Tamas Rudnai wrote:
> As far as I understood solarwind does not want to put a dc-dc in his
> application, and that's where all the problems occur. In other words
> he cannot guarantee stable USB voltage level,
But the point is he can well enough. Four NiMH batteries will produce right
around 5V when full. At 1.05V per cell the voltage drops to 4.2V, which is
just a fraction below the minimum USB voltage. Since Apple is drawing
higher than normal currents, they have to be prepared for a larger cable
drop, so most likely the phone will still charge at 4.2V and a bit lower.
At 1.05V there is still some capacity left in a NiMH cell, but a good
fraction has been used. Also the phone will likely charge with a bit lower..
A simple test with a variable power supply would confirm this, but the OP is
stubbornly refusing to do that.
On Mon, Dec 6, 2010 at 11:25 AM, Olin Lathrop <spam_OUTolin_piclistTakeThisOuTembedinc.com> wrote:
> Tamas Rudnai wrote:
>> As far as I understood solarwind does not want to put a dc-dc in his
>> application, and that's where all the problems occur. In other words
>> he cannot guarantee stable USB voltage level,
>
> But the point is he can well enough. Â Four NiMH batteries will produce right
> around 5V when full. Â At 1.05V per cell the voltage drops to 4.2V, which is
> just a fraction below the minimum USB voltage. Â Since Apple is drawing
> higher than normal currents, they have to be prepared for a larger cable
> drop, so most likely the phone will still charge at 4.2V and a bit lower.
>
> At 1.05V there is still some capacity left in a NiMH cell, but a good
> fraction has been used. Â Also the phone will likely charge with a bit lower.
> A simple test with a variable power supply would confirm this, but the OP is
> stubbornly refusing to do that.
>
I just hooked my $500 cell phone up to a variable power supply and found:
Stubbornly refusing to go higher than 5.5v. This is with both data
lines floating. I was surprised that the current went over 100mA
without any deliberate voltage dividers on the data lines.
This is also through about 5 feet of what appears to be AWG 26 wire so
there will be some associated voltage drop.
It's an HTC phone (Google Nexus One) but there's a good chance phones
of similar vintage do similar things with respect to charging.
I took Martin's data and made a plot of it, which is attached.
This says the phone has just a linear regulator between the USB power and
the LiIon battery. If there was a switcher, the current would not have been
so completely flat from 4.8V on. If the iPhone does this then just 4 NiMH
batteries driving the USB power will work, but leave a significant fraction
of the NiMH battery capacity unused.
part 2 10132 bytes content-type:image/gif; name="phone.gif" (decode) part 3 181 bytes content-type:text/plain; name="ATT00001.txt" (decoded base64)
-- http://www.piclist.com PIC/SX FAQ & list archive
View/change your membership options at
mailman.mit.edu/mailman/listinfo/piclist
>
> I took Martin's data and made a plot of it, which is attached.
>
> This says the phone has just a linear regulator between the USB power and
> the LiIon battery. If there was a switcher, the current would not have
> been
> so completely flat from 4.8V on. If the iPhone does this then just 4 NiMH
> batteries driving the USB power will work, but leave a significant fraction
> of the NiMH battery capacity unused.
>
Olin,
Why would you say that? I suspect that there is some sort of a switching
power supply inside the iPhone from various random articles I read on the
Internet. Of course, they could be wrong.
However, a threshold voltage of 1.15 volts/cell would not allow the pack to
drop below 4.6 volts. At that voltage, the pack would be still be drained
above 90% (if not 95%?) according to the chart here:
He is saying that there doesn't seem to be a switcher because
switchers have a constant power input characteristic within their
valid input operating range and this device does not display this
behavior.
