Jun 04 2015

Batteriser – Cool Tech or Scam?

A company is claiming that they have created a small sleeve costing only $2.50 that will fit over standard alkaline batteries, and boost their life by up to 800%. My initial reaction to this claim is that it is almost certainly total BS. But I am always willing to give a new claim a fair shake and dive into the details.

The reasons for my initial skepticism are many. Energy scams are common. They often promise incredible effects from implausible devices. Usually, by asking one simple question, you can destroy the claims utterly – from where is the extra energy coming? The laws of thermodynamics are called “laws” for a reason, and reality is a harsh mistress.

Bob Roohparvar, owner of Batteroo which is developing the Batteriser, has an answer to this question. He claims that typical alkaline batteries use about 20% of their available energy. The reason for this high inefficiency is that electronic devices are finicky; they need a steady voltage of 1.5. When an alkaline battery drops below 80% capacity the voltage also drops to 1.4 or 1.3. At that point the device will read the battery as dead, leaving 80% of its juice unused.

The Batteriser, Roohparvar claims, is a miniaturized device that boosts the voltage back up to 1.5, and maintains it there until every last drop of energy is extracted from the battery. OK, I thought, at least he has an answer to the most basic question about such claims. But that is about all he has. Even a slightly deeper dive into battery technology and his claims reveals them to be untenable.

Before I deliver the coup de grace, my skepticism about the Batteriser grew the more I read Roohparvar’s claims. First, simple math told me that if batteries normally have a 20% efficiency and he is extracting 100%, that is a 5x improvement, not an 8x improvement. So how does he get to 8x?

Let’s say you buy a new battery. You use it for a month and its voltage drops to 1.4. It’s now ostensibly dead at 1.4 volts, but if you slip on a Batteriser, its output increases to 1.5 volts for another month. That’s already a 2x increase in battery life. Eventually the battery’s natural, unboosted output drops to 1.3 volts—but Batteriser keeps it at 1.5 volts for another month. Now you’ve realized a 3x increase in battery life. And so on, and so on. Roohparvar says Batteriser can continue to deliver a 1.5 volt charge in batteries that have discharged down to 0.6 volts. There are more than eight 0.1 volt steps between 0.6 and 1.5 volts, so, in grossly simplified terms, the Batteriser can extend operational battery life somewhere around a factor of eight.

I call BS on that. His fancy math does not get you around the fact that 100% energy is 5 times 20% energy. He is playing a little shell game. It seems likely, assuming his device even works, that as you extract higher voltage from the battery it will drain faster and faster. In fact, new inefficiencies might be introduced. The way to boost voltage will be to draw more current from the battery, which will heat the battery more, and might make it less efficient (in addition to being a potential hazard). In any case, the 800% claim is almost certainly nonsense, even by Roohparvar’s own logic.

But can, then, the Batteriser extract 500% more life out of those AAs? I don’t think so, and here’s why. While it is true that the voltage of an alkaline battery (and all batteries) drops off as capacity drops, the big drop off doesn’t occur until around 82% of the capacity has been drained, and therefore only about 18% remains. Take a look at the graph – Zn/MNO2 is alkaline batteries, the blue line at the bottom. You can see that over the first 80% of the battery’s life the voltage slowly decreases from 1.5 to about 1.2 volts. Just above 80% it then drops off the cliff.

The question then is – what is the cutoff voltage for most devices? You can look this up yourself. I found a good reference at Popular Mechanics, although it was from 1994. The types of devices that still use alkaline batteries, however, are roughly the same as then. Newer electronic devices, like cell phones and iPads, run on lithium ion batteries. Here is their table:

autofocus camera with flash: 1.3v
halogen flashlight: 1.3v
regular flashlight: 1.2v
headphone/CD player: 1.2v
portable LCD TV: 1.2v
pencil sharpener: 1.2v
AM/FM radio: 1.0v
electric shaver: 1.0v
quartz clock: 1.0v
remote control: 0.8v

You can see that the devices cutoff starting at 1.3v and as low as 0.8v. Except for the high-end devices, most of the devices cutoff at 1.2v or lower, meaning they will extract 80% or more of the stored energy from an alkaline battery. I looked up the cutoff voltage for a couple of more modern devices for comparison. Microsoft cordless optical mice have a cutoff voltage of 0.6v, and Logitech mice of 0.8v. These are therefore very efficient devices.

