Get 5-volts from an exhausted alkaline cell

Every day we throw away usable power alkaline cells. Since the device in which we use them can not work with partially discharged cells, the circuit that I present can raise the voltage of one exhausted cell to 5V to use it in any other application, for example to turn on a led, feed a microcontroller or feed remote sensors that do not demand much energy.

Elements du projet

una-pila.jpg (JPG, 68.28 KB)

Discussion (9 commentaire(s))

petrus bitbyter il y a 5 ans

I have my doubts whether this circuit is working with an almost exhausted battery. Even without analysing it, I see a zener that requires 10mA to do what it has to do. At the 4.7V it means a 47mW, a lot of power from an almost exhausted battery. Did you really build this circuit? I like to see the results.

petrus bitbyter

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Roel Arits il y a 5 ans

The 1N4732 zener has a Izk (minimum zener knee current) of 1mA. That means it starts zenering at 1mA at a voltage "close" to it's 4V7 rating. With 10mA or more, the precise zener voltage of 4V7 is guaranteed.
So lets assume that the zener voltage is about 4V3 with a current of 1mA. Adding up the 0.7V base-emitter junction of the transistor results in an output voltage of 5V.
It is not a very precise voltage regulation, but it is good enough for most applications that require 5V+/-5%.
I've build a similar circuit like this before, and got an efficiency of around 85% out of it.

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M.P.W. van de Kraats il y a 5 ans

One nasty habit of batteries is that they tend to start leaking when totally exhausted.
The chemicals those things leak ususally ruin the device they are in.

Therefore, when one does try to totally exhaust batteries till the last drop, one should add some kind of leak detection as well, to prevent damage to the battery holder, and possibly other...

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Johannes van Kampen il y a 5 ans

if this works correctly, it can also be used with single solar cells or very small ones.

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Marcel Hariga il y a 6 ans

Total uselessness

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Wer Weis il y a 5 ans

in deutsch...
Jeden Tag werfen wir nutzbare Alkaline-Zellen weg. Da das Gerät, in dem wir sie verwenden, nicht mit teilweise entladenen Zellen arbeiten kann, kann der Schaltkreis, den ich vorstelle, die Spannung einer erschöpften Zelle auf 5 V erhöhen, um sie in jeder anderen Anwendung zu verwenden, zum Beispiel um eine LED einzuschalten, einen Mikrocontroller zu speisen oder speisen Sie entfernte Sensoren ein, die nicht viel Energie benötigen.

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David Ashton il y a 6 ans

@MPW Van de Kraats.... good comment. And any batteries which are used in old-fashioned incandescent bulb torches and things with motors in them will have a fair bit of "go" left in them when they are useless in their original appliance. To get them powering a garden LED, or a small breadboard, is a great way to get the most out of them.

I must admit though, anything I can I convert to rechargeables.

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M.P.W. van de Kraats il y a 6 ans

Even if you don"t have any use fo it, it sharpens your electronics skills by going through the design and understand every detail of it.
I would say: ain't that in the line of the Elektor projects ?
They usually ain't to usefull, if not to learn from them, get ideas by understanding new and/or different concepts.

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Isaac Rose il y a 6 ans

If this question has been asked please forgive me, I did look through the most of the comments and did not see this perspective.

There is a fair amount of discussion on using a 1.5v cell. But I have to wonder whether this may be a good substitute for migrating from a 3-4 cell 1.5v battery pack to using a 3.6v LiPo pack. It would be better to redesign the application circuitry to run straight from the 3.6v supply. However that may not be an option due to component selection, design resources, or due to exiting supply chain.

When I have a moment I intend to run this scenario through the simulator in Altium. In the mean time I thought I would mention it here to bring another perspective.

@David Ashton, you make a good point, in that there are COTS IC options, I know maxim semi has some ic’s which can performs the same application. The challenge becomes the return on investment for the ic. Passive components have become so cheap these days, that the cost of the ic can be 2-4 times the amount of the passive in production quantities. For a hobbyist it may be a wash, or the ic may be preferred if you don’t have an adequate stock of passives.

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David Ashton il y a 6 ans

Isaac yes, if you want to run a professional piece of equipment, you would use a Li-battery or something else. But this technique could have many applications - for a hobbyist powering a small breadboard, or maybe even in a wireless IOT sensor somewhere to let it get the most out of an AA or AAA battery. I have an electric roller blind that uses D batteries and i have plenty of them lying around that still have 1.3 volts in them, but are almost uselss for the blind. This circuit could let me use them.

