New answers tagged batteries
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Is this a device for testing lead acid batteries?
The two coils look like low resistance, high power resistors. These could be switched on and off by the relays to see how the battery reacts to having a heavy load applied.
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It says on page 1
• Supports Battery Capacities Above 65 Ahr
• Supports Charge and Discharge Currents
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Most power efficient way would be in software, if you are driving or observing these signals in an MCU.
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A 900 nA static current is ok for your application?
https://assets.nexperia.com/documents/data-sheet/74AUP2G14.pdf
This one has a higher worst case (2uA, 100nA typical), but you would need only one IC:
https://www.ti.com/lit/ds/symlink/sn74hcs14.pdf
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All the answers assumed an external fuse with easy access to both sides.
However, the lithium battery pack you are using is probably made of three or four lithium batteries in parallel.
The final output voltage is created by repeating so many of these groups of 3 or 4 parallel batteries, each connected in series.
The fuse is inside each lithium batteries, at ...
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The logo on that chip resembles Microchip Technology Incorporated.
Here is their website.
They are the makers of the famous PIC microcontrollers and now own the AVR line as well. They, of course, make many other chips.
The logo is usually inverted from that and in a circle, so it may be something else. I couldn't find a closer match.
A website like THIS ...
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Sounds like a USB power bank might fit your needs, though if you attempt to drive all the motors at once under load it might be too much current.
At least the voltage will be regulated to 5V and the battery charge and protection will be taken care of.
The average and peak currents will determine how long a charge will last and whether the maximum output ...
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Your best bet is some battery source and a regulator to put it down to 5V (it seems that all your gear is more or less 5V powered). It was already commented that some onboard regulator could need 7V, for example.
I would, however, worry more about your current budged, powering something like 20 servos will need probably more than 1A (check your datasheet)
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A fuse doesn't protect a battery from overcharging or discharging.
It also doesn't prevent a short circuit, however if a short circuit happens then hopefully the fuse will blow preventing fire or damage to equipment caused by said short circuit. The fuse blows and makes the system safe if correctly installed but the short circuit is still there and must be ...
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This is a bit like asking if you can fill a 42 foot high tank through a spout 12 feet off the ground. No. Your source has to be higher than the destination or you need a pump (a voltage booster in your case).
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Yes it matters. You must make sure that the gate is controlled in reference to the source.
Edit for the OP: You need to make sure that your gate is always connected to a part of the circuit that is connected to the source in such a way as to know the potential difference between the source and the gate. In you circuit, it is unclear that this is true. So I'm ...
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No this is not a viable idea at all.
If you have three batteries, there are only two ways to arrange them. All in series (3S) or all in parallel (3P).
If you want to add capacity to your battery bank, you should add two more batteries so that you have two in parallel and two in series (2P2S).
There are many details to see to when it comes to combining ...
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Not a particularly good idea.. you missed same age/useage? As batteries age their characteristics change, you will eventually over charge one set of batteries and under charge the other. This will shorten the life of all 3 batteries. Better to go 2 + 2 or keep the third separate.
But if you set up the charger(s) to charge each 12V set of batteries ...
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In this diagram it seems left end of the red wire (The blue dot) has highest potential and the PD should drop as we move to the right.
That would be correct if you took battery negative as your reference or "ground" point. In the illustration the author isn't measuring voltage from a ground point, s/he's just measuring difference between two ...
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This circuit should work assuming the fans are 12V. If the unmarked fan is 5V, it might not like having 9V applied. 12V fans won't run at full speed but this will make them safer for little fingers. I'm guessing that is a good thing.
Notice that there is a resistor added to your list of parts... You need that if you use an LED. You can repeat the LED part of ...
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Boost converter single cell AA battery not delivering enough voltage when connected to Wemos D1 mini
I wonder, what I am missing. The specification says 600mA output current.
Well well.
These cheapo DC-DC converters always spec the biggest number they can find in the datasheet with the unit of "Amps" dangling at the end. Usually it's the big number in the title of the datasheet, which means...
Datasheets usually put in the title the biggest ...
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Boost converter single cell AA battery not delivering enough voltage when connected to Wemos D1 mini
What voltage is the battery when operating?
It will also be good to look at the battery voltage with a scope because the voltage may drop significantly when the switch in the converter is conducting but rise during the rest of the cycle. I find that it is essential to use a very low ESR capacitor at the input of the converter - a conventional small ...
