# Tag Info

5

I asked myself much the same question a few years ago. My local very cheap supermarket had an offer on some very expensive name brand AA batteries 'U', so I bought a pack of those, and a pack of their own brand alkaline AAs 'W' at 13% of the name brand price. I also had some own brand cells from a middle-price supermarket 'A', and some rechargable NiMH cells ...

5

The batteries simply last longer, if they are not used up to the full capacity. Maybe the battery pack charge level is not allowed to go fully empty and they leave it 3% charged, and maybe the pack is also not fully charged but only up to 97%. The pack also has to provide a certain capacity over a temperature range, charge cycles, and battery age, so that is ...

4

No. None of the USB specifications involve directly charging the battery, rather they supply power to a battery charge supervising circuit inside the device. Damaging a battery by overcharging, overdischarging, charging after overdischarging, etc, is the fault of the charge supervisory circuit or a defect in the battery itself, not the external power supply. ...

4

a battery creates potential difference by accumulating more electrons on the negative plate. That's true in a sense, but for a battery how the electrons got there (pumped by electrochemical reactions) is probably more important. Therefore the more electrons that accumulate on the negative plate, the higher the battery voltage. Well, yes, but those ...

4

So would it be possible to connect an isolated 1:1 DC/DC converter to one of the batteries and connects its Vout+ and Vout- to the + and - of the other battery? Step in the buck regulator - feed it from your 14.8 volt (series battery) and it should very efficiently step down to provide you with a lower voltage supply. It'll be more efficient than an ...

4

A multimeter's resistance measurement function applies a fixed voltage across the DUT (device under test) and measures the current or applies a fixed current and measures the voltage across the DUT. The second method is simpler as the resultant voltage can be fed to the display's ADC (analogue to digital convertor) and the voltage reading obtained displayed ...

4

If you look up discussions about Porsche Tycan vs. Tesla people are discussing the fact that the Porsche only uses ~80% of the battery while the Tesla allows in the SW to set it to 100% (so I guess its not factory default) Using only a part of the capacity has two effects: more lifetime (wear of the electrodes) faster charging (the curve gets flatter in ...

3

Pretty close but what you describe much more accurately reflects a capacitor than a battery. In a capacitor, you add energy to pull electrons away from where they want to be thus storing potential energy, and that potential energy is released when the electron is released and allowed to pop back. You do move electrons in a similar way in a battery, but the ...

3

I had thought of a $\color{red}{\boxed{\text{buck converter}}}$ module, but in the market I cannot find one that supports 4.2-3.3V in input and output of 1.5V. They are fairly easy to find. For instance, this one springs to mind: - If you need a little more than 500 mA at 1.5 volts then a different device is required. Here's a selection page from ADI ...

3

The L7809 is operating very close to the edge: it has a 2V dropout voltage meaning it needs +11V input to remain stable. You're giving it an unregulated +12V through a diode so very little margin - any spikes on +12V (due to amp current spikes) may freak out the L7809. You can increase your margin by: -making cap at L7809 input larger -running Q1 collector ...

3

A current limiter simply won't work - you can't get energy from nowhere, and causing the supply to "brown out" during transmit is likely to at best yield poor on-air results; more likely cause a crash and hopefully restart of the entire system. To the best of my knowledge, all LoRa radio chips have programmable output power, so if you want to limit ...

3

As others have said, a current limiter is not what you need here. In fact, this is the very cause of your problem: a current limiter will decrease/block the voltage as it approaches the current limit, which is exactly what the battery is doing already. The Tadiran TL-5903 indicates it can supply up to 200mA pulses, but it will not maintain the rated voltage ...

2

Although this is an old question the answer may be useful to others. Summary: The PV panel suggested is of too low a voltage and power rating to be more than very marginally useful in this application. _______________________________ To charge a battery the applied voltage must be at least equal to the highest voltage the battery reaches. In this case either ...

2

The calcium is used as a hardener like antimony was used, a benefit of calcium over antimony is it creates less gassing and that's why batteries can be truly be sealed and maintenance free. The calcium does create higher internal resistance which has to be overcome with higher amperage. The voltage must remain consistently at 12.66 volts to protect vital ...

