In my quest to become an electrical engineer, it seems the more I learn about electricity the more confused I get…
When a capacitor is hooked up to an electrical source it will charge to the same voltage as the voltage applied, over a length of time dependent on the capacitance and the current, right? Does the voltage affect the charging speed?
Okay, where does this current value come from? Does a capacitor count as a “load”? In a pure voltage/capacitance arrangement, how does one calculate the load of a capacitor?
Next question: I have constructed a tesla coil. It has a capacitor bank/spark gap arrangement. The power supply is 15,000v 30mA neon sign transformer. I can adjust the spark gap distance…a shorter distance means the gap will activate at lower charging voltages. Now, as I move the gap farther apart, the sparks become intermittent and presumably closer to the full 15,000v discharge of the capacitors. What if I move them apart so that no spark could possibly occur? My gut tells me this will blow the caps - but since capacitors can only charge to the same voltage as the source, (and assuming the voltage rating is appropriate) would this really put the caps in danger of being destroyed?
Desmostylus is correct. I’ll provide a qualitative description of what goes on when you charge a capacitor.
To model how a capacitor charges you need three things:
Voltage source
Series resistance
Capacitor
Let’s assume the initial voltage on the capacitor is zero. You hook up the battery, resistor, and capacitor in series. What happens to the current and voltage?
Initially a lot of current flows. This is because the capacitor initially looks like a short (i.e. it looks just like a wire). The voltage across a wire is (roughly) zero. Therefore, when you first hook the circuit up, current is at a maximum while the voltage across the capacitor is zero.
But then the clock keeps ticking. What happens a little while later?
After a while the capacitor starts to “charge up.” As more and more charge accumulates on the plates, the capacitor no longer looks like a short circuit. The current starts decreasing while the voltage across the capacitor starts to increase.
After a long while the capacitor charge has built up to the point there will only be a trickle of current. This same current is going through the resistor; if the current is low, the voltage drop across the resistor is low. This means the voltage across the capacitor must be high.
At time = infinity the current finally stops. If there’s no current through the resistor, then the voltage drop across the resistor must be zero (i.e. the resistor looks like a wire). Consequently the capacitor voltage is at a maximum and in fact equals the voltage source.
How fast does all of this happen?
The lower the resistance, the faster everything happens.
The lower the capacitance, the faster everything happens.
I will readily admit I’ve left out a bunch of stuff and glossed over things. If you want more detail just let us know.
For a visual aid, here is the universal charge curve for RC circuits. The basic unit is called a time constant (TC). TC = R×C. 1 time constant is the time it takes for C to charge up to 63.2% of the applied voltage. In this case, the applied voltage is the difference between the source and the charge on the plates of the capacitor.
Let’s assume a 100V DC battery connected to an RC circuit. The values of R and C don’t matter; the curve will still look the same. At one TC, the cap will charge to 63.2% of the applied voltage, or 63.2V in this case. During the 2nd TC, the cap will take on an additional 63.2% of the remaining difference between the charge it has already acquired and the applied voltage, or about 23.3 volts, bringing the charge on the cap up to about 86.5 volts by the end of the 2nd TC. Keep doing the math like this and at the end of the 3rd TC you will have 95.1 volts on the cap, 98.2 volts at the end of the 4th TC, and 99.4 volts at the end of the 5th TC.
Because the charge is always 63.2% of the potantial difference between the cap and the source, the cap will technically never reach a 100V charge, hence the resulting curve is asymptotic.
Maybe, maybe not. Assuming that “aure” = “sure”, and “mass” = “heat”, then yes, they are similar in that there are first order differential equations involved, and yes, solutions to first order differential equations display similar forms.
(Look, Crafter_Man, I already acknowedged that Attrayant had the right idea, but had simply chosen a graph that wasn’t the best illustration.)
Oops, I just did a search on “time constant” and picked the nicest looking graph. Just change Temperature to Voltage and all will be right with the world.
Yes, this all makes sense. We studied RC circuits earlier this year. I was just trying keep the concepts straight.
Now, what about my second question? If I wire a capacitor bank rated at 20,000v parallel to a 15,000v supply and a spark gap (with the spark gap set such that no breakdown will occur), am I in danger of blowing the caps?
Trigonal Planar: I don’t know much about Tesla coils. But my gut reaction is that, as long as you stay below the maximum voltage rating of the capacitor, you probably won’t have a problem.
Just curious… is it just one capacitor? Or multiple caps in series?
what’s been said regarding the (solved) DE’s for capacitors is correct if it’s a DC source, but just remember that, IIRC, capacitors act differently if using an AC source.
Yes, it’s an AC power source I’m using to charge a bank of capacitors, each rated at I belive 1200v wired to give a voltage rating of 22,000v (and appropriate capacitance).
The rest of the tesla coil system isn’t important for my specific question – I assume that since the supply voltage is less than the design voltage, I shouldn’t have a problem directly charging the caps.
I’m really not sure on this since we never really covered charging capacitors with an AC source in class, but we did do a lab on it. We applied a square wave to a capacitor and during the time V=0, the capacitor would discharge. I don’t know about V<0 though, since we didn’t try a sine wave. I do know however that capacitors and inductors do have an impedance when using an AC source, which will have an effect on current and voltage.
I just finished my fundamentals of electrical engineering course (double credit), so hopefully someone more informed can come along.
If you attach a capactior to an AC source, it will charge up as if the source were DC. When the polarity of the source changes (i.e. when there is a negative voltage from the source to the capacitor) the capacitor will start to discharge. This cycle continues.
As such, a capacitor can be used in a filter to help remove AC noise from a DC source.
Charging capacitors with an AC source is futile; you’ll spend as much time charging them as you do discharging them.
