Real world Sanyo 2100 NiMH capacity: Luxeon5W DD (data)

I wondered what is my _real_ battery capacity in a direct drive configuration. So I made some runtime measures on my Lambda Megalite.

LED: Luxeon 5W U4T binned on Lambda heatsink
LED's forward voltage @700mA is 6.40V in operation (when warmed up), 6.5V @ 700mA just 5 seconds after powered on.
This config is a resistored (0.1 ohms) direct drive in a Mag2D using 6 AA cells
Rechargable: 6xSanyo NiHM 2100 (AA)
I put a 0.1 ohm resistor to measure the current.

The measurements were made on a minute resolution. The table below is contains less data.

Continous runtime was 1h37min till current dropped to 700 mA.

[min] Vf(led)[V] I[A] P(led)[W] Cumulated used capacity[mAh]
0 6,93 1,360 9,42
5 6,77 1,200 8,12 126
10 6,72 1,149 7,72 223
20 6,68 1,151 7,69 414
30 6,66 1,162 7,74 607
40 6,65 1,148 7,63 800
50 6,63 1,123 7,45 988
60 6,60 1,084 7,15 1172
70 6,57 1,036 6,81 1349
80 6,54 0,941 6,15 1517
85 6,50 0,878 5,71 1592
90 6,46 0,811 5,24 1662
95 6,42 0,732 4,70 1726
<font color="red"> 97 6,40 0,697 4,46 1749 *** 700 mA limit reached </font>
100 6,36 0,624 3,97 1782
105 6,30 0,545 3,43 1831
110 6,37 0,563 3,59 1881
112 6,31 0,493 3,11 1898

As you can see, usable capacity is much less than advertised. (It is of course not Sanyo's fault, this is true for only this application)
The 2100 mAh theoretical capacity in this case is only 1750 mAh. Sad but true: 83% only.
Not that the "usability" limit was of my choice: till the current dropped to 700 mA.
(If we watch till 500 mA, capacity was 1900 mAh.)

This efficiency of 83% when direct drived is comparable with those current regulators that are able to gain the last energy from the batteries. Of course there may be other factors also.

Humm, there are some current regulators that start out at 96% efficiency at 1 Amp out, and work their way down to 86% efficiency with as little 1.6V in.

http://www.molalla.net/~leeper/effici~1.jpg

(btw, this has been verified by other EE's, and they came up with higher efficiencies than I do)

Is there anything really wrong with their stated capacity?
Do they state they will maintain a certain voltage at 2100 mAH? Have you ran the batteries down and added up the total capacity? You are cutting off a large chunk of the batteries capacity in your test.

Nothing wrong with Sanyo. They are famous of declaring their real data.
Usually companies give their capacity at 0.2C current drain (in this case 2100 * 0.2 = 420mA) until voltage drops below 1.0 volts. (Or 1.1V I don't know). In my case the voltage limit is higher thus resulting in lower "seen" capacity.
If I would test it further I am sure that I reach that 2100 mAh. But the led brigtness is not usable then.

.

[ QUOTE ]
Pinter said:
Nothing wrong with Sanyo. They are famous of declaring their real data.
Usually companies give their capacity at 0.2C current drain (in this case 2100 * 0.2 = 420mA) until voltage drops below 1.0 volts. (Or 1.1V I don't know). In my case the voltage limit is higher thus resulting in lower "seen" capacity.
If I would test it further I am sure that I reach that 2100 mAh. But the led brigtness is not usable then.

[/ QUOTE ]

Even at 200mA, a 5W is using 1.2W of power (6.2V * 0.2A = 1.24W), so it should put out more light than a 1W luxeon driven at 1W (same power, lower current density per die = more light per watt). Personally, I would hardly consider that brighness level unusable, but your needs may differ.

Also note that you are dropping .7 volt across that large .1 ohm current measurement resistor

Nope, he's only dropping 0.07V (0.1 ohms * 0.7A)

Yes, but is mAH the total capacity until it is completely drained or just to your 500mA cutoff point?

Also, did you subtract for the losses of your sense resistor? What guage and length were the wires?
What is the resistance of the battery contacts? What is the switch resistance? I also know the case, cap, battery spring has resistance.

Now if you measure the V at the batteries, then you could have excluded alot of these sources of error.

can some one please explain i realy dont understand does this mean sanyo 2100 is a good batts or not?

