HP graphing calculator power consumption
Eric Rechlin
in order to get some idea of how power consumption has changed over
the years. Going from the 48GX of 1993 to the 50G of 2006 shows
about four times as much (300% more) power used.
The most surprising thing I determined is that the 50G consumes
significantly more power than the 49G+. Not only does the 50G have
one more battery, but it also manages to draw more current, resulting
in an effective 40-50% greater use of power. I do not have any
explanation for the increase.
All measurements were made by measuring the voltage across a 10.0 ohm
resistor in series with the batteries and calculating the current
draw from that. This means that the higher values may be off by a
bit, but the numbers should be accurate for the most part. An Extech
multimeter, obtained at Radio Shack, was used for the measurements.
I repeated a few of the measurements with a nice Fluke multimeter,
and the results were approximately the same, so the Extech can be
trusted.
All tests were done with fresh batteries to ensure consistency.
Tests for each model were done using the first calculator I found
lying around. Tests on subsequent examples showed differences from
unit to unit in the 5-10% range, so more accurate results could be
made by measuring several calculators and averaging them.
All figures below are measurements of current in milliamps. For
power consumption, multiply by 6.0 volts on the 50G and 4.5 volts
on all other models.
Calculator 50G 50G 49G+ 49G+ 49G 48GX 48GX 48GX
# of cards 0 1 0 1 0 0 1 2
---- ---- ---- ---- ---- ---- ---- ----
Idle 15.5 14.9 13.9 12.5 6.9 4.9 5.3 6.0
Blinking 18.8 18.2 16.9 15.6 9.5 8.6 9.0 10.0
Plotting 74.4 79.5 69.3 73.3 20.5 23.2 23.8 25.3
Transmit Serial 89.2 93.1 n/a n/a 21.6 24.3 25.0 26.3
Transmit USB 86.1 89.9 70.4 74.7 n/a n/a n/a n/a
Transmit IR 88.5 92.0 74.1 78.3 n/a 24.3 25.0 26.3
Open I/O Serial 30.7 34.9 n/a n/a 7.4 5.3 5.8 6.4
Open I/O USB 27.7 31.8 25.6 29.8 n/a n/a n/a n/a
Open I/O IR 31.0 35.2 29.9 34.1 n/a 4.9 5.3 6.0
Beep 1 kHz 79.5 83.5 62.7 62.0 21.9 23.6 24.5 25.0
Write flash 90.0 95.2 74.0 73.4 24.5 n/a n/a n/a
Read flash 77.0 82.0 63.6 67.7 24.5 n/a n/a n/a
Write SD n/a 72.0 n/a 68.0 n/a n/a n/a n/a
Read SD n/a 72.0 n/a 68.0 n/a n/a n/a n/a
Off 0.0 0.0 0.0 0.0 0.0 0.0 0.1 0.8
0: Zero cards installed
1: 128 KB HP (or 128 MB Sandisk) card in slot 1
2: 128 KB HP card in slot 1 and 1 MB HP card in slot 2
Running the Meta Kernel on the 48GX drops "blinking" power from 9 mA
to 5.4 mA with only the 128 KB card installed and from 10 mA to 6.1 mA
with both cards installed.
When the IrDA port is open, the 50G and 49G+ briefly show an increase
of about 5 mA about every 5 seconds.
Data transmission tests were performed with no cable connected (for
USB and serial) or no device in range (for infrared).
Regards,
Eric Rechlin
tomcee
Eric:
Very interesting and helpful results. Thanks for your efforts.
Is there any reason to suppose that a 48G (as opposed to a 48GX) would
have a bit less current draw? I also wonder about precise battery
voltage affecting current readings.
I've used 48G's for serial transmission to Plotters, and recall
readings around 16mA. Perhaps in the (not too distant) future I'll
get a couple of the 48G's out and compare.
Regards,
TomCee
richwood
- Show quoted text -
> Eric;
Thank you for the outstanding effort put into gathering this info. It
certainly raises some interesting questions regarding the reasons for
the high power drain of the 50G. Is HP entering the battery
business? ;-)
It is certainly obvious that the 49G+ and 50G are power hogs compared
to the earlier units.
Rich W
Dueño de Monte
power (the maximum) than Read and Write in the SD card ? ! ! ! !
