Substituting Capacitors for a Car Battery?

...what color... That's funny

Moving onto a mechanical course seems like a fun idea .....

Your idea works best on blue cars ........

perhaps a truckload of capacitors?

Yep. Weld-up a trailer or two to carry the capacitors, and see what they do for the power-to-weight ratio. Brilliant idea.

"... I'm an electrical engineering student. ....... I have the formulas to calculate charging and discharging capacitors but is this even possible? ......The capacitors are 350f each and I'm assuming I need 6 or so in parallel....."

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1. "....but is this even possible..."

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-Is what, exactly, even possible?

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--That you have spent enough time studying electrical engineering to claim you are a student....

--dubious.

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--That you have 6 350F capacitors rated to at least 15 volts (since an EE student suggested attaching them in parallel)....

-- similarly dubious.

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--That you 'have the formulas'.

-- This could be in the realm of possibility...as long as 'have' applies strictly to a physical written copy and doesn't imply possessing a sufficiently firm grasp of the concept to run the risk of utility.

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Here, look at some pretty pictures (note some of these are painted blue... which is why a blue car would work best....matching the source to the load)

The last one is probably the best choice. Great Scott!

Wasn't that one the power source for the 'Back To The Future' car?

Well this guy's project is developing at a phenomenal rate. We've gone from replacing a car battery with capacitors, to eliminating the need for gasoline. God Forbid

Watch out the Big Bad Oil Cartels will be arriving soon to cart you off. 'They're coming to take you away hahaha hehehe hahaha, they're coming to take you away'

it is Fall, they perhaps began their journey in Sept

Quite. That's why the original poster will need a pantechnicon of some sort to lug all that capacitance around.

'pantechnicon'...

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I like it! Thank you.

You need one of these....

http://www.gizmag.com/graphene-based-supercapacitor/28579/

Probably have to build it yourself....

Please post for us a picture of your (6) 350 Farad, 15 Volt capacitors.

Did they cost more than your car?

I suspect he is either using six standard 350 farad 2.7 volt super caps in series giving him an actual 58.3 Farad 16 volt capacitor or six 2100 Farad units in series giving him an an actual 350 Farad 16 volt bank.

Retail for the 350 F units new are around $11 each with good quality used take off ones running in the $2 - $5 a piece range and the 2000 F are $52 a piece new with good used take off running between $10 - $20 each.

Maxwell 350F super capacitor. $10.75

Maxwell 2000F super capacitor. $52

Either way I would say he would not be out all that much compared to buying a new good quality automotive battery. The last one I bought new for my pickup set me back around $130.

But don't worry too much about these! I've never detonated them, but these low voltage ultracaps are "safe!" :)

They don't explode like small higher voltage caps. *SUPPOSEDLY* (I don't know for sure!) they just short inside and vent gasses. They don't explode. You can directly short them and they just make sparks. The internal resistance is so low they don't even get warm. If you overvolt them they don't explode then either. They supposedly just melt down inside and vent. The potential is pretty low so it's not like higher voltage caps.

But they are actually pretty great for hybrid cars. I did some math and I'd only need like four of my ultracap packs (so 24 of these individual ultracaps) to give 40 hp for 5 seconds. That's enough to give a realistically usable boost to a hybrid car :)

Not to mention the fact they will outlive the car, have 99.9% charge/discharge efficiency and are MUCH lighter than batteries ;)

Hmm. Throw one into a sink full of water, let me know the results...

Or open the hood in the rain...

You can directly short these and they don't explode :) The ESR is so low they don't generate any significant amount of heat. It's such a small amount of heat that once you discharge them from a direct short, they are supposedly just slightly warm. So throwing them in water would likely do nothing. Remember, each cell is only 2.5 volts max :)

Hmm. What I think is likely and what YOU think is likely seem to differ. OTOH, How bout you try it, and prove me wrong? I have broad shoulders, and in fact, don't really know how multi farad capacitors would behave. I know how snappy motor caps can be if you get them to short across a wedding ring. And they are a LOT smaller.

Hmm. I guess I DID say it was a "stick of dynamite". That would involve explosions, I suppose. But of course, I know better, shorting out any capacitor won't make it explode. I stand corrected from using hyperbole I was just thinking of the energy inside...when you DO short out the capacitor, a multi farad capacitor will discharge, and that energy will normally be dissipated in heat. Its a LOT of energy. And it discharges fast. And it has to go somewhere. Wires will sag and glow. Plastic will melt. Or maybe not. Be interesting to see.

Looking forward to your you tube video of the charged multi farad cap being tossed into a tub of water. That would convince me.

Well, think about it though. Did you watch the video I posted before? it's rather impressive. In case you didn't see it yet, check it out. A tub of water would have a higher resistance than what that guy did. Here's the link again. It's the identical caps that I have.
http://www.youtube.com/watch?v=EoWMF3VkI6U

He's shorting out the capacitor with what looks to be about 10 gauge speaker wire.

A welder for example is in the order of some 100 amps and a low voltage (don't know for sure, but 10-20 volts sounds right!).

I would toss one in a bucket of salt water if I knew that there would be no water ingress. But because of the crazy high taxes and import tax that I had to pay, the value of these to me is quite high now.

The difference with those caps is that the voltage is much much higher. I accidentally shorted out a fully charged camera flash capacitor once and it melted the volt meter leads and actually bent one. That only has a tiny tiny fraction of the energy of one of these ultracaps, but the difference is the camera flash capacitor dissipated all of the energy in the order of a millisecond or less. Ultracaps can't do that (since the potential is so low!).

Yes I did. Sparks flashed. Wires sagged and glowed. 21 minutes to charge at 5 amps. Discharge would be much faster than charging.

Didn't see anything on the video to make me think something like this would not be a hazard in some fairly common conditions. In fact, the narrator says so on several occasions. But then, lots of things are dangerous, and still used.

Carry on. Let us know if it can become a cold weather starting aide. Sounds promising.

Kind of interested in powering a coil gun.

Yeah, sorry for posting that so many times :) But it still remains that high voltage caps "explode" while these just make sparks and burn things. But you're absolutely right! They can be dangerous. But so can car batteries lol. Car batteries can explode and spray acid everywhere :p

I will definitely post something once it gets really cold. :)

Yeah, I've wanted to make a coil gun for fun too :) I'm no expert on that, but it would be a really cool project. Dangerous, but cool :p

'...The ESR is so low they don't generate any significant amount of heat. It's such a small amount of heat that once you discharge them from a direct short, they are supposedly just slightly warm....'

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That isn't good analysis.

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To me, 'direct short' would mean a very low resistance connection between the poles. Assuming very low means a range from negligible up to a value still meaningfully less than the ESR....probably fairly standard....

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In the assumed case, low ESR would be indicative of more heat produced, not less.

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Consider I²R losses when resistance is reduced. The reduction in resistance, is overwhelmed by the increase in current (because the current component is squared while the effect of resistance is not).

