Lithium batter for 3.3V Circuit - Page 1
Forum section: electronics / projects

author: JohnSL
author_rank: Regular Contributor
timestamp: August 03, 2015, 04:03:59 am
content: I'm looking for some advice on how to power a 3.3V circuit from a 3.7V Lithium battery. As I understand, a 3.7V Lithium battery starts out at 4.2V when fully charged and is "depleted" at 3V. Additionally, I want to use a USB 5V connection to charge the battery as well as power the circuit when it's connected. To provide some more background, I started with the ESPToy schematics from Ray's Hobbies and have been adapting it to my specific needs. Here is Ray's schematic:As you can see, he's using diodes to provide a voltage drop. I'm not sure which diodes, by I measure a drop of 0.5V across D1, and the same across D2, while the drop across D3 (when USB power is removed) is 0.7V.The battery charging IC he's using (TP4054) appears to be pin-for-pin compatible with an MCP73831 charger. I found a forum thread about power with the ESP8366 herehttp://forum.makehackvoid.com/t/esp8266-operating-voltage-range-and-sleep-current/286/41. It appears that keeping the board at 3.3V provides optimum performance, as it's power use varies a lot with voltage. However if I use Ray's circuit, I'll get a voltage drop across diode D3 that will bring the range down by .7V so it will be 3.5V fully charged, down to 2.3V depleted. That certainly doesn't sound optimal.It also seems like this would be a common requirement. Is there an IC that provides the battery management, voltage regulation, and dual-power sources (both battery and 5V)?Thanks,-- John

author: Mr.B
author_rank: Unknown
timestamp: August 03, 2015, 04:22:44 am
content: This is the circuit I use.Sorry about the draftmanship... 2 minute rush job...Edit: Quick explanation...With USB power on, power is connected to the LDO via the diode.The battery is isolated by the P Channel FET.The CN30xx charger chip does its job.Disconnect USB.FET connects battery to the LDO.

author: leo_r
author_rank: Newbie
timestamp: August 03, 2015, 10:29:59 am
content: ICs which provide an all-in-one solution for system power management are often called "power path" ICs. Unfortunately all the nice ones come in really hard to use packages (QFN or BGA).I've taken a similar approach to that shown by Mr. B before. The key is the P channel MOSFET which isolates the 5V from the battery when USB is connected, allowing the charging IC to charge it. This is asimilar approach to what Olimex usein some of their microcontroller boards. Using a low dropout linear voltage regulator rather than a diode will waste much less power - the MCP1700T will give 3.3V with a 150mV drop.If you look at a lithium ion discharge curveyou'll find that there's not much more energy beyond 3.45V anyway.There's also monolithic buck-boost switching regulators designed specifically to provide 3.3V from a lithium battery. These generally also come in less useful packages as well. Given the discharge curve of a lithium ion battery I can't imagine incorporating a boost converter gives you much efficiency gain. An LDO might drop an average of 0.4V over the discharge curve, which would mean an efficiency of about 89%. A buck-boost like theTPS6300xapparently peaks at about 96% efficient.

author: amyk
author_rank: Super Contributor
timestamp: August 03, 2015, 12:16:00 pm
content: This should work... it's from an actual production device.

author: technix
author_rank: Super Contributor
timestamp: August 03, 2015, 12:36:04 pm
content: Here is an idea that I have used once, a little bit complex but can be useful: switcher + isolated communication. This will give you a chance to even further the isolation. The gist is use an SMPS to give you regulated 3.3V, and since this usually produces different ground potentials, isolate the communication lines.A friend of mine rolled two versions of this. His first version used the voltage-inverting buck-boost SMPS and optically isolated UART lines. A second version used a non-inverting SEPIC SMPS and forgone the communication isolation.My own version, since it is intended to be an ECG adapter for oscilloscopes (that is, this gadget + your oscilloscope = makeshift ECG. I personally had a bit is cardio issue so I thought this piece of equipment can be a lifesaver, but if power surge goes into the probes it can kill me) I used isolated Cuk SMPS + PWM-based analog isolation.

author: milow
author_rank: Contributor
timestamp: August 03, 2015, 11:42:34 pm
content: Quote from: amyk on August 03, 2015, 12:16:00 pmHow does it work?R114 being 0R renders Q101 useless. Thus the battery is connected to the USB5V rail the switch is turned on. I guess this is not a good idea, so I assume R114 is not used in the finished device. Then Q101 connects the battery to the USB5V rail if and only if the USB's 5V is not present. This protects the battery. But how do you charge the battery?I want to learn from your good design and have many questions.Why do you use two regulators? Can I combine two regulators can it create problems? I could imagine that one slightly higher than the other results in current flowing from one into the other. Does it impact stability?I checked the datasheet of the XC6411. It wants ceramic 1uF, 0.1uF at the each input and each output respectively. I guess the other regulator is similar. You put a 1uF there. Why not (at least) 2uF? Why is there no cap before VIN1?You used a 0.22uF after the combined outputs of the regulators. Did you use 0.22uF because its about two times 0.1uF, which is required by each regulator.How did you design the output filtering? I guess you filtered 3.0V by a LC-lowpass because it has to source more current than the reference voltage, which is filtered by a RC-lowpass. Is this correct? How did you come up with the corner frequencies? Why do you filter after the regulator and not before?Sorry for all the questions. But I think its the best to learn from experts and concrete designs.

author: milow
author_rank: Contributor
timestamp: August 05, 2015, 07:24:47 pm
content: Could anyone else help with the question regarding amyk's design?

author: Mr.B
author_rank: Unknown
timestamp: August 05, 2015, 07:49:01 pm
content: Quote from: milow on August 05, 2015, 07:24:47 pmI don't see how that design could work for the OP or answer his question.There is no charger for a LiPo battery.R114 would need to be omitted for any battery technology.With R114 omitted, you could use the circuit with primary cells... The primary cells would take over when the USB power vanished.Certainly the front end of the circuit does not make a lot of sense to me...I would suggest that the parallel regulators are to provide a manufacturing option. Perhaps the devices have different footprints.I am guessing that only one would be used in production.

author: amyk
author_rank: Super Contributor
timestamp: August 06, 2015, 02:16:56 pm
content: It is not my design, it is a common schematic for"MP4" playersand shows several BOM options.Only R114 is installed, the battery is charged directly from Vbus. A protected cell is used. I believe Q101+R107 is used for the non-rechargeable battery option.Only one of U105 or U104 is installed depending on available parts.

author: PICmonster
author_rank: Regular Contributor
timestamp: August 06, 2015, 02:18:28 pm
content: The intrinsic diode of Q4 is always forward biased in your design, so Q4 will not isolate the battery, Q4 will allays fwd conduct, the MOSFET in practical terms this might just be a schematic symbol that wrong and the MOSFET may not have this diode I have not checked, you might want to try a different MOSFET arrangement

author: JohnSL
author_rank: Regular Contributor
timestamp: August 06, 2015, 03:30:08 pm
content: Thanks everyone--this has been really great, and given me a starting point for a better circuit. I like the MOSFET and LDO approach, as it's very simple and low cost. I love eevblog!

author: Mr.B
author_rank: Unknown
timestamp: August 06, 2015, 10:30:06 pm
content: Quote from: PICmonster on August 06, 2015, 02:18:28 pmIn this particular design it does not matter as the circuit is designed to be always on.If you can tolerate the voltage drop you could replace simply with a diode.The FET is used because it has a very low RDSon.The purpose of the FET/Diode is simply to prevent USB5V connecting directly to the LiPo battery.