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The LEDs are probably in parallel. One AA-battery has about 2000 mAh capacity, so if it lasts about 10 hours the current draw is about 200 mA. If you use a 5V adapter, you need to drop about 2V. 2V/200mA = 10 ohms. Since these are very rough values from rough data, you need to experiment. Buy ten 10 ohm 1W resistors. With these you can get a lot of different ...

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An LM317 linear regulator would be a good solution, since it is cheap and commonly available. It's been around for years. You can see an example circuit below. Instead of the transformer, T1, and diode bridge, D1-D4, you could just use any AC-DC power supply that generates an output voltage of around 3-9V. A 5V supply would be ideal. If the load current is ...

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Happens quite often these days that a product is functional, except one little bummer of a "feature" that makes it unusable by design, like LEDs that dim via a 200Hz PWM. How did that pass quality control? No idea. The usual causes for this kind of sound are: magnetoscriction, aka "coil whine" which occurs as variable magnetic fields ...

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Put both battery boxes in the middle and supply 5V from a phone charger.

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It rather depends what you mean by "connect them in series". A string of multiple LEDs designed to run on 4.5 to 5V will have all the LEDs in parallel, with current limiting resistors as required. That resistance may well be in the control box - white LEDs are intended to run on about 3V. If you connect two LED strings end-to-end, then you will ...

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You can put all the batteries in series and the supercapacitor parallel to them. That way you will get more current and the supercapacitor will always be charging.

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Voltage is a measure of how much energy a unit charge would gain/loose if it moved through the electric field between two points. Remember that 1V = 1 Joule per Coulomb. But its just a measure of how much energy would be lost or gained. That potential exists weather or not you actually move any charges. So voltage can certainly exist without current. ...

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This requires an update in 2020: For most modern batteries, 2.5V is the discharge limit. Older batteries were usually rated at 2.75V or 3.0V, but as I've said, that's not the case in 2020. However, to be completely sure, you do need to consult the battery's manual, as the parameters vary wildly. For example, a typical Sanyo cell will have safe discharge ...

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Parallel connection and increasing total Ah (batteries are from different manufacturers). That's a really bad thing to do. You should never connect batteries / battery packs in parallel unless you REALLY know what you're doing and then the batteries should be of the same model and preferably of similar age and wear level. If you ignore this advice and do it ...

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Search for CR2032 dummy battery or CR2032 battery eliminator. Some have glued a couple cheap-ish custom ENIG PCBs together to make up the 3.2mm thickness.

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“Ground” is always defined in electronics as zero voltage, just like earth-ground. for floating supplies, it is arbitrary yet the convention is still V- rather than V+. For bipolar supplies 0V , it is normally in between V+/V- with 3 terminals.

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Unless you have a reason to do otherwise, it's a modern convention to use the negative lead of the battery pack or power supply as your ground node. Unless you rely on actual physical properties of ground, eg, for radio, or to bond to a conductive housing, in a way it does not even matter. Most uses chose the negative terminal, but a few legacy vehicles have ...

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A few clues: If $P = \frac {V^2} R$ then which way will P go with increasing R? Given that, then what is the worst case for R? What is the worst case for V (the maximum voltage that you can apply to the resistor)? You now have both V and R so you can calculate P.

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If a circuit has a 15 V battery, what's the highest voltage you could put across your resistor?1 How much power will a 1k resistor dissipate at that voltage? 1 Actually a bit of a trick question.

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Although you have resolved the problem, but since the actual cause is still unknown, I will suggest the following; Assuming a defective FET is ruled out, it is very possible that the problem was coming from the LATCH circuit. Considering that your pull-up resistor for Q4 is 100K, any leakage current in the LATCH circuit will result in a Vgs that can keep the ...

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I think the simplest answer is to use an 'off the shelf' MR16 LED lampbulb such as any of these .... https://www.ebay.co.uk/sch/i.html?_from=R40&_trksid=p2334524.m570.l1313&_nkw=mr+16+12v+led+bulbs+-halogen&_sacat=0&LH_TitleDesc=0&_osacat=0&_odkw=mr+16+12v+led+bulbs And step up the battery voltage a bit using something like this ... ...

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Most EVs including bikes already recover 'spare energy' from motion through 'regenerative braking'. The kinetic energy stored is turned into electricity by the motor acting as a motor-generator when braking. No need for anything else at all. There's a video from Bjorn Nyland on youtube showing how he recovered much of the energy used driving his MG ZS EV up ...

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Perpetual motion machines are impossible no matter how clever you think you are and how complex you can build them. In your case, the dynamo will impose an additional load to the motorcycle motor and additional consumption from the battery. And, because you lose some energy as heat at any conversion stage, you will in fact get LESS mileage. Sorry, that's how ...