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Shorting a low ESR capacitor will result in more rapid heating than shorting a capacitor with high ESR.

No it's not good analysis. It's not even anecdotal evidence. But it's true. A "direct short" might even have slightly more resistance than the ESR of the capacitor. You're absolutely right that by the definition of a direct short (significantly less resistance than the ESR), more heat would be generated in the ultracap.

The ESR is 0.0008 ohms. It's hard to get lower than that if you have even a short length of cables going anywhere. No real world condition would allow it to be put in a solid copper vise to directly short it. But even then, the junctions would probably have a higher resistance than 0.0008 ohms. The generated heat within would be a ratio of the ESR and the resistance in the cables.

In the case of any typical battery, the resistance of the wire you're shorting with would be much much less than the ESR of the battery. It's not possible to do that with the ultracaps. Almost all of the heat will be dissipated within the battery.

Let's make an example: take an ultracap, where the resistance of your wire and the junctions is 0.01 ohms and the ESR is 0.0007 ohms (the spec sheet of these caps says 0.7 milliohm). Let's say the current is 250 amps. For that instant in time, 625 watts is being burned up in the cables and 43 watts is being dissipated within the ultracap. A lower ESR ALWAYS means less loss and less heat generated. You'd REALLY have to try to get the definition of a direct short on an ultracap. It would be discharged so quickly (a second or so?) it would barely get warm.

In conclusion, we simulated it in a pspice program. Lower ESR always means lower losses! Charging an ultracap with a voltage source always gave 50% losses, but unless you have a 500 amp DC power supply or you want to for some reason charge them with a large series resistance, that condition won't happen. Charging with a good modern switching power supply to simulate a current source will work much better. In the case of the current source, the losses were proportional to the ESR.

I found the spec sheet if you don't believe my 0.7 mohm claim:

http://www.noodlehed.com/ebay/datasheets/2600f.pdf

The specs say they can be charged/discharged in 0.3 seconds and the maximum current is 600 amps. At first I thought it seemed like magic, but it's actually true! :)

My feeling is that these ultra capacitors are new enough that there could be things I don't know. That makes them fascinating to study both the caps themselves and their effects. Nicola Tesla's was asked once if he had ever studied Ball Lightning, and the answer was..."well, Ball Lighting studied Mr. Tesla for awhile.....

Everybody wants to be the Buck Rogers evil genius as soon as they see some of the frankenstein stuff you can make with coils and capacitors. I did my share of that stuff myself...and anybody who watched the episode where the MythBusters started to shake apart a full size bridge with a five horse lawnmower with a thrown blade knows that old Nick was on to something.

When I was in my teens, I made Tesla coils to make some halloween decorations...fire ladders and such. Sold several of them...they look pretty...sparks go up and up and up and eventually break away, and the next spark rolls up the ladder. Resulted in a visit by some authority to my customer who asked him to cease and desist because he was disrupting communications at the local airport. We were agast because, well, we just hadn't thought of the it. The authority told us, its okay, its pretty common. And showed us how to shield it. (he liked that stuff too!)

So the original OP's question was "could this be a good first year project" for an Engineering Programme? My feeling is that the answer would be both yes and no but mostly no. Yes because it is interesting and has a future and will keep him and his T.A's interested. And no because you risk getting into areas the T.A.s and profs don't have all the answers. ( And I was not just being snarky at noting that your grade is dependent on NOT making the profs look like they don't know what they are talking about! Which come to think of it is true.) There may be things we didn't think about. That coil disrupted communication. This cap may boil batteries with back EMF. A lot of computer tech is driving cars these days, will these result in surges which could destroy emissions controls? (see, you don't know. All you can say is "probably not", it didn't destroy any onboard chips at normal temps....) If anything, I suspect this is graduate study work. Its not the "effect" which we might worry about. Its the fall out.

I didn't mean to sound sour here, I am actually quite stoked about the concept. Its just that engineering is different from science...there should not BE any unknown variables in an engineering project. (Oh lord, wouldn't that be nice!!!! Just once!) A scientific project, well that is another thing entirely. The OP's question was in the engineering realm, not the scientific realm.

I think you should start a new thread just on the topic of using super capacitors. I would certainly haunt it!

link cosmos magazine note 44

link ball lightning note 45

badass of the week

That's a pretty funny story. I've never had an experience like that before :)

About potential unforeseen problems with ultracaps in a car coupled with a normal car battery:

One theoretical problem that we discussed in my research group was grounding the ultracaps on a different location of the battery. My BMW has the battery in the back (as you maybe noticed from the photo).

There was a space under the hood for the ultracaps with easy access to the starter relay connection... which I thought was a better idea. I honestly have no idea if it's a risk, but my colleague said when the starter relay first contacts, the inductance of the car chassis might be enough to cause a ground differential between the front capacitor ground and back battery ground of the car thus frying electronics (he said could be several volts between the grounds of electronics and could fry some computers since they are all networked). I'm not sure if I believe it, but he said it was a risk so I didn't try it :) He said he experienced small inductive ground differentials in cars and that the combination of the capacitance of the large battery being distant from the large capacitor combined with the inductance of chassis could do terrible things lol. On the other hand I don't see how it would be sooo much different than jump starting the car from the + post under the hood (goes to the starter relay) with another running car. That was poorly worded explanation but maybe you get the idea :p

'....Let's make an example: take an ultracap, where the resistance of your wire and the junctions is 0.01 ohms and the ESR is 0.0007 ohms....'

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...oh, come on. You are going to claim that the internal heat is negligible for a short on these supercapacitors, but then use less-than-super-18-gauge-equivolent conductivity for the short?

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Shorts are usually assumed to be direct node to node connections free of resistance. But in the real world of course there is resistance.

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Lets assume 3/0 AWG .... no, that really is overkill. 2 AWG has a resistance of 0.16 ohms for 1000ft. We need about a foot, so, that comes out to 0.00016 ohms for the conductor. Lets be generous and say the connection on each end has the same resistance as the entire conductor.

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3 x 0.00016 ohms = 0.00048 ohms

+ 0.0007 ohms ESR = 0.00118 ohms

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Okay so these charge up to 2.5 volts, right? So initially that equates to a little over 2100 Amps.

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That means a little over 3 KW is being converted to heat at that instant in the capacitors ESR. A little over 700 W is being converted to heat at each connection and across the conductor.

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Next let's look at the claim:

'..... A lower ESR ALWAYS means less loss and less heat generated.....'

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The term 'always' has a history of being problematic. I tend to think 'ALWAYS' (in ALL CAPS, italics, and bold font) is likely to be even worse.

So what would happen in the above scenario if just the ESR were 10 times higher, 0.007 ohms?

The capacitor's ESR would only be converting around 780 watts to heat at that instant.

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And if the ESR were 1000 times higher than original at 0.7 ohms.... the capacitor's ESR would only be converting a little less than 9 watts to heat.

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There are at least two examples that contradict the bold italicized all caps decree of 'always'.

