At 02:29 PM 11/20/2011, you wrote: >Looking at this problem again. > >Would it not make sense that I use 6 volt battery with a LDO regulator for >the 5 V? It looks like the drop out voltage for switching regulators is >kind of large, around 1.5 Volts, so I would have problems. >Using a LDO reg If you're making your own buck regulator, from 6V, yes. The modules can accept 4.5-9V or 9-18V for 5.0V or 12V output. >efficiency wise would be better than boosting to say 7 volts (or 12 volts = as >you suggest) and then bucking to 5 volts. I should be able use most of th= e >battery capacity with a LDO. > >Thanks, > >Gordon Williams Yes, it might. You'll need to sharpen your pencil a bit on the minimum voltage.. take into account discharge rate and temperature effects, and think about what happens when the battery reaches the drop-out point Also, when the battery is being charged you might see around 1W dissipation in the regulator if the load can be connected while charging (and much wors= e were the battery to be dried out or disconnected). Your reverse battery protection has to be low drop too. It's not a bad approach to try. Efficien= cy will be pretty good with this approach. Step-up from 6V to 12V should be straightforward- something like a MC34063 would be the cheapest in volume. The 5V regulator will probably drop out with switching noise from the 12V regulator on top of the output. Converter topologies such as flyback or SEPIC (or those that have transform= ers) don't have the limitations of the simple buck regulators so that's another set of options. There are lots of way to do this.. it depends a lot on the details of your application which is best for you. --sp >----- Original Message ----- >From: "Spehro Pefhany" >To: "Microcontroller discussion list - Public." >Sent: Sunday, November 20, 2011 11:04 AM >Subject: Re: [EE] Battery Configuration with boost or buck converter > > > > At 09:01 AM 11/20/2011, you wrote: > > >Hi, > > > > > >I have to create at dual power supply from batteries: > > > > > >5V @ 500 mA > > > > =3D 2.5W > > > > > > >12V @ 250 mA > > > > =3D 3W > > > > So, roughly equal power consumption on each. > > > > > > >I going to be using 2, 6 Volt 5Ahr sealed lead acid batteries and I'm > > >wondering if it is more efficient/ better to connect them in series or > > >parallel. > > > > Why not start with a 12V 5ah SLA (or 6V 10ah)? You'll have > > less worries about balancing during charge/discharge. > > Anyway, I don't think there's a simple single * answer to this question >for > > switching converters in general. > > > > >Therefore is it more efficient to have a 6V battery supply and step it= up >to > > >12 V and down to 5 V or start with 12V battery and try and get 12 volt= s > > >regulated and 5 volts out of it. > > > > > >When I have a buck regulator, what happens when the input voltage goes >below > > >the output voltage. Is there a dropout voltage similar to a linear > > >regulator? Or, do I just get the unregulated battery voltage at that >time? > > > > You'll get a dropout, but at some point it should cut out entirely so a= s >not > > to damage the switch by causing it to go linear. Step up regulators >typically > > can't output much less voltage than the input, so if the input goes muc= h > > higher than the desired output you lose regulation on the other side. > > > > That's assuming no transformers. If there are transformers you can have > > whatever ratio you want, within reason. > > > > >I'm also looking for a source of pre-made modules around 25 mm x 50 mm= at >a > > >cost < $10 each in ones. Anyone have a favourite source? > > > > There are tons of them (though few are < $10 in singles). Try a paramet= ric > > search at Digikey or whatever. Some of the cheaper ones don't have prop= er > > short-circuit protection, so that should be provided externally if >required. > > Galvanic isolation is nice to have. I've used the TDK CC-E series- they= 're > > cheap and small (but non-potted and electrically noisy). < $11. in > > singles. > > > > >I understand the boost/buck idea, but don't have any experience using >these > > >regulators so any insight of there actual operation would be useful. > > > > You can look at SEPIC converters- they require two inductors each. You >could > > also have a 6V-> 12V converter at 5.5+W then buck that down to 5V with = a >2.5W > > converter, or a 12V->6V at 5.5+W then boost that to 12V with a 3W >converter. > > You lose efficiency both ways, so if each converter is 80% efficient, y= ou > > end up with end-to-end 64% for the portion that goes through both, and = the > > first converter has to be sized for both loads AND the inefficiency of = the > > second converter. This works better if one load is much smaller than > > the other. > > > > * "correct" being assumed > > > > >Best regards, > > > > Spehro Pefhany --"it's the network..." "The Journey is the >reward" > > speff@interlog.com Info for manufacturers: >http://www.trexon.com > > Embedded software/hardware/analog Info for designers: >http://www.speff.com > > > > > > > > -- > > http://www.piclist.com PIC/SX FAQ & list archive > > View/change your membership options at > > http://mailman.mit.edu/mailman/listinfo/piclist > >-- >http://www.piclist.com PIC/SX FAQ & list archive >View/change your membership options at >http://mailman.mit.edu/mailman/listinfo/piclist Spehro Pefhany --"it's the network..." "The Journey is the rewar= d" speff@interlog.com Info for manufacturers: http://www.trexon.co= m Embedded software/hardware/analog Info for designers: http://www.speff.co= m --=20 http://www.piclist.com PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist .