17
\$\begingroup\$

The Batteriser [Edit: removed dead, harmful link] is a crowd-funded product intended to extend battery life by boosting the voltage. It is basically a joule thief that is in a tiny package that slips over the cell.

Dave Jones of EEVBlog did a video debunking the product:

Which the Batteriser people responded to with a video of their own:

And a response from Dave:

The latter two videos primarily deal with the Batteriser promo team's failure to understand how to measure voltage provided by batteries under load versus out of circuit. They believe that a power supply is an "unfair" test because it behaves different from batteries, or that skeptics failed to consider battery internal resistance, etc.

While I think it's obvious that the Batteriser people have failed to grasp some basic concepts, I do question whether a joule thief type circuit is a good way to utilize remaining energy in a cell. (Certainly not the 80% that Batteriser claims we throw away.)

Is there any benefit to using a voltage booster on batteries that are below a device's cutoff/operating voltage?

\$\endgroup\$
  • 1
    \$\begingroup\$ Dave shows that there is so little energy left that it's pointless in this video. \$\endgroup\$ – Matt Young Aug 18 '15 at 17:44
  • \$\begingroup\$ Please see my answer below which also contains information from Dave who referenced a study of >600 discarded batteries, of which >200 were tested showing that more than 33% of the energy remains unused in the average discarded battery. This study and figure were referenced by Dave in an article titled "The Batteriser Explained." \$\endgroup\$ – MicroservicesOnDDD Sep 26 '18 at 1:23
19
\$\begingroup\$

"Is there any benefit to using a voltage booster on batteries that are below a device's cutoff/operating voltage?"

Of course there are benefits in that situation: a battery that would otherwise be dead can still be used for some time. But probably not for long, so it is debatable whether this is usefull.

What DJ (IMO correctly) argues is that the Batteroo claims are wildly exaggerated at best, and the use of their device with batteries that are not yet below the cutoff voltage will lead to some additional energy use, so the overall effect might be negative.

\$\endgroup\$
  • 9
    \$\begingroup\$ ...the use of their device with batteries that are not yet below the cutoff voltage will lead to some additional energy use, so the overall effect might be negative. I don't think this point has been made enough. Considering most devices these days have some kind of DC-DC converter in them, the inefficiencies are compounded, and will result in a loss of battery life. It's unfortunate so many people are buying into this nonsense. \$\endgroup\$ – Matt Young Aug 18 '15 at 17:47
  • \$\begingroup\$ @MattYoung: Putting a switching supply between a battery and a linear supply can greatly improve the efficiency of the linear supply if the switching supply's output voltage is adjusted to match the linear supply's minimum input voltage. I would not be surprised if there are some devices where useful battery life could be improved 50% or more. Unless one knows which devices would benefit from such devices, however, throwing them around randomly isn't apt to offer much benefit. \$\endgroup\$ – supercat Oct 13 '15 at 21:15
  • \$\begingroup\$ I am not aware of many battery-powered devices that use a linear power regulator. \$\endgroup\$ – Wouter van Ooijen Oct 14 '15 at 5:34
8
\$\begingroup\$

Our goal is to keep the load on the batteries running as long as possible. In general, these loads are either fixed resistance (like a basic flashlight) or fixed power (like almost anything electronic beyond a certain complexity). A fixed power load is generally a switching regulator, which has a minimum dropout voltage.

A fixed resistance load doesn't much care what the input voltage is; the power out of the batteries will drop with the square of the voltage. Your bulb gets dimmer as the batteries die, but the dim bulb consumes less energy. You get a little time running bright, and a long time running dim. By putting a boost converter on the batteries into a resistive load, you're effectively turning the lamp into a fixed power load. Now, the lamp runs bright until the dropout voltage is reached, at which point the lamp stops entirely.

If the load was already fixed power, adding another regulator in front of it doesn't change that. The only possible effect you can have is to change the dropout voltage. If you've made the dropout voltage higher than it already was, you've made the device run for a shorter time! If you've made the dropout voltage lower, then you should be able to run the same device until a lower voltage point on the batteries.

However, the total energy you get out of a battery by putting a fixed power load on it is very complex; at lower voltages, you necessarily draw more current, to make up the fixed power (P=VI). The more current you draw, the more the terminal voltage drops due to internal series resistance, the faster the battery dies, and the less total energy you get out of it. So you could only ever be increasing the total energy draw from the batteries by a very small amount, and that amount is almost certain to be consumed by the reduced efficiency from adding another switching regulator to the system.