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^{(upon review, it has become apparent that the bold typeface was only added (by me) when I copied and pasted the text......even so I am leaving my comment as is, both for literary effect and as record of my own fallibility in the matter).}

Yes, you are definitely right on those figures. BUT there's not that much energy in these and they are rated at being charged and discharged in 0.3 seconds. The maximum current is 600 amps (says the manufacturer). You would have to really go out of your way to directly short them to the point where the majority of the energy would be dissipated within to the point they'd become dangerous. The resistance we're talking about is so low it would be hard to measure. Just the oxide layers on the junctions would be enough to throw off figures (at least, that's my thought).

If you short something, even big, the resistance goes up almost immediately. If you used an 18 gauge wire the resistance would go way up and it would vaporize almost immediately so your claim that it's the resistance of an 18 gauge wire is not accurate. Even a motorcycle battery will vaporize an 18 gauge wire. If you used a large battery cable to short it, it would probably get warm in that fraction of a second (and the resistance might even rise to the figure that I gave). That plus the junction resistance, relay resistance or whatever makes my figure at least not unbelievable.

What happens on paper may not happen in real life. You can find a lot of videos of people pretty much directly shorting these. They don't explode, they just get a little bit warm. Even if they did explode when shorted, I wouldn't care. They work well for how I want to use them :p

We just got a $2000 LCR bench meter in my lab. I could try to measure the junction resistance on the caps. It can supposedly measure fractions of milliohms. I'm wondering how much the resistance of the junctions alone would be. It might not even be measurable without specific equipment though because the resistance under load will be much higher than just sitting there.

Also, you're just looking at watts. Since the total storage is finite, lower ESR will mean fewer joules will end up in the ultracap. I haven't actually done the math, but that would have to be the case if we're comparing the same load resistance. Since we're typically talking about discharge times that are fairly short, fewer joules of heat put into the capacitor = better :)

You're a smart guy. You're really good at finding holes in arguments :D It would be fun to talk with you in person :)

Let me reiterate a few things:

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- You application is great. I'm not trying to detract from that (I couldn't if I wanted to... it is a great, reasonably affordable solution for a very common problem).

- I don't doubt the very low ESR of these capacitors. In almost every situation low ESR is a huge advantage.

- My figures for connection resistance as well as my selection of wire gauge were chosen, not to best reflect real world conditions, but to illustrate a point.

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....and that point is: the time to be most on guard to insure you don't make over the top statements is when you are most excited about a great solution.

Your solution is great and that in itself is enough. It is easy to get carried away in the moment and make statements like 'lower ESR always....' when trying to convey your enthusiasm about what is rightfully something to get excited about. Later if some of the statements made in the excitement of the moment are called into question, it will likely reflect poorly, probably unfairly, on your original idea.

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I've just been giving you a hard time when you make statements that are over the top. The criticism of those statements doesn't tie at all to the value of the solution you have devised....except as you decide/react.

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My points of disagreement have been made with the best intentions. I have to tell you I do enjoy the suspense; being a bit worried about what kind of retort I will return to discover you have left. I do have the numbers right, right? Is he going to find a source for typical connection resistance? Am I going to get away with calling an alternator a constant voltage source?

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I appreciate your participation in our friendly disagreement.

Thanks! And I agree, it has been kind of fun. SOOO much better than arguing politics. Usually, politics just turns into completely ignoring what the other is saying and then name calling! That's no fun :p

Yeah, I tend to have a bad habit of saying things just as you described. I do make those generalizations like "things are always this way" and it's not scientific. I am doing my PhD in mechanical engineering so I really have no excuse. Almost all of my knowledge of those mysterious invisible things we call "electrons" that move through my iron beams (as the point of view of a mechanical engineer :D ) is mostly self taught and observational from hobbies, reading stuff and taking stuff apart and burning stuff up :p

So in conclusion, you're a worthy opponent ;) UNTIL NEXT TIME MY FRIEND!!! :D

How do you have time to do this stuff AND a PHD programme at the same time?

BTW, nobody here is your opponent. Every body thinks you are doing some good stuff. I think we are more used to reading text books than enthusiastic magazine style hyperbole, but hey...its all good.

I didn't even know there were capacitors that big. My lab certainly never used them! Ah....the miracles of modern technology. I had always taught my students that the field of electricity has been hijacked by the communications industry. We look at the growth of computers and communications as a huge development. There is just as huge developments coming in the field of linear motor technology, micro environmental controllers, RFID technology, nano-technology, and bio-motors.

There are HUGE major developments coming. Us old pharts know it. You young guys take it for granted. The positive think I take from this thread is that the "young guns" will be able to handle it, and take ideas from everywhere and put them into practice. Hmmm...did we ever get a nail gun big enough to put a satellite into orbit? What happened to all the "star wars" tech?

Anyway, its time for me to unsubscribe from this thread. Please consider contacting the administrators and getting a blog. You have a LOT to contribute to CR4.

So, at thirty below, the cranking amps drops to about twenty percent. (A phenomon I am familiar with here in Canada) Do you think that a parallel capacitor of a mighty size would make up for all those cranking amp losses? If so, that would be more than awesome...it would be ground breaking!

This is a call for "science" to be done.

I DID say I liked the OP's question! Now I see why it had all the right elements. Nick, you didn't replace the battery, but you may have created something really good. Congratulations. Everybody who thinks as I do should give Nick a GA.

Yeah, for sure it would! They have essentially no temperature dependence too so they'll work at -40 the same as they would at +40C :D
The short circuit current of these capacitors is around 1000 amps and are supposedly rated for around 800 amps for a realistic current draw. These are 2600F so 6 in series is 433 farads. The calculations are pretty easy if you know what your starter is rated for (mine is 2.2 kW, and a smaller engine would obviously be less). With 1/2CV^2 I can calculate the joules from 12.6 to 10 volts. That number of joules is enough for three starts at 2.2 kW (minus glow plugs/accessories) at 30 below :D Pretty amazing because these have less energy than a couple AA batteries... difference is it can put out 99% of that energy in a few seconds!

I imagine that it's much better for the battery if you have a "sealed" AGM battery. Every time I start the car, the alternator charges the battery crazy fast (I've seen it charging at 60 amps). Surely it vents some water vapor each time. If the current is smoothed a bit, maybe it would vent fewer gasses.

Really, you could use a small battery in conjunction with these ultracaps and it would be enough since these ultracaps are more than enough to start a car (assuming you don't need to run lights for a long time without the engine running.) That would be smaller and MUCH MUCH cheaper than ultracaps alone. You don't need mega-amps of current haha ;)

I have a second set of ultracaps. I'm going to make a portable battery booster to keep on hand in case a friend needs to jump his car. I'll just buy a small motorcycle battery to put in parallel with it. Given some recovery time, that would be enough to jump start several pickup trucks. There's enough energy for sure! It would be only a few kg and be able to provide a realistic 800 cranking amps.

