Choosing a DC connector for charging - Page 1
Forum section: electronics / projects

author: kian0079
author_rank: Contributor
timestamp: March 26, 2023, 05:56:13 am
content: Hi all,I am building a circuit to charge up to 15 lipo batteries. I have found a power supply brick that supplies 5V 10A output that comes with a 2.5mm diameter DC jack connector. The charging current is set to 560mA, hence if all batteries are charging at the same time, about 8.4A of current to be drawn from the power supply.I am trying to find a suitable DC jack connector. The common (cheaper) ones available are mostly rated for 30V 500mA, 24V 2.5A, or 16V 4A. There are some that are rated for 24V 7A. When choosing a suitable connector, should I be just looking at the current rating only? In this case since my max charging current is 8.4A at 5V, should I be choosing a connector that has a currently rating greater than 8.4A? Or since I am using a low charging voltage, can I use a connector that is rated for 16V 4A or 24V 2.5A?I am thinking, my power dissipation is 5V x 8.4A = 42W, hence would a 24V 2.5A (60W) connector be suitable? Is this the right way to spec the connector? Thanks in advance!

author: mariush
author_rank: Super Contributor
timestamp: March 26, 2023, 06:10:28 am
content: If you're charging all batteries independently, then it would make sense to power your product with a higher input voltage and have one or several dc-dc converters to convert that higher voltage to the lower voltage expected by the charger ICs.For example, design to accept 16.5v... 20v as it's most common range for laptop adapters and you can easily get a 65w-90w laptop adapter , mass produced, cheap in volume.With higher voltage, it's less current, so a plain 2.1/5.5 or 2.5/5.5 barrel jack connector will work with minimal losses.DC-DC Converters would also be able to produce a lower voltage like 4.5v or whatever is closer to the minimum voltage required by the charger IC to correctly charge a battery, and with much higher efficiency.A single battery lithium charger will often be just a fancy linear regulator, so whatever voltage above 4.2v + some margin (usually 0.2-0.5v) will be thrown away as heat.So have maybe 4 dc-dc converters, each powering up to 4 individual charger ICs ... 4 x 560-600mA = <3A is not a problem.Higher input voltage would also open more possibilities, for example it could allow you to make your charger fancier... like maybe add feature to charge a 2S battery (have a JST XH or whatever connector is used by most 2S batteries and a special charger IC that's powered with 9v or higher (through a dc-dc converter from the higher voltage at input) and you could add load balancing on those cells.

author: kian0079
author_rank: Contributor
timestamp: March 26, 2023, 08:05:51 am
content: Thanks mariush!Let me explain and see if I understood your proposed solution.I can get a 16.5 - 20V, 65W-90W laptop charger and this would feed into my DC-DC converter circuit through a plain 2.1/5.5 or 2.5/5.5 barrel jack connector. So its 16.5V going into 4 DC-DC converters. And the output of the DC-DC converter is connected to 4 charging ICs in parallel. I can get a 16V 4A charger at half the price of a 5V 10A charger. Would that make sense?I am using the TP4056 charging IC. It seems to be a cheap and rather popular part that is easily available on LCSC. The data sheet is in Chinese, but from the block diagram provided, it seems to use a 4-6.5 supply connected to a 0.4 ohm resistor and to the VCC of the charging IC. What should the output of the DC-DC converter be in this case? Should it still be 4.5V or it can go even lower to minimize loss? I am not familiar with DC-DC converters, would you be able to recommend a part number that I can take reference from?Thanks in advance!Quote from: mariush on March 26, 2023, 06:10:28 am