I'm not seeing a good argument for this. You'd be better off with rechargeable batteries.

\$\endgroup\$
  • \$\begingroup\$ If a device has a linear supply whose minimum input voltage is excessively high (e.g. a four-cell device that requires 5.2 volts for operation) putting a booster in front can prevent batteries from being incapable of operating it while 80% of their energy remains. If the linear supply uses an excessively large battery stack for its voltage requirement (e.g. using six cells when 5.2 volts would suffice) a buck-mode switcher could substantially reduce current draw. Those scenarios aren't terribly typical, but adding a switcher for devices where they do can greatly improve service life. \$\endgroup\$ – supercat Oct 13 '15 at 21:20
4
\$\begingroup\$

If one has a device which will draw 20mA continuously at any voltage above the minimum required for operation, and will work equally well at any such voltage, a buck-boost switcher that scan scale a battery's voltage up or down so the device always sees that minimum voltage may both reduce the amount of current drawn from batteries which output more voltage than the device would need, and allow continued operation with batteries which produce less voltage. A win-win.

A buck-boost switcher which raises the voltage significantly above what the device would need for operation will waste energy whenever the battery voltage is between what the device needs and what the booster gives the device.

If the useful performance of the device varies with voltage, scaling the battery voltage up may offer enhanced performance at the cost of reduced battery life; scaling it down may offer better battery life in exchange for reduced performance.

If the device draws power intermittently, and the amount of time it requires power will vary with voltage (e.g. it's a motor which periodically needs move something a certain distance) the amount by which scaling the voltage increases or decreases the current drawn from the battery may be larger or smaller than the amount by which it affects the duration.

If the device has a switching supply built into it, adding a second one in front of it may offer little benefit.

In short, there will be some cases where adding a switching supply may greatly improve battery service life; there will be others where it is useless or counterproductive.

\$\endgroup\$
1
\$\begingroup\$

When you test a battery, you have to put a load on it, otherwise the voltage floats up much higher than it should, considering its remaining life.

In the high-demand application, the internal resistance of the battery becomes much more of a factor in what voltage the battery can deliver, leading the battery to reach its cut-off voltage much too early.

Let's use a camera flash as an example, for that is particularly high-demand application.

Especially if you're using your camera in sub-zero temperatures, where the internal resistance increases and the battery chemical reaction proceeds at a slower pace, you'll use up batteries incredibly fast. And those used-up batteries will be considered by the camera to be "dead", for its application, in that cold setting.

But take those "camera-dead" batteries back inside, and let them warm up, and they will indeed still have much of their life still remaining, and will even present a decent voltage as well, even under test load.

There are lots of high-demand applications. Toys or anything motorized, and also poorly designed products, which I see all the time, poorly designed in a variety of ways. But even in the standard scenario, almost everything cuts off at or above 0.8 volts, leaving energy down to 0.5 volts left to be used for the low-power-demand application and some kind of boost converter.

To sum up, the key to understanding this issue is realizing that a cell considered "dead" for the high-demand application will not be considered dead for the low-demand application, but that energy may be inaccessible without some kind of boost converter.

Also key is understanding that low-demand applications may cut-off due to voltage when there is really plenty of energy left in the batteries, which is where the voltage-booster, and I believe the Batteriser product too, if it is quality, will definitely prove useful. So, then, low-power-demand products that cut-off on a low-voltage-basis because they DO NOT have a boost, will DEFINITELY benefit from the boost.

A simple cheap LED flashlight is a good example of both a low-demand application, and a device which cut's off based on voltage, because the cheap LED flashlight uses a resistor and the forward-voltage drop of the LED to decide the cut-off.

So, for a typical 3-cell flashlight, 3x1.5=4.5 volts new. The LED drops about 3 volts. So the natural voltage cutoff for a cheap LED flashlight is actually fairly high, at 3-volts/3-cells = 1 volt per cell.

But lighting those LED's is actually a fairly low-demand application. There is definitely plenty of energy left in those cells.

So, this is the perfect example of when it WOULD be beneficial to use a boost circuit to get the remaining energy out of these cells which have only been used down to 1 volt per cell.