Thanks for the GA vote ;)

As a side note, I bought mine on ebay. They were used Maxwell 2600F ultracaps. It was about $200 for 12 of them. They all operated perfect despite the fact they were used. They can operate for over a million cycles so they will probably never wear out in a car (the cycles are very small!).

"Is this even possible"...? ALL things are possible.....(with the right help).

You might-could start by doing a bit of research ... thinking 'all-around-the-box'.

What cap's are already "in" the automotive market ... what's available for storing energy during regenerative braking...?

Ask the manufacturers a few questions. They just *might* have someone willing to show-off what they know.

Maybe if you step-up to a few 15v 400F units.....

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Click the link to the right of "NO 10"... (go to page 19-of-20 here)

Let the Forum know how it works out --- "Best Wishes" ---

The storage capacity of most capacitor to store energy is poor. But there are some offered that say they can. There are many working on it. MIT working on one. It uses carbon nanotubes as storage.

By the way, in case some of you haven't seen this, here's a great video explaining these exact same ultracaps that I have:
http://www.youtube.com/watch?v=EoWMF3VkI6U

He covers the basics behind the math and makes LOTS OF SPARKS! Sparks are always cool... at least when there are sparks where you want them ;)

I have some really large capacitors that I found in a junk bin at the university I work at. I found six 400 volt 1000 uF capacitors and a crazy 12 kV 1.26 uF capacitor. Despite the high voltages, the total energy storage is tiny in comparison :) The difference is these high voltage caps can discharge their energy in a much shorter time (milliseconds instead of seconds), thus giving a higher power. These were apparently from a high speed camera (think camera flash to the extreme :p ).

I have no idea what I could use these for, but I thought they might be fun to play with ;)

You are likely to incur a loss of efficiency in the electrical system if you install a large capacitor bank... whether in parallel with a battery or stand alone.

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Only 50% of the energy used to charge a capacitor from a constant voltage source ends up as energy stored in the capacitor...the other 50% goes to heat.

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Old lead acid batteries charge far more efficiently from a constant voltage source.

50% losses? Heat where? Shouldn't be because if they had such high charging losses, I would have noticed them getting hot by now. Their mass is relatively small so they would get hot fast. Also, the internal resistance is EXTREMELY low so the losses should be very very low (less than 1/10th of the charging losses in any typical battery as the ESR is 1/10th). I'm pretty sure you can just count on these charging at a 99% efficiency :) . The self-drain rate is also amazingly low. I measured it to be around 5mA or less even including my simple passive balancers.

Extremely low resistance doesn't necessarily equate to extremely low losses/heat....consider the losses/heat associated with very high resistance, like wire insulation......

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They aren't called 'i-squared-r-losses' for nothing... low ESR isn't doing you any favors when it comes to charging.

...anyway at 30 below, are you really taking your gloves off to attempt to sense any fleeting wisps of warmth?

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In fairness, the 50% limit on charging efficiency when using a constant voltage source applies to the full charge, i.e. from 0 V up to the constant voltage.

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Because the capacitor in an automobile especially in parallel with a lead acid battery will not typically be discharged anywhere near 0 V then charging efficiency will be higher.

However, you should get that notion about ESR driving charging efficiency out of your head. The efficiency in charging a capacitor with a constant voltage source is not really that dependent on the resistance (neither ESR nor charging circuit).

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When charging to a constant voltage, the energy converted to heat by resistance (there will be resistance in a real circuit, the amount isn't really critical though) will be:

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E_r = (C/2)(V_final - V_initial )²

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The energy stored in the capacitor will be:

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E_c = (C/2)(V_final² - V_initial ²)

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You can see that when the starting voltage is zero that the energy stored in the capacitor and the energy user by the resistor will be equal, hence the 50% limit on charging efficiency.

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The charging efficiency can be much, much higher for a partial charge, when still charging to a constant voltage but starting at some voltage meaningfully higher than zero. ....

... but not the 99% efficiency you are hoping for in a typical automobile electrical system scenario.

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Assuming (partially for easy calculation) that the alternator charges at a constant current of 14 V and that under the heaviest load (probably when starting) the system voltage drops to 10 V momentarily.

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The highest percentage of the charging energy that could end up as stored energy in the capacitor would be:

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((V_final² - V_initial ²) / ((V_final - V_initial )² + (V_final² - V_initial ²) ) *100%

(196 - 100)/((196-100)+(16))*100% = ~ 86%

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If we look at a more limited charge from 12 V to 14 V the maximum efficiency possible only increases to 93%.

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Efficiency is also hurt if capacitor does not reach the full constant voltage before being discharged again.

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While I disagree with your assertions about 99% charge efficiency, challenging me to support my statement made me look at the numbers and I can see that the addition of a capacitor bank in an auto (if done correctly) is unlikely to lead to a meaningful reduction in efficiency. With the limit of the charging range, even accounting for incomplete charges, the charging efficiency of capacitors isn't that different from standard lead acid batteries.

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If you don't trust me, or just want to see a more in depth explanation, check out this paper.

We're getting a bit off topic but this is interesting anyways ;) But with all due respect, I think you're simplifying it a bit too much. Not many systems would discharge a capacitor bank down to 0V, not even in a hybrid system. Depending on the electronics I'd assume the usable voltage is about 25-50% of fully charged but I have no idea.

I don't think anyone would charge an ultracapacitor with a voltage source :) that would be a bad idea. That's the only condition you'd get 50% losses, because as you're charging it you'd be dissipating a huge amount of energy in wires because the currents would be enormous and half the energy would be lost in the inherent resistance of the wires. Charging a capacitor with an ideal current source would have a practically 99% charging efficiency because there would be very little losses anywhere.

The reason is this: with a voltage source, the ultracap at 0 volts would initially act as a dead short circuit. All of the energy would be wasted in the ESR (tiny) and the resistance of the wires that go to the ultracaps.

Low ESR does a lot for discharging so I'd expect the same for charging though. You get the majority of it back even for high discharge rates. I linked a video in another post "fun with ultracapacitors" in case you didn't see that yet. They compare an ultracap to a AA battery. The AA battery has more energy than one ultracap but if you try to draw a lot of power from it, the battery will probably vent from the heat dissipated within due to the higher ESR. An ultracap can however put out hundreds of amps and not even get warm. When we're talking about the applications in which capacitors are typically used (I just mean quick power cycling) they are more efficient than batteries.

Then that paper is about charging in switched capacitor converters. It's probably a different condition than we're talking about.

The energy losses are given in the paper you linked to. The losses due to ESR are given and would be Iin2*RESR*t which is a component of the total losses. The ESR of ultracaps is about 1/10th of that of any battery that I found (1/100th of small batteries). So automatically, one component due to the losses involved is between 1/10th and 1/100th of a normal battery. The majority of the losses would be due to the circuit that charges it (the resistance that it's in series with).

