Bipolar voltage supply with two XL6019 modules powered via USB - Page 1
Forum section: electronics / projects

author: josar
author_rank: Newbie
timestamp: December 11, 2024, 09:37:05 pm
content: Hello dear community,I am new here and have wanted to register for a long time but I have always lacked the decisive need. Now it's happened and I'm hoping for some expert help.I am currently building a power supply via USB for an old UNISEF Compact Disc Player 1984, for which I no longer have a mains adapter. The idea is actually quite simple, but the devil is in the detail. I'm using a USB-B socket as a 5V connection and two ready-made XL6019 Switching Current Boost DC/DC Converter Modules (incl. circuitry like Joy-IT XL6019). Now the CD player needs a bipolar power supply of +7.5 V and -7.5 V. I therefore take the two modules and use the positive pole of one module and the negative pole of the other module to provide the bipolar voltage for the CD player. I connect the unused poles as common ground (i.e. minus from the first voltage source and plus from the second). And this is where I now encounter the problem that the power supply breaks down completely.This is quite understandable, as I am short-circuiting the two voltage modules to a certain extent, as both modules are connected to the ground of the USB connection. As I understand it, I would have to disconnect the ground somewhere (....?), but then the voltage modules would probably no longer work at all, as they float without a reference.How can I solve the problem without taking a completely new approach? I would like to stick to my idea with the two XL6019 converter modules.Any help, any advice helps me a lot and I thank you very much in advance.Josar

author: trobbins
author_rank: Frequent Contributor
timestamp: December 11, 2024, 11:20:45 pm
content: Yes you need to isolate a supply rail somehow before you can configure a bipolar supply. I've used XL6009 based dc/dc modules and they are very convenient.One option is to use a USB battery 'brick' as an isolated supply rail - I use a brick for ground loop isolation applications, and it's often easy to locate an unused/replaced brick that still functions amongst your friends. The brick may have some 'noise' on it as part of its own conversion from battery to regulated 5V, but that should matter for your application. You may even be able to power the brick through a plugpack whilst using its 5V USB output.I'm not sure what small cheap dc/dc or dc/ac modules are available that provide transformer isolation and allow a low dc output (eg. 12V, but that would then need a step-down after it).

author: Benta
author_rank: Super Contributor
timestamp: December 11, 2024, 11:33:10 pm
content: How about if you tells us about power/current consumption?I'm pretty certain that not many people here know what a UNISEF thingummybob is.

author: RoGeorge
author_rank: Super Contributor
timestamp: December 11, 2024, 11:43:50 pm
content: Quote from: josar on December 11, 2024, 09:37:05 pmOne way would be to use first module between 0..5V USB to get 0..15V, then connect the second module between 0..15V to get 0 .. 7.5V. This way you will get GND, +7.5V, and +15V (relative to the USB GND), which they become respectively -7.5, 0V and +7.5V for the player.This should work, as long as you only listen to the player with a headset or its speaker, but it won't work if you want to plug the line out from the player back into the line in of the PC. (because the GND from the line out of your player might be in fact the +7.5V relative to the PC's GND).Another method would be to use the trick from this video, to get -7.5V regulated relative to the USB GND.The cheapest and simplest way of getting a negative voltage railFilip PiorskiThe module in the video is made with another chip, but the same idea should work for the XL6019 modules, too (though I didn't try, you may want to double check this by comparing the typical schematic between the two chips, search their datasheets and compare them).

author: Ian.M
author_rank: Super Contributor
timestamp: December 12, 2024, 12:18:10 am
content: The XL6019 doesn't have a synchronous rectifier, so a boost module made with it typically uses an external Schottky diode, which means it cant sink current, so RoGeorge's first suggestion wont work, even if you fix it so the second module runs from 5V in not 15V from the first module (because a XL6019 cant be a buck converter - see below). If the load draws more current from the positive rail than from the negative rail, the created 'ground' will rise, which the load will see as the positive rail collapsing together with negative rail overvoltage.You could rearrange it a bit to eliminate the need for a buck converter and prevent the created 'ground' rising - use one module to boost to +7.5V and call that load 'ground' (with USB Gnd becoming load -7.5V), and run the second module from +7.5V (and USB Gnd) boosting to +15V, which becomes load +7.5V. The problem with that is all the power for the second module goes through the first module, including its losses, so efficiency will suck, and I'd be surprised if you can get more than a couple of hundred mA out, even when fed by a 2A USB charger.His second suggestion is only applicable to buck converters, but the XL6019 has one side of its internal switch grounded, so cant be used as a buck converter.You could *try* an inverting charge pump circuit driven by the SW pin, but I suspect the high inrush current will trip the IC's internal overcurrent protection early, severely limiting the duty cycle and thus the output power.The datasheet shows how to get quasi-symmetrical positive and negative rails (Fig.18 on page 11 ofhttps://www.xlsemi.com/datasheet/XL6019-EN.pdf), but only the positive rail is actually regulated + you'd have to replace the inductor with a transformer, probably hand wound on a pot core. That's a lot of effort for an inferior result vs buying a buck module RoGeorge's second suggestion will actually work with. However note that inverting buck circuits typically require a very 'stiff' supply due to their high startup current, so you may run into problems with it tripping the over-current protection of your USB charger wallwart, and repeatedly failing to start.TLDR: with your power rail requirements and the likely current consumption of a mid-80's CD player, I wouldn't start with 5V USB power in!