Say, for example, that I have a boost switcher which takes a voltage
between 2V and 8V input and delivers a regulated 12V output. I then
connect a load to the 12V which draws 1 Amp. Let's also say that the
efficiency of the switcher is a constant 85% from 2V to 8V input (not
really realistic but perhaps not too far off). Then, there is 12W
being delivered to the load and 12/.85=14.1 W being drawn from the
source, independent of the input voltage. So, at 2V, the power supply
will draw 14.1/2=7 Amps. At 5V, it will draw 2.8 Amps. At 8V, it will
draw 1.8 Amps. Notice that the current drain goes down linearly as
input voltage rises. This "negative resistance" characteristic (not
that V/I is negative but that dV/dI is negative) is a good sign that
you are dealing with a switching power supply.
A linear regulator tends to draw a constant current from the source
regardless of the source voltage. This is what Olin saw in the graph.
The values on the low end where the current falls off as the voltage
drops are likely the "dropout" region of the linear regulator, where
the output voltage is no longer being regulated but rather is
following the input voltage minus an offset.
> On Mon, Dec 6, 2010 at 1:52 PM, Olin Lathrop <.....olin_piclistKILLspam.....embedinc.com>wrote:
>
>>
>> I took Martin's data and made a plot of it, which is attached.
>>
>> This says the phone has just a linear regulator between the USB power and
>> the LiIon battery. If there was a switcher, the current would not have
>> been
>> so completely flat from 4.8V on. If the iPhone does this then just 4 NiMH
>> batteries driving the USB power will work, but leave a significant fraction
>> of the NiMH battery capacity unused.
>>
>
> Olin,
>
> Why would you say that? I suspect that there is some sort of a switching
> power supply inside the iPhone from various random articles I read on the
> Internet. Of course, they could be wrong.
>
> However, a threshold voltage of 1.15 volts/cell would not allow the pack to
> drop below 4.6 volts. At that voltage, the pack would be still be drained
> above 90% (if not 95%?) according to the chart here:
>
> http://www.eneloop.info/home/performance-details/discharge-current.html
>
> I think that's a reasonable voltage to cut off at.
> A linear regulator tends to draw a constant current from the source
> regardless of the source voltage. This is what Olin saw in the graph.
> The values on the low end where the current falls off as the voltage
> drops are likely the "dropout" region of the linear regulator, where
> the output voltage is no longer being regulated but rather is
> following the input voltage minus an offset.
>
I thought that graph was *not* from an iPhone, but some Android one? What
also Olin said is that solarwind would need to measure his iPhone and draw
that graphs to go any further from this point.
>
> Sean
>
>
> On Mon, Dec 6, 2010 at 6:13 PM, V G <x.solarwind.xspam_OUTgmail.com> wrote:
> > On Mon, Dec 6, 2010 at 1:52 PM, Olin Lathrop <@spam@olin_piclistKILLspamembedinc.com
> >wrote:
> >
> >>
> >> I took Martin's data and made a plot of it, which is attached.
> >>
> >> This says the phone has just a linear regulator between the USB power
> and
> >> the LiIon battery. If there was a switcher, the current would not have
> >> been
> >> so completely flat from 4.8V on. If the iPhone does this then just 4
> NiMH
> >> batteries driving the USB power will work, but leave a significant
> fraction
> >> of the NiMH battery capacity unused.
> >>
> >
> > Olin,
> >
> > Why would you say that? I suspect that there is some sort of a switching
> > power supply inside the iPhone from various random articles I read on the
> > Internet. Of course, they could be wrong.
> >
> > However, a threshold voltage of 1.15 volts/cell would not allow the pack
> to
> > drop below 4.6 volts. At that voltage, the pack would be still be drained
> > above 90% (if not 95%?) according to the chart here:
> >
> > www.eneloop.info/home/performance-details/discharge-current.html
> >
> > I think that's a reasonable voltage to cut off at.
On Mon, Dec 6, 2010 at 7:45 PM, Tamas Rudnai <KILLspamtamas.rudnaiKILLspamgmail.com> wrote:
> I thought that graph was *not* from an iPhone, but some Android one? What
> also Olin said is that solarwind would need to measure his iPhone and draw
> that graphs to go any further from this point.
>
And Russell said that the iPhone 3 has a linear regulator charge controller..