My car alternator is a voltage source up to 90 amps at idle. Something above idle it's peak is 150 amps according to the Bosch rating on the sticker. The resistance is extremely low because the positive cable going from front to the battery box is about an inch in diameter (it's long but massive!). When it's charging, the alternator goes down to whatever the battery is at until it's charged enough to bring up the system voltage to 14 volts. Otherwise if it were a true voltage source the current would be enormous and something would explode :)

The problem you are facing here is theorists VS the hand on guys.

Theory says a lot of things won't work perfectly but hands on shows they work more than adequately enough to do what they are intended to do!

I for one am a hands on guy as well and I know full well what you are talking about and why it works surprisingly well.

As far as system efficiency I too could care less. It it makes my vehicle start better I could care less about the few watt hours of electrical energy that got wasted in making things work better.

Personally I also do not see lead acid batteries, supercapacitors, starters and alternators as fragile delicate electronic devices either.. I know how much abuse they can take and can survive unharmed.

I've cranked engines with fully charged 12 volt 150 AH batteries until they were dead then hooked them to another vehicle and pull started them letting the stock 75 amp alternator (same Delco 10si model as used in countless vehicles that have a single battery 1/6 the capacity) deal with charging the big pair of near flat dead batteries while still powering the rest of the system without second thought or concern many times as to whether anything would burn up or not.

From my standpoint I know exactly what you are explaining and working with here and I have a very good understanding of why it works regardless of its theoretical quirks and or efficiency limitations.

Personally if anyone is worried about efficiency in their vehicle the electrical system isn't the problem. Its the emission systems and compliance that is what is eating your pocket book alive at the gas station.

I'm not sure you really read my reply or the paper I linked.

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Do you think the alternator in your car is better modeled as a voltage source or a current source?

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BTW, I'm not suggesting what you did isn't a great solution, I was only thinking about the inefficiencies introduced. Upon running some numbers, I don't think there would be a measurable decrease in efficiency (if at all), even though charging efficiency is not realistically going to reach your 99% claim in typical conditions.

Yes I did ;) Did you read my comments? :D Yeah, I'm not taking offense to you at all. I understand you weren't saying it's not a good project or anything like that (and I appreciate that, thank you).

I'm just a mechanical engineer, so maybe I'm wrong about these things. Like tcmtech and yusef1, I've just made some observations on things :) But there is no ideal voltage or current source. An alternator (with the regulator) will act as a voltage source... up to about 90 amps at idle. Beyond that, it acts more like a current source. It will provide up to 90 amps, nothing more, and the voltage will go down if the effective load goes up (like charging this capacitor pack). It's clearly seen if you hook up a clamp meter to your car with a dead battery and all the electronics on. It'll be producing 90 amps until the voltage is up to ~14 volts. If the voltage in your car is less than 14, it probably means it's using more than the rated capacity (or it means the regulator isn't very precise/consistent). I tested that this summer on a camping trip where I drained the battery a lot :) It went up to 89.5 amps until it hit 14 volts, then the current went down.

The paper basically says that a capacitor being charged with a voltage source is inefficient because a cap is a dead short at t=0. At the very first instant in time, all of the energy is being turned into heat in the wiring (in the total circuit resistance). This principle uses the resistance of the wiring to lower the voltage low enough so the current isn't infinitely high. When the capacitor is charged to half the voltage source, half the power is going into the capacitor, half is heat. You get the idea. Modelling with a voltage source would probably only be somewhat realistic with small electronics (in small power supplies etc). If the paper were true, then switching power supplies wouldn't be 90+% efficient these days (of course they use mostly inductors instead though). But that's exactly how they would charge such huge capacitors in a hybrid: switching power supplies to produce an appropriate current to charge the caps.

If the capacitor was being charged with an efficient PWM simulating a current source, 99% of the power coming out of the PWM would be stored in the capacitors. However, I never said anything about the charging circuit efficiency ;) With modern electronics, I'm sure it would be up to 90% total. But 99% of the power going into these capacitors would end up being stored until they self drain.

Come monday, I'll talk with one of the guys in my research group. His background is electrical engineering :)

As long as you have a solution to a problem rather than a solution looking for a problem to solve. Finding applications may be harder than to solve the engineering. So far, we are not doing so bad.

Here.....catch....tosses a cap to you that was charged by the ignition coil.....

you said....

Half the energy would be lost in the inherent resistance of the wires.

Heat is not delivered to a resistance. Heat results from LOSS of resistance. The lower the resistance, the more heat is produced. Measure the resistance in a toaster if you don't believe this. This seems to be a stumbling block with MY students. I worried that you might have fallen into that trap.

Ni-chrome wire (for instance) has the property of increasing resistance as it heats up. This helps to limit the current.

I think we already resolved this. Put 8 AA batteries in series and hook it up to a car starter. The energy is greater than the AA batteries but the potential is the same. Then compare by hooking up the caps to the starter. Which will get hot? The AA batteries will because it's in series with a very low resistance, lower than that of the batteries.

The heat dissipated within something is entirely dependent upon the resistance of what it's in series with. Put a copper pipe across the terminals of your toaster and see what happens. Super low resistance and your house wires will suddenly become the heating element if your circuit breaker doesn't trip.


BTW, I forgot to post a video.

It was on January 19th, which as you can see got down to about -30C

http://marge.campus.ltu.se/temp/old_png/temp_201401.png

That's the weather station right next to my apartment at the university I work at.

This is a link to the video of it starting:

http://nickdittes.phanfare.com/6425066

That's my personal website and I'll leave the video up for at least a year. Then it's probably gone.

I have a leaky diesel injector so it takes several more revolutions than normal to build up 250 bar for the injectors to open (since the high pressure pump is driven by the timing chain). It fires up essentially instantly after that happens. It's a common rail injection system (maximum pressure, something like 1380 bar).

It cranks about as fast at -30C as it does at +30C because of the ultracapacitors.

You call this a resolution? Note....critisism always looks nastier than it really is when written in an email. Please remember, I bear you no animosity. Even though the below is a real flame as I see on my re-read. Well, I am a teacher, so be it.

You have to use the correct terminology. Mostly, I think you do...I just spotted a potential trap which my students fall into.

A copper pipe would break circuit breakers. No question about it. But a nicrome wire is nearly as good. It too is a dead short like the copper pipe, but unlike the mythical pipe, its resistance changes (increase) as it heats up. And besides, the toaster wire is skinny. So there is no difference between the copper pipe and the toaster wire...both draw a huge amount of current, especially at the beginning before the wires heat up. We call this "inrush current".

With nichrome wire, though, the resistance increases as it turns red, which serves to limit the current. You don't feed current to a resistor....you push the current into the resistor with voltage.

WE ALL KNOW this...but the obvious tends to blur as we get into the circuits, and deal with internal resistances of generators, batteries, and whatever. Thevenizing Nortonizing such circuits can give one a headache. and Power is not voltage. But Voltage CAN be part of power. ( E times I. ) Potential is another word for voltage. Energy means nothing...well...maybe power. Hard to fit it into your essay.

