor. So it looks like this:


simulate this circuit – Schematic created using CircuitLab

So when you talk about a voltage supply that provides \$1\:\text{V}\$, it is usually interpreted to mean that \$V_\text{IDEAL}=1\:\text{V}\$. However, nothing is said, just yet, about the value of \$R_\text{INTERNAL}\$.

Now, when you add that the voltage supply is capable of \$1\:\text{A}\$, this usually means that it is safe to use and can comply with a demand for as much as (but not more than) \$1\:\text{A}\$, while still operating close to its nominal voltage.

This idea of "close to" is important. When a voltage supply is providing its "maximum" output current, you expect that the voltage will droop a little from its ideal value. But how much? Well, that depends. A \$9\:\text{V}\$ battery is notorious here. If you pull just \$100\:\text{mA}\$, you might expect the output voltage to be only \$8.8\:\text{V}\$ (fresh battery) because it "droops" a little under this load. But with a mains-powered supply you might expect better. Either way, there will be a specification somewhere that will tell you something about what to expect.

There are several ways to specify the current limit.

One is to specify a range for the expected output voltage when under a specified load. So, for example, your power supply might be rated to provide \$1\pm0.01\:\text{V}\$ at \$1\:\text{A}\$ or it might be rated to provide \$1\pm0.2\:\text{V}\$ at \$1\:\text{A}\$. Any particular one might be on the high or low end. The guarantee simply tells you that it shouldn't be outside that range.

Another way to specify it is to provide the value for \$R_\text{INTERNAL}\$. For example, that \$9\:\text{V}\$ battery is often specified to have \$R_\text{INTERNAL}=2\:\Omega\$. This is just another way of saying that if you tried to momentarily pull \$1\:\text{A}\$ out then you'd expect to see \$9\:\text{V}-2\:\Omega\cdot1\:\text{A}=7\:\text{V}\$ at the output. So the specification might be \$8\:\text{V}\pm1\:\text{V}\$ at \$1\:\text{A}\$ load. (They won't handle that kind of load. But I'm just talking "as if" they might be able to do that.)

This doesn't mean the power supply can't do more than what they are rated to do. Often, they can. It just means that you should not exceed that spec.

So what happens when you directly hook up a discharged capacitor to a voltage supply? Well, if it were an ideal voltage supply and an ideal capacitor using an ideal superconducting wire, there would be an instantaneous current that was infinitely high. And the universe would go boom. But the reality is that wires have some resistance (only a little, but some) and the power supply itself has some limitations (such as imposed by \$R_\text{INTERNAL}\$) and the capacitor itself has its own limits, as well. Reality impinges. So we use models that are appropriate to each situation -- as simple as, but no simpler than, needed.

It's possible that a voltage supply has a built-in current limit control -- many do. In these cases the voltage supply provides the indicated voltage unless the current would exceed some dialed-in current limit. In that case, the voltage supply drops the voltage lower and lower until the dialed-in current limit isn't exceeded. If you hook up a wire across such a voltage supply, expect the output voltage to be close to zero.

If the voltage supply does NOT have a current limit control or a baked-in current limiting circuit, then the \$1\:\text{A}\$ rating just means "what it is rated as safe to use." But it may deliver more than that, too, if you were to try and short it out with a wire.

A discharged capacitor is very much like a wire. It has very little resistance. So it will demand a lot of current if hooked up to a voltage supply. What actually takes place will depend on a lot of details. But in general the capacitor will charge up quickly and the voltage supply will probably exceed its ratings during the earlier part of that time.

A designer simply wouldn't directly hook up a voltage supply to a capacitor. That process isn't manageable and therefore isn't a good idea. Instead, a designer will include a known resistance between the voltage supply and the capacitor; or else they will use a current control circuit. Either way, directly hooking up capacitors to voltage supplies without some kind of management involved is not good practice.

So in your case, the simplest method is to insert a resistor. Now, everything makes sense. Suppose the resistor were \$1\:\Omega\$. Now you have a time constant of \$1\:\text{s}\$. So in about 5-8 time constants, the capacitor should be fully charged. In the meantime, the voltage on the capacitor will follow the equation: \$V_\text{CAPACITOR}=1\:\text{V}\cdot\left(1- e^{\frac{-t}{1\:\Omega\cdot 1\:\text{F}}}\right)\$.

You also know that the peak current will be at the beginning and will start at \$1\:\text{A}\$ but then decline over time.

That's all now very predictable and under management. The internal resistance of the power supply might alter the exact details slightly. But it is, at least, managed. And you know what specification you have to look for, in terms of your power supply's compliance current -- it must be able to handle \$1\:\text{A}\$.


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