Also, keep in mind that many devices have battery packs with multiple batteries. The voltage of each battery is added – so if you have a battery pack with four AA alkaline batteries they will generate 6.0 volts when fresh. Let’s say you are using a device that requires 4.0 volts, that means the batteries will last until they are drained down to 1.0v, which is again north of 80% efficiency.

This means it is largely up to the device manufacturer to design the device to operate efficiently with available batteries – designing the battery pack and the cutoff voltage to meet the needs of the device and optimize the efficiency of the batteries. There certainly can be devices out there that are poorly designed, with high cutoff voltage and/or too small a battery pack that are highly inefficient. It does not seem as if this is typical, however.

All of this means that the Batteriser, if it works optimally and does squeeze every last drop out of alkaline batteries (and does so safely) is probably going to increase typical battery life by about 20%, not 800%. A 20% increase in battery life is not bad, but it probably won’t lead to the successful Indigogo campaign that Roohparvar is counting on. It also may not be cost effective or worth the bother of using the Batteriser (depending on the device). The product may have a niche application for specific devices that have a necessarily high cutoff voltage, but it is not going to revolutionize battery tech or save the world, as the hype promises.

Once again the smell test for overhyped energy tech claims seems to be accurate. The claims seemed too good to be true because they were.

Of course, I am wiling to be proven wrong. I would be happy to test the Batteriser for myself in various devices, and if any of my arguments are not sound please point out my error.

28 responses so far

28 Responses to “Batteriser – Cool Tech or Scam?”

  1. TheDawgLiveson 04 Jun 2015 at 8:55 am

    There is still the huge unanswered question of HOW this device “boosts” the voltage to 1.5v. Also, if this device actually does what Batteroo claims, then it will result in worse performance in devices with low cutoff voltage such as remote controls because it will force more voltage than the device needs resulting in shorter battery life.

  2. Daniel Clementson 04 Jun 2015 at 9:12 am

    The voltage can be stepped up with a very simple circuit consisting of a diode, a coil, and a switch (transistor).

  3. Barryon 04 Jun 2015 at 9:29 am

    The voltage multiplier circuits that Daniel alludes to do work to an extent, using either switching circuitry or big hulking capacitors. Their downside is they tend to produce quite noisy DC voltage and the circuitry itself is bulky (it certainly would not fit in the apparently paper-thin package pictured above). However voltage multipliers don’t increase the total power (the product of voltage and current) that is available from a battery. What is worse, the added multiplier circuitry would present itself as an additional load to the already weakened cell, draining it even faster. I just don’t see where the magic happens.

    Most devices are current-driven loads, and don’t much care about a cell’s terminal voltage (usually anywhere from 1.2 to 1.5 volts will be fine for most devices), but as the available current begins to drop (the cell’s internal resistance changes due to changes in electrolyte chemistry), their performance degrades dramatically.

    So I’m keeping this one in the BS file for now.

  4. eastporteron 04 Jun 2015 at 9:34 am

    Our family keeps track of disposable batteries: the batteries from the camera can go into the wireless mouse> on down to the electric toothbrush. Information on this gizmo may help educate people to the fact that many of the batteries that they throw away still have lots of use.

  5. John Danleyon 04 Jun 2015 at 9:42 am

    Key phrase here is load capacity. The load of BS has reached full capacity.

  6. pdeboeron 04 Jun 2015 at 9:47 am

    Barry, Daniel is not referring to a voltage multiplier, he is referring to a buck boost. They are not bulky, or noisy. However, DC-DC supplies are usually much bigger than the device shown(check digikey).

    Doesn’t matter anyway, because a 1.4V battery is certainly not dead. Most devices can drain down to 1.2V for the reasons Barry said.