You said...."The heat dissipated within something is entirely dependent upon the resistance it is in series with." That is simply not true. The amount of heat dissipated within something is dependent upon the INVERSE of the resistance it encounters. You can argue the opposite and claim you made verbal short cuts, ...but the fact remains...the amount of heat dissipated is an inverse function of the resistance, not the direct funcion you claim. All wires will heat up, that's why they limit the number of circular mils in a conduit. Making the wires bigger so that they dissipate this heat helps a lot to limit the heating. Adding more resistance also limits the heating effect simply because more resistance will reduce the current. You have to figure load (including internal resistances) into your calculations or they simply don't work. Internal battery resistance is a little more complicated. The batteries (in your example) don't heat up because they have more resistance. They heat up because the wires and connectors and chemical electrolytes inside are too small to handle inrush currents that are bigger than they are rated for. Capacitors have blown up in MY lab for much the same reason. They really DO blow up....big bang and all. Inrush currents are short in duration, but they can be devastating. Super caps can cause LOTS of inrush currents when connected to very low resistances.

A final example.... My father melted the wedding ring on his finger when he was messing about with batteries. Not a lot of resistance in a gold band.

Good information in your comment, enough for a GA.

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Seeing as how you are an educator, and criticism is always fair game for criticism, I think you might want to reconsider this particular part of your comment:

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"...A copper pipe would break circuit breakers. No question about it. But a nicrome wire is nearly as good. It too is a dead short like the copper pipe, but unlike the mythical pipe, its resistance changes (increase) as it heats up...."

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That would have to be a mythical pipe, indeed. So mythical that you'd be better off calling it something like 'kopper', since 'copper' has some well defined properties, one of which is an increase in resistivity with increasing temperature. This relationship is fairly linear as long as we aren't too close to absolute zero or the melting point. 'Copper' is not 'unlike' 'nichrome' in this sense.

A big copper pipe would likely not heat up much so would not change resistance much. It was not I who brought the big old copper pipe into the discussion!!!! Grin. Not as much as the wires in the wall and the circuit breakers for sure. A copper wire the size of the nichrome wire in the toaster would also heat up. Not sure if the resistance would increase as fast as the nickel chromium alloy. I could look it up, I suppose. I am sure you did since your comment implied that it does.

Criticism accepted with no problem....grin!

You supposed correctly. The resistance in copper increases with temperature about an order of magnitude faster than it does in nichrome.

Kuul! I did not know that. Thanks for checking.

I'm sorry if I sounded like I was holding animosity towards you ;) It wasn't my intention. I didn't take it that way from you.

And yes, you're absolutely right I do a lot of talking without saying things in the best way. I often type faster than I think.

And I'm perfectly aware of any temperature/resistance effects. I don't think that needs to be considered to the same extent in a car starter system as we hopefully don't have any glowing components in the 1 second that we need to start!

Let's just say the engine will take a certain amount of current to start it. Say 200 amps. So 200 amps is going through the battery and when it's cranking, there's a 4 volt drop (it's normal to see that). The internal resistance is apparently 0.02 ohm. The ultracaps only drop about 1 volt. So 0.005 ohm. That's a difference of 800 watts and 200 watts of internal heat generated within the power source (P=I^2*R). Lower resistance, less heat.

To further what I was trying to say all along, I acknowledge that lower internal resistance of the battery/caps will definitely lead to a higher cranking current (which is exactly the point of doing this all along!!! :D ), up to the point where the starter is no longer capable of pulling more current because of it's own internal resistance based on the available potential. IN PRACTICE with measurements, the current seems to be the approximately the same with or without the ultracaps IF the engine and battery are warm so the 200 amp estimate above perfectly applies, despite the obvious difference in internal resistances of the battery/caps. If the internal resistance is reduced, there will be LESS heat generated within the battery/caps.

That's why I brought up the AA batteries as a comparison. They'd be more or less shorted and do nothing except get hot, because the internal resistance would be significantly higher than that of the starter so that MOST of the energy within would be dissipated within the AA batteries, instead of externally (as in the case of the caps). Do you see my point now? That's all I was arguing all along lol. That's why I said the heat dissipated in a component is dependent upon what it's in series with.

My point with the copper pipe was to say that less resistive heat will be made in the copper pipe than the heating element because it's low resistance because practically, the wires leading to it will have a higher resistance so a higher percent of the power will be dissipated there. Yes, obviously the power would be enormous and obviously trip the breaker.

I really don't think these caps will blow up in any condition. The potential is too low and supposedly they just kind of fizzle out and vent. I haven't seen any evidence of them being capable of detonating. There are lots of videos of people "shorting" them to burn things or make sparks. If you did that with a car battery, they would most definitely explode or at the very least, vent a lot of hydrogen gas and get hot.

I feel no animosity. I am a teacher...its what I DO!

You have clearly fallen into the resistance trap. Most of my students do sooner or later.

Current creates heat.

Resistance limits current.

Increasing voltage increases current though a fixed load.

Lowering resistance allows more current to flow, therefore, more heat.

Current is equal in all parts of a circuit.

Not difficult points, but one that is often overlooked even by people who are familiar with the code book. Why did my Dad's gold ring melt on his finger?

You said...if the internal resistance (of the capacitor) is reduced, then there will be less heat generated. That is simply not true...reducing resistance results in increased current. IN A CIRCUIT. As long as you keep your caps out of a circuit, they won't fail you. The low internal resistance of your ultra caps most certainly WILL create heat. Getting an ultra cap with a lower internal resistance will create even more heat. However, a capacitor sitting by itself is NOT a circuit. Even a capacitor with one function.....supplying a load, is not a circuit. A resonant circuit IS a circuit, and it will heat your caps up just fine. I worry a little (only a little) because your starter motor is a coil, and you MAY get some resonance. A capacitor is a reservoir of electrons. Tesla used to water cool his capacitors because they got so hot, but then, he was creating very powerful resonance circuits. I understand some of his Lyden Jar arrays reached close to one Farad. But if the caps dissipate heat, then, so do the wires and especially the load. How much? Presumably the ultra caps have heat dissipation designed into them, so it hardly matters. The wires are designed to carry sufficient current without heating up too much, so they don't matter much. The load will take the lion's share of the potential. Current will be common throughout the circuit. ....the starter motor would be the most potential drop in the circuit. It would get, what...almost 12 full volts across it. Lots of current, which can and HAS heated series motors up until they fail. Current is common in all parts of a circuit...so what goes across the load ALSO goes across (or rather through) the battery, and presumably the capacitor.

There we go....what more can I say? (wait for it!)

Why did my Dad's ring melt? Too much or too little resistance?. It was a solid gold ring. I think it was too little resistance.

haha, you're such a nice guy! I like Canadians :)

But no I haven't! I acknowledged there is a higher current! More power is getting to the starter! BUT despite the increase in power being delivered to the starter, since the internal resistance of the caps is about 1/10th of a battery, less heat is generated within them for the same power output! BUT it's not going to be 10x the current or 10x the power because the resistance of the starter (and the cables) hasn't changed and is significantly higher than the caps.