It's hard to believe the iPhone 4 would be drastically different,
considering the convenient 5v supply for charging the 4.2v lithium
battery.
-- Martin K
> And Russell said that the iPhone 3 has a linear regulator charge
> controller.
> It's hard to believe the iPhone 4 would be drastically different,
> considering the convenient 5v supply for charging the 4.2v lithium
> battery.
>
<www.tgdaily.com/hardware-features/50344-the-real-iphone-4-teardown>Detailed
reports on D1815a - login and probably for money.
www.ubmtechinsights.com/reports-and-subscriptions/open-market-reports/Report-Profile/?ReportKey=6976
<www.ubmtechinsights.com/reports-and-subscriptions/open-market-reports/Report-Profile/?ReportKey=6976>
But if the material below is both accurate and complete it tells us enough.
Buck regulators mentioned and no boost.
Suggests effectively linear with switching & inductor = buck to improve down
conversion efficiency (as per my prior comments). This gives useful but not
vast gains when battery is low and USB input high.
Re question on LiIon voltage.
Std LiIon charges at constant current to Vplateau and is then charged at
Vplateau with decreasing current until some terget Ilow is reached.
Vplateau is set by design.
Typically 4.2V.
4.3V for the brave and foolish with somewhat more capacity and perhaps
prettyflames.
4.1V for the conservative with notably less capacity and notably longer
cycle life.
4.0V by nobody but less and better again.
As a rule, as Vplateau is reduced the capacity per cycle drops BUT total
capacity/cycle x cycles rises (subject to limits imposed by other
constraints such as calendar life).
LiFePO4 have a 3.6V Vplateau.
DISCHARGE terminal voltage for std LiIon starts at about 3.7V (soon 3.6)
hence the basis for the query.
LiFePo4 starts at 3.3 - 3.2V terminal.
Russell
>From above page:
The following functional blocks will be extracted, analyzed and organized
into hierarchical schematics in the Full Analog Circuit Analysis:
• Various LDO Voltage Regulators
• Buck Regulators
• Battery Management Circuitry
> On 07/12/2010 01:52, M.L. wrote:
> > On Mon, Dec 6, 2010 at 7:45 PM, Tamas Rudnai<TakeThisOuTtamas.rudnaiEraseMEspam_OUTgmail.com>
> wrote:
> >> I thought that graph was *not* from an iPhone, but some Android one?
> What
> >> also Olin said is that solarwind would need to measure his iPhone and
> draw
> >> that graphs to go any further from this point.
> >>
> >
> > And Russell said that the iPhone 3 has a linear regulator charge
> controller.
> > It's hard to believe the iPhone 4 would be drastically different,
> > considering the convenient 5v supply for charging the 4.2v lithium
> > battery.
>
> Is this an iPhone4?
> I was sure I'd seen a teardown of it somewhere online recently, thought
> maybe that would be an idea to get info on the charging circuit if the
> wish is to avoid testing with a bench supply (popular product, sure
> there has to be someone that pulled it apart.
> (I found out below someone has and it doesn't help much anyway...)
> I have access to an iPhone 1 here I would be willing to test, but they
> may have changed things since then.
> I believe they have from what I discovered - I think the iPhone 1 uses
> an LTC 4066 IC - info here:
> tzywen.com/modules.php?name=News&file=article&sid=696
> <http://tzywen.com/modules.php?name=News&file=article&sid=696>
>
> This page might be of interest from Lady Ada:
> www.ladyada.net/make/mintyboost/icharge.html
> Seems the Apple stuff can get very picky about the datalines (similar to
> my Omnia - had to hack a charger for that, would not simply charge from
> any USB source) needing voltage dividers etc.
> I also found the iPhone4 uses this power management IC:
> Dialog D1815A 338S0867
> Info from this page:
>
> www.isuppli.com/Teardowns-Manufacturing-and-Pricing/News/Pages/iPhone-4-Carries-Bill-of-Materials-of-187-51-According-to-iSuppli.aspx
> And this page:
> http://www.tgdaily.com/hardware-features/50344-the-real-iphone-4-teardown
>
> Could not find a datasheet for it though, so no idea about how it works....