I watched the treatment that Dave of EEVblog gave to Batteriser, and I think that he has perhaps overemphasized where Batteriser was wrong, but may not have sufficiently thought out the above things that I relayed, as I have studied the Joule Thief extensively, and I don't think that Dave has done this. I do understand the points Dave raised, and some may still be valid concerns, but I use my Joule Thief circuits all the time, and I can attest that they are definitely beneficial, just as would be any decent boost alternative.

Finally, in an emergency, the boost products, whether Joule Thief or Batteriser, or another product, would come in handy and even may become critical in a Hurricane Florence or other disaster scenario. Sometimes just having a working flashlight is essential, and if having a Batteriser or two lying around allows me to do that, then on that additional count as well, I call Batteriser and the Joule Thief, beneficial.

===========

Edit #1

To answer an asked question, I have absolutely no affiliation with Batteriser, Batteroo Boost, or the company Batteroo, or anyone in it -- just great fondness for the Joule Thief and trying to provide them to the third world, where they can't afford electricity, or batteries, and I don't want the overinflated claims of Batteroo to torpedo the Joule Thief.

To back up what I said, I am going to appeal to Dave of EEVblog and a research study paper he referenced directly.

In his EEVblog post "The Batteriser Explained" (a fairly thorough treatment of the subject in my opinion, and worth reading), Dave states:

HERE is some great research on used batteries. About 33% is wasted based on their data.

I appreciate Dave saying this, because it states that there really is energy left to use in the average discarded battery. He also states the following, which to me, means that the Batteroo product is still useful (just not as useful as they exaggerated):

I’m genuinely baffled as to why Batteroo would need to resort to claims like 8 times life. This thing would still sell like hot cakes if they claimed realistic practical figures. 50% increase in your battery life? – great, countless people would still buy it at the super low price point it’s at...

This study Dave references helps very much to answer this particular stack exchange question, so, to show some of their due diligence, here's the test flow chart:

Research study flow chart showing study test methodology

And here is a scatter plot and curve fit that shows the individual data points and that there is a nice correlation:

Research study scatter-plot with 4th order Polynomial curve fit

This chart shows how much actual capacity was left for many actual discarded batteries.

For their tests, they collected discarded batteries from 19 recycling boxes, then separated the batteries into 5 voltage classes spanning 0.1 volt from 1.1 volt to 1.5 volts. Batteries were randomly selected and discharged using a 120mA constant current load down to 0.9 volts. In the 636 battery study, 265 were discharged down to 0.9v to determine remaining life (mAh). According to their test results for discarded batteries:

  • About 10% can be considered new (see 1.58v data point, fig 4 above)
  • About 30% have more than 50% of their energy left
  • About 40% are fully discharged (defined in study as less than 1 volt)

And lest you think that 1 volt is completely discharged because of their study, they also say:

...all batteries with an initial voltage of less than 1.0V are registered as 0V and assumed fully discharged. Of course this is for most of them not true, they still contain a small remaining capacity which could be used to power low power devices (e.g. clock, or small radio). This was not considered important in our work.

They then go into reasons why people throw away batteries with so much (>=30%) energy left:

  • High Power devices (early cut-off)
  • Ensure Batteries are Good (replace for each use)
  • No (or bad) Battery Tester (unknown state-of-charge)

My most common personal reason is "Ensure Batteries are Good". I have an audio recorder, and don't use it that often, but when I do, I want to make sure that it doesn't fail in the middle of something important (a child's recital). So, my default action is to just put new batteries in.

The bottom line that I want to relay is, don't let Batteroo's inflated claims ruin the truth -- that there really is energy left in discarded batteries. Just watch out for leakage, because the lower the discharge, the higher the pressure.

There definitely is benefit to using a voltage booster (such as Batteroo Boost or a Joule Thief) on “dead” batteries.

\$\endgroup\$
  • \$\begingroup\$ What is your affiliation with Batterizer? \$\endgroup\$ – winny Sep 22 '18 at 8:59
  • \$\begingroup\$ @winny I have absolutely no affiliation with Batteroo, Batteriser, Batteroo Boost, or any related product, company, or employee. I have absolutely no interest in the company or any of their products. I merely am sticking up for the Joule Thief, which would also be "without benefit" if the Batteroo product were found to be "without benefit", since they have the same basic use cases. Please see my edited answer which now supports my position with hard data, and answers from Dave of EEVblog, who, it seems, agrees with me. \$\endgroup\$ – MicroservicesOnDDD Sep 25 '18 at 2:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.