How to interpret battery discharge curves?

Poppy

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How does one interpret battery discharge curves?

Here is a link to the Panasonic Carbon-Zinc D cell discharge curves. It is an Adobe Acrobat PDF file.
It is obvious that the battery will deliver a decreasing voltage for different lengths of time depending upon the amount of resistance to the load. They depict different graphs with different resistance values in ohms.

How can we use these graphs to estimate how long a LED will run with how many lumens?
I know that we will have to check the Cree data sheet for the particular LED to determine Lumens per ma.

Is there a way to calculate what the resistance is to the ma output delivered to the LED?
Is there a way to estimate the run time of a particular LED with one, two, or three batteries in series?
 
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HKJ

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I would not really trust these discharge charts. It is not that they are wrong, but notice the "Daily periode" label, this makes me suspect that they are not continuous discharges curve, but combined from a lot of short discharges.

You can find continuous discharges curve on my website, also for primary batteries. I only have a single D cell and that is alkaline, not zinc-carbon. I do have AA in both zinc-carbon and alkaline and zinc-carbon is very bad.

With curves based on mA I do also believe it is easier to estimate a led runtime.
 

Poppy

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HKJ, Thanks for the response. :thumbsup:

I think that THEIR runtimes are FOUR minutes an hour... allowing a lot of rest time, done eight hours a day for a total of 32 minutes of run-time a day. Definitely NOT continuous. It must take them a month to create that/those graphs.

I am sure that I read that for LOW demand applications that the carbon-zinc is almost as good as alkalines. I don't know what defines something as low demand.

I recently bought an Ozark Trail 300 lumen (lol) lantern for $14.97 at Walmart. With three fresh Carbon-Zinc batteries it pulls 0.41 amps on high, and 0.26 amps on low.

It has a Cree XB-D emitter.
How would I calculate how long it should run in a non-regulated lantern?
Sorry If I am already off topic.
 

HKJ

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Before you can do that you need to make a curve of current consumption vs. voltage for the lamp (Requires a lab. power supply).

The you can print a discharge curve for the battery you want, say this:
EagleTac%2014500%20750mAh%20(Black)-CapacityTime.png


You start at time 0, read the lamp current draw for the battery voltage from the table you did before. This says what curve to start with or more likely between what curves to start. Now you simple has to draw a line between the two curves that are closest to the lamps current consumption, all the time varying the distance to my curves, depending on the lamps actual current draw.
If the lamp has constant current draw at all voltages, there is not much work in it, but with varying current draw it can be a bit of work.

This type of work can be automated, but I do not have time for writing a program (I uses all my spare time on reviews). If somebody wants to play with it they can get a couple of raw battery data files from me.
 

AnAppleSnail

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I am sure that I read that for LOW demand applications that the carbon-zinc is almost as good as alkalines. I don't know what defines something as low demand.

I recently bought an Ozark Trail 300 lumen (lol) lantern for $14.97 at Walmart. With three fresh Carbon-Zinc batteries it pulls 0.41 amps on high, and 0.26 amps on low.

It has a Cree XB-D emitter.
How would I calculate how long it should run in a non-regulated lantern?
Sorry If I am already off topic.

"Low Demand" is a low "C" number. If your battery capacity is C, and your current draw is 1/10th of that, you have "0.1C" draw. That's pretty low. It would take ten hours to drain the battery that way. Note that the XB-Ds forward voltage is 2.9v. Check the Datasheet Here, page 8 for the Vf x Current curve. It wants 0.1A at 2.6v and 1A at 3.3v.

Using three D cells in series, and current draw of 0.41A on high and 0.26A on low, I'd expect the lamp to get annoyingly dim after about 8 hours.

In our case, we'll look on the datasheet for the chart closest to 0.41A current draw, AND assume that the load is exactly a resistor. That is (1.5V = 0.41A x R Ohms). R=3.7 Ohms, pretty close to the 3.9 Ohm figure. We see that figure settle in at (1.29V @ 1 Hour), which gives (V=IR) 0.33A. At hour 9, we see it "cut off" at 0.9V (0.9V = 0.23A x 3.9 Ohms). There is not much visible difference between an LED at 0.33A and at 0.23A. The exact output will be about 30% less, and the apparent lux will not change by much.

