DC-DC converters - Page 1
Forum section: electronics / projects

author: jacobfeder
author_rank: Contributor
timestamp: June 10, 2016, 06:22:35 am
content: Hi,I am designing a mixed-signal system that I want to be battery powered (preferably 4xAAs) and am trying to figure out the best way to step up/down my battery voltage. The different rails I need are:+15V @ 80mA-15V @ 30mA+5V @ 50mAThese are measured consumption values so I probably want about double for safety margin.Obviously the +-15V is for the analog sections. The 5V is needed in both analog and digital sections. I am thinking I might want separate supplies for the analog and digital 5V rails.I want the system to have quite high resolution and accuracy (ADC is 16 bits), so I am a bit worried about the noise from switching DC-DC power supplies. At the same time, the measurement is very low frequency (nearly DC, somewhat like a strain gauge) so the noise may not matter much because it will be averaged out / LPFed. I was thinking about using a combination of a SMPS + linear regulator for the analog section. The digital can have just the SMPS. Any advice?Thanks so much!

author: JPortici
author_rank: Super Contributor
timestamp: June 10, 2016, 08:37:54 am
content: with the correct transformer you should be able to get all three voltages from a single flyback regulator/controller (you can find all info from TI's datasheets)look up the jim williams video regarding minimizing switching regulator's noise.with a 16 bit adc i also recommend a dedicated voltage reference

author: batteksystem
author_rank: Regular Contributor
timestamp: June 10, 2016, 10:08:38 am
content: Quote from: jacobfeder on June 10, 2016, 06:22:35 amFor 4xAA, 1.5V x 4 = 6V1. To prepare for +15V / -15V rail, I will boost the 6V up to 36V, if current is not large you can use a charge pump (hence the integer multiplication), drop it back to 15V with virtual ground in the middle with positive and negative LDOs.2. The current for 5V isn't big, I will just use a good LDO to handle it.

author: sotoszorbas
author_rank: Contributor
timestamp: June 11, 2016, 01:25:14 am
content: Quote from: jacobfeder on June 10, 2016, 06:22:35 amI've made a PSU using as preregulator a step down converter that I design. For the linear part the I used the usual solutions. Let me tell you that if you want a noise free power supply do not use a SMPT . your currents are small enough so the linear option should be OK! If you need to step up the voltage (because of the batteries) use the well known mc34063.http://www.onsemi.com/pub_link/Collateral/MC34063A-D.PDFDaves video:

author: Buriedcode
author_rank: Super Contributor
timestamp: June 11, 2016, 10:48:52 am
content: At those currents the +/-15V supplies can come from discrete charge pumps from a switching power supply. Each one would need a post regulator - unless the rails are not used as a reference, and merely power for the opamps, then you could get away with simple filters. You would need to 'do the math' and simulate them so they can provide sufficient current.4 AA's, at the end of their charge will only provide ~4V, so an LDO for the 5V rail is out.Perhaps a boost converter that has an operating voltage higher than 5.5V (often the max for many converters designed for batteries) set to 6-6.5V. This means the converter will have a very low duty cycle because its only 'boosting' 0.5V or so. Then an LDO for the 5V rail, and then 3 stage charge pumps for ~ +/- 17.7V. - thats (6.5 x 3) minus two diode drops for each stage. A Bat54 will be about 0.3V, so losing 0.6V per stage. With 3 stages, and 6.5V in, thats 19.5V out - (3*2*0.3) = 17.7V.This gives you enough room for a linear regulator.Of course you could just boost to 9V, and use a voltage doubler for 18V, then an inverter after that to get ~-17V from that. Many integrated (and easy to use!) charge pump devices can operate up to 250kHz, some up into the MHz, but I have always used a linear regulator after these

author: Fungus
author_rank: Super Contributor
timestamp: June 11, 2016, 10:57:29 am
content: Quote from: jacobfeder on June 10, 2016, 06:22:35 amYour current draw is very small so maybe the inductor +filtering in your switching supply will be enough.Even so, youalwaysneed decoupling as close to the ADC chip as possible. Put three or four small, ceramic decoupling capacitors in parallel right next to the ADC's power input pin. You can't make these wires too short.The current draw of the ADC will also be fairly constant so if decoupling capacitors aren't enough then another inductor (which is a different size to the inductor on the switcher so they don't resonate) won't hurt. Put an inductor in series and decoupling capacitors on either side, like this.(and maybe parallel the decoupling capacitors on the output side)Quote from: jacobfeder on June 10, 2016, 06:22:35 amThat *might* be enough, who knows? The only way to find out is to try it and see. Take lots of samples ... anddon'ttake them in sync with your ripple (it could happen!).

author: mikerj
author_rank: Super Contributor
timestamp: June 11, 2016, 12:45:23 pm
content: 4xAA to a reasonably low current 5v rail can be done pretty easily using SEPIC type converter which can both raise and lower the supply voltage. This is still a fairly simple circuit, requiring one extra inductor and capacitor over a standard boost or buck configuration.

author: mariush
author_rank: Super Contributor
timestamp: June 11, 2016, 06:41:51 pm
content: QuoteMeasured values, so you really should aim for more than those. Let's say 100mA for +15v , 50mA for -15v and 100mA for 5v ...AA batteries have enough current output at down to about 1v ... you can check batteryshowdown site to see how batteries behave, see low drain (200mA) :http://www.batteryshowdown.com/results-lo.htmlor high drain (1A) :http://www.batteryshowdown.com/results-hi.htmlKeep in mind that there's going to be some efficiency issues when boosting the voltage.. a cheap MC34063 will probably be at around 65-70% efficiency when boosting 6v to 15v so to get your 15v @ 100mA 1.5w you're looking at about 1.9w / 6v = 0.3A or 0.45A in worst case (at 4x1v) it's a significant constant load on just 4 AA batteries.Going with a better boost regulator will matter here .. go for something like aAP3012which will give you over 80% efficiency, orMIC2288which will give you over 85% efficiency orMIC2619(but note this one only handles up to 6.5v at input) orMIC2145with low quiescent current ... and both can work with tiny surface mounted inductors and ceramic capacitors, mc34063 needs big inductors and capacitors to output something reasonable.

author: jacobfeder
author_rank: Contributor
timestamp: June 15, 2016, 03:56:16 am
content: Thanks all for the helpful replies. I will give it a try.

author: jacobfeder
author_rank: Contributor
timestamp: June 16, 2016, 04:24:55 am
content: After doing a bit more searching I'm still in a bit of a pickle. I've decided to go with 18650 lithium cells instead of AA's for capacity reasons. Seems like no matter which circuit / chip combo I look at there is a big downside.Just to clarify a few things:1. The ADC has a dedicated voltage reference so a bit of noise on the rails isn't a huge issue, especially if it can be filtered with an LC first.2. I am interested in measuring the battery voltage with my 5V micro. This is an issue if you use a 30V output regulator and an opamp to generate the virtual ground since the battery voltage will be -15 + 3.7V = -11.3V relative to the micro.3. I am looking for something that is reasonably efficient. Sadly a lot of the converters I've seen (MC34063) are something in the 50-60% range. I was hoping to get more like 75-90%Any more ideas? I'm surprised there aren't more cookie cutter solutions out there considering how common +-15V rails in are.Thanks