Caps v. Batteries

[KilroySmith]

OK, after pulling apart my $15 mini-RC car, I had a question that went back to basic electronics:

How would you test a two-terminal device to determine whether it was a battery or a capacitor? What physical characteristics differentiate them?

- Both are charge storage devices
- Both show a voltage change as the stored charge changes.
- Would a battery display the same reactance as a cap in an AC circuit?

Obviously, the real question was, "Just what the hell is that little cylindrical thing in there? How can it be a battery, if they can charge it in 45 seconds? How can it be a capacitor, if it can run a motorized device for two minutes?". What was really annoying was not being able to formulate a test that would unequivocally determine one or the other.

/frank

 

[CTho9305]

Short it with a low-resistance resistor for a while, and then if you get 0v across its, it is a cap, if you get a low voltage it is a battery? (that is a semi-educated guess.... I'm assuming that a dead battery still has a very weak chemical reaction in it).

I would imagine a cap would have lower internal resistance, so you could see if the terminals spark when it is charged... or get a small resistor and an ammeter.

 

[KilroySmith]

Good try, but some capacitors (electrolytics, for example) will also recover some amount of voltage awhile after a short "should" have drained it. In the real world, there is probably a range of time where a shorted capacitor wouldn't recover, but a battery would. There is no way to determine that theoretically, however.

This is an interesting question for me, because I want to eliminate both the practical, though theoretically invalid, tests as well as the theoretical, though practically invalid tests.

Practical, but theoretically invalid test: Charge the device, then graph it's discharge voltage. In the real world, NiMH, NiCd, or LiIon cells would have a very unique discharge signature compared to a capacitor. Theoretically, however, a battery could have the same discharge signature as a capacitor.

Theoretical, but practically invalid test: Can't think of one at the moment...


/frank

 

[spikemike]

I got one of those too, I opened it up and found mine had a battery in it, you should be able to tell by reading what is on the battery/capacitor, mine said NiMH i think, I don't remember exactly but it was rather obvious it was a battery. I think it would have to be a rather large capacitor to give it a 5-10 minute run time so I bet it is a battery.

 

[dkozloski]

A capacitor that would equal the current storage capacity of a couple of "D" cell batteries would fill a semi-trailer and then some..

 

[CTho9305]

Originally posted by: dkozloski
A capacitor that would equal the current storage capacity of a couple of "D" cell batteries would fill a semi-trailer and then some..

I thouhht so, but wasn't sure enough 😉.

Capacitors that have been drained do not gain voltage by themselves. They are just two metal plates with gunk between them. In a battery, the gunk is different and the metals are different.

 

[KilroySmith]

Originally posted by: dkozloski
A capacitor that would equal the current storage capacity of a couple of "D" cell batteries would fill a semi-trailer and then some..

Not so, young one.

A couple of 'D' size Ni-MH cells might give you 16000 mah @ 1.2v. This is equivalent to (16 A-hr * 3600 sec/hr) = 57600 A-secs. Using the relation 1 A = 1 coulomb/sec, we have approximately 57600 coulombs of charge stored at 1.2V (average). Using the relation Farad = Charge in Coulombs / voltage, this is equivalent to a capacitor of about 48000 Farads.

Now, in the commercial market such things as:
German 50,000 Farad Supercapacitor
are available in a package half the size of a loaf of bread, weighing 2.5KG (~6 pounds). A whole lot smaller than your semi-trailer.

2 D-cells 50000 F cap
325g 2500g
125cm^3 1200 cm^3

So, as a solution, it's roughly an order of magnitude larger in both volume and weight.

I still don't understand how they're charging a battery at 80C.

/frank

 

[JSSheridan]

Perhaps this will work. Connect an inductor across this device, and measure the voltage with an oscilloscope. A capicator-inductor circuit will have a sin wave, while a battery-inductor will have a decaying voltage. Peace.

 

[Shalmanese]

In case your wondering, it is indeed a battery and not a cap. It charges so fast because it doesnt fully charge the battery, just pumps a huge voltage through it for a short time period. Dan from dans data reviewd one here.

 

[Menelaos]

If you had two of them, you could do the following:

Discharge one of them, then connect it to the (fully charged) other (parallel to it). Monitor the voltage across. If it stays about the same, both are batteries. if the voltage drops to about half, both are cap's.

(internal impedance of a batterie is very low, so voltage stays about the same, even for high current. with the cap's the stored charge is devided between both and the voltage is halved (Q = CV))

Hope this helps,

Menel.

 

[MISwannaB]

Originally posted by: JSSheridan
Perhaps this will work. Connect an inductor across this device, and measure the voltage with an oscilloscope. A capicator-inductor circuit will have a sin wave, while a battery-inductor will have a decaying voltage. Peace.

 

[dkozloski]

Kilroy Smith, I have been working with military and industrial electronics for about 45 years. The problem with your theory of the current supplying ability of a capacitor is that the discharge curve would be just about the mirror image of a Ni-Cad battery. The voltage would drop at a high rate until the capacitor trickled off to nothnig with the voltage level being so low it would be useless, your regular RC discharge curve. A Ni-Cad battery maintains the output voltage and thus the current almost to complete discharge. Refigure your capacitor size to allow for a discharge curve like a Ni-Cad and tell us what you get.

 

[KilroySmith]

dkozloski -
OK, good point - with a linear load (a resistor), a theoretical capacitor will show an exponential decay. With a constant-current load, it should give a linear curve. Of course, the supercap I linked shows a constant-current discharge curve that looks remarkably like a Ni-Cd discharge curve, nothing like a theoretical cap would provide.

What does a theoretical battery look like with a linear load? I know that two different examples of real-world batteries (Alkaline and Ni-Cd, for example) can have two different discharge curves with a linear load - nearly linear with Alkaline, extremely non-linear with Ni-Cd.

Is JSSheridan's test valid? I don't believe so, because I don't believe the LC circuit will oscillate on it's own (I should know that, but I have descended into the world of "If a signal doesn't have square edges, it's black magic"). I'm not sure that an LC circuit wouldn't behave similarly to an LB circuit. I might have to break out an O'scope and my old breadboard to find out.

/frank