Buck Converter with Lipo and Charger - Page 1
Forum section: electronics / beginners

author: Lrv209
author_rank: Contributor
timestamp: February 03, 2022, 08:50:10 am
content: Hello,I have developed a buck converter to power a table lamp LED from a LIPO 18650 2600mAh cell. The LED requires 2.9v and 700mA to give the required lumen output and I developed the attached circuit to provide the required power. The IC is enabled/disbaled via the EN logic pin being sent to ground to disable the IC and shut down the output. A switch mounted on the lamp base will be the on/off.I will be charging the LIPO from a charger base (USB connected) that will connect to the lamp base and the LIPO battery via the pins PP1 & PP2. Originally I was just thinking I could use a quality LIPO charger such as the Adafruit 1904 micro charger. I then realised however that the charger could be charging with the lamp on and with no load bearing capability in the charger that could cause problems with overcharging and finding a suitably sized charger with loadbearing has been difficult.I have now added the transistor Q1 as a way of overriding the switch and sending EN to ground when the charger is connected so the IC and lamp are always off when the charger is connected. I have added a Schottky Diode D1 to prevent the battery feeding back in to the transistor.I have already had the buck converter circuit built as a PCB and it has worked just fine so the Buck converter circuit is working as required but I am now looking at the charging/Load bearing problem. My problem is the D1 diode which has a Forward voltage of 435mV at 0.5A. This I think will reduce the maximum charge of the battery to 4.2-0.435=3.76v although the charger will still sense the 4.2v full charge.I am hoping someone may be able to see a better way to to disable the load when charging or to power the Transistor Base other than what I have used or maybe a completely different approach using a mosfet at the IC Vin pin. I can see a how a mosfet could be used but I still require an "event" to change the mosfets status and the only event I can see is the battery charger being connected so I am back to where I am right now.I am very new to electronic design and rying to learn as I go so any advice would be appreciated. The circuit schematic is attached and shows the latest iteration with the Transistor and D1 Diode.Thank you in advance for any ideas, comments, advice or recommendations.- Attachments and other options

author: Peabody
author_rank: Super Contributor
timestamp: February 03, 2022, 03:44:54 pm
content: It's "load sharing". Your circuit would prevent battery current from flowing back into USB, but wouldn't prevent USB power from flowing back into the battery.Below is a typical load sharing circuit. It shows a boost converter, but a buck could be used instead. USB power, when present, charges the battery and powers the load independently at the same time. That's assuming the 5V brick produces enough current to do both. So you don't have to turn off the lamp to charge the battery. The mosfet is needed to prevent the 5V source from flowing back to the battery, but without imposing any voltage drop when the battery is powering the load, as a second diode would. The mosfet needs to be "logic level", with a typical Vgs threshold voltage of 1-1.5V.But I think there is a problem with your on/off setup. When you turn off the lamp, power will flow through the Enable pullup resistor as long as the lamp is off. That will discharge the battery . An alternative is to put a pull-down resistor on Enable, and have the switch, when On, connect power to the Enable pin. Then when the switch is Off, no current flows, and the converter is powered down. Or you could tie Enable high, and have the switch connect power to the converter input when On, as shown in the circuit.

author: Siwastaja
author_rank: Super Contributor
timestamp: February 03, 2022, 03:55:11 pm
content: I fail to see how the extra load could cause overcharging. It could trigger some timeout features in the charger, though, if the lamp is on, and prevent full charge. I would try it without any power path logic. The worst that can happen is the charger is too intelligent and complains somehow, or terminates charging early.For the LED, use a LED driver IC, basically just buck constant current source. It makes little sense to use voltage based buck converter and then lose significant part of the energy in series resistor to regulate the current to prevent LED destruction. LEDs are current driven components, and current mode buck without voltage feedback is actuallysimpler, with less parts. Look at the LED driver section of your favorite distributor.The battery is li-ion, not LIPO. Sorry for nitpick, I can't resist.

author: Peabody
author_rank: Super Contributor
timestamp: February 03, 2022, 04:20:05 pm
content: Quote from: Siwastaja on February 03, 2022, 03:55:11 pmThe typical charger terminates charging when charging current drops below 10% of the full constant-current level. But if there's any significant load current, and there is no power path circuit, then the charger will see the load current as charging current, and termination will never happen. So the risk is not terminating early. The risk is never terminating. It's not good to continue to apply 4.2V to a fully charged battery.

author: Siwastaja
author_rank: Super Contributor
timestamp: February 03, 2022, 04:31:06 pm
content: Oh, floating at 4.20V isn't a big thing. It results in like 101% SoC. Just adjust the charger to output 4.15V if you are worried.

author: Peabody
author_rank: Super Contributor
timestamp: February 03, 2022, 08:40:37 pm
content: Well, I'm certainly not an expert on this, but every lithium charger I've ever seen terminates charging. None of them continue to trickle at 4.2V, or at any other voltage. So I assumed there was a reason why they do that. Also, the popular charger ICs all seem to be fixed at 4.2V, with no provision for any adjustment to a lower voltage.

