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Tripping a 20A breaker Answered

Hello. I need help with charging a capacitor bank I have. It consists of 10 capacitors: 4 4700uf capacitors and 6 6800uf capacitors. All are wired in parallel. I have been charging them with 110AC rectified to DC. The only problem is that whenever I flip the switch to start charging, it trips our shop's 20A breaker and you have to flip it back to start charging again. It trips our breaker every time and dims the lights making my dad angry. So I need some tips on what wattage resistor i need to use to keep from tripping the breaker. I experimented with a 10W resistor and that was way too much resistance. I am not allowed to experiment anymore until someone credible can tell me what to do. Thanks for your help.



7 years ago

Well, using ohm's law we can see that 110V/20A = 5.5ohms. So, your capacitor bank must have a total resistance that is less than 5.5 ohms. Your resistor should probably be in the 5-10 ohm range. But be aware, it will drop between 100 and 200W.

So, where do you get a low resistance, high wattage resistor? A light bulb, of course! Wire up one or two 100W light bulbs in series with your capacitor bank, and it should charge up without blowing the breaker.

...except the cold resistance is an order of magnitude lower than the hot.....

Suggestion for a low resistance high wattage resistor....Electronics Goldmine, or Allelectrionics.

Either buy them individually or get the Alelectrionic's Genormous surprise box....they normally have a few in there (but that is a bit of a hit or miss thing).

Many of the ceramic "block" style resistors are on orders of 10 ohms or lower (some as low as .005 ohms).


7 years ago

As jeff-o says, lightbulbs are easily available high-wattage low value resistors. Helpfully, as your supply voltage is roughly 100, a 100W lightbulb will (under normal circumstances) pass 1A. If you want your capacitors to start charging at 1A, wire a 100W lightbulb in series with the bank. I assume you are using a bridge rectifier, so wire the bulb in series with the AC supply to the rectifier.  If you want to charge at 2A or 3A, use two or three bulbs, wired in parallel with each other, all wired in series to the rectifier.

You could alternatively go below 1A by using several bulbs in series, but at this point the bulbs aren't seeing the entire supply voltage so working out their resistance is harder (bulbs have lower resistance when underpowered due to running at a lower temperature).


Reply 7 years ago

Incidentally, with a 100W lightbulb in series, your time constant is around 6.5 seconds so the bank should be mostly charged within a minute. You can double check this calculation with a calculator like the one at http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capdis.html and plugging in the relevant numbers.

BTW you do know that working with this much capacitance can possibly KILL YOU! (or at best give you temporary Tourretts Syndrome)
There was a Navy technician who thought of the bright idea to test his own body's resistance. He used a standard Ohm meter powered with a 9v battery. Desiring to get the best connection possible he clamped down on the leads with both of his thumbs. This resulted in his breaking the skin with the probes & allowed the 9v battery to connect through his blood stream across his heart muscle. This caused the heart to go into defibulation & he died. BE CAREFUL OUT THERE!

.  I have serious doubts about a 9V "transistor"/PP3 battery (or any other 9VDC source) being to kill someone in that manner. If you use 1K ohms as the resistance of a human from hand to hand (on the low side, even for punctured skin), that's only 9mA. DC usually takes at least 250mA to kill someone (unless you run wires straight to the heart).
.  Some electrocution info at http://en.wikipedia.org/wiki/Electric_shock.

I am NOT an Electrical engineer, just an Electronic Technician. My degree is 10 years old & I haven't worked in designing systems ever. So, I don't think I qualify to meet your dad's definition as credible. Disclaimer said; this question intrigues me & I wish to talk through the details & see if this can help to understand a solution.. What I remember from my education is: Capacitors take an in rush of current (# of electrons, AKA Amps) to charge them. You are obviously exceeding 20 amps & the breaker trips. Once they are partially charged with the initial attempt then they take less current to fill them to capacity. So that is why they don't trip the breaker on the second switching attempt. I understand your attempt is to define the needed rating for a "current limiting resistor" Your aim is to slow down the flow of electrons so it will take longer to charge the capacitors.
What was your result of adding the 10 W resistor? Did it take too long to fill the capacitors or simply blow the resistor?
This is the same as adding a 10 W light bulb on the positive side of the capacitors. Intuitively I think this would burn out the resistor because those capacitors are still hungry for the charge. They would still suck those electrons through the resistor. It might have been too little resistance. Maybe adding a big resistor would slow down the electrons. An example of a big honking resistor would be a hair dryer or heater coil. But that would take an in-rush of current as well.

Have you thought of wiring them (capacitors) in series starting with just one & adding them one by one. They would definitely take less amps initially and should take longer to charge that way as in the end you have created the equivilent of one large 59,600uf capacitor. You could then add a switch to rewire them to parallel when charged if needed.

You might be way ahead of me on this having attempted these suggestions already but I am hoping to help by just "thinking out loud" here. I'll be watching the forum for any other comments & maybe get my own thinking corrected in the process. Best regards

I was using a 10w ceramic resistor and it just limited the current flow too much. It did not get too hot or explode and I am well aware of the risks in using these. Thanks.

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