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[on:]

I'm building a solar powered LED watch and I'm trying to calculate how much current the display will draw so that I can find out what kind solar panel I need. The display is a 4 digit, 7 segment display.

This is what I have so far:

LED current should be ~500uA or more preferably.
At any given time, when looking at the watch there will be ~16/28 segments lit.
The watch will be checked 50 max. in a day (random figure I found online, can't be bothered to actually count.  )
The display will be on for 3 seconds when checking the time.
Only one digit will be displayed at a time. Each digit will be on for 2ms before switching to the next digit, with no dead time in between digits and/or wrap-arounds (last digit back to first digit)

Since ~16 segments will be displayed, this means that on average 4 LEDs are on at any time. 16 segmets/4 digits = 4 segments.

4 LEDs for 3 seconds is the same as 12 LED's for 1 second.
12 LEDs * 500uA = 6000uA/s
6000uA/s * 50 times a day = 300000uA/s = ~83 uA/h

Assuming the watch can charge for 12 hours a day.

83uA/h = ~7 uA/12h
so the solar panel will need to provide at least 7uA continuously for 12 hours. And if I want each LED to get 1mA then the solar panel will need to provide at least 14uA for 12 hours.

Does any of this sound correct? And yes, I know I'm not taking into account active and sleeping MCU current draw, charging circuitry losses, etc.

[Reply #1 on:]

Why don't you just try to wire everything up first, on a battery or ac adapter and measure the current and voltage needed to run the watch. Then once to find the current, you can find a solar panel(s) to power your watch.

[Reply #2 on:]

Are you going to see anything with just 0.5mA going to an LED?
I'd start with a simple test there before getting too much farther.

"Assuming the watch can charge for 12 hours a day."
Doesn't sound realistic.

[Reply #3 on:]

@HazardsMind That's not the easiast thing to do when the current draw changes every 2ms.  But, yes, I have already wired it together and the average current draw that my multimeter gives me seems to be reasonable.

@CrossRoads Yes, even at 250uA with the 25% duty cycle it's running at, it's comparable to an LED watch that I bought. It's amazing what LEDs can do. What would say is a reasonable charging time? I sit in a classrooms for 7 hours a day and at my desk for at least a few hours. With a small, low quality panel at my (not so) well lit desk I get a minimum of 20uA.

Anyway, It was supposed to be more of a check that my math was correct (hence the title), rather than finding other methods to check current consumption.

[Reply #4 on:]

The math may be correct, I didn't really run thru it, you seem to be on the right path.
What are you using for overnight charge storage so you don't lose the time?

[Reply #5 on:]

I'm using a 3.0v ML1220 (http://www.digikey.ca/product-detail/en/ML-1220%2FF1AN/P295-ND/447504) lithium battery that I had laying around. It's sort of overkill I think. Even If the watch pulls 1mA (which is about 1000 times what it should when left alone) it should still get through the night if fully charged. I'm not entirely sure about charging though. I'm going to use a 4v solar panel (don't actually have it yet so I can't check voltage under load and different light intensity). Initially I was going to drop the voltage with diodes so that even in max. light I would only get 3.1v after the diodes, but then I can't charge the battery unless the panel is putting out it's max voltage. My second idea was to use a 3.3v switching regulator and series diode on the output to drop the voltage a bit, but I'm not sure which regulator to pick. Does this one look OK? (http://semicon.njr.co.jp/eng/PDF/NJU7261series_E.pdf). It says it's a step-up but the input voltage can be higher than the output so I'm guessing it can work as a step-down. What do I need other than a 100uH inductor and decoupling caps? Could I use a 3v regulator and skip the diode? The datasheet doesn't mention reverse current draw when Vin < Vout which is why I thought of using 3.3v + diode. Will these regulators even work at 20uA input? My latest idea was to use a comparator to compare solar panel voltage and battery voltage and somehow switch in 0, 1, or 2 series diodes to drop the voltage form solar panel to battery. Not sure how I would implement that though. Sorry for the eruption of questions, I've pondering all of this for a quite a while and appreciate any input

[Reply #6 on:]

On average, 5 segments are lit -> 2.5ma per digit. 2.5ma * 3 seconds * 50 times = 375 ma*seconds.

