New answers tagged

1

The capacity of the capacitors can be back-calculated. Let's do some math. The capacity of the battery is 2900mAh. Let's say that the phone will run for 6 hours on this. Roughly the phone is then consuming 480mA. Also, a typical Li-ion battery goes from 4.2 ish volts down to 3.4 with a hard cutoff at 3.0v. Armed with these facts suppose we want the phone to ...


0

First thing would be to confirm the output voltages of whatever the existing power supply. If it only outputs a single 12V supply, then you can connect a 12V battery to the output of the power supply section. But remember, typical lead acid batteries can be charged up to 14.4V and your amplifier should be able to withstand that. Also, it would be better if ...


0

I would recommend against your approach and use the solution that the camera manufacturers have already made. Every camera brand already has a solution to this assuming you can get AC power to the tripod position. The solution is a Camera AC Power Adapter, these work by having a dummy battery that is connected to an AC powered regulator. This will ensure you ...


2

The item in question appears to have a color-changing LED and a water pump. The LED will be, likely by far, the lesser of the two energy-consuming items in the device. The easiest way to determine the power consumption maximum is to refer to the nameplate, typically on the bottom, which should give an upper bound on the power consumption. If is is powered by ...


1

Yes, an analogue voltmeter, by itself, could disturb the precise measurement of a voltage by loading the source and changing its voltage. The characteristic of an analogue voltmeter, which would decide how much it, by itself, could load the source and change the voltage being measured, is known as its 'sensitivity'. The sensitivity of an analogue voltmeter ...


2

The disturbance should cancel out as far as the voltage reading goes because both terminals on the battery (the exposed terminals where you connect the meter) are the same metal, and likewise both of your probes are the same metal. On a car battery the terminals are lead or a lead alloy. Your meter probes are, let's say, nickel plated brass. As you follow ...


-2

No it can't. Hmm the voltmeter is connected in parallel to the branch we want to measure its voltage. A perfect voltmeter has infinite resistance, and if you connect an infinite resistance in parallel to the load then the total resistance doesnt change at all and we have the same voltage drop on the load.


-3

Charge is always taken from the voltage source. In gold-leaf electrometers, that charge will be a transient event. Thus the source can re_stabilize. However, standard voltmeters demand a steady flow of charge, a current, to stimulate voltage_dividers (voltage_ratiometric measurments). Some (old HP meters) may be AC_coupled, but that just means a AC signal ...


4

is it possible? No. 5 V < 8 V so no charge will flow into the battery. is it safe? No. Lithium batteries are explosive. any other way? Yes. Buy a proper charger. This will monitor the battery's state of charge and limit voltage and current to a safe value to prevent fire and explosion. ... lithium iron battery ... Are you sure it's not lithium-ion?


0

The datasheet provides an example application diagram which shows it can output 8 amps at 3.3VDC which is 26.4W and it is far from 100W. But other than that this is my only concern. Buck converter power levels don't work the way you think they work. If a buck converter can output 8A, it pretty much do that anywhere in its operating voltage range. The ...


1

Yes, you can. I do not see anything wrong with it except for the extra junk of communication cables and apparent higher cost. In fact, that is what happens in some big energy storage systems. It is easier to replace a single BMS instead of the single BMS for multiple smaller modules. It also helps with the integrity of the individual cell. Basically your ...


0

As mentioned in the comments, trying to estimate the SOC with only a current-based method is not going to be precise since there are other variables you're not accounting for. Coulomb counting relies on integration and that is its main drawback -- any measurement offset, error, or otherwise uncertainty grows without bounds over time until you reset the ...


0

The regulator you listed may continue working down to 1.7 volts - that is its UVLO level so, adding a diode in series with the power rail to the device is not going to work AND, the diode will represent a big power loss in normal operation because it's conducting power supply feed current. This defeats the object of using a switching regulator in many ways. ...


3

That depends on the chemistry. For non-rechargable lithium manganese dioxide batteries that would be theoretically 3.4V. However, you can also buy rechargable lithium cobalt oxide button cell batteries, which is theoretically 4.2V, just like any other lithium-ion battery.


2

To answer my original question, yes it is possible for an old battery to sink large currents. I believe it is caused by plates disintegrating and pieces falling and accumulating on the bottom, eventually creating a short. But this condtion usually persists past the first few minutes of charging, so it can not be called inrush current. Now that I have thought ...


0

I am using IRF3205 to switching GND, and the gate is controlled by the microcontroller for the delay, so when the connector attaches to the load circuit, the GND is already disconnected, so the spark not happens. Solution: - When the power In signal detected by the microcontroller, it turns on the gate after a few seconds to avoid spark. (Note:- Here I can ...