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Charging a lead-acid batteries is rather complex matter. First, the safety: Even a discharged car battery can output enough current to melt your wires and/or start a fire and/or burn your hands, if you make a small mistake and short the battery. An over-charged, over-discharged or simply damaged battery may produce hydrogen. It is explosive when reaches ...

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"Flat" discharge curves are, first of all, a matter of scale and discharge conditions. During discharge, two things happen: The open-circuit voltage decreases. (Valid for Li-ion, Pb-acid, NiCd, NiMH and most other chemistries, but not all - there are some batteries that don't do that, at least not in their useful discharge range.) The internal ...

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You appear to be asking for a (rechargeable) battery power pack for mobile use. Note that your mixer is highly unlikely to draw 1A from it's power inlet. I'd measure the current draw. It'll most likely be less than 100mA. The 1A A.C. supply is simply used for convenience. There's little demand for 1 watt AC adaptors. Similarly, it's unlikely that exactly 9 ...

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What you are currently doing exactly opposite of what the article you linked to suggests charging lead-acid batteries. The article says to use CCCV method which means constant current constant voltage. It means that the charger first limits the charge current to constant predefined level and the charging voltage is at the battery voltage level, until the ...

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I have a 9,6v 2000mAh Ni-mh battery ... I need 30 to 60 minutes of battery so if I remember my lessons from high school I have approximately 18 W of power available. Not quite. You have 18 Wh of energy available. (Energy = Power × Time.) So, for a discharge to 50% of capacity (safety margin) you have 9 Wh available. To use that in one hour means your lamp ...

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You should get a 12 volt battery for this. There are many sizes of 12 volt batteries. As Chris pointed out in his comment: A resistor is not suitable for this. Even a linear voltage regulator would quickly result in dead batteries.

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This would be the battery protection circuit of the Li-Po battery. It prevents over-current over-voltage and under-voltage as a minimum in most Li-Po battery systems. This circuit and its parameters are set by the manufacturer, for the particular Li-Po it is meant to protect. Generally it is not a good idea to tinker around with this as it could render the ...

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Why could this be happening? Is my TP-4056 module bad? My TP4056 was also doing same. I used a new battery and tried changing switches. Lastly I found the problem was with the TP4056 itself. I exchanged it for a new one and voila. No more power drain. As I had the same problem as you and as the replacement 4056 and almost all others do not have this problem ...

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If your motors are 24v, and you are placing the pairs of motors in series, doesn't your supply need to be 48v? Your existing circuit provides approx. 12v to a paralleled pair of 24v motors.

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9V batteries have a very small capacity. Most of them are 200mAh, some more expensive ones are 650mAh. It would be better to use a 12V, 18V, 24V charger. If you really want it to be remote, you will need a laptop battery at 19V, 3000 mAh or a car battery at 55 000 mAh. There are smaller car batteries, which are called motorcycle and electrical bike batteries....

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Please read the datasheet correctly, there is 7mv of hysteresis for Standby comparator. If the battery voltage slightly goes up after shutdown and the STBY pin input voltage exceeded the upper STBY threshold of 1200mV+7mv than your regulator will start and again after some battery drainage the voltage crosses the lower STBY threshold it will shutdown. But ...

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First one, of we assume the 18A is accurate, the peak-to-peak voltage is 18mV/m$\Omega$, average is half of that. An average-reading multimeter (not a true-RMS type) will read about 11% (just $\pi$/$\sqrt{8}$) high so you’ll get about 10mV/m$\Omega$. Divide the reading in mV by 10 to get m$\Omega$. Whether the 18A is accurate depends on the actual ...

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The internal resistance of a battery is generally calculated from its open circuit voltage Vo, load voltage Vl, and the load resistance Rl: Ri = (Vo/Vl - 1) * Rl. It's that simple and you can manually make such a circuit easy. There can be many alternate methods but in a general sense the easiest one is usually recommended. Some basic info with example can ...

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The AH (Ampere-hour) capacity of a battery is specified at a specific rate of discharge which is typically 10-hour or 20-hour rate. Thus a 9 AH battery capacity specified at a discharge rate of 10 hours means that it can supply a constant current of 0.9 Amp.for 10 hours before it gets fully discharged. As the discharge rate is increased its AH capacity gets ...

1

I was able to solve this by using an H-Bridge wired as an active rectifier. By using the recommendation here as inspiration, I bought a MOSFET Array 2 N and 2 P-Channel H-Bridge (DMHC3025) and wired it up as shown in the image below. This removed the voltage drop across the diode bridge and fix the inconsistent performance of my charging circuit.

