I am a student who has a lot of interest in electrical engineering, so I look for a lot of basics and go through various trials and errors. Of course, I'm a beginner who doesn't even know the basics...
I don’t have good writing skills, so I’ll ask you in short sentences.
There is a lighting fixture with 3 1.5v batteries
(Small light with 12 to 15 LED mini bulbs)
Trying to replace it with secondary batteries such as lithium-ion or polymer batteries.
(I know the purchase route)
Usually small batteries are going to use a boost circuit from 3.7v to 4.5v.
Battery and PCM protection circuit and converter already exist
This is the introduction...
This question is
1. I was thinking of a lithium polymer that I heard that lithium-based kids have a lot of advantages, but because of their stability, it is not lithium-ion. But because of the large price difference... Is the stability of using a can-type lithium ion such as 18650 in the presence of a protection circuit poorly??
2. It's a very basic question, but I'm confused... The current value.
Does the current automatically change according to the load depending on the nominal voltage of general batteries and the resistance value of the device to be used??
Or is there a fixed section?
For example, if you insert 4.5k ohm to a 4.5v battery, it takes 1mA load, and if you insert 450 ohms, will it take 10mA automatically??
If so, is there something like the maximum discharge rate the battery can withstand? (To prevent overheating and explosion hazard..?)
3. This is the next question. I'm confused because it's the same battery...
The auxiliary battery is set to 5V 2A in the specification column. Is this listed under the assumption that it is a mobile phone?
Is it possible to output up to 2A depending on the resistance?
Or what kind of resistance (such as a light bulb) will be delivered to 5V2A unconditionally and the rest will leak as heat, causing breakdown or explosion
I know that the battery discharge adjusts as the voltage drops as it discharges, but I'm confused.
4. This is the end.
If I use a polymer cell converter from 3.7 to 4.5, is the battery not affected? Is the fever worse or not??
There is a product, but I don't know the principle.
Is the principle simply increasing the voltage and decreasing the current value...?
Please long answer.
I'm afraid of exploding various parts while making a fuss so I want to put safety first.
Answer B
1. In terms of stability, they are equal. There is no difference, so don't worry.
2. The description of the load current is correct, and there is a discharge curve depending on the battery. In other words, there is a curve for how the discharge time comes out according to the discharge capacity. Normally, you can use up to the indicated capacity. From 5V/2A to 2A...
3. Same as in #2.
4. Are you using a converter after the battery? It’s no problem because it’s originally used for this purpose.
Answer A
1. When comparing lithium-ion and lithium polymer, use lithium-ion (18650). Lithium polymer is easily swelling (swelling), and its characteristics are slightly inferior to lithium ion. And even in terms of stability, lithium-ion is better. The previous Samsung cell phone explosion was also a lithium polymer. If you use it without force, you do not have to worry about an explosion accident.
2. There is a maximum discharge rate value for each battery or rechargeable battery. In the case of lithium ion, it is usually 2C~3C. For example, if the current capacity of 18650 is 2600mAh, 1C = 2600mA, and 2C = 3200mA. In the case of a high discharge rate (non-protected) that enters an e-cigarette or charging drill, it can be up to 10C~20C. In other words, it can produce up to 20~40A instantly.
In general, the maximum discharge rate of rechargeable batteries is much higher than that of general primary batteries.
3. In the auxiliary battery, 5V means output voltage and 2A is a warning to use max. 2A or less.
If more than 2A flows, the overcurrent protection circuit operates in the boost converter (converter that makes 3.7V -> 5V) inside the auxiliary battery and cuts the output. So it's not a breakdown or fire, but it's not good because it puts too much pressure on the converter.
Since it is through a DC-DC converter, the output voltage always maintains a constant 5V even if the internal battery voltage drops. Then, when the battery is almost discharged, the converter cuts the output.
4. If your product originally used three 1.5V batteries, no matter how high the current consumption is, it will only be around hundreds of mA. That way, the battery or converter does not generate heat.
There is nothing you can do if you are scared by battery explosions like someone saying, “I’m scared of maggots, and I can’t get it.” Of course you have to be careful, but lithium rechargeable batteries don't explode as easily as you think.
The 18650 with built-in protection circuit will never explode unless you stab it with an awl or hit it with a hammer.
When you get used to it, the fear goes away. So don't be scared, keep touching, experimenting and making.