0

As I learned, the arrow of the voltage is commonly drawn from + to - like this: - + | | | | 1.5V | | | |_| - ˅ As the technical direction of current is defined as from + to - it totally makes sense to label the Cathode the higher potential because the energy of a positive charge is higher there. Edit: Image of a complete schematic: Note that all ...


1

Parallel connection of lead-acid batteries is done routinely in a lot of cases - including almost all UPS devices, small boats, offroad cars, etc... The more identical batteries are, the better. They ABSOLUTELY must be the same voltage. They MUST be of the same type (flooded/gel/AGM, starter/traction/standby), it is good if they are the same brand and even ...


0

The battery positive terminal is called the cathode: - Picture from Battery university


2

Conventional current is "upside down" with regard to electrons, due to some arbitrary choices of labelling made centuries ago. That's why I normally suggest ignoring them and focusing on conventional current and voltage analysis. Electrons are negatively charged. Each electron has a charge of minus 1.602176634×10−19 coulombs. You could label the &...


1

0v point of a circuit is completely a matter of choice. In the above drawing, yo can as well put 0v at point A and then you'll have -1.5v at point B. It is only the difference that matters.


2

9 volt batteries are designed for low current long life applications. Even if you could draw 1.5 Amps at 9V in an ideal setup, with a typical 500 mAh 9V alkaline battery, you would get less than a 3rd of an hour life on it. In reality the battery chemistry will reduce the voltage and current capacity as the current draw increases. From https://www....


2

Most lead-acid batteries charge at a constant 14 4 volts, so charging several in parallel is really just a charge-current issue. If the charger cannot supply enough current it will likely lower the charge voltage to protect itself. As the batteries charge up the voltage will rise, but should NOT go over 14.4 volts, or you could "cook" the batteries,...


5

Batteries don't often fail low voltage or short circuit, but if they do, then a 'very parallel' arrangement could be bad as all the good batteries gang up on the bad one to force a high current through it. Protect each battery with a fuse in series. Before connecting them in parallel, make sure one or more of them aren't duds, check each individually into a ...


2

9V/6 ohms= 1.5A which will kill the very cheap low power batteries extremely quickly. Why kill the battery with a parallel resistor? Rayovac does not show performance graphs so I show one of an Energizer cheap old 9V Zinv battery like what you asked about.


1

Power adapter are rated by their maximum current and voltage, a 9V, 1.5A rating means it will output 9V max and 1.5A max. The exact current that will be output depends on the circuit. You can think of it like this: the adapter outputs a specific voltage and then the circuit draws whatever current it wants till the adapters reaches its maximum. Kinda like ...


1

There are generally three classes of issue here: First, as pointed out, you've misapplied an NPN device as a high side switch (switching the power), and mixed up the terminals of the transistor, too. To switch the power rail, you need an actual high side switch, like a PNP transistor or better yet a PFET. Or, if you must use a NPN or N-channel device, ...


1

You can't. A starter battery specs list both 20-hour discharge capacity (in Ah) AND CCA in Ampere, which is also a capacity, but for some pretty high current discharge where lead-acid batteries are profoundly non-linear. The capacity in both modes usually differs 10..30-fold. Both of them can only be tested and not really measured (except by testing multiple ...


0

No way to calculate it properly afaik. However, some of the cheaper Chinese battery testers on the market (Biltema, Topdon etc) simply seem to estimate it by measuring the battery's internal resistance (R), then present the CCA as: CCA = 2985,6204 / R (with R in mOhm), CCA rounded to nearest multiple of 5. Very strange. I have tried to find out why they do ...


0

You are correct in interpreting the spec of 5V for 4.2 CV with an input power dropping resistor to bring the input to 4.3 or more at max current. Thus a series connection is not possible from a single source. However, nothing indicates a problem duplicating the layout in parallel if your source can handle the current such as from a PC PSU using 6A max for 6 ...


0

You can use one TP4056 module per cell, but you will have to use fully isolated power supplies for each one. This would require one USB charger per cell and you would have to verify that each charger is isolated (put a 10K resistor between the grounds of the chargers and then measure the voltage across the resistor, is should be 0V if the power supplies are ...


2

As well as wanting very low voltage drop in battery backup mode, in the comments you say the mains supply's output is 9V nominal, and I want to have as much of that available as possible The obvious solution is to combine the power sources through 2 'ideal diodes' which have very low voltage drop. If a dedicated IC is not acceptable then you could use ...


0

It seems like there is a delay during switching of the relay. The relay is a "break-before-make" type switch. This results in a momentarily cut of the load's supply. Could this be improved by designing a transistor-based switch? You have a few options. These two came to my mind: Have a circuit that is a type of "make-before-break" -- ...


0

Backwards could cause it, but it can also just happen. I had one cell from a pack of four loose voltage well before the others. this triggered an alarm so replaced them all. If I had continued with them it could have reverse charged the weak cell and caused a leak. after a second duracell destroyed my maglite I stopped using that brand.