So I should run all the programs from the SD insted of the Flash, and
install all libs on port 1 (not Flash) to be more power effective.
I´ll try it.
Daniel
>
> Calculator 50G 50G 49G+ 49G+ 49G 48GX 48GX 48GX
> # of cards 0 1 0 1 0 0 1 2
> ---- ---- ---- ---- ---- ---- ---- ----
richwood
>
> - Show quoted text -
Daniel;
Let us know the results as running from SD memory may slow down
operations due to reading speed. Also I note that the power on time
is greater with a large SD card installed so you might try running
speed with various sizes of SD cards to see if there is any noticeable
difference.
Rich W
Raymond Del Tondo
the storage space on an SD card is also Flash...
The high values for internal Flash access can indicate either
a higher access speed than to external Flash,
or the built-in Flash chips aren't optimized for low power
Also note the significant difference between the FHB (49g)
and the ARM-based machines
From an energy-oriented viewpoint the ARM-based
calcs are major energy-wasters,
compared to the FHB (49g) or the real HP-48 machines.
Raymond
"Dueño de Monte" <duenod...@gmail.com> schrieb im Newsbeitrag
news:1187571718.8...@22g2000hsm.googlegroups.com...
Dueño de Monte
> Hi,
>
> the storage space on an SD card is also Flash...
> The high values for internal Flash access can indicate either
> a higher access speed than to external Flash,
> or the built-in Flash chips aren't optimized for low power
>
> Also note the significant difference between the FHB (49g)
> and the ARM-based machines
>
> From an energy-oriented viewpoint the ARM-based
> calcs are major energy-wasters,
> compared to the FHB (49g) or the real HP-48 machines.
>
> Raymond
Yes I notice the ARM machine consumes much more, but it worth it, the
speed in calculus in between my HP48S (Saturn 2-4 Mhz) and the HP49g+/
50g (+75 Mhz) is huge.
I´ll pay for this !
Storage is what now confuse me !
Internal Flash should me more efficient than SD, but is a little bit
the opposite, based on Eric measurements. So a little more research
should be done here, and I will try a little bit on it.
Now I move all the programs to SD, and will test the battery consume
(BatStatus application).
On first impressión do not notice any execution time difference on
must of aplications used (I have a 2 Gb SD card now) and the start up
time is just the same it was (really short).
So, will tell you in few days if notice any change.
Daniel
John
"The most surprising thing I determined is that the 50G consumes
significantly more power than the 49G+. Not only does the 50G have
one more battery, but it also manages to draw more current, resulting
in an effective 40-50% greater use of power. I do not have any
explanation for the increase."
Well, I have to wonder what at all is surprising about that.
No offense, but a second year EE student would have known that in advance.
You increase the voltage , you increase the current. While it is not
a purely resistive load, to some degree it is.
Heres an explanation: Ohm Law
As for power consumption, batteries really only care about how much
current they are suppling.
And if you jack the voltage 30%, even if the the current was the same,
guess what happens to the "power consumption".
(only comparing 50G to 49G+ - apples to apples)
Now comparing the current draw, with the exception of the beep and flash
and I/O, for the most part it is a constant 2-5 ma difference.
As for the flash, I would like to see the unit to unit variation.
It would be much more meaningful to state your test conditions and get
10 or 20 of use to try it. Also, it would be meaningful to see what
happens as the batteries drain.
A.L.
>No offense, but a second year EE student would have known that in advance.
>
>You increase the voltage , you increase the current. While it is not
>a purely resistive load, to some degree it is.
Hahahahahaa! The best joke ever!
A.L.
Eric Rechlin
> Eric Rechlin wrote:
> > The most surprising thing I determined is that the 50G consumes
> > significantly more power than the 49G+. Not only does the 50G have
> > one more battery, but it also manages to draw more current, resulting
> > in an effective 40-50% greater use of power. I do not have any
> > explanation for the increase."
>
> Well, I have to wonder what at all is surprising about that.
>
> No offense, but a second year EE student would have known that in
> advance.
>
> You increase the voltage , you increase the current. While it is not
> a purely resistive load, to some degree it is.
I completely understand that. I do have an EE degree. No need to
explain Ohm's Law. :)
Perhaps I was not clear, but what I do not have is an explanation for
what changed from the 49G+ to the 50G that resulted in the higher power
consumption. The only differences I know are the following:
1. Ability to be powered by USB.
2. Wires from the CPU to an extra connector for a rudimentary serial
port.
