axi0m said:I have been testing this equation with various different capacitors and capacitor bank values for V, R and C. I have found that when combining two capacitors in series (V doubles, C is divided in half, R is doubled) the following equation yields the same I-curve over t as a single capacitor. Wouldn't the area of the I curve be equal to the capacitor's (or bank's) total energy? In other words, shouldn't U=It?
I = V/R * e^(-t/RC)
berkeman said:Just putting two caps in series does nothing to the voltage and resistance. you must mean something else in your question?
axi0m said:Ahhh, I guess I've been gravely mistaken. I was under the belief that when you add two capacitors in:
series, you double the ESR (because two resistors in series double the resistance,) divide capacitance by two, and double the voltage
parallel, you double capacitance, voltage remains the same, and you divide resistance in half
Would it be too much to ask for you to kindly provide me with the proper values of said properties when combining two capacitors in parallel and in series?
berkeman said:Ah, you didn't say that the R in your question was the ESR of the cap(s). Yes, series connecting two identical caps will halve the capacitance and double the ESR. By saying that it will double some voltage, I assume you mean if they are charged up initially and then placed in series still charged. Yes, that will double the voltage.
Are you then shorting them out to use the equation that you posted? The only "R" is the ESR values?
axi0m said:By doubling voltage, I meant that if you place two 450V max. caps in series, you can now charge the bank up to 900V rather than only 450V.
Yes, they will be shorted-out and there is no additionally resistance in the circuit, other than ESR.
berkeman said:Oh, that's different. In general you want to use a single cap with the full voltage rating. If you are placing caps in series to try to get a higher overall voltage rating, you need to do something additional to ensure that the total voltage divides equally across the two lower voltage capacitors. Otherwise, one could have its voltage rating exceeded and blow, and then the other will blow.
You would generally ensure the equal voltage division with large-value resistors that are placed around each cap. Something like 100kOhms around each cap.
Can you not obtain a single cap with the full voltage rating?
The problem with two caps in series is that during discharge, the voltage on one could drop faster than the other, you could overvoltage one, or even reverse voltage one.. If you do have series caps, put bleeder resistors across them, or if you have many cap-pairs in series, tie the inter-cap connections together.axi0m said:Well, I'm designing a cap bank for a rail gun so relatively high ampere currents are desired. Though not yet exacted, said current will probably require a higher voltage than I can find in a cap of sufficient capacitance. So, at this point, I'm trying to configure the best cap bank with regard to series, parallel or a combination.
Bob S said:The problem with two caps in series is that during discharge, the voltage on one could drop faster than the other, you could overvoltage one, or even reverse voltage one.. If you do have series caps, put bleeder resistors across them, or if you have many cap-pairs in series, tie the inter-cap connections together.
The Capacitor Discharge equation represents the relationship between the voltage across a capacitor, the capacitance of the capacitor, and the time it takes for the capacitor to discharge.
The Capacitor Discharge equation is derived from Kirchoff's Voltage Law, which states that the sum of voltage drops in a closed circuit is equal to the sum of voltage sources in the circuit. This law is applied to a circuit containing a capacitor to derive the equation.
The voltage (V) is measured in volts (V), the capacitance (C) is measured in farads (F), and the time (t) is measured in seconds (s).
The Capacitor Discharge equation is commonly used in the design and analysis of electronic circuits, particularly in applications involving capacitors such as power supplies, filters, and timing circuits. It is also used in experiments to measure the capacitance of a capacitor.
The Capacitor Discharge equation assumes that the capacitor is ideal and has no resistance. In real circuits, there is always some resistance present which can affect the discharge time. Additionally, the equation assumes that the voltage across the capacitor is changing continuously, which may not always be the case.