But wait! As the voltage drops, the current to the LED drops. at 0.9V per cell, three batteries gives 2.7V - The bottom of Cree's charts for the XB-D. The typical XB-D will pass 100mA at 2.7v. This would be only 25% of 'high.' The LED will present a lower load to the batteries than a resistor - So at "9 hours," the 3.9Ohm load would quit. The LED would still be passing about 150mA, with the batteries a bit over 1.0V per cell.

What's next? Extrapolation is not so accurate as measurement, but at ten hours we'll have about 50% of the original 'high' output. At 20 hours, about 20%. At 40 hours, about 10%... You may see where I am going with this. When do we stop a runtime test?
 

Poppy

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THANK YOU BOTH!!!

In our case, we'll look on the datasheet for the chart closest to 0.41A current draw, AND assume that the load is exactly a resistor. That is (1.5V = 0.41A x R Ohms). R=3.7 Ohms, pretty close to the 3.9 Ohm figure. We see that figure settle in at (1.29V @ 1 Hour), which gives (V=IR) 0.33A. At hour 9, we see it "cut off" at 0.9V (0.9V = 0.23A x 3.9 Ohms). There is not much visible difference between an LED at 0.33A and at 0.23A. The exact output will be about 30% less, and the apparent lux will not change by much.

I am pretty much following all this, and will re-read it a few times, but I am wondering if you made an error in math. Considering that batteries are in series, shouldn't it be (4.5V= 0.41A x R Ohms) ?
 

AnAppleSnail

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THANK YOU BOTH!!!



I am pretty much following all this, and will re-read it a few times, but I am wondering if you made an error in math. Considering that batteries are in series, shouldn't it be (4.5V= 0.41A x R Ohms) ?

We're looking at current from the battery, and we have test charts with one battery. What R value makes the battery give pretty close to 0.41A ("High" mode in your lantern)?

1.5V = 0.41A x R, R is nearly 3.9 Ohm.

Let's use one-battery values for current, and keep in the back of our mind that we have 3 cells in series for the lantern. The 3 cells in series will be identical to one in series for current, and deliver 3x the voltage.

Edit: Don't forget that you can answer this question FOR SURE with about $6 in batteries and measuring once or twice hourly for several hours. Estimates are good, but measurement is king.
 

Poppy

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We're looking at current from the battery, and we have test charts with one battery. What R value makes the battery give pretty close to 0.41A ("High" mode in your lantern)?

1.5V = 0.41A x R, R is nearly 3.9 Ohm.

Let's use one-battery values for current, and keep in the back of our mind that we have 3 cells in series for the lantern. The 3 cells in series will be identical to one in series for current, and deliver 3x the voltage.

Edit: Don't forget that you can answer this question FOR SURE with about $6 in batteries and measuring once or twice hourly for several hours. Estimates are good, but measurement is king.

Two questions:
1. Measure battery voltage, or amperage drawn?
2. Will we be able to calculate emitter lumens output at each measurement point?

I already did some measurements of voltages, for example, I noted that after 11 hours on high I got 1.189 volts, but I let it rest for 5 hours and the batteries recovered to 1.358 volts.
I let it run for another 4.25 hours (I forgot to measure the voltage then) but after resting 8 hours, they measured 1.312 volts.

EDIT: The batteries were Panasonic Carbon-Zinc two for a dollar at the dollar store.
 

AnAppleSnail

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If you measure voltage + current, you'll get more information about battery life and output over time. Current will give you output information, while voltage x current will help you characterize battery life.

You can estimate emitter lumens, but don't forget that not only is the bin unknown, but the brightness bins are 7%. You can calculate the relative output change by current - So if you get a lumen output you trust, you could then calculate other outputs based on the different current levels. Use the % output chart from Cree's datasheet, or see if a CPF member has a better one for your use.

For your measurements, try to measure "under load." This will take two multimeters - One measuring voltage, the other current. This gives the best information about the light system. If your 1.189v per cell was under load, the lamp saw 3.567v - Plenty of headroom to maintain LED output. If you also had current, you could estimate the light output. Without more information about the circuits inside, we can't estimate output very well.
 

