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# Tag Info

1

WD40 displaces moisture but allows contacts to penetrate the thin film for a reasonable resistance ~ 100 mOhms, if there is sufficient force. It will eventually dry out and leave a thin residue so it is not permanent. If you had liquid Tin, after cleaning off oxide, dip then rinse in water. That is better but you have to buy bulk (2L?) so this is not ...

0

The bigger the battery in a single cell, the lower it's internal resistance. Of course battery chemistry like electrolytic caps has a wide range in ESR (effective Series Resistance) So shorting its voltage across it's internal ESR will produce the short circuit current. This is dangerous on Lithium cells and lead acid cells due to the energy stored. Voc/...

2

Just so this actually has an answer: yes, it is because the battery has internal resistance and because your wire is not an ideal short circuit. Information about the capacity of a given battery can be found in the battery's data sheet, which should be available from the manufacturer. However, batteries are generally not designed to used with an extremely ...

2

Because in the real world there is nothing such as a perfect wire or a perfect battery. There are parasitic resistances, inductances, and capacitances everywhere. Plus there is the issue of the chemical reactions in the battery that is taking place to produce that electricity. That can only happen so fast even if you had a battery with zero parasitics. ...

0

You can use technical Vaseline (best solution), or - if you haven't - any grease or machine/engine oil. And don't use any vegetable or animal fat, or vegetable oil - these tends to hardening because of oxidation, or to decomposing, and can cause corrosion.

0

Any spray is likely to be an insulator, negating the function of a contact. Most solders, whether pure tin, older (non-RoHS) Pb/Sn, or newer Bi-RoHS types, are more corrosion-resistant than plain steel. Having had the same issue, I sand the actual contact area to remove any oxidation and cover just that area with solder. Making solder adhere to steel or ...

-1

You discharge too deep. Don’t go below 3.65v deeper discharge overloads the cell. You may also consider increasing the battery bank to reduce discharge rates. Then if you fast charge back a cell overloaded at discharge time, you heat it up further. Damaging it. You may need two banks and a switching circuitry, in order to prevent deep discharge and fast ...

4

Yes, current will flow into the battery, and in a sense you are trying to charge the battery. The possible danger is that the battery is not designed to be recharged, or that the value of $R$ is not high enough to limit the current to a safe level. In either case there is the possibility of damaging the battery, perhaps causing a fire.

1

Can I replace it with ANY 12 volt battery of comparable size? No. In the first instance, is it a rechargeable or non-rechargeable battery? -- you can't swap really swap them. Rechargeable batteries come in a number of different kinds which have different charging circuits and properties. You can't use a battery of one type where the charging circuit is ...

2

Yes, this is perfectly normal. A battery has an internal resistance. When you add a load on to it, it becomes a resistor divider. For example, assume you have a 12V battery and it read 12V exactly when measuring it. Imagine this battery has a 0.2 Ohm internal resistance. You then add a load of 10 Ohms. Use the voltage divider equation [Rload/(Rbat+Rload)]...

3

The further we discharge lithium batteries, the more damaged they get (a deeper discharge is equivalent to a higher cycling count). This is particularly true for a very deep discharge, where battery voltage decreases rapidly: Image source: batteryuniversity.com Keeping the user from discharging the battery to this extend increases battery life time ...

0

how about a standard boost converter, that uses the battery to absorb the energy of the inductor? and a low-power comparator to shut off the boost converter once charged to 14 volts you might call this a flyback converter, with some shutdown.

2

The current will be shared equally by the batteries only if they are exactly identical to each other. That is almost certainly not the case. In a real voltage source, like a battery, there is a small resistance in series with the voltage source. If the value of this resistance is different for the two batteries then the battery with the smaller resistance ...

0

I add 6pcs 2.7v 500f supercapacitors in series with protection board and that drain my car battery in 6 hours, so I decide to remove all the protection boards and not use it, now I have 7pcs 2.7v 500f supercapacitors in series and paralleled with my car battery, everything working perfectly, the capacitor not over charge since my alternator maximum charge is ...

2

Imagine that all the NPN transistors behave as a controlled switch. The switch contacts are between collector and emitter. When the switch closes, the LED turns on. That switch is controlled by the Base-Emitter voltage $V_{BE}$. $V_{BE}$ needs to be more than 0.7 V for the switch to close and the LED to turn on. That $V_{BE}$ input of the transistor ...

0

As long as your ebike controller doesn't check if the "correct" (original) battery is installed you shouldn't have an issue replacing the 10ah battery with the20ah battery. If however it does check you could theoretically swap the BMS from the 20ah to the 10ah battery but this will most likely not be needed. The difference in capacity shouldn't be an issue (...

-2

I have just put 5 volts right on the + and - of were the battery would go. They do have another pin I guess for sense. I did it on 2 old cellphones that the battery went bad on. But they only stay going for about a day. It thinks the battery is draining down very low so it turns off. I guess have to put something on the 3rd pin but I don't know what. I ...