3. An extra battery, and any changes in voltage regulation circuitry
needed to handle this.
The only one that sounds likely is that number 3 was poorly implemented.
Given other examples of Kinpo engineering, this does not surprise me.
But I have not compared the schematics of the 49G+ and the 50G, nor do I
have them, nor have I opened both to compare their circuitry, so I
cannot answer the question.
> And if you jack the voltage 30%, even if the current was the same, guess
> what happens to the "power consumption".
Yes, I know, and I already explained this in my post.
> It would be much more meaningful to state your test conditions and get
> 10 or 20 of use to try it.
I did state my test conditions. If you have any specific questions, feel
free to ask and I will answer. I believe I gave enough information for
anybody with the time and ambition to reproduce the tests. The only ones
that were a bit questionable were the flash tests, because it doesn't
take long to fill 128 KB of flash, and the numbers jump around a lot.
> Also, it would be meaningful to see what happens as the batteries drain.
I did not see any appreciable difference when experimenting with both
fresh and partially-used batteries. I did not try any nearly-spent
batteries, however.
Regards,
Eric Rechlin
Joel Kolstad
news:bq3zi.11768$7e6....@bignews4.bellsouth.net...
> You increase the voltage , you increase the current. While it is not
> a purely resistive load, to some degree it is.
Actually, umm... for the past decade or thereabouts, it's been more common to
find switching power supplies than linear power supplies in electronics, so as
you increase the voltage you *decrease* the current (since the load is trying
to draw a constant power).
Paul Schlyter
John <dead...@hex.com> wrote:
> Sorry to start a new thread on this, but that M5 thing go in the way.
>
> "The most surprising thing I determined is that the 50G consumes
> significantly more power than the 49G+. Not only does the 50G have
> one more battery, but it also manages to draw more current, resulting
> in an effective 40-50% greater use of power. I do not have any
> explanation for the increase."
>
> Well, I have to wonder what at all is surprising about that.
>
> No offense, but a second year EE student would have known that in advance.
>
> You increase the voltage , you increase the current. While it is not
> a purely resistive load, to some degree it is.
>
> Heres an explanation: Ohm Law
>
> As for power consumption, batteries really only care about how much
> current they are suppling.
>
> And if you jack the voltage 30%, even if the the current was the same,
> guess what happens to the "power consumption".
So one way to increase the battery lifetime of the 50G would be to
remove one of the four batteries and replace it with e.g. a nail or
other metal object with an appropriate length? The 50G ought to work
well on 3 batteries, since the 49G+ did so .... right? <g>
> (only comparing 50G to 49G+ - apples to apples)
> Now comparing the current draw, with the exception of the beep and flash
> and I/O, for the most part it is a constant 2-5 ma difference.
>
> As for the flash, I would like to see the unit to unit variation.
>
> It would be much more meaningful to state your test conditions and get
> 10 or 20 of use to try it. Also, it would be meaningful to see what
> happens as the batteries drain.
--
----------------------------------------------------------------
Paul Schlyter, Grev Turegatan 40, SE-114 38 Stockholm, SWEDEN
e-mail: pausch at stockholm dot bostream dot se
WWW: http://stjarnhimlen.se/
John
increase the power by 30%, so I think your statement should have been
more like:
"The most unsurprising thing I determined is that the 50G does not
consume significantly more power than the 49G+. Since the 50G has one
more battery, it's expected that it will draw more current, resulting
in an effective 10-20% greater use of power after subtracting the 30%
power increase from the extra cell. This pretty much explains the power
increase completely, except for the flash which does not follow the
pattern."
As for your test conditions, several of the tests require that you run
some code or set some conditions to measure it, and I think that for
meaningful results you should explain in a repeatable way how you
performed those tests.
Paul Schlyter
John <dead...@hex.com> wrote:
> Well, ok no one seems to dispute that the 30% voltage increase will
> increase the power by 30%,
Actually, according to Ohm's and Joule's laws, a 30% increase in
voltage will increase the power not by 30% but by 69%, assuming
that the resistance remains unchanged.