    There are several different ways to do a DC-DC boost. A voltage multiplier is not one of them as it is an AC to DC and is only typically used in particle accelerators.

  7. Banzai Otison 04 Jun 2015 at 11:20 am

    Small typo? I believe the company’s name is spelled ‘Batterwoo’. 😉

  8. Jim Shaveron 04 Jun 2015 at 11:37 am

    pdeboer: The gist of what Barry wrote is true — A DC-DC converter circuit will drain the battery faster, due to its own unavoidable power dissipation. So even if the device works (which it probably doesn’t, given all the identified technical issues), you would have to be careful to not install it until your battery-operated device stops working. Even then, assuming you could really access the remaining 20% or less of energy from the battery (which you probably can’t), the DC-DC converter would turn a significant fraction of that 20% or less into waste heat. So you’re left with a slim chance of pulling a fraction of the residual energy from the drained battery and turning that fraction of energy into useful work.

    Also, I have to say that Roohparvar’s pseudo-technical explanation of there being eight 0.1V steps down to 0.6V and therefore an 800% energy boost is truly befitting of the crappiest of sci-fi B-movie dialogues.

    If it quacks like a duck…

  9. Karl Withakayon 04 Jun 2015 at 12:23 pm

    You can’t get something for nothing, and you can’t cheat Ohm’s law. V=IR You can’t just increase voltage with no offsetting cost.

    This isn’t particularly innovative tech here. If it worked as billed, not only would it have been done before, but manufacturers would have been building it directly into the devices themselves.

    Reading through the patent, it’s hard not to conclude deliberate fraud here, but I suppose I shouldn’t absolutely rule out Hanlon’s Razor: Never attribute to malice that which can be adequately explained by stupidity.

    The patent claims section reads like a marketing brochure. The technical details in the patent read like a patent for a black box. It covers many different means of applying the black box, but it never describes what’s in the box. For all I can tell from skimming through the patent, the DC-DC convertor may consist of micro-leprechauns used to maintain the voltage.

    From the patent: [Although specific embodiments have been illustrated and described herein, it will be appreciated by those of ordinary skill in the art that any arrangement that is calculated or designed to achieve the same purposes may be substituted for the specific embodiments shown. Many adaptations of the disclosure will be apparent to those of ordinary skill in the art. Accordingly, this application is intended to cover any adaptations or variations of the disclosure.]

    That’s pretty rich coming from what is essentially a novel use patent for application & combination of mundane prior art technologies of battery sleeves and dc-dc voltage convertors.

  10. Karl Withakayon 04 Jun 2015 at 12:30 pm

    From the credulous PCWorld articly:

    [“There’s actually no IP [intellectual property] in the boost circuitry. Our technology is really a miniaturization technique that allows us to build the sleeve. We have some IP in some of the IC circuits that are in there, but the key is we’ve been able to miniaturize the boost circuit to a point that no one else has been able to achieve. “]

    Except that degree of miniaturization is only needed post hoc. What haven’t electronics manufacturers been incorporating “boost circuitry” into the devices all along to increase battery life without needing aftermarket sleeves? The world wonders.

  11. ThatCarlon 04 Jun 2015 at 2:24 pm

    Karl & TheDawgLives:

    Many devices already do have voltage regulators. This device would be unnecessary for them.

    Another catch is that increasing voltage does not mean the battery can put out any more current, so for some power-hungry devices without a built-in regulator this wouldn’t help.

    They don’t take up a whole ton of space, but it might take some custom manufacturing to fit in the proposed design.

  12. Pete Aon 04 Jun 2015 at 2:51 pm

    Karl Withakay— I really enjoyed your comment 🙂

    To the readers who may not understand why, I ask: How the heck do you think electronic flash, which requires many hundreds of volts to operate its xenon tube, derives this high voltage from a battery consisting of one to four cells each having, at most, only 1.5 volts?

    The inverter technology used in modern state-of-the-art flashguns is awesome because it works really well and the manufacturers refuse to masquerade it as Batterwoo [many thanks to Banzai Otis for this word].

