LTC3786 3V to 18V boost converter - Page 1
Forum section: electronics / projects

author: Hamed_ta
author_rank: Contributor
timestamp: November 14, 2019, 04:28:30 am
content: Hi,I'm designing a li-po battery booster with LPC3786 which is able to convert 3-4.2V battery voltage to 18V 5A, for this purpose I tried to customized one of the sample circuits from the datasheet, (3786 F13a ) I have changed a few of components and entered the values in the LTpowerCAD project file and it seems to work, the only problem I can see is that as the original sample circuit's output voltage is 5V the output of the circuit is connected to the output pin of the LTC1754 pin (through the body diode of mosfet Q), which has the maximum rate 6V of the pin voltage so now with the output voltage of 18V is there any other solution for that ? maybe just replacing mosfet with a simple diod ?is there any other problem you can mention in the circuts ?

author: SiliconWizard
author_rank: Super Contributor
timestamp: November 14, 2019, 03:49:06 pm
content: Quote from: Hamed_ta on November 14, 2019, 04:28:30 amGood catch! You could have just overlooked that and be in for some nice magic smoke.There was no problem in the original design as the output voltage was 5V and the output of the LTC1754 -5 was also 5V.The issue here is that the VBIAS pin requires a min. voltage of 4.5V, which is why they added a charge pump to generate 5V from the battery. Directly connecting VIN (thus the battery) to VBIAS, which is the simplest way of using this converter if VIN >= 4.5V, and what is suggested in the datasheet, was not an option here.I'm just having a quick look, but am I right thinking that the whole idea around D2 and the PMOS controlled by PGOOD is for bootstrapping (meaning the VBIAS will be taken from VOUT through D2 once the LTC3786 has reached regulation)? I'm not sure I see a big benefit power-consumption-wise, as the charge pump will still be running all the time? So couldn't you just 1/ remove D2 and 2/ directly connect the VOUT from the LTC1754 to the VBIAS pin, remove the PMOS, and call it a day?Edit: after reading a bit more closely the datasheet, it appears there can be significant current drawn from VBIAS (whereas I had assumed it was negligible), thus this explains why they added this switching over circuit. (I still thnk it could be further improved by automatically shutting down the external charge pump when VOUT is regulated, but I see the point.)So yeah, you have two options here: you can either replace the PMOS with a diode (VOUT of LTC1754 to VBIAS), which I think would be good enough for bootstrapping as long as you use a schottky diode (remember you must provide at least 4.5V, thus you're allowed MAX 0.5V drop with the diode...). Not the most efficient, but since it would only provide current while starting, I don't think it matters. The other option would be to replace the single PMOS with two PMOS back-to-back for a true bidirectional switch.As I said above, it could be a bonus idea to find a way to shut down the external charge pump (/SHDN pin) using the PGOOD output of the converter, so you'd save some power once the converter runs regulated (always a good idea when running off batteries!) (Admittedly, with a typ. operating current of 11µA, it may not matter much, depends on your overall power consumption and the battery's capacity of course!)

author: Miti
author_rank: Super Contributor
timestamp: November 14, 2019, 04:20:22 pm
content: You may be dangerously close to the Vdss max of the power transistors. It is only 20V.

author: Hamed_ta
author_rank: Contributor
timestamp: November 16, 2019, 12:29:00 pm
content: Thank you for your analyzing it was awesome,I've already tested the schottky diode instead of PMOS but something strange happened, still I have about 13 volt on LTC1754 out pin! which I don't know where it comes from !? maybe it the leakage current of diod? do you think it can be harmful for the LTC1754 ?and in general do you think this circuit will work ?

author: Hamed_ta
author_rank: Contributor
timestamp: November 16, 2019, 12:31:24 pm
content: Quote from: Miti on November 14, 2019, 04:20:22 pmyes thank you butI'm going to limit the output voltage to 18 volt do you think still it can be dangerous ?

author: Hamed_ta
author_rank: Contributor
timestamp: November 23, 2019, 04:36:11 pm
content: any comment ?

author: Hamed_ta
author_rank: Contributor
timestamp: December 20, 2019, 02:13:06 pm
content: I designed a PCB for this project, can you please check that and let me know if there is any design problems in that ?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 25, 2019, 01:42:33 am
content: Yikes, cut planes!? No no no no!!Tim

author: SiliconWizard
author_rank: Super Contributor
timestamp: December 25, 2019, 01:48:51 am
content: Quote from: T3sl4co1l on December 25, 2019, 01:42:33 am+1

author: NiHaoMike
author_rank: Super Contributor
timestamp: December 25, 2019, 04:46:21 am
content: If you just need about 20V or so, go with a 5S pack rather than trying to parallel a bunch of cells and then boosting the voltage. It will be cheaper and more efficient.

author: Siwastaja
author_rank: Super Contributor
timestamp: December 25, 2019, 12:40:36 pm
content: Quote from: Hamed_ta on November 16, 2019, 12:31:24 pmNo matter how perfect your layout will be, you will have parasitic ringing on the switch node which will exceed the 20V rating. By how much? Hard to say. As some kind of rule of thumb, a very well laid out converted might achieve 15-20% overshoot voltage, while a decent one gets around 40%, and a poor beginner project could closer to 100% overshoot.I strongly recommend on designing so that the ringing won't exceed the Vds rating, ever.Still, it may work: it depends on the avalanche rating of the MOSFETs, and the ringing energy in the LC parasitics. This energy is now converted as extra heat in the avalanching MOSFET. You will see it in the scope trace as a flattening SW node voltage somewhere around 22-23V or whatever the avalanching voltage of your 20V MOSFET is.It's easier to analyze when youdon'thit the avalanche rating. You don't necessarily need to addmuchextra voltage headroom, but do leave a little. Then the avalanche capability of the MOSFET works as your safety margin. Use a high bandwidth scope (100MHz as a very bare minimum, preferably more) to look for the maximum voltage of the SW node.An RC snubber may be used to reduce the overshoot, but a properly sized MOSFET that can handle the overshoot is likely simpler.Vds max = 25V would be a good starting point, and even that doesn't have much excess, only some 38%.