author: Lrv209
author_rank: Contributor
timestamp: February 04, 2022, 09:04:45 am
content: Hello Siwastaja and thank you for your reply and advice and especially thank you for mentioing the constant current LED driver as a alternative to the buck converter circuit I have drawn. I was unaware these items existed but am certainly very interested in using an appropriate IC to create what I think will be a better and more efficient circuit to power my one LED.Taking your advice I have looked for an IC that might work and so far the nearest I have found is the TPS92360https://www.ti.com/product/TPS92360but as I am very inexperienced in this sort of design I am hoping you can advise if I am on the right track with this item. Hope you can advise.I also have now learnt the difference between Li-Ion and Li-Polymer and as you very correctly stated my battery is a Li-Ion battery so thank you for being pedantic it has helped me learn again.regards

author: Siwastaja
author_rank: Super Contributor
timestamp: February 04, 2022, 02:02:19 pm
content: That chip is a boost converter, which is a very good topology for an LED driver, but it has one big limitation: the LED Vf must be significantly larger than the maximum possible input voltage, otherwise the LEDs are always on (through the diode which you can see on the front page of the datasheet!), and current can't be regulated or limited.This means, for a li-ion battery with Vin_max=4.2V, the load must be at least two white LEDs in series. If you have just one high-power LED with Vf=2.9V, this won't fly. For larger string of LEDs, this is excellent. Do you have just a single LED, if yes, then look atbucktopology LED drivers, and pay attention to the minimum voltage drop-out (you would want to use the battery down to, say, 3.2V, and if the LED requires 2.9V, the controller needs to go nearly full duty cycle at the end. Possible with buck regulators based on P channel MOSFETs, but not necessarily every chip on the market.)Remove lithium polymer and LiPo terms from your mind. They do not exist beyond research efforts in 1990's, despite the fact that a few manufacturers use the term for pouch form factor li-ion cells, purely for marketing reasons only. (You know, even if a technology fails to emerge, nothing prevents you from starting using the name if it sounds cool, right! Similarly, you can call a gasoline car an EV because it has electric windows. In a pouch-type li-ion cell, there is polymer in the outer package (aluminium-plastic vacuum bag), but the electrolyte is not polymer, which was the original point of the LIPO research.)

author: Lrv209
author_rank: Contributor
timestamp: February 09, 2022, 08:07:25 am
content: @peabodyHello Peabody,I think I am following your comment regarding the location of the on-off switch and the current flowing constantly when the switch is OFF. As a result I have made a change to the circuit as shown in the attached schematic and would appreciate your comment on whether I have correctly interpreted your suggestion. It certainly makes sense to me to restrict the unwanted current flow and loss.In the circuit I have made R6 10k but am wondering if that is the best size. I calculate the current at EN will be 0.037mA when the switch is closed (ON) so 10k R6 should work OK but I am really only guessing in this case.Hope you advise and would appreciate any suggestions.Cheers

author: Peabody
author_rank: Super Contributor
timestamp: February 09, 2022, 10:33:21 pm
content: Yes, that's what I had in mind, but with current coming from R5, the voltage will be barely above ground when you close the switch. So I would sugggest you change R6 to 1Meg, and just eliminate R5. Then the voltage on EN would be at the full rail voltage when the switch is on, but the current through R6 would be very low.

author: Lrv209
author_rank: Contributor
timestamp: February 10, 2022, 12:48:36 am
content: @peabodyThank you for the reply and advice i will go with something similar however the switch i am intending to use and the only one i have found that suits the device i am building is only rated to 500mA and i am using a 2600mAh li-ion battery so i am concerned the current will exceed the switch capacity if not reduced by R5. If i change R5 to 20ohm i will reduce the current through the switch to 200mA well within its rating and then keep R6 at 1meg or slightly lower i think i should achieve a similar result. Hope you can comment.Thank you again for the helpRegards

author: Peabody
author_rank: Super Contributor
timestamp: February 10, 2022, 03:37:28 pm
content: The only current that will flow through the switch is what flows out through R6. No current will flow into the EN pin. But there's no problem with having R5 there. Just make sure it is a low value relative to R6 so the voltage at the midpoint of the resistor divider, which is where EN connects, will be high enough to turn on the regulator when the switch is on. The datasheet will tell you what that voltage needs to be.

author: Lrv209
author_rank: Contributor
timestamp: February 11, 2022, 06:33:51 am
content: @peabodyAgain thank you for your help. While i have looked at the circuit for many hours i was too naive to see that when i include R5 & R6 on each side of En i have a voltage divider. Lack of experience showing again.So using the standard formula for VD calcs and a low 100k R5 and large 1M R6 i have a voltage at EN of 3.364v well within the range required of 0.3v-6v. So all clear now and so obvious with your help.And so obvious also when i consider your other comments re current flow which with the resistors i have listed will be so small it will not be any problem for any switchRegards and again thank you.