>
> Looks like the bench supply test may be the only way to find out the
> desired information...
>
>
>
>
> I registered and requested the info anyway, as I figured some figures
> would still be nice for (useful) Vin range - not been asked for any
> payment so far. Will update if I receive anything.
UBM TechInsights has a number of standard reports available to help
you address your own specific needs. These reports have standardized
content but a variety of options are available. Our advanced
technical reports range in price from $8,500 - $26,000 (see notes on
pricing, below). Samples of our advanced technical reports include:
....
Rolling together parts of my prior answers to this and another question.
- LiIon and LiPo = Lithium Polymer (as opposed to LiFePO4 which is
different) typically need 4.2V at the battery for full charging.
- I guesstimate that you'll be liable to drop 0.2V in the iPhone and
0.2V in the charger when you are at limiting low battery condition at
full power.
- If so, you will need a minimum NimH battery voltage of 4.2 + 0.2 +
0.2 = 4.6V to fully charge the LiIon battery.
- 4.6 / 4 = 1.15 Volt.
So IF you can achieve only 0.2V drop in charger and 0.2V drop in
iPhone you *may* 'get away with" batteries as low as 1.15V to provide
a full charge.
- YMMV.
I'd expect that it probably would.
0.2V is easily "eaten up"
A quick flick through the LTC4066 data sheet doesn't seem to provide a
Vin-Vbat saturation figure (or I missed it) but based on other
figures, something in the 50 mV - 150 mV range seems likely. At say
100 mV that alone is half your allowed 0.2V for my above assumptions
to hold.
A way to check the iPhone charger drop is.
- Open iPhone (Gargoyle knows) and attach a wire or wires to battery
AFAIK iP3 has soldered in battery (non replaceable, sealed for death,
200 charge cycles lifetime :-( )(be evil).
- Charge from approved or whatever USB charger
- Measure voltage drop on each charger lead between charger and battery.
This will give you some indication of capabilities. If you use a
variable bench supply as charger (as has been suggested)( and which
has a MUCH more controlled voltage than a USB port does) you can
measure voltage and current at various battery voltages.
This allows you to determine minimum voltage drop at peak desired
current (1 A in your case AFAIR).
It apparently had a USB A socket on the outside that was marked "12VDC
out"! I didn't get to test it so don't know if it lived up to its promise,
nasty if it did!
I just tested the open circuit output voltage of the little 1 Amp AC charger
that you get with the phone. It has a USB host socket so you can plug in
your USB cable directly.
The +5 V line is reading 5.10 volts
One data line is reading 2.70 volts.
The other data line is reading 2.00 volts.
A question: Since the data lines have to be set at a certain voltage (with
resistor voltage dividers), and the battery pack voltage will vary between
4.4 - 5.6 volts, how would I get a steady voltage on the data lines to tell
the iPhone to draw 1 Amp? Would I need to just stick a low power voltage
regulator at around 3.0 or 3.3 volts and create a divider from there
> A question: Since the data lines have to be set at a certain voltage (with
> resistor voltage dividers), and the battery pack voltage will vary between
> 4.4 - 5.6 volts, how would I get a steady voltage on the data lines to tell
> the iPhone to draw 1 Amp? Would I need to just stick a low power voltage
> regulator at around 3.0 or 3.3 volts and create a divider from there?
Consider:
-Read the article/s referenced here (by me and others) that told what
they did and provided circuits.
On Thu, 2010-12-09 at 08:44 -0500, V G wrote:
> I just tested the open circuit output voltage of the little 1 Amp AC charger
> that you get with the phone. It has a USB host socket so you can plug in
> your USB cable directly.
>
> The +5 V line is reading 5.10 volts
> One data line is reading 2.70 volts.
> The other data line is reading 2.00 volts.
