New answers tagged batteries
1
I have an RV that I've done quite abit of electrical work, so I know where you are coming from. To start, let me make sure I understand what you are trying to do. Take a look at this block diagram:
I believe the only information I left off is the 7A max for the charge controller(s). To start, let me bring up the topic of this related question: Charging ...
2
Your 75W fan required 220 Volts, it will not operate at typical battery voltages such as 12V.
A 75W fan operating (indirectly) from a 12V supply would draw 6.25 A not 0.34 A.
Even with a theoretical 100% efficient invertor, the current on the 12V side is greater than the current on the 220V side to deliver the same power across the system from battery to ...
0
The other answer that said you don't have to worry about current is correct but not for the right reasons. On a DC motor, from the perspective of the equivalent circuit, the current draw is going to be a function of the internal resistance and also the back-emf. The back-emf is a voltage of opposite polarity as your battery that will work to reduce the ...
3
You don't need to worry about the current input, the resistance of the motor (which can be obtained using V = IR : 12V = 1A*R => R = 12 Ohm ) will limit this. What ever 12V battery you use will only supply 1A at that resistance (V = IR again). A larger battery will last for a longer amount of time because it has more stored charge (measured in Amp Hours). ...
1
That circuit uses the CN3083 LiIon charger IC <- datasheet.
It can work with batteries with or without temperature sensors.
If the IC specs suit your need the circuit will work as well as the CN3083 does.
Horrendously drawn Charger shield cct diagram here
1
Schottky diodes don't make great blocking diodes for the purpose of power rails. They have often have high reverse leakage currents.
Your initial in-rush current is probably high causing the battery voltage to dip below the part's UVLO. To verify this put a scope on your battery and set it to trigger on it dipping more than 100mV below it's nominal value. ...
1
Legal Disclaimer: Li-ion batteries can catch fire if overcharged, or improperly charged, or exposed to heat so be very very careful. I don't guarantee this advice won't cause damage or loss, or hurt you or anyone else.
Yes, it looks like you should be able to. As long as your circuit doesn't overcharge your battery or exceed it's charging rate you should be ...
1
I would NOT design anything that would require direct AC power if I were you. Really. You can design your circuit to take 12/24VDC or whatever you need from a wall wart and then use that to charge your D cell.
Seriously. Do not go near the AC outlet. There are lots of design safety issues to consider, the first of which is using an isolating transformer.
...
0
Using a different charger for what the battery is rated for is generally a bad idea. the battery is designed to be used with a specific charger that is rated for the same voltage and Amp Hours (Ah). You should not use this charger with this battery.
0
Useful battery life is a function of two main parameters:
How quickly the chemicals in the battery will be depleted
The level of depletion at which the device becomes unusable
The capacity figures for batteries indicate how much current they can supply before they are depleted to the point that the manufacturer does not expect them to be useful for ...
4
The stated requirement can be met relatively easily. Whether that will treat the batteries as they should be treated is unknown to me, ...
A warning: "Modern" AA NimH above about 2000 mAh capacity will not tolerate even a whiff of long term trickle charging. Whereas C/10 rate was once a commonly accepted level, this is now a fatal overcharge long term.
...
1
Here is a low power LDO voltage regulator (TPS79733) that consumes less than 2uA quiescent current whilst delivering up to 10mA to keep your MCU ticking over when it's asleep.
It has a fixed output of 3.3V and I think the clincher would be to arrange for your TPS63061 switcher to generate maybe 50mV more output (3.35V) so that when it gets activated, it ...
1
The data sheet you link to says (on page 6) that the battery is CR123 (not AA), which is a non-rechargeable lithium battery.
0
Depending on the battery voltage variation while charging and the tolerance required on the 1A you may need a constant current 9A sink.
Good question.
2
Depending on the charger you won't know until you examine it, but generally I would say yes, it hurts your battery. It doesn't take too much current (a few mA) depending on battery, to cause permanent damage after a few months.
3
If it's a duty cycle of 4%, then the average load is 4% of 1.95mA, or 78 uA. The battery should last 2610 / 0.078 hours, or 33461 hours. Over 3 years.
Battery capacity changes with temperature, time, and discharge rate (and the 2610 mAh rating isn't too precise either, in my experience.) These calculations are probably within ±20% of the real answer.