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Malcolm Watts il y a 6 ans

I concur with comments about leaking batteries. In recent times currently manufactured alkalines appear to be more susceptible to leaking than they were in the past.
Commenting on the circuit (which I have not analysed too closely) I would suggest that ideal active devices that should be considered for this idea range from specifically designed I.C.s for energy scavenging (high-efficiency boost converters for using wireless power which is otherwise unused) to the Zetex range of low Vce(sat) transistors which give superior efficiency to bog-standard types. I have used these in a range of projects and can attest to their excellent performance.

Malcolm Watts

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KENT SWAN il y a 6 ans

Converting 1.5V to 5V from a boost circuit, regardless of power source and regardless of its type is governed by the standard rules of conservation energy where Power out = power in * conversion efficiency.

Pin * efficiency = Pout
Power = I *E
(Ein * Iin) * efficiency) = (Eout * Iout)
Iin = (Eout*Iout)/(efficiency * Ein)

Assuming you wany 50ma at 5V and can convert at 90% efficiency then

Iin = (5V *0.050ma)/( .9*1.5V)
Iin = 185ma where 18.5ma is lost by the conversion circuitry

The current required to maintain the output voltage increases as the source voltage drops

Iin = 277ma (battery 1.0V)

An AA high capacity primary (non rechargable) battery typically has a capacity of lets say 2500 mah. At end of life it will have used approximately 90% of its charge and the voltage will have dropped lets say to 1.0V This means that there's approximately 250mah remaining. The low voltage switching regulator will have a minimum voltage input typically around .8 volts. We can then only extract current from 1.0V to 0.8 volts which because of internal resistance in the battery means that such an extractor is conceptually valid for normal lifetime use but is in fact a wortheless use of silicon to recover useful energy from used batteries as the runningtime will be measured in a handfull of minutes

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Johannes van Kampen il y a 6 ans

You know the solar garden lights? They do the same thing using 1 NiCad cell to feed a 5V white Led. I found many different IC's inside from Chinese makes in a transistor hous with usually 4 pins. Have fun experimenting with them!

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KENT SWAN il y a 6 ans

Side note. The above discussion does not account for any issues which may be caused by an imbalance in the stack caused by different end state capacities and voltages. For non rechargable cells this is not likely to cause any significant problems but that is another discussion.

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KENT SWAN il y a 6 ans

An out of the box refinement which takes note of the law of conservation of energy would be to stack a number of 'used' batteries to achieve a net voltage greater than or equal to 5V (most likely 5 or 6 cells) lets say 6 cells. Instead of a boost regulator use a highly efficient buck boost device with a very low voltage cutoff. 6 new cells = 9V where as 6 used cells would have a net voltage of 5 to 6 volts.

Lets say that the cells are 1 volt. The rules of battery capacity say that if you stack a number of cells which have a specific capacity, then the stack will still have the same capacity but the voltage will be the sum of the voltages of each cell.

Going back to the 'debunk' response, with six cells we would have a battery with 250mah at 6V. With the input and output voltages being roughly the same then the 250mah would become availible. This means that, at 50ma draw you would get roughly 4 hours from the reclaimed stack. If you started with 6 new cells the buck-boost would be in buck mode for most of the stack's life but as it switched to boost mode, the difference in input to output would allow you to get the most out of the batteries.

The net of this discussion is that it's best to match the battery voltage with the use voltage and that buck is preferred (eg input voltage > output voltage).

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Peter Williams il y a 6 ans

Duracell alkali batteries probably best to avoid leakage

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M.P.W. van de Kraats il y a 6 ans

Yes, I strongly agree there,
However, mentioned circuitry is most likely to please the cheapgoat, that tries to save at each and every corner.
Not quite the type of public that is likely to buy Duracell.....

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M.P.W. van de Kraats il y a 6 ans

The concept sounds like a nice idea to me, especially the combination with the generated 5V output level.
However, what jumps to mind when reading this, is the image of batteries that start leaking because of being left in a device for too long.
When using such a circuitry, we should be carefull not to squeeze out the chemicals together with the last drop of energy......
Need for an additional leackage sensor ??

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petrus bitbyter il y a 9 ans

I did not analyse your circuit in detail but a quick scan make me doubt it will work. An almost empty 1.5V battery will not provide much energy anymore and I expect that all it can provide will be consumed by the circuit itself, leaving nothing for the load. Nevertheless, if you want to be sure why not building the circuit. Cannot be that difficult, can it?