In other words, a smaller percent of energy is lost in the power source! That's all I was trying to say all along!!!!! :D

Thats what you have been saying all along. I get that. I still have a problem with it.

If we had a starter load of say, 200 amps (not unlikely) that same amperage is going through the capacitor. Or the battery. The formula is dirt simple, and no doubt known to you. Power (to heat things up) is equal to current times voltage. Voltage is equal to current times resistance. Current is equal to the inverse of the resistance. (Amps equal voltage over resistance., if voltage stays the same (at 12 volts which it will NOT for long be the case in a capacitor), then as resistance goes down (say by substituting a less R cap or dropping in an ultra cap instead of a high R battery), current has to go up. More current equals more heating of whatever it is going through. I think your capacitors are designed to handle huge but short term currents, and that would be expected. In other words, the cap will not heat up because it is designed to heat sink or otherwise dissipate the heat. It is practically a dead short to a changing current. Cranking a battery will change the current as it discharges. A dead short heats up.

Again, I ask, why did my father`s gold ring melt on his finger

I just don't see how 200 amps which go through the starter is not going through the capacitor. But you cannot deliver that 200 (or whatever) amps without letting it leave the capacitor. So 200 (or whatever) amps will be going through the capacitor. The battery will heat up a LOT more because it is not a capacitor....it needs to reverse a chemical reaction...which is why it heats up...NOT because it has a higher internal resistance. The batteries chemical reaction is a load.

Thats what I have been trying to say all along as well. Shall we agree to disagree

But you're forgetting that all along I've been talking about starting a car, not anything else like shorting a ring. It takes *approximately* the same amount of energy to start a car whether it's from an ultracap or a battery (especially with a diesel, since it pretty much needs that energy to heat the air within the cylinders to ignite the fuel). When very cold, an excellent battery might need 1 or several seconds additional to crank compared to these caps, because it will be cranking slower and the compression heating will be less (more cylinder losses at slower cranking speeds).

I said before in #61 that the current going through the starter is also going through the caps/battery. I used the 200 amp example to show the heat buildup in the caps and battery was 200 and 800 watts? I already said that. I pointed that out that if it's 200 amps (I measured that with a 500 amp DC shunt meter that's now mounted within the car) there would be LESS heat within the caps than a battery, and as you just added now, it's not only because the internal R but of the electrochemical reactions as well. The ESR (or whatever we want to call the internal resistance) is responsible for the heat generated within a battery. Less ESR = less heat within the source, assuming roughly the same power output.

According to a simple calculation, the ultracaps have enough juice to crank the starter for 12 seconds (assuming if it's still a 2kW starter as I read in a technical brief for this engine) and assuming it's from ~12.5 volts down to 8 volts. That's a lot of cranking! That math assumes the battery isn't contributing. As I'm sure you saw in the video, it was really only a couple seconds of cranking at most even at -30C. So the voltage drop is minimal. I'm neglecting the glow plugs, fuel pumps etc but I have a 92 Ahr battery in parallel to take the "small" loads (even though I realize, the glow plugs are A LOT of power!!!)

I also pointed out that with measurements, the starting current is only higher when the battery isn't capable of pushing that many amps when it's cold. The internal R of the caps ARE less than 1/10th of that of the battery, but as I stated before, the current WILL NOT be 10 times more because of the starter resistance.

The current due to the caps may be marginally more when it's warm out (as my measurements showed). The impedance of the starter is high enough it will ONLY take a certain maximum current due to the fact the impedance of the battery and caps are both very low! The internal R of the battery (when warm) is low enough that even with an order of magnitude change to the caps, it doesn't seem to increase the total current (again stated, unless the battery is really cold when the battery can't do it! This is the purpose!).

I appreciate your comments because you have made me think about what I'm stating but I'm sorry, I still don't see how I'm possibly wrong in any way.

If used properly, and assuming the same power output, lower cap/battery ESR equals less loss. It's EXACTLY the same reason why you'd want bigger power cables going to the starter, or lower loss transmission lines in power grid distribution.

Dead shorts equal heat

Your super caps are nearly a dead short. To crank they must pass about 900 watts through them for a few seconds. Good thing its only a few seconds.

Current creates heat, and that heat must be dissipated somehow. The caps are clearly designed to do that.

Why would you assume a higher Resistance would result in more heat?

The battery heats up because it has other things going on.

Not that it matters. Clearly you got it working. The super caps are clearly capable of dissipating the 8 or 9 hundred watts going through them. But like I said, but them into a resonant circuit where they will get exposed to both inrush and outrush currents three or four times a second, and they WILL heat up.

What would be the difference between my dad's gold ring and the tiny resistance in a super cap? You have STILL not answered that. But you can't dismiss that question. It is key.

Your last paragraph is spot on. Bigger power cables create more path for current. Current is still there.

I'm sorry, but you are wrong. Now, I respectfully want you to consider the following calculations below as proof of what I was saying :) You will see, nearly HALF as much heat is dissipated within the capacitors/battery pack compared to the battery alone, DESPITE THE EXTRA CURRENT DUE TO THE VERY LOW RESISTANCE! :) :)

Yes, I obviously get that dead shorting something will result in a huge amount of power being dissipated in a very short time. But we're not doing that! Copper can vaporize and explode like a bomb... I have never stated otherwise and never said you were wrong about a short circuit and high current/power... but a cap is not a short circuit.

...and we're not dead shorting things here. The cables/starter limit the current. If you put a gold ring, or better yet, a pure solid silver block IN SERIES with the starter, there would be very little heat coming from that little silver block because the resistance is so low. The cap's ESR is something like 0.0008 ohms (as I stated very very early in this discussion). Even if it changes by one order of magnitude in ESR due to some magnificent temperature change, it's still 0.008 ohms, or an order of magnitude less than a typical battery. So, ideally 32 watts of heat for 200 amps per cap (and there are 6). Realistically, maybe more, but not much more!

Where are you getting that 900 watts figure? According to the manufacturer, these ultracaps can be charged/discharged REPEATEDLY in several seconds at up to something like 600 amps continuously. These specific caps were apparently taken (used) from a hybrid electric city bus and used for that purpose. 900 watts would vaporise these. They have a small mass.

You have to remember that the starter motor and cables aren't changing. You change the ESR of the source and the load (cables/starter) will still be the current limiter. The current will only rise marginally.

LET'S DO SOME MATH!!! :) Starter (including cable drop), 8 volts drop @200 amps = 0.04 ohm

typical battery: 4 volts drop @200 amps = 0.02 ohm

Total series resistance of the whole system = 0.06 ohm

Now let's change that system to what it MIGHT be with the caps in parallel with the battery. The published ESR of the caps when new was 0.0008 ohms each plus the junctions which for an estimate's sake, is probably equal to the ESR of each cap. So let's just say 0.0048 ohms for the caps, 0.0048 ohms for the junctions and 0.0096 total.