Also, adding a 4th cell to 3 cells will increase the voltage
by 33.333.....% rather than by 30%, and this will increase the
power by 78% (77.77777......%)
elly...@gmail.com
the hp 50g best? alkaline? Ni-Mh? Ni-Cd?
elly
Eric Rechlin
> Well, ok no one seems to dispute that the 30% voltage increase will
> increase the power by 30%, so I think your statement should have been more
> like:
>
> "The most unsurprising thing I determined is that the 50G does not
> consume significantly more power than the 49G+. Since the 50G has one
> more battery, it's expected that it will draw more current, resulting
> in an effective 10-20% greater use of power after subtracting the 30%
> power increase from the extra cell. This pretty much explains the power
> increase completely, except for the flash which does not follow the
> pattern."
No, I disagree entirely. First of all, the 50G effectively consumes
something like 50% more power than the 49G+. In other words, it will use up
36 batteries (in 9 sets) in the time it takes a 49G+ to use up 24 batteries
(in 8 sets), though the exact ratio does depend on use. Second, your
"explanation" isn't taking into account anything but the most basic
solution.
It would make sense if the 50G used something like a diode or a resistor to
cause a voltage drop because of the extra battery. You seem fairly
confident that is what they do. You could certainly be right, but I don't
want to make any assumptions without looking inside. To be honest, I never
really considered that option because of the inherent inefficiency, as you
point out. Maybe I'm missing something, but wouldn't it make much more
sense for them to use a switching regulator? Then there would be no reason
to expect it to draw more power than the maximum 10% or so overhead of a
regulator.
In my second post I did say that the most likely reason for the increase was
a "poorly implemented" voltage regulator. A passive component, which you
describe, would be just that, in my opinion.
> As for your test conditions, several of the tests require that you run
> some code or set some conditions to measure it, and I think that for
> meaningful results you should explain in a repeatable way how you
> performed those tests.
It probably isn't going to make much of a difference, but for the "blinking"
test I pressed a number while at the stack and let the cursor blink. For
the plotting test, I plotted 'X^2+Y^2' in Fast3D mode (I used wireframe mode
on the 48GX, so in hindsight I should have picked something else that worked
equally well on all models, but my goal of the particular test was simply to
be CPU intensive). The other tests just consisted of copying ~100 KB files
to and from flash and the SD card. The calculators had stock settings
except for RPN mode and soft menus.
Regards,
Eric Rechlin
John
Why don't you see that 30% of that power is just due to the extra battery.
Batteries supply current - if the current draw was 50% more you would
have a case, but it's not so you don't.
Lets look at these cases:
50G/0 49G+/0
Idle 15.5 13.9 +1.6ma +11.5%
Blinking 18.8 16.9 +1.9ma +11.2%
Plotting 74.4 69.3 +5.1ma +5.2%
As you can see, the difference in current draw is pretty minimal.
Thats all the battery cares about - how much current you draw.
Last I checked they were rated in mah, not watth.
I assume most of the other case are infrequent, but what ever you think
is the proper mix of the test cases would allow you to calculate the
average change in battery life.
For me idle is the most frequent case, as I make a calculation then
look at it for a while. So I should expect about a 10% decrease in
battery life. It's just the physics of how batteries work.
John
BTW If you point was "Hey they totally blew it going to 4 cells" I
agree. If they had use a good quality switching regulator they could
have increased battery life by a good amount.
And, actually I over looked the fact that since there is an extra cell,
at only 10% more current draw but 30% more voltage, perhaps the battery
life is longer.
From posts here tough it sounds like may unit draw significantly more
that current than your figure.
richwood
>From looking back at early posts here after the 50G came out I get the
impression that part of the problem is that the 50G turns on the 'Low
Battery" indicator at a significantly higher voltage per cell than the
three battery units did. Reports indicated a low power indicator turn
on at about 1.05 volts per cell or less on the 3 battery powered units
and at about 1.2 volts per cell for the 50G. With turn on at the 1.2v
level a alkaline cell is only about half discharged. A good part of
the observed decrease in battery life can be directly related to this
change I would expect.
The only reason I can see that makes sense to me for the change, from
an engineering standpoint, is the expectation that users of the 50G
are going to use NIMH batteries. At 1.05V a NIMH battey is so close
to fully discharged that unless it is immediately changed there is an
excellent chance of program disruption and/or memory loss. At 1.2V
the NIMH battery is still about 85% to 90% discharged but has some
reserve capacity left so that immediate change is not so critical.