    Both the voltage drop and the leakage current across all semiconductor junctions will render the device negatively efficient throughout the working life of chemical cells and ionic cells (e.g. lithium). Inverters rapidly decrease in efficiency as the number of cells in the battery approaches one!

    I clearly remember the “metal balls in a wire cage” for insertion into fuel tanks and the magnetic fuel pipe clamps that were being vended as ‘scientifically proven’ to increase the miles-per-gallon of road vehicles. Without the vendors of quackery, humans would have little incentive to learn essential critical thinking skills.

  13. _Arthuron 04 Jun 2015 at 3:31 pm

    Joule thief circuit:

  14. Karl Withakayon 04 Jun 2015 at 3:42 pm

    Pete A,

    Don’t forget the Vortec Cyclone air intake system to simultaneously boost horsepower and increase fuel economy.

    From the Vortec Cyclone web site:

    A: You can find skeptics and negative information about any product. The bottom line is, our tests show that the product works, and independent dyno tests confirm our horsepower gains. We have hundreds of testimonials from satisfied customers, and our return rate is less than 5%. We all use Vortec Cyclones in our personal vehicles.”]


  15. Damloweton 04 Jun 2015 at 4:23 pm

    Another angle on this also, if all these devices have issues with with 1.2v being a flat battery, how do they manage to operate at all on rechargeables? A rechargeable cell has a starting voltage of 1.2v, and if the initial claim of ‘devices”refusing to work on anything less than 1.4v is silly.

    As has been mentioned, most devices seem to be current driven and don’t have issues between a quartet of 1.5v alkalines for 6v, or a quartet of 1.2v NiMH/NiCd ect at 4.8v. One noteable annecdotal exception is my first digital camera, the thing just would not fire up with anything less than 2 new 1.5v alkaline batteries, and then destroy them within what seemed like 20-30 photos!


  16. DietRichColaon 04 Jun 2015 at 4:44 pm

    Welcome to another installment of NeuroLogica Chemistry Lessons by DietRichCola:

    I am not an electrician, electrical engineer, or even particularly well-practiced in my electrochemistry anymore. However, from a chemistry perspective my understanding is this:

    The voltage of any galvanic / voltaic cell is an intensive property of the electrochemical cell at standard conditions (1M or 1 mol / 1 L). The voltage for any electrochemical cell is determined by the net chemical reaction. If a AA battery and a AAA are based off of the same reaction (they aren’t always), they will have the exact same voltage, even though the AA is a “larger” battery. It is true that the AA battery will have a larger store of energy, but this is not reflected by the voltage.

    On a purely chemical scale, any change in voltage of the cell from is technically based on the RELATIVE ion concentration difference from standard conditions in the anode and the cathode as the battery is being depleted. This is related to total chemical amount, but only in a relative sense AND it is a logarithmic relationship, not a linear one:

    Ecell = E°cell – (0.0592/n)log([anode ion]/[cathode ion])

    Ecell is the actual cell voltage, E°cell is the maximum cell potential at standard conditions (25 °C and 1M), n is the number of electrons transferred in the redox reaction, and the ion concentrations in the anode and cathode. As the anode (where all the electrons come from) gets used up, the anode ion concentrations will increase (since this is the metal having lost its electrons). Meanwhile the cathode ion concentration will decrease as the ions gain electrons and become neutral metals.

    All of this is to say, on a purely chemical level, you won’t see drop of 0.1 V from any cell potential involving a 1 electron process, until you have a 48:1 ion ratio, which means you’re at 98% of the reaction completion. In a 2 electron process you need a 2,389:1 ion ratio, which means you’re at 99.06% reaction completion. This is also assuming they start at standard conditions, when in fact they usually start with an increased cathode ion concentration relative to the anode in order to utilize this effect effect in reverse (i.e. to increase the initial voltage of the cell and give it longer before it depletes). But all this means that for much of the battery life, the voltage should be nearly constant, as reflected in the diagram Dr. Novella posted.