2

Possible reasons: Many of those adapters are unregulated which means it might give 9 V when fully loaded but a higher voltage when lightly loaded. You used an AC adapter. The alternating current alternates the polarity and that or may have caused damage. You used one with polarity reversed. Check the label. Normally they are centre-pin positive. You ...

2

You will be connecting all devices in parallel, so the voltage requirement stays constant at 5V. To get the needed current just sum up all device input currents. If you have 4 devices with 2A each you need 8A. The power is V*I, in this case 40W. If you want to power the devices from battery for a longer time, you obviously also have to keep in mind the ...

0

Car batteries are starting batteries - their internal resistance is low since approx., 450 amps (for petrol engines) and 800 amps (diesel) and even more need to flow to start the engine. Internal resistance goes up (high) with age and use - eventually even if the battery goes up to a decent 12.5 volts, it may still not charge enough due to high resistance ...

3

I don't know who designed that board, but it has several questionable connections. First, CH340C chip datasheet requires V3 pin connected to VCC if VCC is 3.3V. On the board it is permanently connected to decoupling capacitor, which corresponds to 5V configuration. On the other hand, the LEDs are connected to 3.3V supply, so they will (maybe) operate in ...

5

The best report on this subject https://batteryuniversity.com/learn/article/how_to_prolong_lithium_based_batteries The lowest aging rate is known to be 0'C 66% SoC which is the approx. SoC charge level used by OEM's to ship/store LiPo batteries. Depth of discharge %DoD, greatly affects total lifetime Ah capacity. NMC LiPO4 100% DoD ~...

2

You'll find yourself in a situation where both are true. Point is: the aging processes in your Lithium battery are are dominated by usage (especially: decomposition of the electrolytic carriers, and forming of metal dendrites at high currents). Obviously, not using the battery will avoid these. Lithium Batteries have an excellent shelf life. Leaving them ...

0

Replace the batteries with fresh alkaline ones. What you describe is normal behavior for batteries that are either capable of supplying the current draw or are close to the end of their life. The "recovery" is due to polarization.

0

My thoughts were if sizing a AC sine wave inverter, go with larger than needed for the current solar panels installed but not push it to the limits, but still have the option under peak demand. But of course still wondering the efficiency of doing so or more importantly how to size the battery array feeding the inverter to provide optimum output but minimize ...

0

Cell means a single encased electrochemical unit (one positive and one negative electrode) which exhibits a voltage differential across its two terminals Battery means two or more cells or batteries which are electrically connected together and fitted with devices necessary for use, for example, case, terminals, marking and protective devices. Cell ...

0

In addition my concurrence with @hobbs I support your endeavor to reduce DoD to 50% but also prevent overvoltage. The stronger battery has lower ESR and will provide most of the surge currents and possibly the older battery will age slower as a result of tandem connections with a new battery. (unless it develops a cell short and drains the other battery) ...

1

Ah is Amps x hours, so in theory a 65 Ah battery can produce 65 A for 1 hour, or 32.5 A for 2 hours etc. If we naively also assume that the battery produces 12 V throughout the discharge then the 12 V 30 W device would draw 30/12 = 2.5 A, and the battery would last 65/2.5 = 26 hours. In practice a typical lead-acid battery's rated capacity is for 20 ...

3

Assuming that you are directly connecting the batteries in parallel, and the cables you use are adequate, the two batteries will always have pretty much the same voltage, and if they're the same chemistry they will discharge to the same level, and basically the whole combined capacity will be usable. If you take 50% of the combined capacity, each battery ...

0

It would be a bit hard to accurately predict that without knowing a lot about the battery for example: The depth of discharge allowed by battery chemistry Battery internal resistance and its true AH rating which may not he same as claimed by different suppliers. A good way would be to measure this could be by using simple energy meter that measure total ...

1

26 hours if everything behaves ideally as you described. 12V/30W is 2.5A, so 65AH/2.5A = 26 hours.

1

Details are missing so an accurate calculation cannot be made. A "12V" battery starts at about 13.8V then its voltage drops as it is used. You did not say what is the minimum voltage of the device that you are powering. You also did not say the minimum voltage of the battery when it has provided 65Ah.

0

It's called buck-boost battery charger ( if reversible ) Renesas has a portfolio with this family, ISL9241.

1

Typically these are not done in the typical series or parallel method that simple projects use. When you need a large number of leds and low power, you use one of two (or both) methods. The first is multiplexing/charlieplexing/scanning. This is also used for addressable leds or displays, to display a picture/video/individual pixel, not just backlights, but ...

2

If keyboard components aren't rated for LED light through them, don't buy them. They are likely to be opaque, which means you could put a searchlight down there and not get any light through. Anyway you would be unable to power it off USB, and you'll need to get under 2.5W for that, and will want to get under 500mw if possible. How to size Let's take ...