Unless the designers had their heads inserted in their hemmoroidal
tissue, or I do, this seems to me to be the only logical explanation
for the observed design change.
Rich W
Eric Smith
> Well, ok no one seems to dispute that the 30% voltage increase will
> increase the power by 30%,
I do. For a resistive load, the power increases with the square of the
voltage, so increasing the voltage by 30% will increase the power by almost
70%.
The electronics in the 49G+ and 50G isn't a purely resistive load, but
it still may have a more than linear increase.
If the 50G had used a switching regulator, the current would have decreased
with the voltage increase. Oh well. Maybe in the next model...
Eric Smith
> Thats all the battery cares about - how much current you draw.
> Last I checked they were rated in mah, not watth.
That's how they're rated, but it doesn't tell the whole story. The
reality is that if you draw a constant current, the voltage will decline
on a curve, and the shape of the curve varies depending on the battery
technology and the specific cell design. Sometimes the battery specs
include such a curve at a particular load current.
However, most devices do not draw a constant load current. They usually
do one of two things:
1) draw less current as the voltage drops (resistive load)
2) draw more current as the voltage drops (switching regulator, constant power)
This makes predicting the battery behavior rather complicated, and is one
of the reasons why so many devices that try to provide a battery guage to
the user do such a poor job of it.
Rich
>
> Why don't you see that 30% of that power is just due to the extra battery.
>
> Batteries supply current -
No, batteries supply voltage. Current depends upon the load resistance.
Any constant current device varies the voltage to achieve the desired
current. Current is a result or voltage and resistance, not a cause.
> if the current draw was 50% more you would
> have a case, but it's not so you don't.
The IC's in the calculator are most lokely CMOS. CMOS draws current only
when changing state, that is, as a function of clock speed. I'm not clear
that the same CMOS circuity would draw 50% more current at a slightly higher
voltage. If this were the case, then one poster's suggestion, short out one
of the battery slots, would work. It's a simple test, but I'd think that
it unlikely to produce the desired result. My guess is that there's more
circuitry in the 50G.
I don't have one yet. Been out of the picture for a few years. But as the
kids get older, I may be able to spare some cash. :-)
> Lets look at these cases:
> 50G/0 49G+/0
> Idle 15.5 13.9 +1.6ma +11.5%
> Blinking 18.8 16.9 +1.9ma +11.2%
> Plotting 74.4 69.3 +5.1ma +5.2%
>
> As you can see, the difference in current draw is pretty minimal.
>
> Thats all the battery cares about - how much current you draw.
Batteries don't even care about that. :-)
> Last I checked they were rated in mah, not watth.
My latest rechargeables are rated at 2100 mAH. They are hybrids.
Cheers,
Rich
Rich
>
>
> BTW If you point was "Hey they totally blew it going to 4 cells" I
> agree. If they had use a good quality switching regulator they could
> have increased battery life by a good amount.
>
> And, actually I over looked the fact that since there is an extra cell,
> at only 10% more current draw but 30% more voltage, perhaps the battery
> life is longer.
If the batteries are in parallel, then adding one more battery will increase
current capacity by, in this case, (3+1)/3=1.33 times.
If the batteries are in series, then current capacity is that of the weakest
battery, or basically every battery if you start with fresh batteries, that
is, it's unchanged. I don't have a 50G, but I'm pretty sure the batteries are
in series.
Cheers,
Rich
John
>> Well, ok no one seems to dispute that the 30% voltage increase will
>> increase the power by 30%,
>
> I do. For a resistive load, the power increases with the square of the
> voltage, so increasing the voltage by 30% will increase the power by almost
> 70%.
Last I checked P=I*E.
** Assuming the current remained the same **, P=I * (1.30 E)
John
> No, batteries supply voltage. Current depends upon the load resistance.
> Any constant current device varies the voltage to achieve the desired
> current. Current is a result or voltage and resistance, not a cause.
No one said as battery is a constant current device.
Quite the opposite.
My point is when you have a voltage source, the variable, and what
determines battery life, is the current draw.
Rich
>
> > No, batteries supply voltage. Current depends upon the load resistance.
> > Any constant current device varies the voltage to achieve the desired
> > current. Current is a result or voltage and resistance, not a cause.