3
Your device needs 1.95 mA × 35 ms = 68.25 µA-s per transmission.
Your battery is good for 2610 mA-h / 68.25 µA-s = 137.65 million transmissions.
To turn this answer into a time value, you would need to know how many transmissions occur per some unit of time.
1
"Real" batteries are capable of maintaining their voltage while delivering significant current.
However, the potato batteries are not capable of delivering much current - they have a relatively high internal resistance, so the voltage will drop significantly when they are delivering current. Putting the potatoes in parallel doubles the available current.
...
2
The Duralock battery has improved alkaline cells (improved purity of actual chemical ingredients) not only this but the Duralock battery protects the anode and cathode with separators that limit power transfer when the battery is not in use. On top of this, it has a sort of triple corrosion protection that surrounds the contents in an acid resistant, ...
-1
From www.elprocus.com:
Lead Acid battery is a heavy current device and it contains sulphuric acid. It is necessary to take precautions during battery charging to avoid hazards due to sparking, acid spilling, explosion etc.
Following are some tips to be followed while charging Lead Acid Battery:
Always keep the battery in a well ventilated ...
3
If what you are saying is that your transformers are each rated at 12V (with a center tap) @ 0.5A, you can put a full-wave rectifier on just one of them and pull 1A @ 6V from the output. The average current through each half of the secondary won't exceed the 0.5A rating.
simulate this circuit – Schematic created using CircuitLab
0
Because you will need to rectify any secondary voltage of a transformer in your application and, you might also need a regulator circuit to provide the correct DC level, I propose that you wire the secondaries of the two transformers in series and use a single bridge rectifier and a step-down (buck) SMPSU. Primaries in parallel.
It's not good practise to ...
1
If the LEDs are to be directly powered by the batteries, the blue, white, and possibly the green LEDs (depending on which type of green LED it is) will not work at all on 3 Volts: They require a forward voltage of 3+ Volts, 3.4 Volts as specified in the question.
Also, once the voltage starts dropping due to depletion, even the marginally lit ones (e.g. the ...
0
You don't really give useful information in your question. What are your requirements, how long do you need the leds to be on? Do you need them to be at a semi-constant brightness? Do you need them all of them to be on (because once the battery voltage starts dropping, the blue, green and white will turn off first since they need more voltage, and leave the ...
0
You might be out of luck. Batteries can't just charge when you apply a voltage; there are special current and voltage conditions that must be fulfilled in order to charge the battery, and these are met using special charging circuitry. If the battery charging circuitry is located on the board that you may have damaged and is damaged, the charger itself isn't ...
2
I've built several products around different chemistries. I have found LiPo the easiest to use since there are already specialised charger IC's that do all the work however, like you said, having them in series is not that straight forward. I believe the NiMH is a better solution if you don't have a lot of experience with chargers or if you don't find a good ...
1
Your daughter must invent a more efficient solar panel, some other, more efficient means of converting solar energy to electrical energy, or make a much more efficient car. I'm assuming charging the battery already in the car with a solar panel doesn't count.
Of course the energy in sunlight varies by latitude, weather, and time of day, but \$120W/m^2\$ is ...
0
What I haven't seen much of is this: SIZE MATTERS! Not for voltage, that will remain constant, all else remaining the same, but size DOES affect AMPERAGE! More plate surface area, (and electrolyte) equal more amps, all else remaining the same. So, the 'bigger' the battery, assuming the identical construction, compared to a smaller battery, will have the ...
0
A definition of potential difference ( what is measured in volts and sometimes called voltage ) may be useful: it is the work the battery does on a unit charge moving it from one terminal to the other. Thus since the same charge is moved twice each time thru the same voltage the total work done is 2 * V. In a parallel connection the charge moves thru one ...
6
Fortunately, there is. You can find diagrams of the discharging process of a battery type, search for "lead-acid discharge graph" or so. This returns graphs like this one:
In the datasheet of your battery you can find the exact graph for your battery.
Say you want a circuit to detect 80% discharged, just check in the graph, the voltage of a lead acid ...
1
Taking the question at face value, ALL the answers given are wrong [ :-) ] and your question and a subsequent comment do not match.
You say you have 4 x 12V batteries to make a 48V supply.
