New answers tagged

2

You need to measure the output current and determine what the charge is with that current. You could use a look up table with the curves you posted. You could also try to make a fit for each curve. There is a part that can be seen as "internal resistance", so you might do a fit for the low current curve and try to add a "resistance" ...


1

I see 2 options how you could estimate the state : you can use a "Coulomb Counter" : it is a device that measures (rather accuratly) the current, and integrates it over time. Basically, it gives you the number of Ah flowing in/out the battery (more precisely, the difference between the Ah flowing in and those flowing out). you can measure the ...


0

Sometimes battery packs have a protection module, sometimes they don't. Most likely they don't just put a diode in series with 5V, as that would damage the battery quite quickly and it might become dangerous. If that device is dangerous, then you can inform your local electrical safety authorities about it so the sellers have to make a product recall.


1

That is a very nice SOC chip with DSP for low cost (<<$1) and small instruction set (RISC) that does it all including 16 bit ADC to measure voltage and current to regulate LiPo battery charge. https://github.com/RDA5981/RDA5856/blob/master/Datasheet/RDA5856TE_Datasheet_V0.03.pdf Tear-down (perhaps a fine example of IP theft, RDA does not hold ...


0

The batteries you've linked are two-terminal 12V lead-acid types. These consist of six series-connected cells internally - with no means to balance (or equalize) them. It doesn't matter if you add more series cells; the charging current will be the same through all cells and the overall situation for each cell won't be any different to the single-battery ...


0

In short, you are right, batteries can discharge faster than charge. But in terms of efficiency of use, the matter is not the same. The way the question as initially done does not consider some very important topics, as battery is essentially a (rechargeable) source of electrical energy. These are some relevant points to consider: How much is the charging ...


2

When charging, the lower-capacity cell will reach its maximum voltage (say, 4.2 V) first, but the pack as a whole will not have reached its maximum voltage (8.4 V) yet, so constant-currect charging will continue and the lower-capacity cell will be overcharged. So no, not safe. Some (of the better) BMSes will stop the charging before something goes really ...


0

I know question is quite old, but the topic is still useful to several readers, so here are my answers and a circuit I built that charges a lead acid battery as you originally wished: For example, I could attach something that would charge it, keeping it at 14.6V until the current goes down to .1A, and turn off. When this device would turn off, the charge ...


0

A more nuanced answer than you received so far is that 4.27 V is not a significant problem as long as the charger is turned off (disconnected) after the cell is full. Then, the cell voltage will relax down to a safer value. Compared to charging to 4.2 V, the advantage of a higher charging voltage is that it stores a tiny bit more energy in the cell. The ...


2

Charging a Li-Ion battery to 4.27V probably won't cause a fire, but it would make me uncomfortable. Your batteries will likely suffer from a reduced lifetime however. Fully charging and discharging the battery puts stress on it. That stress will reduce the lifetime of the battery, and the capacity will continue to reduce as time goes on. To limit the stress ...


0

Can I use a resistor in series, and if so, what value? Yes, you can use a resistor in series. The value must be chosen to limit current to less than the specified 'trickle' charge rate, which is typically capacity / 10. In your case this is 600/10 = 60 mA. Under charge the cell voltage will quickly rise to ~1.3 V, then gradually rise to ~1.5 V after 12 ...


1

Short answer : no, you cannot use a 5V supply to charge a NiMH cell Long answer : Whatever resistor you put between you NiMh cell and the 5V supply, when the battery is full, there will still be a voltage diferrence, so there will still be current charging the already full battery : so you will distroy it. Buy yourself a NiMh charger (it costs only a few $). ...


0

Let’s see your wish-list: Charge a 12V car battery from the “main battery”. <=> Assumed here the main battery is the battery connected to the car starter engine and alternator. Use of thin cables, to not draw to much power in case “aux” battery is empty. Here is a problem, as thin cables should not be used to present a high resistance to limit the ...


0

Charging Lead-Acid batteries in parallel is not a problem, but a larger battery charger that feeds up to 6A, when equally divided by your 3 batteries of 9Ah, would give 2A/battery, in the best case. Sealed-LA batteries maximum charging current usually is 10-20% of rated capacity. For a 9Ah, I_max = 1A ~ 2A. In the present case the above 2A (or more) is a tad ...


1

That seems unnecessary. Better to use a single boost regulator to charge the LiPos in series, and use a single FET to disconnect the load. The technique of cell balancing is maybe what you're looking for. Here's an Analog appnote about that: https://www.analog.com/en/technical-articles/active-battery-cell-balancing.html Figure 6, "High efficiency ...


2

Different battery chemistries have different nominal voltages and different discharge profiles. Unfortunately nickel-cadmium and nickel metal hydride Cells have a nominal voltage of 1.2V which is somewhat lower than the 1.5V of traditional zinc carbon and alkaline cells. Most equipment would tolerate this, but some would not. There were attempts over the ...


14

How do they work? I think, maybe they have some integrated DC-to-DC transformer circuitry? Correct. There's been multiple types over the last decade that tries to work like a ubiquitous battery. These are just the latest, thanks to Miniaturization of electronics. Yesterday's 4 inch pcb is today's dime size single chip IC. While not the one you are asking ...