author: mariush
author_rank: Super Contributor
timestamp: March 26, 2023, 09:05:22 am
content: TP4056 is like a linear regulator, with a bit more intelligence.It can work with as little as 4v, to charge very discharged batteries, for example from 3.0v - 3.2v up to around 3.9v or so, but the battery would not be fully charged this way.If you want to fully charge a lithium battery, the input voltage must be a bit above 4.2v because the final step of charging a battery is keeping the voltage at around 4.2v for a few minutes.High current power supplies can be more expensive than high voltage, low current power supplies. It often makes more sense to go with a mass produced power supply available in a higher voltage and use dc-dc converters to produce your desired voltage.I don't have time to filter what LCSC has, but basically you'd look at the input voltage, frequency, ease of implementingfor voltage I'd pick at least 20v or higher, but there's good chips that support maximum 14-16v which can still be used if you limit your input power supplies to 12v)for frequency, i'd try to pick something higher frequency like 600kHz or more - higher frequency means smaller cheaper inductors, smaller capacitors, but the conversion efficiency drops a bit the higher the frequency.You may be able to find 1.5-2mhz chips that support up to 14-16v input, but only up to 1mhz for higher input voltage rangefor easy of implementing, I'd look for chips that require as few components as possible, for example I'd avoid chips that need extra compensation networks (resistors and capacitors with values that must be calculated with formulas in datasheet), I'd give priority to chips with datasheets that have good instructions and sometimes even example layout circuitsFor example AP62300 and AP62301 can work with up to 18v input voltage, and output up to 7v at 3A ... and runs at 750kHz and they're 6 pin chips, easy to solder and use.Besides the chip, only an inductor and a few ceramic capacitors and resistors are needed and datasheet even contains example circuit for 5v output.So such chips could be used with a 12v input and possibly even a 16.5v laptop adapter power supply.Here's the LCSC link :https://www.lcsc.com/product-detail/DC-DC-Converters_Diodes-Incorporated-AP62300TWU-7_C1883519.htmlIf I'm reading the datasheet correctly, AP62301 is an AP62300 without PFM, pfm is basically a mode in which the regulator outputs a "cleaner" voltage and is more efficient at very low output currents. So AP62300 is better and you can use whichever of the two.With up to 3A of output current, you could power 5 tp4056 at the same time, or you could configure your TP4056 chargers to limit charging current to around 450mA and then you could safely power 6 TP4056 at same time.So it would cost you around 2-3$ in parts to make 3 such dc-dc convertersFor higher voltage, there's plenty of optionsRT8250 can do up to 23v , outputs up to 3A current :https://www.lcsc.com/product-detail/DC-DC-Converters_Richtek-Tech-RT8250NGSP_C48415.htmlIt does require a compensation network but the values are pre-calculated and listed in datasheet so other than being a few extra capacitors and resistors to order, it's easy. It even suggests part numbers for the inductor required and there's layout example for the circuit board.In theory you could also buy ready made step-down/buck converter boards from eBay with those super common chips, but they can be less reliable.For example see this 1$ "module" :https://www.ebay.com/itm/175569869721The chip supports up to 24v and can do 3A for short periods of time. Continuously, without extra cooling, it should not be used for more than 2A of current.You can set the output to a fixed voltage like 5v by adding a blob of solder on the bottom or you can put a blob of solder on ADJ and then use the small trimpot to set the voltage.Another example, 3.5$ for 5 "modules" :https://www.ebay.com/itm/163320251988This one uses either MP2307 which does up to 23v, outputs up to 3A of current... or MP1584 which is 28v max input, 3A out ... both chips show up in board pictures on the eBay listing ... the MP1584 version is a bit better but no idea what you'd get.

author: mariush
author_rank: Super Contributor
timestamp: March 26, 2023, 10:16:22 am
content: Quote from: kian0079 on March 26, 2023, 08:05:51 amForgot about this...That 0.4 ohm resistor is not needed, and most likely it's there in the datasheet to reduce the input voltage.For example, at 1A charge current , the voltage dropped on the resistor will be V = I x R = 1A x 0.4 ohm = 0.4vSo you have 5v input, but when charging the battery with 1A current, the regulator will get 4.6v ... but you'll dissipate P= IxIxR = 0.4 watts into the resistor (you'd need a 0.5w or 1w rated resistor), while you could use a dc-dc converter to produce 4.5v and avoid the resistor and the heat it would generate.

author: jonpaul
author_rank: Super Contributor
timestamp: March 26, 2023, 12:18:12 pm
content: for fire safety use only the industry standard quick disconnect polarized UPS battery connectors, see APC, Tripp-Lite replacement batteries.Available in capacities for 24V battery 7,12,35 AH.Made by AMP.Jon

author: kian0079
author_rank: Contributor
timestamp: March 26, 2023, 12:26:06 pm
content: Thank you Mariush! Very insightful comments from you and I learnt a lot today!Quote from: mariush on March 26, 2023, 09:05:22 am