3

A reverse-inserted cell in a series chain of cells effectively is being charged by the other cells. And it says very clearly on the battery "Do not charge - may leak or explode" ;) Take a look inside a young-age-grade toy's battery box and you'll find there's extra plastic or battery contacts added specifically to make it impossible to reverse-...


1

Difficult to get to the bottom of this with unknowns but I guess the jist of it is: Can a reverse polarity cell, in a series arrangement, be damaged by other cells around it? Most definitely if there is a load which will pass more than trivial current completing the circuit. Consider something simple like a 4-cell flashlight. You basically have 4 cells in ...


0

I believe this would work, but you may want to consider a protection chip like https://www.ti.com/document-viewer/BQ294533/datasheet/abstract#SLUSAJ36867 as it requires less external transistors, or one with internal transistors, which may save space in a constrained application like a cubesat. Test it thoroughly, simulate various failures.


1

The negative terminal on a battery connect directly to the body of the car, usually the frame or engine. So when you connect your jumper cables to some arbitrary piece of bare metal, you're in fact hooking it to the negative terminal of the battery. It's so much easier hooking up jumper cables when you can just hook up the negative to any bolt on the engines....


2

Does that mean the negative pole of the battery is connected to all of the metal in the car, and thus the current flows through the whole car body during the process? Yes. It's cheaper and more convenient than running a copper ground wire everywhere.


0

Using a booster can have an effect on the battery. Say you need to get 1 ampere at 12 volts from your 3.7 volt battery.. Your booster has to put out 12 watts continuosly (1A x 12V = 12 watts.) To put out those 12 watts continuosly, the converter has to take in 12 watts continuosly. Given 12 watts and 3.7 volts, the converter will draw 3.24 amperes (12 ...


1

My suggestions, MOSFET based inrush current limiter: https://github.com/msglazer/Anti-Spark_Switch (Generally used in octocopter) Anti-Spark XT60, XT90 connectors explanation: https://youtu.be/X71Suakve6A (Video) In Anti-spark connectors, they are using small value high power resistors approx 6 ohms. This Resistor helps in reduce inrush current due to ...


1

Basically your capacity will reduce in accordance with the efficiency of the boost converter and the voltage transformation ratio of the boost converter (if you are measuring the current at the 12V output. Assuming the efficiency of the converter is 80%, then 2100mAh will be effectively reduced to: $$ 2100 \text{mAh}\times 0.8 \times \frac{3.7}{12} = 518 \...


4

You won't get as many mAh from the 12V supply as you do from the battery, but you'll get almost as many mWh. A boost supply can't provide free energy, so when it generates a higher voltage than at the input, it must draw more current from the input than it puts out. In the ideal case, Po=Pi, meaning Vo * Io = Vi * Ii. In real life, the power supply will ...


0

First of all, you need to convert 9V to 5V. The easiest way is to find a linear voltage regulator or LDO (low dropout regulator, essentially the same thing). Its advantage is low noise. And for your low-current application it would be the best choice. Besides, if you look for fixed voltage output versions, you can find just a 3-pin linear voltage regulator ...


2

I see two potential problems with your circuit. The first problem is battery protection. Li-Ion batterys need a proper charger and protection circuit or they can be a fire hazard. There are Li-Ion cargers that have a wide DC input range an should be used here. The battery pack should also be protected from overdischarge. There are Battery Management Systems (...


0

Depending on the laptop, the task may be as simple as 150W/24V solar panel + voltage limiting shunt for 150W/19V or 20V (where most laptops work) + the proper jack to get the power into the laptop. No MPPT, no inverter, shunts are way cheaper and simpler than MPPT or PWM controllers and the laptop will use whatever it can, depending on the available sunlight....


1

Your plan is generally reasonable. Needed current ratings all depend on the voltage of the batteries; I'm guessing that you're thinking of a 12 volt battery, but for a solar power system that isn't connecting to any other DC equipment meant for auto use, you should consider 24 or 48 volt nominal voltage to get more efficient operation and not need quite as ...


1

You can get a spark hooking up 12V jumper cables between cars because you are making and breaking a circuit in the many-millisecond time it takes to connect. The inductance in the cables or receiving circuit is surprising in its ability to keep current flowing in a collapsing magnetic field when a circuit is broken. It's nothing to worry about. If you want ...


0

Charging and discharging batteries in series can lead to voltage imbalance over time. You can read more about it here. To get the most lifetime from your batteries, you can charge them in series but then you should have a charge balancing circuit.


1

It is a good safety practice that you should always have at least an extra layer of protection when dealing with Li-ion batteries. A proper design shall have: A well designed, robust charging and discharging circuit A battery protection circuit A battery with protection circuit built-in #3 would be the last resort in case both 1 & 2 failed. You can ...


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