| Electronics > Projects, Designs, and Technical Stuff |
| Implementing USB-C in project |
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| Surajg:
Hello Everyone, I am newbie to electronics and this is my first post. Please forgive me if I have posted it in incorrect forum :) I am working on a project planning to implement USB-C in my project which will eventually be product someday. There will be USB-C slot on the product which will power the product and charge the lithium ion battery. Here are basic parameters of the project: Input parameter: 5v,500mA : Considering worst case were user can use any adapter maybe a mobile charger 5v, 1A-3A or 9V-3A, or 12V-2.25A, 20V-1.35A Considering USB-C PD options available. Battery specification: 1s3p battery pack 3.7V, 2550mAH 14550 cells Datasheet : http://www.tenergy.com/30001_datasheet.pdf Output requirement: 12V, 0.7A maximum requirement. I have attached rough block diagram for the same. Block diagram explanation : Planning to use USB-C CC controller IC which will handshake information on higher current requirement in case adapter is capable. There is charge controller IC (https://www.ti.com/lit/ds/symlink/bq25703a.pdf) which is Buck-Boost Charge Controller With System from Texas instrument which can support input and output requirement of project. Charge controller IC is connected to battery for charging. Further the battery provides output to boost converter (3.7V->12V) as final output. As per my understanding : Batteries can be charged at 0.2C nominal : which is 0.5A (Will it take roughly 5hrs to charge?) 0.5C : 1.275A 1C : 2.55A charging current (Will it take roughly 2.5hrs to charge?) Battery can discharge : 3.7V and 0.5C - 1.275A current 1C - 2.55A current 3C - 7.65A current (Are the current mentioned correct?) Questions: 1. If I connect a boost converter which will provide 12V and rated at 0.7A at output what should be input current of boost converter IC? I followed this document of TI instrument (https://www.ti.com/lit/an/slva372c/slva372c.pdf) (90% efficiency, 30% p-p ripple current) the input current required comes out to be 2.62A. Is this correct? 2. Am I missing anything in this circuit? 3. Can 5V, 500mA mobile charger charge the batteries? What will be charge time? 4. For above mentioned battery pack, can it power 12V, 1A product without any issue? What will be discharge time? 5. If there is USB PD adapter at input, can it provide power to system with boost converter in between? 6. Should I consider 3S combination of lithium ion batteries instead of 1s3p with boost converter at input? Will it be more efficient system? Thank you very much in advance :D |
| jbb:
--- Quote from: Surajg on April 23, 2020, 04:11:53 pm ---Hello Everyone, I am newbie to electronics and this is my first post. Please forgive me if I have posted it in incorrect forum :) --- End quote --- The technical details I think fall into the Projects category. This could be quite a hard project to start with, so good luck. Challenging areas: * switch mode converters (charger, boost) * USB C PD (not as easy as we might wish) * LiIon battery charging; there are safety concerns --- Quote ---I am working on a project planning to implement USB-C in my project which will eventually be product someday. There will be USB-C slot on the product which will power the product and charge the lithium ion battery. Here are basic parameters of the project: Input parameter: 5v,500mA : Considering worst case were user can use any adapter maybe a mobile charger 5v, 1A-3A or 9V-3A, or 12V-2.25A, 20V-1.35A Considering USB-C PD options available. --- End quote --- OK, this is the sort of thing USB C PD is intended for. Looks like your maximum power is 27W. I think the USB C PD spec says that if you can meet the power at a lower voltage, you shouldn't go up to the next one (i.e. you can do it at 12V, so no need for 20V). Also, and I could be wrong, I thought the spec preferred 15V to 12V? --- Quote ---Battery specification: 1s3p battery pack 3.7V, 2550mAH 14550 cells Datasheet : http://www.tenergy.com/30001_datasheet.pdf --- End quote --- OK, 800mAh cells. You might do better with a single 18650 cell as they're very common and made in huge numbers. The cell voltage range will be around 3.0 (flat) - 4.2V (at end of charge). 3.7V is the average. A 1S3P pack would be 3.7V avg * 2.4Ah = 8.9Wh. --- Quote ---Output requirement: 12V, 0.7A maximum requirement. --- End quote --- Any tolerance information on that? E.g. is it 12V +- 10%, or 9V - 15V, or what? Also, do you know average current (useful for estimating battery life). --- Quote ---I have attached rough block diagram for the same. (Attachment Link) Block diagram explanation : Planning to use USB-C CC controller IC which will handshake information on higher current requirement in case adapter is capable. There is charge controller IC (https://www.ti.com/lit/ds/symlink/bq25703a.pdf) which is Buck-Boost Charge Controller With System from Texas instrument which can support input and output requirement of project. Charge controller IC is connected to battery for charging. Further the battery provides output to boost converter (3.7V->12V) as final output. --- End quote --- Thank you for block diagram! I've evaluated that chip before, and it works, but PCB layout is quite important and could be challenging. Having separate SYS output is great, having external MOSFETs is inconvenient. With a 1S pack, you could look for a buck charger (still having separate SYS output) with internal MOSFETs and save yourself some trouble. --- Quote ---As per my understanding : Batteries can be charged at 0.2C nominal : which is 0.5A (Will it take roughly 5hrs to charge?) 0.5C : 1.275A 1C : 2.55A charging current (Will it take roughly 2.5hrs to charge?) Battery can discharge : 3.7V and 0.5C - 1.275A current 1C - 2.55A current 3C - 7.65A current (Are the current mentioned correct?) --- End quote --- Nope, those cells are only rated to 1C discharge: i.e. 800mA each. Thus total pack would be rated for 2.4A max (I suggest you don't push all the way to the edge). Maximum charge is also 1C, i.e. 2.4A. Towards the end of charge you'll be spending 4.2V * 2.4A = 10W max on charging. --- Quote ---Questions: 1. If I connect a boost converter which will provide 12V and rated at 0.7A at output what should be input current of boost converter IC? I followed this document of TI instrument (https://www.ti.com/lit/an/slva372c/slva372c.pdf) (90% efficiency, 30% p-p ripple current) the input current required comes out to be 2.62A. Is this correct? 