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I have spent much time in the back shop of one of our local battery pack builders / re-builders. They are often required to remove individual cells from packs. They use a large box-cutter type knife and a hammer to cut the existing nickel or nickel-steel strip from the individual cells. This is the kind of knife with snap-off blade segments. You want to ...

1

In series cells, the weakest has the largest voltage mismatch error prone to over/under voltage damage during CV charge or low SOC. In parallel cells the “strongest” battery has the largest current mismatch due to $$ΔI = Vc / ΔESR [mΩ]$$ Therefore all cells must be perfectly tested and matched before any strings series or shunt or both. Balancing extends ...

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The drawback in connecting the two MPPT chargers is that you have two contradictory purposes: (a) you want to use the wind power, not dump it and (b) you want to monitor the energy production of the two systems. These contradict because MPPT controllers are only truly MPPT when there is adequate demand. Otherwise they throttle back, to protect batteries ...

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The big unknown here is that you have two MPPT chargers, one for the solar and one for the wind. These are fairly complicated devices with a microprocessor inside making decisions, both with regard to finding the maximum power point of the input and assessing the state of charge of the battery on the output. The good news is that they should themselves act ...

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Dilute sulfuric acid forms two ions: $H^+$ and $SO_4^{2-}$. So half reactions for Charging: Positive Plate: $PbSO_4 + 2H_2O \to PbO_2 + SO_4^{2-} + 4H^+ + 2e^-$ Lead sulfate becomes Lead peroxide. Negative Plate: $PbSO_4 + 2e^- \to Pb + SO_4^{2-}$ Lead sulfate becomes Lead. Your full reaction is correct but your half reactions were wrong (ions ...

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By symmetry, the current through each cell is the same at 20/12 = 1.66A per cell. There would be no current through the lateral connections (assuming all cells are matched). The current through each of the lengthwise connections would be the same and each would contribute half of the current. The current through each successive leg of the interconnect would ...

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It's the battery the manufacture specifies for that UPS, and CyberPower is a reputable brand. I expect it'll work just fine - as long as you keep its specifications in mind. The UPS itself is rated for 2.5 minutes of run time at full load. That is, it will run for 2.5 minutes while delivering 1000 watts. That's a 9Ah battery. Just on the mathematics, that'...

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The earthing symbol, in an automotive electrical system schematic, does signify actual connection to chassis. The path from the power source to the loads, in an automobile, is through copper cables whereas the return path is through the low resistance steel chassis. The advantages of such a system are: Cost savings in copper cables to the extent of 50%. ...

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The chassis, metal body on a monocoque car is used as the negative wire, or ground. This saves on needing two cables (supply and return) to each device or light etc. The accountants save on the cost by reducing the number of cables and often by minimizing the cable size used to each device. So each battery negative is connected to the chassis, you could ...

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As comment said: slightly off topic here I found this in less than 30 seconds searching the web: https://en.wikipedia.org/wiki/List_of_battery_sizes#Cylindrical_lithium-ion_rechargeable_battery plus of course you can mount almost any smaller type of cell with some handicraft work.

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So first of all there are two ways the battery can produce heat. Due to Internal resistance (Ohmic Loss) Due to chemical loss Your battery configuration is 12S60P, which means 60 cells are combined in a parallel configuration and there are 12 such parallel packs connected in series to provide 44.4V and 345AH. Now if the cell datasheet says the Internal ...

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The second solution with lead acid would be the easiest solution, bare in mind that lead acid need to be refreshed after deep discharge, and maintenance charged. Li-ion need extra care when charging and discharing, at least least dubble protection for OVP and over current i recommended preferable with a hard fuse aswell. Besides the charging circuit. ...

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Normally one would expect the negative electrode to get smaller and the positive to get bigger, as the technical defined direction of current from plus to negative is wrong. In reality its via versa, charge is phyically transported from minus to plus - so ions are transported from the negative towards the positive electrode. "dry" Effects: https://...

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Put 4 NiMH cells = 4.8V or fully charged 5.6V

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As a general rule, it is not adviseable to mix batteries' or cells' chemistries. (edited after a comment) And any kind of mixing should be avoided: don't mix cells of same chemistry, but different capacities; don't mix cells of different brands; don't mix cells of different age; don't mix cells of different charge levels... In short: if there is any ...

1

A battery is "powered" by a chemical reaction that creates a potential difference between the terminals by "pushing" electrons from one terminal to the other. The reaction proceeds until the voltage between the terminals reaches some value that depends on the chemistry, and then it stops. If you complete a circuit (letting charge flow ...

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Could you drain a battery with only one terminal connection? No. "A current must always return to its source" This statement is a simplification, which ignores the category of phenomena we call “electrostatic”. It would be better to say that a continuous current must return to its source. Current is the flow of charges. Therefore, any current ...

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