    Now, very few chemical reactions have an exact voltage of 1.5 V. Most real batteries often have multiple cells set in series to get the desired voltage. This means that if individual cells experience small drops in voltage, if they are all being depleted at the same rate the cumulative effect of this could get you to a 0.1 V drop a lot faster. But I still have to call bullshit on the idea that when the voltage has dropped by 0.2 V the batter is still at 80% capacity.

    So, if the Mr. Roohparvar is talking about the battery capacity in terms of the total amount of chemical reaction that can take place, and he’s claiming that the voltage is dropping because the chemicals are getting used up and that is making the voltage drop… then this guy is dead wrong. That’s not how voltage works in comparison to battery capacity IF you mean capacity in terms of total chemical reaction.

    Now, the truth is, real batteries die for a lot of reasons, not typically related to the electrochemical cell potential I described. From what I remember these include the electrodes eroding, the salt-bridge breaking down, the housing of the battery breaking down. But based on everything I know, I believe the PRIMARY reason a battery dies is due to internal resistance. Over the life of the battery, the internal resistance of the battery increases due to several factors (such as the ones I just mentioned). It does manifest itself as a drop in voltage because:

    V = E – I*r

    Where V is apparent voltage, E is the maximum cell voltage, I is the current, and r is the internal resistance. In this case, the “I” represents the maximum current the battery can produce. It is the current that drives most devices, so when the current gets too low, the effective voltage will approach zero, and that means the device won’t run. We can rearrange the equation thusly:

    I(max) = E / r

    Since E is fairly constant over the life of the battery (as I described in the section about eletrochemical potential), as the internal resistance increases, the maximum current the battery can produce will drop too low to be useful. I believe this is what Barry was eluding to.

    It is possible that the internal resistance of the battery goes up fast enough that the effective voltage is too low even when there is still “80%” of the available chemical reaction left. This would mean that the apparent voltage drop is not from the chemical reaction running out, but that the act of converting the battery to 20% products has already increased the internal resistance to the point that the battery is useless. IF this is indeed true, then some of Mr. Roohparvar’s statements do seem to hold water:

    From the article:

    “The time it takes for the battery voltage to drop by 0.1V is longer at lower voltages versus at higher voltages. That means that if a constant current was drawn from the battery, it would take the battery a lot longer to discharge from 1.2V to 1.1V than it would from 1.5V to 1.4V. This means that the extent to which the battery life is increased could be even higher.”

    At the very beginning of a battery’s life, if indeed the cathode cell has a high concentration of ions versus the anode, this would be true. Voltage change is logarithmic, so you could see an initial steep drop, and only a minor drop in the “middle” of the battery life when the anode:cathode ratios are still 1:10 to 10:1, which is effectively the 9% to 91% depletion of the battery’s chemical reaction.

    Is it possible his device is decreasing the internal resistance? I have no idea. It’s possible… but then why isn’t he explaining it in those terms so that electrically literate people can understand what he’s claiming, instead of using language that seems to convey a misunderstanding of the chemistry, physics, and electrical properties?

    So I do believe part of this comes on the definition of the capacity. Battery companies measure the capacity of a battery based on how long it can be used before the load bearing current is useless. So is it true that your battery still has chemicals left to continue reacting even after the battery is effectively dead? Absolutely, otherwise the dead battery voltage would be zero, not 1.2 V.

    Now is there still 80% of unreacted chemicals left so that you could view this as still being at “80% capacity”? I have no idea. Based on electrochemistry alone the answer is no, you won’t get that big of a voltage drop until the chemical reaction is nearly depleted. But how quickly does internal resistance rise relative to the available chemicals? That I’m not sure of. Even if this is true, by exactly what process is the company correcting this problem? Since batteries primarily die due to internal resistance, how is this device correcting that?

    So… I’m not calling outright bullshit yet, but I need a lot more info.

  17. DietRichColaon 04 Jun 2015 at 4:57 pm

    Holy crap, I wrote another long one.

    I will say, I started off writing with my mind set that this was a huge scam. However, as I wrote it, I did see some potential ways that what this person said *could* be true. It has to do with internal resistance. Also, with us being careful about how we are talking about voltage, because it is not just the voltage that matters but the available current based on the circuit load.

