Good video. Just a small correction to the title - capacitance decreases when you put them in series, so the total bank capacitance would be about 42F at 30V rather than 500F.
Hello, and thanks for posting the video, however the entire bank appears to be connected in series! So...you would have 2.5v x 12 = 30v (voltage adds in series)....and 500F /12 = 41.7F (capacitance divides in series--with equal valued caps!). this gives you a total energy of 18750J. The energy is conserved, so if you calculate the energy that can be stored in ONE cap (1562.5J) and multiply that by 12, you get the same energy as the bank of 12 has. There is still a lot to learn, but judging from your other videos, you'll get there FAST!
@@SheikhN-bible-syndrome OK, I will do this as I am writing...the formula for the energy 'E' in joules stored in a capacitor is 0.5 x C x V^2. So for ONE of your caps, where 'C=500F' and 'V=2,5v', we get 0.5 x 500F x (2.5 x 2.5) = 1562.5J. The total bank would store 12 times this energy...12 x 1562.5 = 18,765J Another way: consider the energy of your entire 12-pack. using C=41.7F and V=30v, gives 0.5 x 41.7F x (30 x 30) = 18765J of energy. Divide this by 12 and you get 1563.75...the charge on EACH cap
I added a rather lengthy reply, but didn't see it go through....so, here is a similar version... Actually it's 500F REGUARDLESS of the voltage the capacitor is charged to....or they couldn't print 500F on the label...this 500F is constant at any voltage! Don't confuse capacitance (C in Farads) with energy (E in Joules), or charge (Q in Columbs). And, of, course, Voltage (V). They are VERY different,. What is important is to know how thay are related in the capacitor--HERE GOES.... A LARGER capacitance (C) capacitor simply means that that capacitor can hold more charge (actually, DISPLACE more charge, [Colombs (Q)] with less voltage (V) across it. The relationship between charge (Q), voltage (V), and capacitance (C) in Farads (F), and resulting stored charge(Q) in a cap at this voltage is: Equation (1): Q=CV. A handy thing to know is that one Coulomb of charge is 6.24EE18 electrons (6,240,000,000,000,000,000 electrons! This sort of gives you a 'feel' on an atomic scale. If you charge a 500F cap to 1 volt, the amount of energy stored is twice that of a 1000F cap at the same 1 volt (see eq. 1). For capacitors in series, the capacitance divides, and the voltage adds. Take your exaample, 12 500F caps charged to 2.5v each, in series had a total capacitance of 500F/12 = 41.7F at 2.5v x 12 = 30v (your video title should actually be: 42F at 30v). NOW, concering the energy, E, (which is conserved): There is another simple equation: Equation (2): E = (1/2) C V^2 [note V^2 is V squared....so the E rises faster with increasing V than with increasing C] Using equation 2 with a SINGLE 500F cap at 2.5v gives a stored energy of 1562.5J. If you have 12 of these in series, that is, 41.7F @ 30v, then plugging in C=41.7F and V=30v gives (using eq. 2): 18765J of energy. This is exactly 12 x the 1562.5J of ONE of the caps. "energy conversation", verify you get these values using eq 2. For caps in parallel the max. voltage is that of one single cap, but the capacitance (C) adds. In your case, A bank of 12 500F, 2.5v caps in parallel will be: 6000F @ 2.5v. AGAIN if you plug this C=6000F and V=2.5v into eq 2 you get E = 18765J. This works even if they are in some sort of series/parallel configuration. Energy is always constant, but C and V can vary. IT ISS VERY IMPORTANT THAT YOU KNOW THE MAX ALLOWED VALUE OF V, OR OVERVOLTAGE COULD CAUSE A FIRE/EXPLOSION!!! One more interesting thing that is non-intuitive: When you charge (or discharge) a capacitor, you add ZERO electrons to it!!! The charge-voltage placed across them only SEPERATES the + and - charges between plates. ...probably left out part of whwt I wanted to write, but hope this helps some ....always glad to talk capacitors.
Couldnt you just put a cheap active balancer on it. The start correcting at a few millivolt difference. Some even have Bluetooth and you can check each voltage on your phone.
Man, forget the solder. You could stick welding rods on there. That is exactly what happened. You melted the wires together. Had the leads remained in contact, the caps would survive, but the wire woud get so hot it would probably melt and the insulation burn.
500 Farad =500 Amp/Secs @ 30 Volts = 15000 Watts - I would say be very careful indeed, 42 Farad @ 30 Volts = 1260 Watts, still very dangerous current especially if hands are wet
@@maliknaveedislam Hi Duke, I'm not an electronics engineer, but as a guide. You have 3000 Watts at 48 Volts, the Amps would be Watts divided by Volts so 3000/48 = 62.5 Amps. You would want some headroom in there say 80 Amps. Beyond that I would get advice from a pro to be safe, also use an appropriate fuse to keep things safe. Cheers
@@johnlannikk2701 this isn't going to harm anyone at 30 volts, regardless of the potential energy release. Human body resistance is too high to be harmed by this.
If you want to get the full capacity, or if you want to make your bank safe to charge without your attention, IE practical operation. He stopped when the first cap was at 2.5v or whatever. Others would have been at a lower voltage, so less charged. There was a portion of the full capacity of the bank which was unused, but the charge state (voltage level in this case) was kept at a safe level by stopping the charge process. If you have a way of balancing the caps (most banks use a dissipative mechanism to burn away the excess power on a cap at rated voltage) that gives you the chance to fully charge all caps in the bank, without taking some caps to a dangerous (IE over rated spec) charge voltage. Of course you're then turning power into heat just to waste it. Lose power to gain stored power. Do you gain significantly more stored energy than the power you wasted to keep the balance? Is there a less wasteful process to keep the caps balanced? Does practicality overrule possibility when you're designing a product for commerical production? If the capacitors all had identical characteristics (IE perfect capacitors in a perfect world) then charging would be even for each cap. But we do not live in a perfect world. Limiting the total voltage to ~90% of rated voltage times number of capacitors didn't help in this case, so balancing would be something you might do to eak out maximum capacity, or as a safety mechanism to be sure that the caps are always passively (IE no human monitoring or intervention) kept to safe charge states.
The Duke Of Swat when capacitors are in series they gain volts but loose capacitance. 1 500f capactor has 2.6 volts but 2 will have 5.2 volts (double) but loose half capacitance so only 250 amps
The Duke Of Swat 6 of the caps in the video is the best you can get. It will be 15volts 80 amps in series or 3000 amps 2.6v in parallel. Problem is 2.6 can run 3000 through anything but gold and copper
The Duke Of Swat tbh i think maybe 4 car batteries and a circuit to limit it to 80amp would be better a 48v 80amp capacitor set up isnt sold prebuilt and putting one together would be very tough