1

I realize this is an old question. I decided to answer it anyway. OP is probably gone but others may benefit. Fuse size cannot be discussed without discussing wire size. The purpose of the fuse is to prevent wiring from getting too hot. If the wire gets too hot it may cause burns to people who touch it or even cause the insulation to smoulder or burn. It ...

0

The most common symptoms of a "dead" battery, such as an 18650 cell are that the effective capacitance reduces from 10k Farads to < 1% so that it overcharges without a balancer and quickly undercharges and even reverse charges under load unless protected such as with a balancer which is now powered by the batteries. The other characteristic is the ESR ...

2

LEDs do have a maximum current rating, but it does not mean they must be run at full power all the time. For this application at least two things can be used. Buy better, more efficient LEDs so that they light up brightly enough at lower current, say 1 mA or less. You can also divide the LEDs into multiple groups and multiplex them so that only one group is ...

1

I'm looking at using white 3mm LEDs with a max current of 20 mA and a forward voltage of 3-3.4v. Don't run them at their max current. That's not good for reliability anyway. Run them at the minimum current needed for visibility under the conditions you want them visible. For many of today's LEDs that could be somewhere in the range of 1-5 mA. 77 LEDs at ...

0

Normally batteries are rated in terms of multiple of capacity discharge; a 5C 2.8Ah battery would be dischargeable at 14A. There are loads of 18650 cells that can achieve this. I've seen up to 30C advertised. You can buy 18650 cells for like 2$for a decent brand name one (Panasonic, Samsung etc), and you'll probably find the power to weight and power to ... 0 The Lipo batteries used in drones are usually rated for at least 10C, which means the maximium safe current is 10 times the Ah capacity. Your main pack has a capacity of 2Ah, so at 2A it is only being drained at 1C. Thus it would seem to be quite safe. However if current is being drawn in pulses the rms current will be higher than the measured average ... 1 Barring better information: If you know that someone is using safely using the battery packs for 7 minutes at 4.5 A, I think it's pretty safe to continuously draw 2 A – modelling the internal resistance as mostly ohmic, that's not even ¼ of the heat that gets generated at 4.5 A. But you ask to verify the safety of your system, and that's impossible without ... 0 Current (Amps) is not the total power storage of a battery. Cell phone batteries are all 3.7V so they are often compared by the mAh rating of the battery. In reality to calculate run time you need to use Watt Hour (Wh) which is the total amount of energy stored. To find total Energy storage of a battery (assuming battery capacity given in amp hours): $E[... 1 According to some youtube videos, once up to temperature the Fridge used 27 Wh and 2 Amps in 4 Hours, averaging 6.75 Wh and 0.5 Amps The 0.5 A and the 6.75 Wh seems not consistent. the voltage has to be$\frac{6.75 \text{Wh}}{ 0.5 \text{A}} = 13.5 \text{V} $. The voltage at the output of the powerbank is$ 12 \text{V} \$. consider only the Wh ...

1

If you place two batteries in series but keep the load the same, the battery life is halved? Battery life implies the number of battery cycles before the battery is no longer useful. "Run-time" would be a better description. Yes, the run-time would be halved. In practice it would be even worse because battery capacity decreases at higher discharge rates. ...

1

The measurement results you obtained are not enough to determine the internal resistance. simulate this circuit – Schematic created using CircuitLab You measured the voltage across BATT and Rinternal, and the current through them, (green coloured variables below), but it still gives you 1 equation with 2 (red coloured) unknown variabless (which has ...

5

Alkaline battery capacity depends too much on load, it's too complex for the average consumer to understand and the difference between brands is small. Primary (non-rechargeable) lithium AA cells have more stable capacity but they are oriented to the same market. Capacity for NIMH batteries actually is quite stable under load. There is a trade-off between ...

18

cIt's complex, and some of the answers are "soft" and some of you assumptions are (reasonably enough) inexact. Li-ion are measured in mWh LiIon cells are frequently measured in mAh capacity. LiIon batteries (1 or more cells) often have mWh and mAh markings. Neither is a certain measure of what a user will receive. Both are useful. (cf George Box's "All ...

7

Not all loads are constant power or constant resistance or constant current or constant in any other category as the voltage drops at 20% and the double layer effects in all batteries as well as the double C (Farads) and ESR (milliohms) products affect the high drain capacity reductions more. Squeezing out the last 10% SoC might result in 30% drop in ...

32

It's marketing, like PMPO watts. Big numbers sell! If the battery is 1.2V nominal, there won't be a big difference between W.h and A.h numbers. But if the battery is 3.6V nominal, the W.h number will be 3.6 times more bigger than the A.h number! So a 1200 mWh 3.6V battery sounds a lot better than a 330 mAh battery... even though it's the same battery of ...

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