>
> A question: Since the data lines have to be set at a certain voltage (with
> resistor voltage dividers), and the battery pack voltage will vary between
> 4.4 - 5.6 volts, how would I get a steady voltage on the data lines to tell
> the iPhone to draw 1 Amp? Would I need to just stick a low power voltage
> regulator at around 3.0 or 3.3 volts and create a divider from there?
The iPhone isn't looking for a "steady voltage", it's looking for a
ratio between lines, exactly what using resistors gets you, so the
obvious solution? Use resistors.
Almost every commercial high volume USB device I've seen manufactured in
the last 2 or 3 years has been "charge-only" adapter aware in one way or
another.
A lot of device ports have the "charger only" connection detection circuits
built-in now (think SOCs used by smart phones, etc.). If not, the units
I've looked at roll their own. Regardless, the conditions for detecting an
external "charger only" connection are clearly spelled out in the above
specification. Also note that the specification covers input voltage ranges
and other pertinent details.
Matt Pobursky
Maximum Performance Systems
On Fri, 10 Dec 2010 03:58:34 +1300, RussellMc wrote: {Quote hidden}
>> A question: Since the data lines have to be set at a certain voltage
>> (with resistor voltage dividers), and the battery pack voltage will
>> vary between 4.4 - 5.6 volts, how would I get a steady voltage on the
>> data lines to tell the iPhone to draw 1 Amp? Would I need to just stick
>> a low power voltage regulator at around 3.0 or 3.3 volts and create a
>> divider from there?
>>
>
> Consider:
>
> -Read the article/s referenced here (by me and others) that told what
> they did and provided circuits.
>
> - Do what they did.
>
>
> The iPhone isn't looking for a "steady voltage", it's looking for a
> ratio between lines, exactly what using resistors gets you, so the
> obvious solution? Use resistors.
On Thu, 2010-12-09 at 11:45 -0500, V G wrote:
> On Thu, Dec 9, 2010 at 10:11 AM, Herbert Graf <hkgrafEraseME.....gmail.com> wrote:
>
> > The iPhone isn't looking for a "steady voltage", it's looking for a
> > ratio between lines, exactly what using resistors gets you, so the
> > obvious solution? Use resistors.
>
>
> Where did you get this information from?
Have you ever tried googling before asking questions here?
>> The iPhone isn't looking for a "steady voltage", it's looking for a
>> ratio between lines, exactly what using resistors gets you, so the
>> obvious solution? Use resistors.
> Where did you get this information from?
Even if you don't follow Herbert's advice, you could consider using my
already completed Googling of several days ago.
I provided links to discussion of the IC used, how to use resistors
to set charge rate, circuit diagram of how to use resistors, values to
use to achieve desired result, photo of inside of phone with charging
IC identified, link to data sheet of IC. The data sheet discusses how
the resistor method works and I have pointed out, perhaps several
times, that that information is in the IC data sheet (LTC4066).
If you always followed all the links that I provide you with you'd be
an international expert* in no time :-)
Russell
* Area of thus acquired expertise cannot be guaranteed
Solarwind - what model and submodel of iPhone is it?
Innards vary. Methods may not.
For some reason I thought you had said iPhone 3 but I see you didn't
specify. Maybe then a 4?
IF resistors are used on data leads et al then a "genuine" charger and
an ohm-meter and some playing and thinking may well provide an
appropriate answer.
> Solarwind - what model and submodel of iPhone is it?
> Innards vary. Methods may not.
>
I have an iPhone 4. Not sure exact "submodel".
>
> For some reason I thought you had said iPhone 3 but I see you didn't
> specify. Maybe then a 4?
>
> IF resistors are used on data leads et al then a "genuine" charger and
> an ohm-meter and some playing and thinking may well provide an
> appropriate answer.
True. I'm going to use Ladyada's resistor values to get a 1 Amp charge. My
biggest concern right now is hoping that the initial high voltage (5.6
volts) of the 4-cell NiMH eneloop battery pack wont damage the phone, since
5.6 volts is above the USB spec
> True. I'm going to use Ladyada's resistor values to get a 1 Amp charge. My
> biggest concern right now is hoping that the initial high voltage (5.6
> volts) of the 4-cell NiMH eneloop battery pack wont damage the phone, since
> 5.6 volts is above the USB spec.