    I really don’t know enough about how fast internal resistance increases relative to the “reaction coordinate” or “extent of reaction” in an electrochemical cell. Those are both fancy chemistry terms for how complete the reaction is. But I guess it very well could be possible that by time the reaction is 20% done the internal resistance could have already increased enough to make the battery effectively useless. That would just mean that disposable batteries are way more inefficient than I thought.

  18. BillyJoe7on 04 Jun 2015 at 5:34 pm

    “and the magnetic fuel pipe clamps that were being vended as ‘scientifically proven’ to increase the miles-per-gallon of road vehicles”

    This was perpetrated by Australian racing car driver and fellow Victorian, Peter Brock. After being such a hero amongst the petrol heads, he lost a lot of credibility over that scam. He died a few years ago…in a car accident.

  19. Sylakon 04 Jun 2015 at 6:26 pm

    The Red flag was as soon as I read the “reason”, : “The reason for this high inefficiency is that electronic devices are finicky; they need a steady voltage of 1.5. When an alkaline battery drops below 80% capacity the voltage also drops to 1.4 or 1.3. At that point the device will read the battery as dead, leaving 80% of its juice unused.

    So why they work with NiMH batterie that output 1.2v then? I got some here, reading 1,29 Volt on my Amprobe.

    V = RI . This guy clearly is not aiming for people with electronic/electricity knowledge. ( I’M a TI guy, but the technical degree I’m currently doing in TI include lot of electronics class, plus the degree I failed in electronic 17 years ago. failed but he basic knowledge is still there).

    The source cannot create energy. The resistance ( or impedance since is it the resistance at the input of the whole device) would probably be constant. If a battery push to maintain voltage, at one point, there’s less electron physically in the source, so less currents, voltage will drop, whatever you do. Also those chemical have a maximun of “speed” they can release the current, so even if you try to push them, it won’t work, they will probably blow. That’s why there a lot of different type of batteries, Like Deep cycle marines ones, Car battery that can give Supoer high current for short period, some that are super stable etc.

    Also I thought it was ridiculous to think companies would sell stuff that is 80% inefficient. It’s not cost effective to nobody. Also devices have a range ( like you pointed out) of voltage they can work within. Like a mouse that work until 0.6V, it is because it work a 0.6V, it probably has a regulator inside, using only .6V, when the source go under that, than the regulator can’t do is job anymore. Man, I just had a big class ( I passed but it was hard) On Dc power supplies this last semester. I hope this guy don’t encounter on of my Alternative Current teacher of last year. I loved to talk sh** about scammers, like people selling speaker with “RMS watt” and stuff like that ( watt ARE rms always, because power in alternative current need to be calculated with the RMS voltage ( P= V*I), if you don’t you are not getting the real power drain. This batteriser guy would have suffer LOL

    On a side note, going back to Cegep ( it is between high school and universities here in Québec,) at 36 year old, is great you should try if you hate you job. LOl I recommenced electronic or Network/computers)

  20. flowinon 04 Jun 2015 at 7:30 pm

    The guys are Jewel thieves! its a scam.
    or maybe not:


  21. Oadzielon 04 Jun 2015 at 7:38 pm

    Well for one you wouldn’t be increasing the ‘Voltage’ as that is the wrong term to use in this context, The correct term being E.M.F.

    However it IS true that it is possible to increase the potential difference across the terminals of the battery, the easiest way to do this would be to increase the temperature the battery is operating at. Increasing the Temp will in turn increase the resistance which when applied to V=IR the potential difference therefore is increased. However just adding a sleeve to the battery will only yield a negligible effect.

    The other thing to note is that Kirchhoff’s Second Law states that the E.M.F equals the sum of the Potential Differences in a closed loop. Therefore if the P.D across the battery is increased the P.D and therefore energy available to the rest of the circuit is actually DECREASED.

    In conclusion, due to my above arguments if the batteriser works in the way I theories then using one would actually DECREASE the performance of the Battery.