author: Hamed_ta
author_rank: Contributor
timestamp: December 25, 2019, 02:18:59 pm
content: what does it mean ?

author: Hamed_ta
author_rank: Contributor
timestamp: December 25, 2019, 02:20:57 pm
content: it is designed to work with only single cell batteries to avoiding balance charging and discharging trouble.

author: Hamed_ta
author_rank: Contributor
timestamp: December 25, 2019, 02:22:06 pm
content: yes good point, thank you I will change the Mosfets, so there isn't any other problem in design and layout ?

author: Hamed_ta
author_rank: Contributor
timestamp: December 25, 2019, 02:29:38 pm
content: I was thinking of adding precise output current adjustment, is it possible you think by just adding another voltage divider in the circuit of inductor current sensing to be able to controll inductor peak current to adjust output current precisely?

author: NiHaoMike
author_rank: Super Contributor
timestamp: December 25, 2019, 03:55:36 pm
content: Quote from: Hamed_ta on December 25, 2019, 02:20:57 pmA balancing BMS would be way cheaper than such a big boost converter.

author: Hamed_ta
author_rank: Contributor
timestamp: December 25, 2019, 08:39:28 pm
content: Does it discharge batteries balanced ? And is it efficient?

author: NiHaoMike
author_rank: Super Contributor
timestamp: December 25, 2019, 08:45:31 pm
content: Those BMS modules typically balance on the charge cycle and the decent ones lose little within their ratings. (It's worth noting that a good quality pack will be well matched and the balancer rarely comes into play.) To further lower losses, you can get a BMS with a higher current rating than you need and add a fuse for overcurrent protection.

author: Hamed_ta
author_rank: Contributor
timestamp: December 26, 2019, 02:14:13 am
content: exactly the problem is that I could charge the batteries in series well balanced but for discharging under high current the unbalanced discharging problem was still there, also other reason is that in parallel, it is possible to have batteries in different capacities, right?

author: Siwastaja
author_rank: Super Contributor
timestamp: December 26, 2019, 08:57:51 am
content: Quote from: Hamed_ta on December 26, 2019, 02:14:13 amA proper BMS takes care of all that.The #1 task of a BMS has nothing to do with balancing, it's cell-level cutoff for both high and low voltage threshold. Meaning it controls your charger to turn off when any cell hits full, and controlsallyour loads to turn off when any cell hits empty.Balancing is a bonus. Very little balancing is needed with li-ion. The prerequisite for most BMS's to be able to balance is that youdofull charge cycles every once in a while. If you don't, nothing catastrophic happens, but slight imbalance may develop and reduce the pack capacity in the long run. This may be an actual issue if the user partially charges for years, for example, always running between 80% and 20%.Note that the market is full of different products tagged "BMS", and some of them won't do shit, or are outright dangerous. Even with a proper BMS, you need to make sure you let it do its job: turn off the charger, and turn offallthe loads. Some have integrated MOSFET power switches that actually disconnect the battery, and are hence easier to use, but include extra price and extra losses in these power switches.The market is saturated with BMS ICs for 5s or 6s li-ion, so you have a lot to choose from; and a lot of appnotes to base your design on. Some have dangerous catches, but your alternative of designing a simple 1s BMS + boost converter has traps as well, so if you actually need high-power unregulated 18Vish, I would go with 5s li-ion instead of 1s + boost.

author: Hamed_ta
author_rank: Contributor
timestamp: December 26, 2019, 01:23:03 pm
content: still I think the esiest and most flexible way to design a battery powered power supply is single cell battery approach because how many batteries with diffrent capacities you would have you can be sure that always all of them will charge and discharge fully balanced, but it has it's own drawbacks too including1 - by only using boost converter you can't make voltages less than input voltage for example 3.3V output when battries is fully charged.2 - the VGS voltage for biasing common Mosfets is higher than input voltage so you need use logic level mosfets with low maximum VDS which limits output voltage.3 - the current will be really high so you should take care of high current issues (huge inductor with high I sat).4 - the boost converter doesn't disconnect output from input when it's disable you should take care of that with extra switch.5 - the voltage for supply IC is too low so same as my case a charge pump extra booster is needed.despite all the concerns that I mentioned as the most important and expensive part in a battery power supply is the battries, still I think using one cell battries and boost the voltage is the more rational as the charge circuit design is simple and straight forward.now what approach do you have in mind for adjust output current ?however please let me know if you know any BMS design which be able to take care of all concerns in series battery maybe in next design I put it in mind.

author: NiHaoMike
author_rank: Super Contributor
timestamp: December 26, 2019, 04:13:51 pm
content: Even in laptops, where the majority of the load is on the order of 1.2V or less, it's very uncommon to find a 1S battery configuration. My guess is that it's easier to deal with lower current at a higher voltage.

author: Hamed_ta
author_rank: Contributor
timestamp: January 03, 2020, 09:16:39 am
content: this converter should be which be able to produce voltages from 5v to 18v and current 0-5A,but when I design in LTpowerCAD by reducing output voltage and current the inductor and sense resistor values which were ok for maximum voltage and current output, get out of range ? what is the solution ?