Show me the numbers!!!!

2 theoretical situations.

Luxeon III or K2 running at 4V and 700ma, what is the regulated runtime on a single 123?

Same situation running at 4.2 volts and 1000ma, what is the regulated runtime on a single 123?

How do you calculate it, do you have to know the driver, or is there a general rule of thumb?

Bill

You know the output power (Pout) of your converter (V times the I you stated). If you know your converter efficiency* (say 80%) then you calculate Pin = Pout/Eff. So for this example Pin is 1.25 times Pout.

Pin is the power the battery has to supply. If you know the Watt-hours rating of the battery (at the current draw you are using) then Time = Watt-hours rating / Pin. The trick is knowing the W-hr rating of the battery. At high currents the rating goes down because power is wasted inside the battery.

The Duracell OEM page has graphs that show run time for various power draws. For an alkaline AA see this: http://www.duracell.com/oem/primary/alkaline/mx1500.asp
Unfortunately it appears you have to ask them to get the equivalent data for their DL123 cells. But, just for education, if you look at the top graph on the linked page you can see that you estimate the run time for various powers (Pin). And you could interpolate between two lines if your power wasn't on the graph.

The run time depends somewhat on the cutoff voltage you choose. Basically this would be the voltage that the driver drops out of regulation at.

So if you can find a 123 graph like this you are all set.

Greg

(* the driver efficiency is probably not constant over a wide range of input voltages so the results may not be perfectly accurate.)

Tom Poast has done the graphs for 123's here:

https://www.candlepowerforums.com/threads/67078

The best of the 123's have about 3 watt hours at 1 amp and 4 watt hours at .5 amp, so figure 3.5 watt hours at 700 ma.

I'm working on it, but someone else post to show me how I screwed up

Bill

Bill

Found it: http://www.duracell.com/oem/Pdf/new/Li123_US_OS.pdf

The graph of interest is the first one "Delivered Energy vs. Power Drain". The graph is quite crude but at 1W it appears you get 3.5 W-hours, at 2W you get about 3 W-hours, and at 4W you only get (about) 2 W-hours.

Your first example Pout is 2.8 Watts. With a 90% efficient driver the Pin is 3.1 Watts.

Your second example Pout is 4.2 Watts. At 90% efficiency Pin is 4.7W and is off the graph.

This is why the HDS lights' runtime is 3x as long at bright settings and 2x at low settings when you use the 2x123 tube.

Greg

Dont forget regulator efficiencies vary greatly. You can have from 87%-over 95% depending on the design and is differential as the output of the cell changes.

His graphs are at a constant current draw. With a regulated boost driver the current will increase as the cell voltage drops as it tries to keep Iout or Pout constant. Thus even if you start out with 1A draw the current will increase and you will drop out of regulation faster than you would expect from the 1A (or whatever current) curve on these graphs.

Greg

bwaites said:

Tom Poast has done the graphs for 123's here:

https://www.candlepowerforums.com/threads/67078

The best of the 123's have about 3 watt hours at 1 amp and 4 watt hours at .5 amp, so figure 3.5 watt hours at 700 ma.

I'm working on it, but someone else post to show me how I screwed up

Bill

Bill

Click to expand...

I mentioned that it varies in my first post. You either have to estimate it or perhaps (if you have the driver already) measure it over the voltage range you expect to use (then use that to come up with an average efficiency).

Greg

Luna said:

Dont forget regulator efficiencies vary greatly. You can have from 87%-over 95% depending on the design and is differential as the output of the cell changes.

Click to expand...

Lets assume a 90% efficient driver.

So as I understand it, runtime on 1 cell would be approx. 1 hour at 700ma and 40 minutes at 1A?

These just need to be ballpark on a 90% efficient driver, I'm still VERY preliminary.

(And I don't intend to design the circuit, I'm on the other end, but need to have some gross idea of runtime!)

Bill

Less I think.

From the graph, at 3.1 Watts draw the energy is about 2.4 Wh. So that works out to .77 hours.

Your 1A scenario is off the graph but at 4 Watts the energy available is 2 Wh (at 1.8V cutoff). So that is 1/2 hour. At 4.7 Watts you might get 20 minutes.

Now this is assuming perfect regulation. If the regulation is not perfect then you will draw less power as the cell voltage drops so the runtime before "moon mode" will be extended.

For example if it linearly drops from 100% at the start to 50% brightness at dropout, the area under the curve is 75% of the perfect 100% regulation case so the runtime may be about 1.33 (= 4/3) of the above numbers.

Greg

Greg,

I was using the numbers from SilverFox's graphs and when I actually put them on paper instead of trying to do them in my head I got about 45 minutes and 31 minutes, so that is close enough for the current situation.

If you think of anything else, please post!!

Bill

If your current draw is constant (at 700 or 1000 mA) over the life of the battery then SilverFox's graphs are fine.

If the current to the LED is constant then the current drawn from the battery won't be constant. Then the Duracell graph is better.

If the regulation is less than perfect you will get a run time somewhere between these two methods (correction: it could be longer than the constant current case since the current draw my drop as the voltage drops).

Greg

Bill,
The first graph below is a run time on a LuxIII at 900 mA using a NexGen converter that is a reasonably efficient converter, IMHO. I don't recall the LED used but am certain the Vf is below your K2:



Now, below is another graph using a NexGen set at 1100 mA and using a UX1J which has a Vf below your K2:



I think your K2 will proably show some regulation at 700 mA on a single CR123 but at one amp, I think you will just see a power curve based on the efficiency of the converter and the capacity of the battery tested.

4.2 watts demand on a CR123? I hear Scotty calling cap'n Kirk......

Let me make up some numbers for different types of drivers: battery voltage, current draw from battery, power in to driver.

1) Perfect regulator:
3.0V, 1.0A, so P=3W
2.5V, 1.2A, so P=3W
2.0V, 1.5A, so P=3W
1.9V, 0.1A (moon mode)

2) Constant current draw circuit
3.0V, 1A, P=3W
2.5V, 1A, P=2.5W
2.0V, 1A, P=2W

3) Less than perfect regulation (current draw goes up but P goes down somewhat)
3.0V, 1.0A, P=3W
2.5V, 1.1A, P=2.75W
2.0V, 1.25A, P=2.5W (power is down 17%)

4) Worse regulation (current draw drops as V goes down. This is still semi-regulated if the current goes down slower than the voltage {on a percent basis})
3.0V, 1.0A, P=3W
2.5V, 0.9A, P=2.25W
2.0V, 0.8A, P=1.6W (power is down 47%)

Don't know if this helps any It was fun to think about anyway.

Greg

Greg, thanks for all the help!!

Don, I used your old IS info on LuxIII's and K2's trying to work through the possibilities.


I appreciate you jumping in and helping here!!!!

The problem begs a one cell solution, but may require the addition of another or maybe 2 CR2's.

Bill