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Don't.
The battery manufacturers specifically tell you not to do this, as the batteries are quite heat-sensitive. If you manage to set one on fire, you can't extinguish it either: it reacts explosively with water, so you have to bury it in sand or powder.
The correct solution is to get spring-loaded contacts or "pogo pins", attach them to a PCB or piece of ...
38
[Should I] connect the fuse to the negative terminal of the battery since it's where the actual flow of electrons originate which is opposite to the conventional flow of current from the positive terminal?
Forget about electron flow. It only causes confusion such as in your thinking. Electrical current flows in a circuit in the same way that a bicycle chain ...
37
I did my PhD thesis on energy management in wireless sensor networks and was working with sensor nodes using CR2032 batteries. We designed the nodes ourselves (my supervisor designed the PCB and I designed the firmware and all the energy-related testing).
I can confirm what people above say that you can draw 100mA peaks from a new CR2032 cell. But as they ...
30
If the batteries are not perfectly balanced they would still have some net voltage. And that is assuming that the batteries are connected in series.
If the batteries are connected in parallel, flipping one of them the other way around will basically create a dead short between them. At best, that is going to drain them.
In either case, storing batteries in ...
29
First, let's see how much energy or capacity you need. 70 µA for 10 years comes out to 6.1 Ah, and you say this is at 3 V. That's a lot more than a CRxxxx battery can do.
Another problem is that you need a battery that is good for 10 years regardless of discharge. Many batteries aren't specified for such a long shelf life, let alone service life. ...
26
What could explain such a short battery life? Could it be the 5v regulator?
As mentioned, the 7805 has about 4mA of quiescent current. You need to find a data sheet for the battery (Eveready has nice battery datasheets, if you're using an alkaline cell). It's probably no more than 100mAh -- 100mAh / 4mA = 25 hours, so that should say something to you.
...
26
You can, but as per @Dampmaskin's answer, it's not really the best plan.
Why not do as often seen in new electronic devices, and inserst a plastic slip between a battery and one of the terminals (or between two batteries). If the slip is long enough to be seen outside the closed battery box, and a clearly visible sign says something like "remove before use"...
23
Long lasting (very low self-discharge) batteries usually doesn't have a very good discharge characteristics.
For example Tadiran SL-360, 14500 form factor offers 2.1AH at 6mA discharge but can not be discharged above 20mA. In the datasheet, you can see it can discharge 10 years at 20uA.
One way to get around it is to use a energy buffering solution: ...
23
There will always be some variation in capacity between batteries. In toy use the current is typically drawn at a high level compared to many other applications, so the internal resistance of the batteries is very important.
In that service the voltage under load drops quite rapidly as the end-of-life approaches, so with minor differences in capacity the ...
22
The sensor (typically an ionization chamber with some radioactive Americium 241) is normally specified at 9V. I think it would work at a lower voltage, but with less sensitivity.
More information can be found here.
Since the battery drain is very low, it's not necessarily true that cells with higher ampere-hour capacity would last significantly longer- ...
20
Step one is trying to avoid wanting to solder to any kind of pre-fab battery, as you don't know 100% for sure what's in them in ways of protection.
Step two is really thinking hard if you can't stick to step one.
Step three would be: Very, very carefully.
To expand on step three:
Many single-purpose pre-fab lithium cells for such things as phones will have ...
20
There are Li-ion batteries and battery packs designed specifically to be soldered, which usually look like this:
I'd strongly suggest you buy one of these instead of trying to reuse a cell phone battery. Your device will most probably be badly soldered and thus unreliable if you do. There's aslo a danger of shorting or overheating the battery while holding ...
20
Toys are often supplied with the cheapest battery cells which may be lower capacity, but also have the widest tolerance. This tolerance and the "weakest cell's fails 1st" is the general rule.
To understand this, model each battery as a charged capacitor, say 1000 farads +/- 20 %. Then put them in series with the same initial voltage and load them with, say ...
answered May 19 '17 at 12:39
Tony Stewart Sunnyskyguy EE75
94.3k2 gold badges34 silver badges139 bronze badges
16
Use conductive glue. Here's an example. A few caveats, taken from the reviews:
It isn't tacky.. so either use a clamp, or be prepared to hold the object (for a long time.) It's thin & watery..apply sparingly in layers. ...and even after it's dried to the touch, it won't conduct electricity very well until it's sat for at least a few days to thoroughly ...
15
How about using an alternative circuit like this:
Q4 is a P-mosfet, kept off when there is a voltage in the main supply (USB in this case). When the main supply is disconnected the gate is pulled down and the mosfet conducts and provides output from the battery source.
The diodes are Schottky type for low Vf and the mosfet should apparently be ...
15
In addition to the notes on battery manufacturing tolerances, I have seen (at least one) device that would split off battery electrical connections to different parts of the circuit.