>
> No one said as battery is a constant current device.
That was added to illustrate that to control current you have one option,
vary the voltage.
> Quite the opposite.
The battery model used in my BSEE classes was as a voltage source with
an internal resistance.
> My point is when you have a voltage source, the variable, and what
> determines battery life, is the current draw.
My point is that shy of having an active device, voltage determines
current draw, current is a result, not a cause.
Cheers,
Rich
Rich
> Eric Smith wrote:
>
>>> Well, ok no one seems to dispute that the 30% voltage increase will
>>> increase the power by 30%,
>>
>> I do. For a resistive load, the power increases with the square of the
>> voltage, so increasing the voltage by 30% will increase the power by
>> almost
>> 70%.
>
> Last I checked P=I*E.
> ** Assuming the current remained the same **, P=I * (1.30 E)
Last I checked, if you increase voltage, you also increase current.
So P=I*V and I=V/r, put them together and P=(V/r)*V=V^2/r.
So if we increase the voltage by 33%, from V=4.5V to V=6V, and let's
make R=1 for convenience, we get the following results.
P(4.5)=(4.5^2)/1= 20.25 watts.
P(6.0)=(6^2)/1 = 36.00 watts.
The power consumed has increased by 77.78%, according to the delta% button
on my HP-15C. Below it I have my 6S and 6S solar, but they are so cumbersome
to use in comparison. I may get a 33s when they are back on the market.
>> The electronics in the 49G+ and 50G isn't a purely resistive load, but
>> it still may have a more than linear increase.
I suspect that it is more or less a resistive load, with the resistance
being a function of what the operation(s) the calculator is performing.
And from the table posted, while turned on, the load does not seem to
vary a great deal.
Cheers,
Rich
The Phantom
posted on March 22, 2006 and the thread, "Re: 4 batteries vs. 3 batteries",
posted on July 30, 2006.
I mentioned in the second one that I had powered the HP50G from a variable
bench supply and found that the current drain doesn't vary with applied
voltage; it's constant. This means that it doesn't have a switching
regulator supplying the CPU, just a simple linear regulator, and the extra
voltage supplied by the 4th cell is just wasted, as far as powering the CPU
in concerned. The extra voltage may be needed for something else, but when
that something else isn't running, the extra voltage is wasted as heat in
the linear regulator that supplies the CPU.
Linear voltage regulators just drop the excess voltage. They don't
maintain a constant *power* drain if the input voltage is raised; they
maintain a constant *current* drain as the input voltage varies. To
maintain a constant *power* drain for a higher input voltage to the
regulator (which would be a lower *current* drain in this case) would
require a switching regulator, and apparently they didn't implement that.
There may be another regulator to supply whatever it is that needs the
extra voltage, or that other function may run on the unregulated voltage
from the 4 cells.
Saturn rising
> Go back and have a look at the thread, "Oxyride AAA's are rechargeable!",
> posted on March 22, 2006
http://groups.google.com/group/comp.sys.hp48/browse_frm/thread/3646a1cfd585d0e6
> and the thread, "Re: 4 batteries vs. 3 batteries",
> posted on July 30, 2006.
http://groups.google.com/group/comp.sys.hp48/browse_frm/thread/96e4695744e66f37
> I mentioned in the second one that I had powered the HP50G from a variable
> bench supply and found that the current drain doesn't vary with applied
> voltage; it's constant. This means that it doesn't have a switching
> regulator supplying the CPU, just a simple linear regulator, and the extra
> voltage supplied by the 4th cell is just wasted, as far as powering the CPU
> is concerned. The extra voltage may be needed for something else, but
> when that something else isn't running, the extra voltage is wasted as heat
> in the linear regulator that supplies the CPU.
>
> Linear voltage regulators just drop the excess voltage. They don't
> maintain a constant *power* drain if the input voltage is raised; they
> maintain a constant *current* drain as the input voltage varies. To
> maintain a constant *power* drain for a higher input voltage to the
> regulator (which would be a lower *current* drain in this case) would
> require a switching regulator, and apparently they didn't implement that.
>
> There may be another regulator to supply whatever it is that needs the
> extra voltage, or that other function may run on the unregulated voltage
> from the 4 cells.
A very definitive answer!
We "Rodger" that; over & out!
-=-=-=-