Capacitor plus battery in parallel = 0.0064 ohms and the total system resistance in series is approximately 0.0464 ohms.

12v/0.0464 ohm = 258 amps and from what I measured, the difference is maybe a little bit smaller! The current is only slightly higher with the caps than with the battery. In my calculation, maybe I underestimated the ESR of the cap/battery pack.

Furthermore, 258 amps through 0.0064 ohms (battery+capacitor) is 426 watts (remember, only half that power is in the caps... the other half is assumed to be in the junctions/cables).

Starter power = 2662 watts.

If I ONLY had a car battery, 200 amps is 800 watts through the battery and ONLY 1600 watts going through the starter. You see what I mean? The % of the power going to the starter is a ratio of the source/load impedance. A MUCH higher % of the energy is going towards starting the car, not heating the battery. I said that many times.

Despite my ranting style of typing, this was a good discussion! Now I feel even safer having these in my car! :D

EDIT: I'm sure you can find fault with my estimates, but I did show that a lower ESR will lead to less power being dissipated within the battery/caps. More gets to the starter. That was my point all along. Nothing else.

Okay, I'm wrong. Ohms law is incorrect. Got it.

Fine.

Unsubscribes.

The problem with capacitors is they discharge all their energy in a single burst; then need to be recharged; unlike a battery that can be discharged at a controlled rate. Capacitors work great for ignition systems, but I don't see much use if a constant flow of energy is needed, like to operate lights, radio, etc. This is not my field of expertise, but it sounds like a worthwhile project. How would you get an initial charge to start a car? Small 12V battery?

When I ran the battery shop back in the day, I learned a lot about batteries and the way the extreme cold can make things difficult. They were aircraft batteries, and sometimes they would arrive in my shop covered in frost. Flying at altitude can do that to a battery. But I didn't do engineering studies, this is only observations.

The batts would arrive and be put on a shelf to be load tested. A failure there would simply mean take it out of service and replace it. (Mallory designed the load banks and testing procedure we used...which struck me at the time to be a designed to ace the batts, but at least the military batteries got renewed a lot.) This involved some pretty heavy load banks. SOP was to let them warm up first. But I was not about to stay an hour late just to let the batt warm up when I could test it, and get it out onto the "delivery" rack. So once in a while I ran cranking amp tests on those frost covered batteries.

Not surprisingly, the chemical reactions of a lead acid battery failed to keep up with the load bank, and its voltage plunged. The cranking amps rose alarmingly as a result because, well, its a heavy load, and if it can't take it in volts, it'll take it in amps. (over simplification of the I times V but makes for a nice turn of phrase)

If there were a parallel capacitor big enough to keep the V topped up for the few seconds of cranking loads, the current would not have to peak, the fuse links would not fry, and the starter windings would not go crispy. After the first three seconds, of course, the battery would be nice and warm inside, and internal cell V would rise to near normal, even under load. Now, you would need to isolate the capacitor because you have not finished spooling up the engine, and you don't need the parasitic parallel capacitor load sucking energy from that.

If this were a Ni Cad, at this point we would be worrying about intercellular membrane overload breakdown and possible thermal runaway due to the overload resulting in recharging the capacitor as well as starting the engine. They run pretty close to the edge you know. And as for the Lithium Ion ones...well, just forget it. They won't take start loads now! (useless damned crap batteries..oh did I say that out loud?)

So that would be "normal" operation. I don't think we would be looking at a simple "drop in" parallel installation, at least not without a cargo plane load of research to take of things which can go wrong. I spotted a couple just because they resemble the experience I had with the stuff. For sure, just to begin with, I would have some serious current limiting on the power supplies. You are going to have a drop in voltage regardless.... its a series load going into near dead short! That will drag ANY voltage down! The caps will react to that V drop even when things are warm. They shouldn't fry starter leads....but remember, they and the starter were made to be run by a battery and its known internal resistance. The cap will throw off that calculation.

So, yes, it really does need some research. As you pointed out, its not as easy as it might seem at first. But then, the original question was about a car battery. As long as my alternator can keep up, a simple drop in parallel cap might well be a saviour on one of those minus thirty five days up here. Probably you don't have this problem where you hang your hat though. But I am totally stoked about the idea up here in God's country.

But why in blue cars only?

It was a joke. He wasn't serious.

Hello great people, l'd like to find out if l could use 3 nos 400 volt capacitor at 2500 farad each to jumpstart my Camry car.Thanks for your kind consideration.

Oladeji Williams

Did you perchance bother to read ANY of the posts here, paying particular attention to the one with the highest "Good Answer" score (#8)...by "mjb1962853".

or , did you follow the link to THIS thread, linked above, to learn a bit more on the subject?

The simple answer {that you seek} to your question is "No" ...

(unless you merely wish to "jumpstart" a whole-lotta-sparks)

Dont dispair, there is a couple of old and wanna be funny guys here that would not offer help that easy or dont know much about the subject or would not give it away in a hurry.

But since you indicate you are willing to learn and get hands on I am inclined to at least offer you a snippet of life experience.

Nothing fancy, only the other day my car battery died and I jump started the car. Last I measured there was less than 8 Volts over the terminals.

While the car was running I noticed that the ABS light came on and would not go away until the new battery was installed.

This made me think that some safety related system will always require enough Voltage which is provided past the alternator directly from the battery system.

You might want to consider this in your attempt to replace the battery.

Good luck with tinkering but stay safe!

I've had extremely good luck with my ultracapacitors. I've had them in for 5 or so years now. Definitely no replacement for a good old lead acid battery, but is ideal for supplementing it.

It doesn't create a massive surge in current draw by the starter motor, as was predicted by another member here. The inductance of the chassis and the resistance of the wires limit the current enough it only makes an insignificant difference in current draw. The impedance of a 1000 CCA lead acid battery is similar the capacitors, but the capacitors will provide that at any temperature on the surface of our planet lol.

There ya go, sounds like a solution for weak batteries in cold regions of this earth!

One use for early primitive versions was in Siberia (Russia) for augmenting the lead acid battery of vehicles (for starting purposes) which really struggle in the cold climate.

It's more than just weak batteries, but with diesels it's important to crank it fast enough to reduce cylinder heat loss so it ignites better. It'll crank at 0F like it does at +40F lol. The standard battery in this car is massive. It's 60 pounds, and it's an AGM so a standard battery would be even heavier. But even that cranks slower when it's really cold.

I can cold start my diesel BMW when it's 5F (-15C) without waiting for the glowplugs and it'll start without hesitation. Much colder, like 0F, it'll take ~2 seconds of cranking to make it start without the glowplugs. The glowplugs take about 200 amps for a couple seconds, but levels off to ~100 amps as they are fully hot, so it really takes a massive battery to start it. I actually jump started an industrial 100 kW generator that hadn't been run in over a decade. It was a 5.8 liter Perkins industrial engine. That actually took fewer amps to start than the car does! lol. Although the 16mm^2 jumper cables limited the current somewhat, it cranked just fine.