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The 560 Ohm resistor feeding gate current to the thyristor via the 1n4002 diode, is almost certainly too high a value leading to the device not being turned on properly. This in turn will increase the dissipation and cause over heating. You need to reduce the value at least by 10 to about 56 Ohms. One simple way to arrive at the value is to monitor the ...
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Many electronic appliances with motors require several times more current when starting up (for the first few seconds) than they do to run continuously.
To accurately calculate your power requirements, use an amp-hour or watt-hour meter to determine how many watt-hours the machine uses per hour.
Multiply that number by the desired runtime in hours to get the ...
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The direction of current through the battery determines whether it is charging or discharging.
The battery is trying to push current in a particular direction. If the current flows in that direction, the battery is discharging. If the current flows in the other direction, the battery is charging.
It is a little bit like a spring or a clockwork toy. When you ...
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Normally laboratory power supplies do NOT tolerate reverse current, they would get damaged. As Sphero mentioned, ideally you would use a electronic load if you have it, but this can be a relatively expensive equipment.
A cheaper solution is to build a current sink circuit, using one opamp and a power transistor as explained here, the supply in this case ...
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I just realised that ohm's law doesn't apply to a fan since the fan isn't a resistor. The fan takes what it needs in power. When I connect the fan to the 12V battery (which I just tested), the fans spins and draws the amount of current it needs.
If I want to turn the fan on or off using a microcontroller, I just need to switch the voltage supply to the fan. ...
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The fan you show is a 4-wire PC case fan. Very common, nothing special. Being 4-wire, it includes the ability to do PWM control. If you feed it a PWM signal the power will be reduced in proportion to the duty cycle.
Here's a link to how to make a PWM control using the venerable 555 IC: https://www.electronics-tutorials.ws/blog/pulse-width-modulation.html
...
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You have linked to an eBay ad instead of a datasheet. The fan has four wires so it probably has a PWM speed control input on it and electronics to efficiently control the speed without the high heat loss of a resistor.
Internally the fan is not a simple brushed motor but is a miniature brushless DC motor with a chip generating a three-phase supply to the ...
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No you generally cannot. Otherwise there would be no need for "Electronic Loads".
These devices can be set to constant current, constant resistance or constant power draw to emulate whatever your real load is.
They are available from a number of sources, generally cost will increase with increasing power capability (and with the accuracy and ...
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That depends by what you mean power supply. DC-DC converter in constant current mode with load in output and battery as input can serve your purpose. But quiscient current of converter should be taken in account. LM317 comes in mind. Heatsink may be needed.
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Calling these `relays' is probably the issue. You'll need contactors and be careful of derating your contacts to the kind of load, especially if it's DC (DC arcing is harder to quench!).
Another option is, as suggested in the comments, a starter relay. In some country they are called `solenoids' which is mildly incorrect, but whatever.
These things are huge, ...
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for a beginner how do you explain this?
Assume a tank is full of water, and there's a faucet attached to the tank. You can adjust the flow rate of the faucet: If you open it more, the tank will drain more quickly.
Now the Amp-Hour (i.e. charge) is the amount of water in the tank, the flow rate of the water is Amperes.
7 Amp-Hours means the source can ...
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Yes, of course. When you redirect some of the electrons so they go through your laptop, the laptop pushes back against them, which decreases the speed of their looping (i.e. decreases the current). Eventually the current stops completely, and then you have no more energy. Your superconductor battery is now flat.
To recharge it, you have to push the electrons ...
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Energy stored in a magnetic field is extracted when you use it, and the field weakens. This is a bit of a simplification but close enough.
The energy in an inductance L is E = \$0.5I^2L\$. To store much energy that way relative to a battery you would need a lot of inductance and/or a lot of current. All superconductors have limits in current density and ...
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T is the terminal for the battery charging device to measure battery temperature with a temperature sensor. The blue part is the temperature sensor. The sensor resistance changes according to the temperature.
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You won't likely be able to see any noise from the battery on an oscilloscope because the noise floor of the oscilloscope is too large.
To actually make a useful measurement, you'll likely need to use (eg. SR560) or make a low-noise AC-coupled amplifier that covers the bandwidth of interest.
Provided you stay within the limits the oscilloscope is rated for (...
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Would a cracked sealed lead acid battery be a safety issue?
YES! It is a safety issue.
The corrosive gel inside has started to leak (hence that white crust, similar to what can form on car batteries, especially the older non-sealed types).
I recently had the misfortune to drip some of the corrosive gel from an old, small SLA battery similar to yours (I didn'...