2. Am I missing anything in this circuit? 3. Can 5V, 500mA mobile charger charge the batteries? What will be charge time? 4. For above mentioned battery pack, can it power 12V, 1A product without any issue? What will be discharge time? 5. If there is USB PD adapter at input, can it provide power to system with boost converter in between? 6. Should I consider 3S combination of lithium ion batteries instead of 1s3p with boost converter at input? Will it be more efficient system? --- End quote --- * The input current you calculated matches the values you provided, but I think you should calculate like this: * Output 12V * 1A = 12W max out * Efficiency 90%, so 12W / 0.9 = 13.3W max in * Pack voltage at end of discharge: 3.0V * Hence max current draw = 13.3 W / 3.0V = 4.44A This is twice what your proposed pack can support! * 5V * 500mA = 2.5W. Assuming some cabling, converter losses, you can only count on 2W. If the average load is less than 2W, you can charge the batteries. Could be quite slow; (8.9Wh / 2W = 5 hours approx. * Nope, that pack can't give you enough current * Yes, if you get a charger with a separate SYS output * 3S combination is tricky, because you need a more complicated protector for safety (must monitor voltage of every cell). Also output voltage would be from 3*3.0 = 9V to 3*4.2 = 12.6V, which might not be enough for your load. Also, 1A load is more than the cells can supply (1C = 0.8A). Efficiency in discharge would a higher because there's no boost converter. |
| jbb:
Ah, here’s something... the LiPow charger. It doesn’t do exactly what you want but could be a good start. It’s open source so you can have a look at the hardware and code. |
| Surajg:
Thanks alot for your reply. --- Quote ---OK, this is the sort of thing USB C PD is intended for. Looks like your maximum power is 27W. I think the USB C PD spec says that if you can meet the power at a lower voltage, you shouldn't go up to the next one (i.e. you can do it at 12V, so no need for 20V). Also, and I could be wrong, I thought the spec preferred 15V to 12V? --- End quote --- Correct. I was planning to limit it to 12V. --- Quote ---OK, 800mAh cells. You might do better with a single 18650 cell as they're very common and made in huge numbers. The cell voltage range will be around 3.0 (flat) - 4.2V (at end of charge). 3.7V is the average. A 1S3P pack would be 3.7V avg * 2.4Ah = 8.9Wh. --- End quote --- That perfectly make sense. Using 18650 single cell would also provide roughly same capacity as 1s3p pack with 14500 cells 850mAH. --- Quote ---Any tolerance information on that? E.g. is it 12V +- 10%, or 9V - 15V, or what? Also, do you know average current (useful for estimating battery life). --- End quote --- Sorry I forgot to add some details. In this project there is an electromagnet (rated at 12v, 0.5A for max pulling power), and other electronics which take 5-6V, 0.5A. Most of the time electromagnet will be inactive, so overall system requirement will be 5v, 0.2A. Electromagnet will be ON for 5-15 second when there is trigger from user input, and OFF until there is next trigger from user input which can happen anywhere between 30s-3minutes. Electromagnet will be ON at max power which require 12V,0.5A. Hence the requirement of system 12v, 1A as explained in previous block diagram. So battery should be capable of delivering enough input current to boost converter such that maximum output current of boost converter meets the system requirement of 12V,1A. --- Quote ---Nope, those cells are only rated to 1C discharge: i.e. 800mA each. Thus total pack would be rated for 2.4A max (I suggest you don't push all the way to the edge). Maximum charge is also 1C, i.e. 2.4A. Towards the end of charge you'll be spending 4.2V * 2.4A = 10W max on charging. --- End quote --- Ohh. Then I need to check cells with higher discharge rate, else batteries will be unable to deliver such high current to boost converter correct. --- Quote ---The input current you calculated matches the values you provided, but I think you should calculate like this: Output 12V * 1A = 12W max out Efficiency 90%, so 12W / 0.9 = 13.3W max in Pack voltage at end of discharge: 3.0V Hence max current draw = 13.3 W / 3.0V = 4.44A This is twice what your proposed pack can support! --- End quote --- Understood. That's way high current than what battery can provide. But as per my above explanation of electromagnet, it will be ON for few seconds, so can battery pack provide the enough power to the system with such requirement? --- Quote ---5V * 500mA = 2.5W. Assuming some cabling, converter losses, you can only count on 2W. If the average load is less than 2W, you can charge the batteries. Could be quite slow; (8.9Wh / 2W = 5 hours approx. --- End quote --- Agreed ! --- Quote ---Nope, that pack can't give you enough current --- End quote --- I was considering 2.62A current input as per TI calculation and 3C cell discharge could cater the required input current. Thanks for the explanation on point 1. --- Quote ---Yes, if you get a charger with a separate SYS output --- End quote --- Ok! --- Quote ---3S combination is tricky, because you need a more complicated protector for safety (must monitor voltage of every cell). Also output voltage would be from 3*3.0 = 9V to 3*4.2 = 12.6V, which might not be enough for your load. Also, 1A load is more than the cells can supply (1C = 0.8A). Efficiency in discharge would a higher because there's no boost converter. --- End quote --- Correct. I assumed that usually battery packs comes with inbuilt protection circuit, and for charging I was planning to use CC-CV charger IC for safe charging. Should I find some battery pack with higher discharge rate? Will that solve the issue of required high input current? I will continue to read more and update you on progress, thank you very much. |
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