Interesting, found my Mag 2D has 0.25 ohms in the switch mechanism from the battery plus spring to the point where the base of the PR bulb hits in the socket. Thats just in one portion of the flashlight.

Jarhead,
Resistors and other losses have no direct effect on the capacity. Capacity means how much charge (current) can leave the cells at given circumstances. Losses affect only the _efficiency_ of the circuit.

Nominal capacity (let's suppose it is 2100 mAh) is calculated by drawing 420 mA-s continously till voltage drops to 1.0V (at 20C degree - room temperature). It is only a number for comparison. Seen capacity can be larger when using these rechargables for many short period, and can decrease if you pull e.g. 2A from them. Also if they get too hot (internal resistance heats internal chemistry at large currents) performace is decreased.

And yes, I can say that Sanyo is a good choice. I wonder how other brands declared to be 2300 mAh can fulfill such a test.

Am I missing something or what? I think I see two glaring errors in your runtime test for mAH, and not accounting for the resistance of the flashlight to take the batteries down to 1.0V each.

First:

If I pulled 1000 ma per hour for two hours I would have a 2000 mAH battery.

If I pulled 1000 ma per sixty minutes, and I ran the test for 120 minutes, I'd have a 2000mAH battery.

Think about it, milliamps per/for hour(60 minutes)...not milliamps for 97 minutes. You have not converted into mAH.

Okay, I see you had 1749 for your mAH rating to 700 mA, and I note 97 minutes not 60 minutes. Note the rating is mAH
Now we have to go back and state this in hours, for mAH. In other words you put out a total 1749 milliamps for 97 minutes.

Right now you are at mA97minutes, not mAH. So, if I take 97/60 = 1.617 hours that your test ran. So I still need to get to hours equivalent, not 97 minutes. To get there, I need to multiply your 1749 milliamps total by 1.617 to get to mAH. Which turns out to 2827.55 mAH. Well above and beyond their rating.



Second error:

Now, you cut off at 97 minutes to 700mA.

Your luxeon at this point has an effective resistance of 6.40/0.697A= 9.18 ohms. You stated at this at your cutoff point being 700mA. Your MagLite has probably a good 0.5 ohms if I run from the bottom of the bulb base through everything and back up to the bulb side. So the total circuit resistance is 9.68 ohms. So the voltage on battery terminals was .697A x 9.68 ohms = 6.747V when you hit 700mA.
I take 6.747V and divide it by 6 cells, we get 1.125 volts still on each cell, well above the runtime to 1V.

If I were to take out the .5 ohms of the flashlight, at this point, I'd still be getting 0.735 A out of the battery.
Battery voltage 6.747V/9.18 ohms = 0.735 A

I also need to run the test down until each battery is to 1V each, you ran it down to 1.125V each. This would buy me more time too.

Seems to me, that their batteries are under-rated.

Found the datasheet for this battery yet?
This is an older 2100mAH one:
http://sanyo.wslogic.com/pdf/pdfs/HR-3U(2100).pdf

I noticed they say something about a 1H rest period in there too.

From the press release: http://www.sanyo.co.jp/koho/hypertext4-eng/0307news-e/0710-e.html

"*2 Actual Battery capacity based on charge/discharge levels as established by JIS C8708 1997(4.2.1)
"

[error1]

>>Right now you are at mA97minutes, not mAH

I think the calculation is correct. Ok, I did not mentioned the formula I used:

Used_capacity[mAh] = sum(I1*t1 + I2*t2 + .. In*tn)
where
I1 is the measured current in between 0..1 minutes
...
In is the measured current in between (n-1)..n minutes
In my case t1=t2=...tn= (1/60) hours.

Measure unit of "I" is [mA], of "t" is [h], so their product is [mAh]

I made an assumption that the current is constant during the whole minute (that is I measured at the beginning of each minute). I guess this approximation does not give significant error.

[Error2]

While it is true that @700mA the effective resistance is 6.40/0.697A= 9.18 ohms, this is not applicable for lower (or higher) voltages.
When reaching 700mA, I made the test further.
After 20 minutes Vled=5.99V I=0.246A (Rled=24.3 ohms)
*After 28 minutes Vled=5.85V I=0.144A (Rled=40.6 ohms)
After 80 minutes Vled=5.50V I=0.018A (Rled=305 ohms)

*Perhaps this is where batteries are at 1.0V each (assuming all other losses (switch+battery+0.1ohm resistor) are 1 ohms.

