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Old 06-27-11, 01:36 AM   #1
jsdavis
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Energy use by lights

I have a PDW Dangerzone and a Planet Bike 2W Blaze that I run on Sanyo Eneloops. I charge the batteries up every Sunday night so there is no doubt in my mind whether or not I have good batteries for the commute week. My commute is during the day usually, but I run the lights in blink anyway because in urban environment, the lighting is always changing between light and shadow.

The amount of charge applied over the past few weeks has been fairly consistent. My AAA are charged with about 380-410 mAh for the PDW and the AAs for the PB Blaze get 330-350 mAh.

What I'm curious about is how my PDW is using more energy than my PB when it has two 0.5W LEDs and the PB has one 2W LED. Does this mean that the PDW is on more of the time during blink function compared to the PB?

PDW claims 50 hours run time, this seems very optimistic considering that the battery seems to be half-depleted after about 6 hours. The PB claim of 18 hours flash does seem realistic, and perhaps underrated since the batteries were only depleted by 1/5 after 6 hours.

Also please check the below calculation for cost to operate...I think it's right but need someone to do a look over.

Charge applied x voltage
1520mAh * 1.2V = 1.8 W*h

Cost of electricity
12 cents per kWh -> 0.012 cents per W*h

Charging a NiMH battery is about 67% efficient, and assume power adapter and charger are about the same so about 44% efficient.

Amount of energy needed to charge the set of batteries taking efficiency into account
1.8 W*h / 0.44 = 4.09 W*h

Total cost:
4.09 W*h x 0.012 cents / W*h = 0.049 cents

So for me to operate my lights is about 0.05 or 1/20 cents for electricity per week.
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Old 06-27-11, 08:15 AM   #2
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those wattage numbers for leds are very approximate. Your post made me realize they probably aren't using high power emitters for rear lights, they have 3.3-3.7v forward voltage drop, so they'd either have to have a boost driver or go to 3 batteries.
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Old 06-27-11, 11:19 AM   #3
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Originally Posted by jsdavis View Post
What I'm curious about is how my PDW is using more energy than my PB when it has two 0.5W LEDs and the PB has one 2W LED. Does this mean that the PDW is on more of the time during blink function compared to the PB?
Could be. Certainly, the rating of the LED has little bearing on how much power it's actually fed.

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PDW claims 50 hours run time, this seems very optimistic considering that the battery seems to be half-depleted after about 6 hours.
They might measure run time on the lowest power setting with the most expensive non-rechargeable battery available. PB does seem to give reasonable, real-world values, but maybe PDW doesn't (I don't have any PDW lights.)

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Also please check the below calculation for cost to operate...I think it's right but need someone to do a look over.
It's minuscule. That should be close enough.

But your figures look accurate enough, once one realizes that "1520 mAh" is the four batteries added up.

Also, we really don't know how efficient the charger is. The applied voltage is more like 1.4 volts, but does your charger have a wall-wart (like my BC-900) that sucks power even when not actively charging? It will depend on how quickly you unplug it.
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Old 06-27-11, 01:46 PM   #4
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I doubt that there is any way to determine the power consumption of lights in blinking mode. There are a number of reason why, including the circuit efficiency itself.

The whole "charger" total mA measuring is bogus. The display only reflects the current drawn into a tube of electro-chemical paste - not an actual expression of power consumption - as in light.

The larger costs of of using these products lays with the energy consumed during manufacture and the chemical wastes associated with the technologies surrounding them. I would suspect using a generator could be an increase in efficiency - but that's a guess as well.
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Old 06-27-11, 02:53 PM   #5
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I doubt that there is any way to determine the power consumption of lights in blinking mode. There are a number of reason why, including the circuit efficiency itself.
Of course there is -- measure it.

Considering that the circuit goes on and off quite quickly, a multimeter won't work well in many cases, but you could charge a battery, then discharge it and measure it's capacity (how to do this is left as an exercise for the reader, though it's not difficult) -- then charge it again and see how long it would power the light until it reaches your chosen cutoff voltage and calculate the power used -- this would be pretty accurate.


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The whole "charger" total mA measuring is bogus. The display only reflects the current drawn into a tube of electro-chemical paste - not an actual expression of power consumption - as in light.
It sounds like he's trying to calculate how much it costs to use the light -- so measuring the power used by the charger is exactly the right thing to do. (But measuring how much power is put into the battery is wrong -- the charger is likely very inefficient if powered by 110v AC and charging small batteries. Instead, measure the power consumed by the charger. Perhaps with one of these (though it'll cost more than many years worth of electricity to charge your lights!)

