NimH AAA to 5V, suggestions?
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From: dagmargoodboat on 6 Aug 2010 12:28 On Aug 5, 6:17 am, Jan Panteltje <pNaonStpealm...(a)yahoo.com> wrote: > On a sunny day (Wed, 4 Aug 2010 20:36:09 -0700 (PDT)) it happened > dagmargoodb...(a)yahoo.com wrote in > <703ed962-238a-4025-a4b7-bebc334b4...(a)14g2000yqa.googlegroups.com>: > > > > >> PS > >> I build your circuit, and it works perfectly, even with almost empty NiMH= > >s. > >> Will make a little PCB for it, and use that, it seems more efficient then= > > my version. > >> No need to order some chip :-) > >> Here is the setup, I use the box to switch between full and almost empty = > >NiMH batteries. > >> ftp://panteltje.com/pub/1.2_to_5V_converter_test_img_2249.jpg > >> This is the collector waveform: > >> ftp://panteltje.com/pub/1.2_to_5V_converter_Vce_img_2250.jpg > >> The 7805 on the LED PCB is disconnected, I go in after that. > >> Project page: > >> http://panteltje.com/panteltje/pic/sign_pic/index.html > > >> Thanks :-) > > >The switching transistor isn't saturating very well--that's costing > >you a few % in efficiency. FMMT617 is quite good. What kind are you > >using? > > Just a BC547. > > >If the output's 5.5V, then that collector waveform must be about 1.2V/ > >division. What's the deal, is the output set low, or are those metric > >volts? > > I think in that version it was 4.85 V output. > See my other post for the improved version, different coil. I saw the other post--70% efficiency is decent. I get 80% commonly, 85% with great care in similar circuits. You have to use a low Vce(sat) transistor--otherwise 150mV out of 1.2Vcc in a discontinuous- mode converter works out to about a 8-10% loss. -- Cheers, James Arthur
From: dagmargoodboat on 6 Aug 2010 12:35 On Aug 6, 9:53 am, Jan Panteltje <pantel...(a)gmail.com> wrote: > On Aug 5, 11:45 am, "Tim Williams" <tmoran...(a)charter.net> wrote: > > > > > "Jan Panteltje" <pNaonStpealm...(a)yahoo.com> wrote in message > > >news:i3c59s$2r4$1(a)news.albasani.net... > > > > On a sunny day (Wed, 04 Aug 2010 16:31:50 GMT) it happened Jan Panteltje > > > <pNaonStpealm...(a)yahoo.com> wrote in <i3c4lv$1u...(a)news.albasani.net>: > > > >>This is that coil in detail, I added a 5 turns secondary for feedback > > >>:-) > > > Forgot lthe link: > > >ftp://panteltje.com/pub/1.2_to_5V_converter_coil_detail_img_2256.jpg > > > That's like the constant current buck switcher I made,http://myweb.msoe..edu/williamstm/Images/CC_Buck1.jpg > > but I only needed two turns. BJTs sure have high transconductance. > > > Tim > > Yes, I just monitored Vce on the scope and kept adding turns until I > no longer did see any improvement. > Also tried a synchronous detector with a BJT, but the Schottky was > better and simpler. > Even tried it as a switch 'upside down'. > About 70% efficiency is OK at 1.2 V I think. > What I am thinking about now is a simple low battery detector to > switch it off when battery voltage drops below say 1.05 V. > At .95 V the circuit no longer hacks it, and the output drops to 3V, > and The PIC stops working correctly. > It is easy to detect with a BJT using the 0.7 V Vbe as reference, , > but it needs some flip-flop action too. > To protect the expensive rechargeables. No need to protect a single NiMH cell--you can't damage it here. Full discharge is fine--damage comes from reverse charging the cell. This circuit can't do that. In practice you won't fully discharge the cell, because the PIC will quit before you ever get there anyway. -- Cheers, James Arthur
From: Jan Panteltje on 6 Aug 2010 13:08 On a sunny day (Fri, 6 Aug 2010 09:35:10 -0700 (PDT)) it happened dagmargoodboat(a)yahoo.com wrote in <48030cf6-4982-4bc9-a33f-5e979bb69cbf(a)f6g2000yqa.googlegroups.com>: >No need to protect a single NiMH cell--you can't damage it here. Full >discharge is fine--damage comes from reverse charging the cell. This >circuit can't do that. That iwould be good, that would simplify things. I was conidering using the analog comparator in the PIC, combined with some I/O and RC time. >In practice you won't fully discharge the cell, because the PIC will >quit before you ever get there anyway. Yes it will stop working normally, but all LEDs seem to go on then :-) A brown out detection and / or watchdog would simply keep restarting it. and the power conversion circuit will keep running and consuming some power of its own. Just looked up NiMH : Wikipedia seems to disagree with you: http://en.wikipedia.org/wiki/Nickel_metal_hydride_battery <Quote> A fully-charged cell measures 1.4?1.45 V (unloaded), and supplies a nominal average 1.25 V/cell during discharge, down to about 1.0-1.1 V/cell (further discharge may cause permanent damage, and the risk is increased with multi-cell packs). <end quote>
From: Jan Panteltje on 6 Aug 2010 13:11 On a sunny day (Fri, 6 Aug 2010 09:28:00 -0700 (PDT)) it happened dagmargoodboat(a)yahoo.com wrote in <cfbeee0b-9301-410a-997d-e61e355b605a(a)f42g2000yqn.googlegroups.com>: >I saw the other post--70% efficiency is decent. I get 80% commonly, >85% with great care in similar circuits. You have to use a low >Vce(sat) transistor--otherwise 150mV out of 1.2Vcc in a discontinuous- >mode converter works out to about a 8-10% loss. It is just a BC547, 'made in China' nothing special. You also lose .5V in the Schottky, at 5V that is an other 10%.
From: Jan Panteltje on 6 Aug 2010 13:52
On a sunny day (Fri, 6 Aug 2010 09:35:10 -0700 (PDT)) it happened dagmargoodboat(a)yahoo.com wrote in <48030cf6-4982-4bc9-a33f-5e979bb69cbf(a)f6g2000yqa.googlegroups.com>: PS I just had a cool idea how to do the auto shut off: -------------> to converter | + 1.2V NiMH _____| - | d === g ---------------------------< from Vout (5V nom) --- s IRLZ34N MOSFET |C1 | /// /// The MOSFET is in series with the battery minus connection. At switch on C1 is discharged, and the circuit will produce 5V output for a few milliseconds before C1 charges. That is enough to make the MOSFET conduct. When output voltage drops below Vgs of the MOSFET (say 2.5 V, else use a divider), the MOSFET Rds increases, output voltage falls further, and zero current will flow. I mean really ZERO. Now that is a minimum component solution of the ultimate kind :-) The Rds-on of this MOSFET is a few milli-Ohm, so 90 mA current causes millivolts drop, no losses to speak of,. I should patent this :-) Still have to try it though :-) But I *have* this MOSFET. |