I imagine that OL will note that you declined to use a power supply
which could be precisely configured but "hope" that batteries won't
hurt your phone :-).
I'd say that there is a good chance that it will be OK, for a wide but
incomplete range of good. Apple may be evil but they are less often
technically stupid. Has been known to happen :-).
> I imagine that OL will note that you declined to use a power supply
> which could be precisely configured but "hope" that batteries won't
> hurt your phone :-).
>
> I'd say that there is a good chance that it will be OK, for a wide but
> incomplete range of good. Apple may be evil but they are less often
> technically stupid. Has been known to happen :-).
I don't have access to a bench supply at the moment, but in any case,
shouldn't the effect of over voltage from either a bench supply or batteries
be the same
> I don't have access to a bench supply at the moment, but in any case,
> shouldn't the effect of over voltage from either a bench supply or batteries
> be the same?
Potentially.
(Pun noticed).
A reasonable bench supply can be current limited as desired - and
while you don't don't want to connect an above spec voltage regardless
of current drain, it current limiting may provide some protection.
As you said you wanted to be able to terminate charge on low battery
you will need a series pass element (transistor usually) so can limit
maximum output voltage extremely easily. An eg CES2310 FET, anyold
[tm] opamp (say LM358 / LM324) a cheap reference diode (TL431) and a
few resistors will allow an acceptable solution.
> As you said you wanted to be able to terminate charge on low battery
> you will need a series pass element (transistor usually) so can limit
> maximum output voltage extremely easily. An eg CES2310 FET, anyold
> [tm] opamp (say LM358 / LM324) a cheap reference diode (TL431) and a
> few resistors will allow an acceptable solution.
What does low battery protection have to do with maximum output voltage? I'm
not understanding what you're saying
> I think Russell means the same circuit can also limit maximum voltage,
> creating a "window" for acceptable battery voltage.
Hmm, I dont see how that would work. Are you saying I should use two
comparators, one for shutting off charging when voltage is low, and the
other for routing power through a linear regulator when voltage is too high
> > As you said you wanted to be able to terminate charge on low battery
> > you will need a series pass element (transistor usually) so can limit
> > maximum output voltage extremely easily. An eg CES2310 FET, anyold
> > [tm] opamp (say LM358 / LM324) a cheap reference diode (TL431) and a
> > few resistors will allow an acceptable solution.
> What does low battery protection have to do with maximum output voltage? I'm
> not understanding what you're saying.
To turn the output off you need a "switch".
I've mentioned a very nice FET that would do that, but there are many available.
A FET can be used as a controlled resistance.
Once you have it in series with the battery you can use extremely
simple electronics to monitor the output voltage and to turn the FET
on somewhat when required when/if the FET reaches some upper
threshhold.
So, the FET serves two rioles - low voltage turn off switch and over
voltage output limiter.
Power dissipation needs to be watched in the latter role.
At 1 Amp out you get 100 mA per 0.1V drop.
Well heatsunk that FET is good for maybe a few 100 mW.
If you want to drop more & often you need a larger pkg.
On Fri, Dec 10, 2010 at 12:39 AM, Oli Glaser <EraseMEoli.glaserspamspamBeGonetalktalk.net> wrote:
> If 5.6V managed to hurt the phone I would be amazed - as noted above
> Apple are many things, but not (usually) technically inept. If worried
> I'm pretty sure you could find out the absolute maximum specs from
> Google/Apple, or voltage limit your design (transistor/zener?)
> Current limited bench supply would be safer for testing if available.
It doesn't seem like it should, however, most chargers and USB ports
are going to be putting out 5.0 - 5.2v
And then the relatively high current will be passing through 3-5 feet
of the thinnest wire they could manage...
If it is a linear regulator, as it is in my phone, then the
dissipation will be significantly higher if you supply it with more
than what it was designed to normally operate at.