2

Can we eliminate CV mode and use it? Yes. If use like this Any battery degradation/life cycle/capacity issues? You will be at about 70% of maximum capacity, and have a huge increase in cycle life, see the third table at https://batteryuniversity.com/learn/article/how_to_prolong_lithium_based_batteries . Not discharging them below, say, 20% will also ...

2

The batteries have a higher specific heat capacity than the heatshrink and the heatshrink should shield the hot air from the PCB while it is not tight. So don't keep pointing the heatgun at it once the heatshrink has closed around the battery and don't point the heatgun at anything until it has warmed up. Heatshrink will also continue to shrink a bit after ...

2

As others have previously stated in the comments, there is no inherent issue with using a battery protection IC unless the overarching mechanical design of your enclosure is very poor. A more fragile lithium-ion battery with a ribbon cable will still break if somebody rips it out with extreme force. A battery will still catch fire if you puncture it. The ...

2

So would it be possible to connect an isolated 1:1 DC/DC converter to one of the batteries and connects its Vout+ and Vout- to the + and - of the other battery? Trying to do so with an off-the-shelf converter would just lead to trouble and frustration -- without a specially-built circuit either the top battery would just go along for the ride, or the bottom ...

2

If your PLC takes an AC input, give it an AC input. UPS are there to accomodate any fluctuations and surge. Cause when AC mains comes back? Bye bye PLC. I own a UPS business and I have seen plenty of people make a silly mistake like this and it will cost you dearly. The price of your UPS will be the least of your worries. My two cents worth.

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No it will not, you have to have circuit charges and monitors voltage, current and temperature within the batteries specification you also need to have at least one over current protection circuit at least, preferable with a fuse aswell.

2

By the way, why do batteries even puff up? Lithium-polymer batteries 'puff up' when some of the electrolyte decomposes into gas. This can be caused by over-voltage, over-discharge, high temperature (ambient or self-heating at high current), poorly formulated electrolyte or mechanical damage. Combination of high temperature and high voltage exacerbates ...

2

Yes, exceeding the spec is not good. It will probably work for a while. Or even forever. But a) no way to say for sure and b) if you build many circuits like that, you can never tell how many will fail. The dumbest and, yet in this case still acceptable option I see is 2 series diodes before 1117 that will drop 1.2-1.6V. You could get away with one, but I'm ...

2

If the barrel jack isn't connected to anything, it is just an open circuit - it will not drain the battery.

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The cutoff voltage is 4.4 V. It's a high voltage lithium polymer accumulator. You can hook it up to a standard charging circuit (that means 4.2 V cutoff I assume) but you won't get the claimed capacity out of it. But you will drastically increase the cycle count of the battery, so that's a plus.

2

You don't want to charge lead-acid from a constant voltage, and if you do, you need to change the voltage depending on the state of charge of the battery (switching from constant voltage "second stage" charging to trickle charge). Google "lead acid charging algorithm".

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Here they seem to be working on robots which digest insects: Domestic robots with a taste for flesh But the robots also have a taste for flesh. They can gain energy by chomping on flies and mice, an idea inspired by researchers at Bristol Robotics Lab, UK, who built a fly-powered robot and have also suggested that marine robots could feed on plankton. The ...

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This answer is to some extent a work in progress. Check back if you care. As a first step it outlines the requirement and overviews the extreme cycle life capabilities in selected situations. What is the longest lasting (number of years or cycles) commercially available battery technology? I asked this question because I felt that an answer would be useful ...

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When you fill the battery, it will start showing a voltage across its poles; this voltage will push a current into your PSU and power parts of it. Most PSUs don't like this very much. If you connect things wrong, a large current may flow into the PSU and fry it. Also, the PSU will generate a voltage of 12V, and will only be able to charge a (nominally) 12V ...

2

I know that there are modules like the TP4056 for Lipo batteries and 18650 cells, but I don't know that there is something similar for an AA battery. it's not about the form factor (AA), but about the chemistry (NiMH/NiCd?). Yes, charge ICs for such batteries are available. You'd want to check the offerings of the major power IC manufacturers (Maxim, Texas ...

2

The individual cell capacity of 3400 mAh is only a nominal value at the given discharge rate. The actual value of individual cells in the pack would vary within the tolerance limits. As the strength of a chain depends on its weakest link so also in a battery pack its total capacity will be less than that which is calculated on the basis of the nominal value ...

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