>> If I were to take out the .5 ohms of the flashlight, at
>> this point, I'd still be getting 0.735 A out of the
>> battery. Battery voltage 6.747V/9.18 ohms = 0.735 A

Again, the Vf and I is determined by the LED's characteristics. At Vf=6.74V I should be 1170 mA. Unfortunately linear formulas do not apply here.

Oh, I see your mAH column now, sorry.

Right:

At Vf=6.74V I should be 1170 mA. Unfortunately linear formulas do not apply here.

Right, so you see how taking the flashlight resistance out, you'd pull more power.

But you had 0.5-1 ohm of flashlight resistance.

The point is, at your 700mA you measured 6.4V at the LED, after the drop of the flashlight resistance. The cells still had plenty of capacity before they hit 1V (they were alot higher). So you cut the true capacity of the cells rather short. If I were to go bypass these resistances, at the point you hit 700mA, your current level would jump back up, thus extending the run time test.

Also, you measured Vf on the LED side of the resistor, so we can toss in another 0.1 ohm for the sense resistor...

So, when you run it down to the to the 1V per cell level, what is your capacity, in mAH, do you figure then?

If we measured at the battery terminals, then we'd know right when the 1V point was hit.

Okay, so this is what I see.

Before you cut it off at the 700mA point, which was 97 minutes. But we have to keep going to 1V per cell.

The last data in the list, 112 6,31 0,493 3,11 1898mAH

And you figure it hit 1V at 5.85V
*After 28 minutes Vled=5.85V I=0.144A (Rled=40.6 ohms)
Thats where you figure you reached 1V per cell

112 minutes + 28 more minutes =140 minutes

So I see now 140 minutes run time to 1V per cell.

So, when you use all your data to this point, what is the mAH rating then?

Of course with your direct drive, there will be a large difference in light output, from the beginning, to the end.

You start off by sending 1.360A into the poor LED, and it's Vf is 6.93. W=V*I So you are hitting the LED with 9.4248W
at the beginning, and it will be plenty bright, possibly color shifted due to this.

Then at the end, which you chose as 700mA, you have, Vf= 6.40 and a current of 0.697A. W=V*I which is 4.4608 Watts into the LED.

Start at 9.4248 W end at 4.46W, you are going to have a very wide variation in output brightness.

Where a properly designed constant current boost will hold the intensity the same for nearly all the "life" of the battery. And we should be seeing CC converters that are 90-96% efficient hitting the shelves soon, that will run a 3W at 1A which is about 3.9W (allowed by the datasheet to do this for the Luxeon III). That alone would give you alot more runtime beyond your 83% "efficiency" of the direct drive. Plus NiMH can be discharged to about 0.9V before risking self reversal of the cell. So tack that on top.

I've been testing boosts that will run a 5W, the best I've seen is 90% so far. I may go to a external low resistance dual fet sychronous design which would push this efficiency level alot higher.

The perfect world would be a buck and boost which is +90% efficient when running a 5W, and I can do that, but I'm trying to find parts to shrink the footprint to where I'd like it. Basically I have two separate circuits rigged with a comparator, with some hysteresis, so they smoothly switch over.

>>>So, when you run it down to the to the 1V per cell
>>>level, what is your capacity, in mAH, do you figure then?
I will check it in a few days, tables are another pc.

Yes, CC converters have two advantages at this area.

1. initial phase over 1 Amps is a complete waste when direct drived. Even the brightness is diminished because of overheating. Plus too much power is lost on the internal battery resistance and on the switch.
Even if the circuit has only 90% efficiency, it helps to save more than its losses.

2. Pulling less current from batteries while they are fresh and their voltage are high. AA rechargables are performing better when current draw is smaller (not 0.5-0.6C)

I ordered uflex drivers from George and will make the same tests again.

I don't think BB, MM, or the Uflex is the benchmark for converters anymore....

Give it a month or so.

; )

pint, i added up the A measurements until you hit the 97min cutoff, divided by number of readings, got about 1a average.

since 97 minutes is about 1.6x1hour, i came to the conclusion that you have pulled ~1600mAH from the batts after 97min.

You got 1.7AH...? whats up with that?