Assuming 100% storage efficiency, and 12hr sun light on the solar panel, it has to provide 375 ma * seconds / 12 hour = 8.7ua.

or 87ua at 10% storage efficiency; or 270ua assuming 3hr sun light, etc.

[Reply #7 on:]

What I think you should do is to provide a means to keep time but not displaying it, when the battery is low, or on backup battery.

Like what some calculators do.

[Reply #8 on:]

You are following the charging recommendations here?
http://industrial.panasonic.com/www-data/pdf/AAA4000/AAA4000PE17.pdf
Looks  like you don't take it much above the output voltage at all.

[Reply #9 on:]

Yes, I'm following those charging recommendations. It's too bad the battery requires such a precise voltage. Maybe I can cheat and take it up to 3.5 or even 4v at low current? I'd rather not though, and of course it says right there that I shouldn't.

@dhenry Like I said, the solar panel will be under artificial light for about 12 hours in a day. I planning on using a monocrystalline cell (http://www.digikey.ca/product-detail/en/KXOB22-01X8/KXOB22-01X8-ND/2754274) so it should work inside just fine. My only gripe is that the cell is expensive. There's also this one (http://www.digikey.ca/product-detail/en/CPC1824N/CLA290-ND/1485717) which is much cheaper, but I don't know if it will provide enough current. Either way, I need to figure out a charging circuit.

[Reply #10 on:]

How about using a super capacitor instead of a battery? As far as I know their capacity is more linearly proportional to their voltage than a lithium battery so they don't require a specific voltage to be charged. I could just buy a 4+ volt super cap, and simply use 1 diode to stop reverse current draw back through the solar panel. The watch will run fine anywhere from about 2.5-5v anyway. I just discharged a 3600uF capacitor from 4v at ~6mA/s for 1 second to simulate the watch while active, and it was at 2.2v afterwards. So really a super cap with over 4mF of capacitance should work; at least until it gets recharged. I could get something like this (http://www.digikey.ca/product-detail/en/EEC-S0HD334V/P11065-ND/300481) for way cheaper than the cheapest 4+ volt battery.

[Reply #11 on:]

Supercaps self-discharge in hours/days, no use for a timepiece.  They are best for very medium/high
currents for short times, totally opposite of lithium batteries!  The varying voltage requires a suitable
SMPS to manage one too.

Back to your maths, if the displays are taking 0.5mA per segment, you need to look at what the
rest of the circuitry is consuming, the display might not be the dominant factor.

4 LEDs for 3 seconds is the same as 12 LED's for 1 second.
12 LEDs * 500uA = 6000uA/s
6000uA/s * 50 times a day = 300000uA/s = ~83 uA/h

You don't mean "uA/h" you mean "uAh" - no division anywhere.

You are over-complicating your approach here - charge = current x time, simple:

4 LEDs = 2mA, 3 seconds 50 times a day = 150 s, therefore charge is 0.002 x 150 = 0.3C

1 mAh = 3.6C  (because 1Ah = 1A for 3600s), so 0.3C = 0.083mAh per day  (so you got the right answer).


What you need to also measure is the current when the watch is off (presumably there is a RTC of some
sort (these take about 0.5uA on battery-backup or so, which is 12uAh per day - that is typically running
from a separate button cell though.

Someone mentioned:

Are you going to see anything with just 0.5mA going to an LED?

You will need high-brightness LEDs at that current level, and you shouldn't expect it to work
in daylight.

Good polycrystalline silicon solar cell yields about 0.5V and 30mA / cm^2 in full sun.   Derate by a factor of
10 for overcast skies, 4 for winter, so expect 0.8mA/cm^2 averaged at 0.5V, and you'll need about 6 to 8
in series to get the voltage, so 100uA/cm^2 at 3 to 4V...  (But check my arithmetic yourself!)