If you really want to be through, take the cost of your batteries, divide by the number of cycles they're likely to survive, and add that to the cost ...

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The larger costs of of using these products lays with the energy consumed during manufacture and the chemical wastes associated with the technologies surrounding them. I would suspect using a generator could be an increase in efficiency - but that's a guess as well.
Possibly.
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Old 06-27-11, 07:01 PM   #6
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The cost was just out of curiosity; not really important. If it's close, then good enough.

My charger is the Maha Wizard charger with a brick that outputs 12V @ 2A. I do not know the efficiency of this. I usually only have it plugged in for 2 or 3 hours at a time.

Last edited by jsdavis; 06-27-11 at 07:05 PM.
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Old 06-28-11, 07:56 AM   #7
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Good for you doug -I need a few people to be more conceited than me. But again - this is mostly a waste of time - all you're doing is measuring a circuit and battery charging numbers - that not measuring the light's true current consumption. But I digress - you can continue playing.
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Old 06-28-11, 08:47 AM   #8
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Good for you doug -I need a few people to be more conceited than me. But again - this is mostly a waste of time - all you're doing is measuring a circuit and battery charging numbers - that not measuring the light's true current consumption. But I digress - you can continue playing.
Conceited? Perhaps, but I certainly did propose how to measure *discharging* numbers too. Perhaps you didn't understand me.

Modern LED lights control how much power they use by turning the LED on and off quickly. Multimeters don't always handle this well, making measuring the power at the light problematic. (But certainly not impossible, with the right gear.)

However, if you take a set of batteries, fully charge it, then discharge them down to 1.2 volts (to pick a figure) at a low but constant discharge rate (most R/C chargers will do this for you) you can easily determine the actual capacity of the batteries. Since you're measuring the discharge rather than the charge, you aren't measuring the "charging numbers". R/C people have been doing this for decades to know when it's time to discard their old batteries (if your receiver batteries fail in flight, any warning you get is usually very subtle, and then suddenly it's too late and the plane crashes.)

You measure the discharge current rather than just the charging current because when you charge some of the current is lost by the battery -- and old batteries usually lose more, so you don't want to guess at the difference when your R/C plane is on the line.

So you measure the capacity of the battery, then fully charge it again, then time how long the light stays on until the voltage drops to the same point (you'll have to occasionally measure the voltage) and do some simple math -- you'll know how much current the light averages in whatever mode you just measured. Yes, there's some small errors in here -- the self discharging of the battery (use eneloops to minimize this) and the voltage sag of the battery while it's on the charger is probably higher than it on the light (as the discharge rate is likely higher, unless it's a powerful light) -- but these errors are likely small if this is done right.

Though there is an easier way -- just put a fairly large capacitor (like a farad, though that's likely bigger than needed) in parallel with the battery pack (move the batteries outside of the light if they're stored internally.) Give it a minute or so to fully charge the capacitor, then use your multimeter (cheap one are OK for this) to measure the steady DC current between the batteries and the capacitor. The capacitor will almost completely smooth out the spikes in the current used allowing any multimeter to give an accurate reading.

And really, the way that the OP was simply measuring the amount of current put into his batteries as given by his charger -- the accuracy of that is reasonable. Yes, it overstates the energy used a bit, as some is lost, but it's on the order of 10% or 20% -- probably an acceptable error for his purposes.

I did also talk about measuring the power consumed by the charger itself (if you're worried about just how big the fraction of a penny it costs you to use is) if you somehow confused the two.

And I certainly do appropriately appreciate your permission to continue playing!

Last edited by dougmc; 06-28-11 at 09:16 AM.
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Old 06-29-11, 10:45 AM   #9
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Conceited? ... to continue playing!
Kudos on your patience to explain this after the prior inappropriate comments. I can confirm that the measurement of capacity required to replacein and out is indeed how many R/C'rs get a handle on the energy requirement of their power systems. Average voltage and time are often not considered (to ultimately calculate power and energy) but this can be estimated or with the right equipment, energy can be measured directly. Is this capacity approach extremely accurate? No. But it is legitimate within it's known limitations...
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