Poppy

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I can get access to the pos and neg terminals of the led.
With the light ON, if I use my DVTM, set to amps, and test across those two terminals, I get a reading, and the LED goes out. Is that OK? OR do I have to cut one of the wires to the LED, and have the current travel through the amp meter?
 

AnAppleSnail

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It sounds like you shorted the batteries through your ammeter. Amp-meters have to go in series, and volt-meters have to go in parallel to the batteries.

You'll want a non-conducting spacer between the (-) connection in the battery box. I suggest a paperclip with paper tape on one side. Wedge it in between the battery and the (-) input spring.

You then connect your voltmeter to this (-) and the farthest (+), and your amp-meter to this (-) and the battery box input which you have isolated.

Here it is in a line: The @| is my paperclip contact isolated from the battery box.

(+ Input)(+ Probe)[+D CELL-][+D CELL-][+D CELL-]@(-Probe)(Amp-probe +)|(Amp-probe -)(- input)

Clear as mud? Good luck!
 

Poppy

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It sounds like you shorted the batteries through your ammeter. Amp-meters have to go in series, and volt-meters have to go in parallel to the batteries.

You'll want a non-conducting spacer between the (-) connection in the battery box. I suggest a paperclip with paper tape on one side. Wedge it in between the battery and the (-) input spring.

You then connect your voltmeter to this (-) and the farthest (+), and your amp-meter to this (-) and the battery box input which you have isolated.

Here it is in a line: The @| is my paperclip contact isolated from the battery box.

(+ Input)(+ Probe)[+D CELL-][+D CELL-][+D CELL-]@(-Probe)(Amp-probe +)|(Amp-probe -)(- input)

Clear as mud? Good luck!

The way the box is set up, the box lid has to be closed to carry the final positive to the switch. It seems like it is a mechanical switch, adding a resistor for the difference between high and low. I didn't test it but... maybe.

OK, so I have two meters, one set on DC volts, the other amperes. I used your idea of a paperclip to break/split the circuit and connecting the ammeter across the split so that I could measure the current flow. That meter has an On/Off switch, and it must be ON in order for the current to flow. I hope that it doesn't add additional resistance and muck up the results.

I also connected the Volt meter so that I can pull voltage readings (of all three batteries in series) at any time. We'll just assume that they each deplete at the same rate, and divide the total voltage by three if that is needed for calculations.

I won't start tonight, but hopefully in the morning, and hopefully I will take consistent readings with consistently spaced time intervals.
 

Poppy

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I started this morning and this is what I have four hours later.
Looking at the cree data sheet's graph of % lux vs ma, I multiplied the ma% by 122 lumens to get an estimated lumens
Does that seem right?
These by the way are voltages under load, and amperes under load.
The starting voltage without load was 4.62 volts. (actually the batteries were at 4.79 volts fresh out of the package, but I jerked around for 15 minutes getting crappy readings due to a poor connection at the amp-meter.)

TimeAmperageVoltageCalc VxA WattsEstimated Lumens
00.74.222.954210
50.684.112.7948
100.674.082.7336
150.654.052.6325
200.644.022.5728
250.6242.48
300.613.982.4278
350.63.962.376183
400.593.942.3246
450.583.932.2794
500.573.912.2287
550.573.92.223
600.563.882.1728
650.553.872.1285
700.543.862.0844
750.543.852.079
800.533.842.0352
850.523.821.9864
900.523.811.9812
950.513.811.9431
1000.513.81.938
1050.53.791.895164
1100.53.781.89
1150.493.771.8473
1200.493.771.8473
1250.493.761.8424
1300.483.741.7952
1350.483.741.7952
1400.473.731.7531
1450.473.721.7484
1500.463.721.7112
1550.463.711.7066
1600.453.711.6695
1650.453.71.665
1700.453.71.665
1750.453.691.6605
1800.443.691.6236
1850.443.681.6192
1900.433.671.5781
1950.433.671.5781
2000.433.661.5738
2100.423.651.533
2250.413.641.4924
2400.413.631.4883136
 
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AnAppleSnail

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Wow! I'll check the math another day.

The math looks good. You've traced it to about 7.3 watt*hours consumed from each battery.
 
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