So while the potential is unlikely to blow out any parts, the
dissipation could (possibly.)
>True. I'm going to use Ladyada's resistor values to get a 1 Amp charge. My
>biggest concern right now is hoping that the initial high voltage (5.6
>volts) of the 4-cell NiMH eneloop battery pack wont damage the phone, since
>5.6 volts is above the USB spec.
If I can make a suggestion: select a lower charge current if possible.
If you look at the discharge curves for the Eneloop cells, you will see that you get far better performance at lower discharge currents. Either drop the current or use multiple cells in a series / parallel arrangement so that the individual cells see a lower discharge current.
Regarding over-voltage: the cheap Chinese-made portable charger that I have has several 5.1V 1W zener diodes in parallel right across the output terminals. Their purpose is to deliberately discharge the battery to below the zener voltage. This ensures that the unit does not damage an overly-sensitive device - 5.1V is within the USB spec.
You could do something similar if you wish.
Or: if you are going to add something to shut your unit down when it becomes discharged, you already have most of what you need to provide an automatic voltage switch.
What I mean by that is: your end-of-battery-life voltage detector needs a comparitor and an accurate voltage reference. Its child's-play to add another comparitor driving a MOSFET that puts a short across an ordinary silicon diode being used as a 0.6V drop. The MOSFET would remain OFF while the terminal voltage was above 4.3V or so, then would turn ON so as to put a short across the diode for the remainder of the discharge time. Note that the diode is in series with the negative end of the battery - you can use a really inexpensive N-channel MOSFET as the switch across the diode.
If you already have the components for the end-of-life battery disconnect, the very few extra components needed for this extra option would cost almost nothing.
>As you said you wanted to be able to terminate charge on low battery
>you will need a series pass element (transistor usually) so can limit
>maximum output voltage extremely easily. An eg CES2310 FET, anyold
>[tm] opamp (say LM358 / LM324) a cheap reference diode (TL431) and a
>few resistors will allow an acceptable solution.
Russell - you are suggesting what I think are completely inappropriate component choices.
Its important that the battery voltage monitor circuit consume as little current as possible. The reference and op-amp mentioned in your post are, relatively speaking, current hogs when compared to modern equivalents.
It should be possible to get the quiescent current down to 10uA with only a little extra effort spent in component selection.
It occurred to me this morning that a 'new' student, like the OP, might
benefit from a little engineering theory discussion before going into
component selection. Maybe get him to model the battery first, et al, as
you would in engineering school. It would provide insights and maybe a
little more reasoned thought. Nothing too rigorous or theoretical - like
a first or second 'circuits' course perhaps.
There are a few very knowledgeable persons here who could do that, and
in all humility I'll defer to them. It seems it may shorten the
roundabout discussion when someone "doesn't get it" and the ensuing
frustrations. Even thinking about a real battery as a source and series
resistance could have answered a few of his questions I think.
Even in formal education, it's gotten very easy to be or make someone
into a "parts picker". It has an instant gratification component, but as
in software engineering, knowing how to drag a visual component onto a
window (ie, VB or Delphi) doesn't make one a software engineer or even a
programmer.
Once you understand the model for the cap, or battery, or FET, you can
understand the significance of the various specs and prioritize them for
your needs, as well as understand what happens and why. THEN the
discussion of which part number makes sense.
It seems he wants to learn, and wants to build, but if he's not taught
the basics (and these projects are perfect simple playgrounds for that),
then he'll build a habit of hacking things together rather than
understanding design IMHO. Others (lurkers) might benefit from the
wisdom of senior members too.
> If you already have the components for the end-of-life battery
> disconnect, the very few extra components needed for this extra
> option would cost almost nothing.
>>As you said you wanted to be able to terminate charge on low battery
>>you will need a series pass element (transistor usually) so can limit
>>maximum output voltage extremely easily. An eg CES2310 FET, any old
>>[tm] opamp (say LM358 / LM324) a cheap reference diode (TL431) and a
>>few resistors will allow an acceptable solution.