author: josar
author_rank: Newbie
timestamp: December 12, 2024, 08:53:25 am
content: Thank you very much for all the constructive contributions. I would like to respond to some of the ideas raised in order to understand them better.Does USB battery brick mean a portable USB power bank? I would use this instead of the XL6019 or in front of it for ground isolation? Could it also work with an AC coupling using capacitors at some point? Just as an idea.The solution with buck converter is interesting, but I would have to deviate from the XL6019. I can still do that in the future. At first I wanted to investigate all the possibilities to stay with the XL6019 (investment is investment). But it looks like that's probably not going to work, at all.It is probably also not possible to replace the XL6019 with a pin-compatible buck converter IC on the ready-built M-SU-XL6019 module, as it still requires external circuitry.Incidentally, I assume a current requirement of 0.8 - 1.0 amps. USB should therefore be sufficient, I guess. I use a 1.8 amp USB charger.

author: trobbins
author_rank: Frequent Contributor
timestamp: December 12, 2024, 09:49:45 am
content: Yes, if you use a USB battery bank then that is isolated, and likely even isolated with the bank being charged by a wall-wart.You could also try and use two wall-warts, each into a XL6019 module, with one pos 7.5V output to the other's neg output. Perhaps try that first, as it may be the simplest approach. A concern may be switchmode related noise coupling into the disc player.

author: Ian.M
author_rank: Super Contributor
timestamp: December 12, 2024, 10:44:03 am
content: Quote from: josar on December 12, 2024, 08:53:25 amYou cant directly compare currents at different voltages when assessing the input supply requirements. Due to the action of switching DC-DC converters you need to compare power.A 1.8A 5.25V USB charger theoretically outputs 9.45W of power. A 2.1A one would output 11W However in practice you get less as the losses in the cable, and often the USB PSU design, typically cause the output voltage to droop at full load.If your load draws 0.8A at 15V (+/-7.5V) that's 12W, more than a normal single USB charger (non USB C) can deliver. However the DC-DC converters (XL6019 or any other) aren't 100% efficient so you need extra input power to make up for the converter losses.I hope you now understand trobbins's suggestion to use two USB chargers is because of *TWO* factors:One USB charger cant provide enough power.To be able stack the output sides of two XL6019 (or any other non-isolated) boost modules in series, their inputs need to be independently powered each with its own floating (non-grounded) supply.As I mentioned earlier I wouldn't start from there, especially not with two USB chargers as two wallwarts for one device with a 'soap on a rope' lump to combine their outputs is just plain fugly. However if you don't have a bipolar lab supply for testing, it would be a good 'quick & dirty' way to power the CD player to test how much current it actually needs at max. volume so you can design a proper power suppply for it.

author: josar
author_rank: Newbie
timestamp: December 12, 2024, 03:08:35 pm
content: Okay, got it. Then I'll fiddle together something like a USB power bank for one of the power rails and then measure the power requirement. Once that's clear, I can get to work on the development of a "clean power supply".In the latter case, one could theoretically take a 2A 7.5V transformer-based power supply, remove the bridge rectifier and insert 2 half-wave rectifiers (diode and electrolytic capacitor) for plus and minus. That already makes a "quick and dirty" +/- 7.5V power supply, doesn't it?

author: Ian.M
author_rank: Super Contributor
timestamp: December 12, 2024, 03:19:29 pm
content: Yes, and 'back in the day' many types of equipment that needed positive and negative rails got them that way from a simple AC output wallwart with a two wire output lead. The problem is the ripple from a half-wave rectified supply is 50 Hz not the 100 Hz one gets from a full wave rectifier so for the same load current it needs double the reservoir capacitance. With a vintage wallwart and modern capacitors, you can *probably* cram enough reservoir capacitance in there . . .Also, as the CD player in question ran from 10 1.5 V batteries, it may not tolerate an unregulated linear power supply (which depending on the transformer can rise significantly when unloaded - some small transformers, the voltage can double), so you may need to regulate a higher voltage positive and negative supply down to +/- 7.5 V eg using LM317 and LM337 regulators, which at 1A load current *will* require heatsinking, which cant be done effectively inside a typical wallwart.

author: trobbins
author_rank: Frequent Contributor
timestamp: December 12, 2024, 11:12:34 pm
content: It may be easier to locate two plugpacks with 12Vdc output - as they are somewhat ubiquitous for a large number of commercial products. If they don't have an output dc that is galvanically connected to mains protective earth (eg. mains 2-prong only), then likely they can each be linear regulated to 7.5Vdc. If the disk player load is fairly constant, then a dropping resistor can be located before the linear regulator so that the regulator dissipates somewhat less power and hence alleviates heatsinking. The two 7.5V rails can then be connected in series, and the mid-point used as 0V for the disk player. If you already have 5V or 6V positive linear regulators then they can be used as circa 7.5V regulators with a LED or a few 1N4148 etc in the ground leg. If the disk player was battery powered originally, and it used 5x 1.5V cells for each 'side', then the player likely can be operated from any dc voltage between say two by 6-8V.