Your diagram implies (but does not state) that the panel is 24V rated.
In a comment you say
I don't think I know the PV array's knew voltage. I know it produces ...
4
There are quite a few different methods in use, all of which have their pros and cons.
One of the simplest is to just measure the voltage out of the battery. As the battery is being charged or dischared, the output voltge will vary, as demonstrated in this graph from digikey:
Of course, the curve will vary wildly by types of batteries as well as ...
3
ACPI is the biggest practical example I'm aware of. It serves as a standard for power monitoring, control, and everything in-between. It replaced older standards that placed responsibility for power monitoring on the BIOS, which in turn required dependence on chip firmware for reporting the necessary information to the BIOS (charge levels, depletion rates). ...
5
For a hobbyist project, I'd go with a NiMH option. They are significantly easier to charge (you can charge them with a commodity linear regulator (LM317) and a few passive components), and safer. The weight difference will likely not matter much- your robot doesn't seem to need to change directions very quickly, and it's not something like a quadcopter where ...
3
A cell or a battery is essentially a charge "pump". Now, to help form an intuition for the answer to your question, fall back to the hydraulic analogy.
Two water pumps in parallel can produce twice the water flow of one (ideally).
Two water pumps in series can produce twice the pressure (or head) of one (ideally).
3
TL;DR version:
Use two alkaline AA cells, a 2.7 kOhm resistor and the LED in series, such that the LED is provided the bare minimum current that lights it up sufficiently.
Summarizing from chat discussion around this question.
Determine empirically what minimum current through your specific LED provides sufficient light to be acceptable as "staying ...
2
Charging the battery safely is a primary concern.
I typically like to DIY; however when safety is a factor, it would probably be better to buy a product from a quality company. So to give you some ideas, this is a "smart" charger you can set for various current and voltage settings (up to 9.6 volts); it will also turn off automatically. It will also charge ...
-2
You could use a regulator, but a cheaper alternative would be to just use a diodes in series. Although it's not very reliable (voltage goes down as the batteries drain), it should get the job done. If you use 1N4148, you will get a drop of 0.6 volts per diode. The peak would be 5.5 volts. (6.1 = 0.6 volts)
Crude, but works!
2
(Question subsequently fixed) First off, to get 6 Volts from the alkaline AA cells, they would need to be wired in series, not in parallel. Paralleling batteries will increase current delivery capacity but keep the voltage the same as a single cell.
While pre-built buck regulator modules are available at the same source (eBay.com) as the boost regulator ...
1
I don't know how efficient the joule thief is so I'd go for the LTC3531 and an op-amp to control the FB pin on the device. The intention of the op-amp is to monitor LED current and therefore keep the drive to the LED constant at 20mA. Here is a picture that hopefully demonstrates the efficiencies: -
The lower picture is the adjustable version and with a ...
2
(This answer summarizes Anindo Ghosh's suggestions on the topic made in EE chat plus a few of my own observations. Please note that I am not considering this answer definitive: just want to add something else to the mix of suggestions)
This answer assumes (without justification) a LiPO battery-based design. Some background information:
LiPO battery cells ...
0
Using AA batteries is not practical if you need an ampere for several hours. Based on your level of experience (I'm guessing) a small 12-volt sealed lead-acid battery sounds like a good choice. There's no point in trying to develop a high-efficiency regulator for the microcontroller, so just use a linear voltage regulator. Keep in mind that the voltage from ...
1
How about 4 NiMH C-cells in series? They are rated at 5000mAH and 4 of them will give you about 4.8V. Should give you at least 2 hours @ 2 amps power draw, but you'll want some sort of over-discharge protection to prevent damaging the batteries by over discharging them. They should be able to stay above 4.5V for most of their usable capacity, but if you need ...
1
Easiest method is to observe the voltage with an oscilloscope.
WHY does the equipment fail with high ripple voltage?
How do you know?
Who says that it does?
Please provide links to manual and other details.
.
1
Quick first pass - more anon:
What colour LEDs?
What model/brand and how many.
With modern LEDs 10A ~+ 3000 lumen+
VERY bright.
Why do you need so much light?
If you are using ye olde junk because they are chaep then a modern high efficincy version may greatly reduce current needs.
If you are using modern LEDs you're stuck with providing it if you need ...
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