> Russell - you are suggesting what I think are completely
> inappropriate component choices.
>
> Its important that the battery voltage monitor circuit consume as
> little current as possible. The reference and op-amp mentioned in
> your post are, relatively speaking, current hogs when compared to
> modern equivalents.
Yesish. Not id3eal choices for ultimate residual battery life.
Good for availability and cost.
For cheap and available could have said TLV431 which is about the same
price and about 50 uA.
Even a TL431 and 358 should be about 1 mA combined AFAIR.
With 2500 mAh cells if you terminate at say 2% of capacity you have 50
mAh left so the monitor circuit would consume that in ~~~ 50 hours.
Not ideal but OKish in many cases.
Even swapping to TLV431 should almost double that.
One solution is to have the low voltage trip turn the circuit off
completely - needing manual reset or much higher battery voltage to
restart. Quiescent current could be << 1 uA.
For lowest reasonably possible quiescent an LM385 and a package of
CMOS Schmitt inverters as per functionally similar thread today. The
reference becomes almost the only drain,
On 10/12/2010 17:06, Bob Blick wrote:
> On Fri, 10 Dec 2010 05:39:13 +0000, "Oli Glaser" said:
>
>> I'm pretty sure you could find out the absolute maximum specs from
>> Google/Apple,
> Technical information from Apple? It's been thirty years since you could
> get technical information from Apple. All we get now is press releases
> :)
>
> I can't even find out what the current firmware version is for an iPod.
>
> Cheers,
>
> Bob
>
Ah yes, I see what you mean, maybe I was a bit optimistic there...
I avoid Apple stuff in general so it was a guess - I thought they would still give some basic technical info about the various interfaces (to the outside world) of their products, min/max temp, voltage etc, that sort of thing.
I suppose based on their assumption that you never use anything but the "proper" Apple accessories then they can avoid even doing that though.
What if you wish to develop a pluggable accessory though? (we are toying with an idea for one, not done any real research yet though) e.g. like the radio plugin (low power transmitter) that came out for the iPhone 1 to listen to your music on the car radio. I guess there is some tightly controlled Apple protocol and IDE? Or maybe everything non Apple is "unofficial" and not guaranteed for use with the phone?
On Dec 10, 2010, at 7:14 PM, Oli Glaser wrote:
>
> What if you wish to develop a pluggable accessory though? (we are toying
> with an idea for one, not done any real research yet though) e.g. like
> the radio plugin (low power transmitter) that came out for the iPhone 1
> to listen to your music on the car radio. I guess there is some tightly
> controlled Apple protocol and IDE? Or maybe everything non Apple is
> "unofficial" and not guaranteed for use with the phone?
>
> Or: if you are going to add something to shut your unit down when it
> becomes discharged, you already have most of what you need to provide
> an automatic voltage switch.
>
> What I mean by that is: your end-of-battery-life voltage detector
> needs a comparitor and an accurate voltage reference. Its
> child's-play to add another comparitor driving a MOSFET that puts a
> short across an ordinary silicon diode being used as a 0.6V
> drop. The MOSFET would remain OFF while the terminal voltage was
> above 4.3V or so, then would turn ON so as to put a short across the
> diode for the remainder of the discharge time. Note that the diode
> is in series with the negative end of the battery - you can use a
> really inexpensive N-channel MOSFET as the switch across the diode.
>
> If you already have the components for the end-of-life battery
> disconnect, the very few extra components needed for this extra
> option would cost almost nothing.
>
> dwayne
I know how to make the circuit if I have a voltage reference. I already drew
it out using FETs and comparators and regulators to handle over voltage, but
I don't know where I would get an adjustable voltage reference that can
supply a voltage ABOVE the battery voltage. I would rather not make a
switching step up regulator or use a transformer. I want to keep part count
low and the circuit board tiny.
It's hard to find low part count switching regulators that can supply a
little over 1 amp. That's why I'm going through all the trouble and just use
a battery directly
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