  22. Willyon 04 Jun 2015 at 9:30 pm

    Gosh, it must be true! I see the drops in voltage are suggestive of incremental “quanta”. He’s not claiming that the drops are continuous; note they drop by tenths of a volt! QM=TRUE!

  23. ccbowerson 05 Jun 2015 at 12:25 am

    That comes across as an embarrassingly credulous article from an editor-in-chief of PC World. There are a few sentences of token skepticism, but even these are tempered. The overall effect is a lack of skepticism given the claims put forth.

  24. Sylakon 05 Jun 2015 at 12:55 am

    Any person who know a little bit of tech should spot that it is BS. pc-world… believe in this? wow.
    it has all the stuff of pseudo-science.
    Make over the top claims : Check
    Violate multiple laws of the universe : Check
    On indiegogo : Check
    sciency-sounding mechanism to explain how it work : Check

    we have a winner.

    It make sense that if a trick like that would have work, Big Battery would have use it. Duracell and Energizer are always trying to sell use new “quantum power ultra 3000” battery and stuff them with more power. I don’t know how much they spend in R&D, but they would have figured out a simple trick like that is worth 2.50$.

    If a short circuit happen inside the circuitry of the devices, would will also short circuit you battery straight like putting a wire between the pole.

    Indiegogo is really bad at vetting what get founded, P.M machines/free energy. ridiculous devices that can read a lot of biometric and make no sens etc.

  25. Boodaon 05 Jun 2015 at 2:37 am


    If the problem with battery life is internal resistance, then how can an external sleeve decrease resistance if the current has to pass through the internal resistance to get to it?

    I’m a layman when it comes to electricity and chemistry. I tried to understand the converters and boosters other people have mentioned, and all I could conclude from it was, “that seems terribly inefficient.”

  26. Pete Aon 05 Jun 2015 at 3:59 pm

    Distilled water is an electrolyte that has a very high internal resistance. By placing graphite electrodes in distilled water it is trivial to cyclicly charge and discharge this weak electrolyte. The potential difference of the electrodes during the discharge cycle V = E (the electrolytic electromotive force) – I (the load current) x Ri (the effective internal resistance of the source).

    Caveat: All electrolytes exhibit non-linear behaviour with changes in their electromotive force. If they didn’t then they wouldn’t be electrolytes; they would be better described as being non-ionic dielectrics — possible candidates for manufacturing capacitors, coaxial cables, etc. The very strong non-linearity of some electrolytes makes them ideally suited for cells/batteries to power electronics that has a narrow range of working voltage.

    Think of a 12 V lead-acid car battery: circa 16 V at the end of a full charge, which it will never receive from the car’s alternator because this would induce gassing of hydrogen and oxygen at a potentially dangerous level. The alternator voltage regulator limits the battery electromotive force to 14.4 V as a safe compromise. Once its electromotive force drops to circa 10 V its effective internal resistance has become too high to turn the starter motor. No amount of Batterwoo can overcome or delay the inevitable problem that we all refer to as “having a flat battery”.

    Many thanks to the commentators who pointed out the “Joule thief”: I was fully aware of the concept, but not its highly apt name 🙂

  27. DietRichColaon 05 Jun 2015 at 11:53 pm


    “If the problem with battery life is internal resistance, then how can an external sleeve decrease resistance if the current has to pass through the internal resistance to get to it?”

    I have no idea. This is where my knowledge is at an end. All I was trying to say is that some of what is described in terms of how much energy might be left in the battery could actually be plausible… if internal resistance increases that fast. How this little sleeve would correct that, I have no idea.

    “that seems terribly inefficient”

    I agree. But maybe it’s just an issue of function and necessity. We want batteries to power our devices… Maybe that is just the trade off we make.

    I still think the language of the article is suspect, and seems like a scam, but I need more info to make a decision

  28. DaveK23on 08 Jun 2015 at 6:33 pm

    Poohrarvar is telling the truth: both 20% and 500% *are* “up to 800%”. Of course, so is 0%.

Trackback URI | Comments RSS

Leave a Reply

You must be logged in to post a comment.