Your idea is not brand new. If you like to read more about this idea, google for 'joule thief'.

BTW were did you find this circuit?

petrus bitbyter

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purple-bobby il y a 6 ans

How does Q2 switch off and charge C2 through D1?

Assume cell is giving 0.8V at 20uA or 600uA or 20mA. V_BE is 0.65V and transistor gain is 30. V_CE(sat) is 0.25V

At switch on there is 0.15V = (0.8V - 0.65V) across R4 and the base current of Q1 -22uA; its collector current is -660uA.
Q2 base current is 660uA and its collector current up to 20mA, however the conduction path is initially through C1 and Q1, providing positive feedback. C1 will charge up to 0.15V which is 22.5pC at 22.5mA it lasts 1ns.
There is a brief spike in current through Q2 before returning to ~20mA.
The inductor current will increase while there is a voltage across it. The voltage is 0.55V = (0.8V - 0.25V), it will reach 20mA in 9ms.

I cannot see a mechanism to switch Q2 off and force the 20mA inductor current through D1 and charge up C2. If Q2 does switch off the rise in voltage accros Q2 will discharge C1 and switch Q1 then Q2 off. However to me, it seems L1 takes too long to get going. In the 1ns while C1 charges, the current in L1 reaches 1.6uA, the gain of Q1 times Q2 would be less than 0.1.

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David Ashton il y a 6 ans

Not to pour cold water on your circuit but I recently saw a "Joule Thief" IC that only needs one inductor to function, here is a link to the datasheet:

https://www.ledsales.com.au/pdf/zxsc380.pdf

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Florian Menke il y a 6 ans

Petrus: I think this circuit is more a "Boost-Konverter", than a "Joule-Thief". Florian Menke

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petrus bitbyter il y a 9 ans

I agree about simulation. It has its use but never can replace the real thing. Finding a 330uH coil, I buld the circuit using BC550 and BC560 transistors and a 4.7V/250mW zener. Connecting warn out batteries I got some voltage but not as high as 5V. Only somwhere between 2.5V and 3.5V depending on the batteriy. Using a new 1.5 battery did not work at all. Q2 was conducting and was never switched off. Apparently real electronics also require some startup conditions. Adding an extra 150pF capacitor did the trick. Using a new battery I get 5V without a load which drops down to 4.2V with a 1k resistor on the output. I ordered some 2N3904, some 2N3906 and some 4.7V/1.3W zeners. I'll go on when they arrive. petrus bitbyter

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genuensis il y a 9 ans

hello, I built the circuit on a breadboard and I can confirm that it works fine, I tried to change the inductor and, provided it is not big, nothing changes (but 1 mH is already too much), I also tried to replace transistors and again it is always ok. It looks very reliable. Efficiency is around 60% for my try, I worry what will happen with old germanium transistors (considering such a low voltage). If you try to simulate it, put a current generator giving a short negative pulse on q2 base to start it.

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Juan.Canton il y a 9 ans

Petrus; This type of circuit simulation has tricks, due to start its operation by the inestability of the components, and this is not easy to simulate, since we would have to establish certain initial conditions that some simulators do not include. As for the inductance you can use either between 100uH and 470uH, it is not a critical component. For testing purposes you can feed the circuit with more voltage without exceeding the output voltage because this, through the network that regulates the output voltage, will stop the oscillation, I hope I helped. Juan Canton

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petrus bitbyter il y a 9 ans

Reading back my comment I realised it looked like I accused you to have copied i.e. stolen the schematic from somebody else, which was not my intention. I was just interested in some more background then you provided. You did not tell how the circuit is supposed to work and I did not make up from the text that you build the circuit. Other important details are the properties of the coil you used. So if you designed the circuit from sctratch, just say so. If you based it on some existing design, just say so. There's nothing wrong with that. (If you don't want to tell a thing, so be it.) Meanwhile I simulated the circuit using LTspice but no joy. I'll build the circuit once I find a suitable 330uH coil. Other components can be found easily but I currently have no coils in that range available. I'll report here the results (or lack of it). petrus bitbyter

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Juan.Canton il y a 9 ans

Petrus: Obviously I already built and use this circuit, and you can get 5mA @ 5V with a power of. 8V. There may be other similar I am not enlightened to make this idea happen only to me, just want to share my experiences. I recommend you build it so that you can confirm its performance. If there are any exactly equal, I invite you to give us the exact reference to show that it is copied.

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