In my case, it was a cheap alarm clock with 3 AAA batteries, 2 were in series to run the electronics, and then a third was in series with the first two to run the LED/Alarm.
...
14
LiPo is MUCH easier to manage well than NimH.
Energy densities for top capacity NimH are about the same as LiPo nowadays.
NimH is a relatively hard battery chemistry to manage well. Charging at low rates is not usually advised and negative voltage deflection under charge or temperature rise are the usual end of charge detection methods. In contrast LiPo is ...
13
Your series resistor is the current controlling device, but you made a calculation error. The resistor value is not 6 V divided by the current, but the voltage across the resistor divided by the current, and that's only 2.7 V, being 6 V minus the 3.3 V of the LEDs.
If you would now take a 150 Ω resistor, for (6 V - 3.3 V)/ 150 Ω = 18 mA, that ...
13
Very often, multiple batteries are series-connected. And the total voltage is used to power "stuff".
In that case, flipping a battery to turn off current works just fine.
However, the less-usual case of parallel-connected batteries won't allow you to flip one: the resulting failure could be spectacular.
It would be unusual, but possible that the series-...
12
From the data sheet:
Nominal capacity indicates duration until the voltage drops down to 2.0V when discharged at a nominal discharge current at 20 deg. C
So, if you discharge at a steady 0.2mA, you will get 220mAh capacity, or 1100 hours. Discharging at a higher rate will reduce the capacity of the battery. Pulsing will also affect this, as BLE ...
12
All those parts can run from 3 to 5V so use a battery that doesn't need a regulator, a 16500 Li-ion cell, or a 3xAAA battery pack are about the same size as the 9V and produce voltages in that range. (or even a Li-po cell)
Without the regulator the microcontroller can shut down and the circuit will only need a few microamps.
11
Looks like a self-oscillating converter, probably a primary winding, a feedback winding and a secondary winding (6 pins). It would be similar to this, but with many more turns on the secondary:
I think this is a blocking oscillator with the transformer primary in the emitter of the transistor and the feedback winding blocking the base voltage.
The ...
11
I have practical experience with PC power supplies. Pretty much any modern ATX PSU contains a main section (that delivers full horsepower when the PSU is on) and a tiny section to deliver +5VSB. In a decent PSU, both are SMPS, i.e. do not contain a classic iron-core AC-only transformer (which would surely blow if fed by DC mains).
There are still some cheap ...
11
Lead acid batteries are fantastic at providing a lot of power for a short period of time. In the automotive world, this is referred to as Cold Cranking Amps. From GNB Systems FAQ page (found via a Google search):
Cranking amps are the numbers of amperes a lead-acid battery at 32 degrees F (0 degrees C) can deliver for 30 seconds and maintain at least 1.2 ...
11
The idle current of a 7805 regulator is around 4 mA so, armed with the ampere hour capacity of your battery, work out how long it will last with a continuous drain of 4 mA.
If you establish that is the problem, you will find that there are plenty of regulators having a significantly lower quiescent current.
Once the battery drops to about 7 volts you are ...
11
Since the negative terminal of the battery is normally considered "Ground" or "Zero Volts", a fuse in the negative lead would leave the rest of the circuit "hot" - usually Not a Good Thing.
Recommended practice is to place the fuse near the positive terminal of the battery, so the whole circuit will be dead if the fuse blows.
(Of course, if the positive ...
11
A 9V battery is a poor choice for providing electrical energy for heating.
An Alkaline 9V battery can typically provide maybe a few watts of heat energy for a short while.
If your wire resistance is too high the heat per wire length will be too low to notice.
If your wire resistance is too low the battery will be excessively "loaded down" and will not ...
10
The Arduino Uno isn't fit to run off a coin cell.
The Uno isn't exactly low-power. It contains 2 microcontrollers running at high clock frequencies and consumes a couple tens of mA
A coin cell will give you 3 V, while the Uno is designed to run on 5 V.
Most important: a coin cell has enough energy to power the Arduino for a couple of hours, but can only ...
10
A CR2032 has a capcity of ~ 200mAH. For 2 weeks, this gives you an average current budget of ~ 500 uA. A plain 555 has a supply current of a few mA, so you'll need to use a cmos 555, which has a supply current of (max, @ 5V) 250uA. Use the highest value you can use for the timing resistor.
Assuming your LED current is 2mA, and the LED is on for 3/10 of a ...
10
The idea behind the Vbat input on chips such as the STM32 is that the RTC and other low-power peripherals will normally be powered from the main Vdd, automatically switching to Vbat when Vdd is absent. On-chip, this can be implemented with something as simple as a diode-OR:
simulate this circuit – Schematic created using CircuitLab
To save pins (I ...
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