27

The batteries you show use Li-ion cells internally and include a step-down and a USB 5V to Li-ion charge controller. They’re essentially power banks that output 1.5V with a step-down instead of 5V or 12V with a step-up. (Why must the battery be Li-ion? For this specific cell form factor [AA], a comparable NiMH battery is only 600-1000mAh (720 ~ 1200mWh.) So ...


3

There are a number of USB-charged AA Lithium-ion batteries from China. They use a switched inductor to regulate the charging with an IC. Since the USB plug and charging circuit are inside the battery and take up space, then the cell itself is smaller than an ordinary AA battery. Therefore the battery capacity of these rechargeable Li-ion batteries is less ...


0

While it is technically possible it would require careful design. Regenerated AC from motors is invariably converted to DC and then smoothed to provide voltage suited to operating conventional LiIon chargers. In the case of highly variable voltages a boost or buck boost conversion scheme is liable to be used. It may be hard to sensibly distinguish between ...


3

Lithium batteries need careful attention to current, voltage and temperature to ensure safe charging. Same also applies to discharging - an EV that's thrashed hard will heat its battery and sometimes will have to engage 'limp-home' mode until its batteries cool down. At any rate, electric vehicles include controllers for both grid charge and regenerative ...


1

Voltage controls the speed of the fan. But you cannot use a higher voltage than the fan is designed for, only lower voltage. If a fan is designed for 12 volts and you apply, for instance, 24 volts (two 12 volts cells in series) you will probably burn the fan. But you can slow down the speed of the fan by applying lower voltage (to a certain limit). I have ...


0

I want to use 12 volt cell connected in parallel will it increase the amp? Maybe. I'm assuming that your fan is a simple device consisting only of a motor and a switch and a battery—no fancy (expensive) electronic speed controls. If that's the case, then as user_1818839 said, the speed of the fan will be proportional to the Voltage. But note! It's not the ...


-2

It's the voltage that increases speed. To increase wattage You will have to align your 12 V batteries plus on plus, minus on minus, so you will have more amps. More amps means more power per hour which in turn means more Wattage, as long as you are not losing it in... f.i. thin wiring, heat, and many more possibilities. Your main course would be, building up ...


0

Please understand that a power-path circuit is not always necessary. Often, the much simpler circuit that places the charger, the battery, and the load all in parallel is sufficient. However, if the charger is able to provide far more current than the battery can safely accept, then this simple circuit won't work.: if the load is off, all of the charger's ...


0

What about a 2-transistor constant current sink? Use an NPN as the control transistor. If you put in a 0.2 Ohm resistor, the NPN will choke the pass transistor at about 3-4 A. If current is lower, the pass transistor will be fully on and can be a low resistance N-MOSFET. This is probably a little bit more efficient than using only a resistor of around 1-2 ...


0

The way everyone and their dog would do it is a boost regulator and a single 3.7V cell. Series cell combinations require a charge balancer, so that’s way more complicated than just having a boost regulator. The cells can be connected in parallel without much trouble if you need to – it’s the series connection that’s problematic. The parallel connection ...


2

I want to implement fast charging in this system. How can I go about achieving it? This requires an external circuit to be designed that sets the charging current. The BQ76940 is not a charger circuit; it is a cell-balancer circuit. The charging circuit is external to this device and, when designing a charger circuit, you must consider that the BQ76940 can ...


0

Fast charge may cause the battery overheating. The battery temperature should be supervised.


2

Assuming the phone is meant to charge via a USB cable, then the charging speed is defined by the relevant USB standards, and possibly proprietary standards defined by the manufacturer. Charging speed is also determined by charging circuit in the phone itself. If the charging circuit in the phone only draws a little current, then the battery won't charge ...


0

Because some chargers adhere to the standard for fast charging a phone with the same standard. Like to fast charge an older iPhone required resistors on the data line and to fast charge an older galaxy phone you needed Samsung Adaptive Fast Charge. Or newer phones require QC2 then QC3 or USB C PD with the right profiles or Apple certified MFi. The max charge ...


0

Don't reinvent the wheel. They sell a ready-made solution: 12 V to 12 V battery charger. https://www.google.com/search?client=firefox-b-1-d&q=12+V+to+12+V+battery+charger


1

No, it is not safe if you don't have a charger. Please understand that a regulator is not a charger. Also the BMS is not a charger either. A buck or boost regulator is a power supply and batteries can't be connected to a power supply. Batteries can be connected only to a charger which safely charges the batteries with the correct current and voltage and ...


2

As long as the available input power minus the boost regulator efficiency exceeds your required output power, then generally speaking, yes. A 5V 2.1 Amp usb charger would be just about topped out if your charging the batteries at 8.7 V 1 Amp at 90% efficiency. You seem to be missing a lithium charging circuit (unless your bms includes that). In that case I ...


0

This has been addressed in a practical manner here: https://whiteboxcellphone.com/psd/rpi-supercap.html Basically, a supercapacitor across pins 2 and 6 of the GPIO will charge while the Pi is powered and discharge for a few seconds when it is not. However, you have to balance the benefit of a large capacitor against the ability of the charger to feed the ...


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