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Yes. It is a safety issue. Buy a new battery. Put that one in a plastic garbage bag to contain the powder. You can drop it off at the place where you buy the new one. They will make sure it gets recycled. Lead acid batteries are very recyclable (people will even pay you for old non-functional lead acid batteries). If you just want to get rid of it, you can ...
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Probably fine. Wiring details may matter if you discharge at a very high rate. Because each wire acts like a small resistor. If the resistance is not equal in all pathways then the cells may not share current equally during charge and discharge.
Also, don't connect batteries together in parallel unless and until you somehow get them both at the same voltage (...
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That series-parallel arrangment is legal but could give some issue since cells will not be discharged in the same way. It will works but you'll have less than 4 times the capacity, probably.
Usually is preferrable to put all the cells in series and step down from the resulting voltage with a converter; even in that case there are cell balancing issues at ...
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Since you have not mentioned any resistors, the current is mainly limited by the internal resistance of the AA cells. Such large currents can overheat wires, and can start a fire.
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It looks like a thermal protector. This is likely a self-resetting bimetal type, unlike the cylindrical type that has a low melting point metal inside with a spring- they open once and then have to be replaced.
Here is a typical product (photo from Alibaba):
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Thermal protection.
A sane battery is expected to have one for fire safety.
May be either a fuse (single-use) or a bimetal disconnector (reversible - reconnects the circuit once cooled).
Most of these have a temperature written on them.
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Datasheet
"If the positive terminal and the negative terminal of the battery pack are short, during the delay time of short limit detector, large current flows through the FET. Select an appropriate FET with large enough current capacity to prevent the IC from burning damage."
"Output Delay for Short Detector vs. Temperature " (p24 18.) ...
answered Feb 11 at 17:12
Tony Stewart Sunnyskyguy EE75
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For the circuit itself I don't know that part but these protectors are meant to be used in only one way, so you just have to keep the sample circuit and recalculate some component about your battery.
For the mosfet you simply needs two logic level n-channel power mosfets; drain rating and rdson depending on your current (they are not switching so you should ...
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Try this, a 2-transistor constant-current sink that is driven through a zener diode as the detection threshold. Here is an innergoogle schematic grab that is close:
http://lednique.com/power-supplies/simple-constant-current-driver/
To this circuit, add a zener diode in series with R1, and connect its anode to Vbb. Not counting the LED, 5 components ...
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Could a circuit with positive feedback be an initial idea?
It borrows your idea of using the 68k resistor to polarize the BE junction, but the feedback causes an abrupt increase in the current:
It is not perfect, since the brightness will be reduced gradually before the LED abruptly turns off. If you adjust the hysteresis and the max. LED current maybe it ...
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LiPo batteries have a nonlinear behaviour with temperature and varies widely with quality of construction, although there are some generalizations.
Assumptions
Battery model specifications take precedence over generic expectations
test with controlled constant current sink vs voltage must not stay under/over voltage for any significant time.
if specs define ...
answered Feb 11 at 1:17
Tony Stewart Sunnyskyguy EE75
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The problem is simply that you have too much load for your batteries.
Battery voltage drops when you draw too much current (when the load is too heavy.)
Boosting the voltage draws more current from the battery, making the battery voltage drop even more.
Every voltage booster you add just makes the problem worse.
You need a battery that can deliver more ...
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Power quality is a problem and servo motors are very demanding when there is any position error. You must have a very low ESR in your power source.
Voltage boosters are a poor solution for servo motors from batteries because the output impedance is inherently increased and both voltage boosters and motors demand huge surge currents starting up.
You need low ...
answered Feb 10 at 18:28
Tony Stewart Sunnyskyguy EE75
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Add a resistor south of Q2 (in series with the collector) connect the south end of R3 to Q2 collector instead of ground.
That will give hysterisis boosting the switching speed of the led. try 22 ohms to start.
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tl; dr: consider a 4s RC hobby pack+charger solution. All the problems are solved for you.
I can't recommend 3s + boost converter for running a high-current motor load as this adds losses and increases the current demand on the battery pack, shortening its life. Using 4 cells (4s) is better, and as it so happens this is a popular pack for RC cars anyway. ...
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I decided to start charging the battery if it's <30% charge and stop charging it once it reaches 90% charge.
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The simple way is to use zener diode, for 9V battery is 6v8 or 7v5 good choice. When the voltage across R2 is higher than 0.7V, i.e Vbatt is higher than Vzen + 0.7 ,the led is gloving.
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