Solar battery charge for time lapse photography...

[powerfulpup]

I want to set up a camera for time lapse photography at a remote site and leave it unattended for extended periods of time. This camera is fairly easy on batteries to begin with, two AAs last for 250+ photos, and if I kill the LCD screen can probably get a bit more. Anyway, what I want to do is get a cheap battery (I'm thinking SLA), and hook an inverter up to it which the DC adapter for the camera can then plug into and voila, more power than the AAs provide. The camera DC input says 3V, FYI, but I'm not sure how to figure out what each picture actually draws, and it may not be possible given the variations in autofocusing and light correction, etc.

On those same battery terminals that have the inverter attached, can I latch on a solar battery charger? This doesn't have to be pretty or fancy, and it can't be expensive or the wife won't think my pictures are worth it... I just want it to be functional. Will the camera basically drain the battery with the solar panel topping it off maintaining a constant charge, assuming the camera doesn't draw more than the solar panel provides? Is it okay to connect the charging cables and draining cables (for lack of better words on my part) to the same terminals on the battery?

Excuse the crude image, but this is what I was envisioning. Is the theory of what I described correct, or am I way off and setting myself up for a meltdown of equipment?

Solar.jpg

I searched around the interwebs and on here and found things for security cameras, cameras with a laptop, cameras with remote control... nothing quite as basic as what I'm trying to do. Any suggestions or thoughts are appreciated.

[daz]

This camera is fairly easy on batteries to begin with, two AAs last for 250+ photos, and if I kill the LCD screen can probably get a bit more.

Hi there!

Unfortunately no-one will be able to help you design a solar system for your application, as you have not told us what your solar insolation and camera's power consumption is.
You say that 2xAA batteries last 250+ photos, but that doesnt tell us what the power consumption is (in Watts). To design the system, we need to know how much power you require first, and then you work it backwards!


As an example; your camera has a power consumption of 10W while taking photographs, and a standby power of 5W, etc etc. The camera will be active during the hours of 6am - 10am, and 3pm and 6pm, for a total of 7 hours.

--to calculate power requirements-----
So 7 hours (active) x 10W = 70W, and 17 hours (standby) x 5W = 85W, which gives you a sum total of 102WH per day.

--to calculate battery requirements----
@50% DOD (depth of discharge)

102W / 12V = 8.5A x 2 = 17AH ---> round this up to the next available capacity

@30% DOD

102W / 12V = (8.5A/30)100 = 28.33A ---> round this up to the next available capacity

--to calculate solar PV array----------
You have not told us what the solar insolation is for your area, so I am going to assume 4 solar hours. You will need to find out what the exact solar insolation value is for your area, and then insert that number (replacing my 4).

If we use a PWM charge controller then.....
102W x 2** = 204W required,
204W / 4 solar hours = 51W solar panel required ---> round this up to the next common wattage

**the number "2" figure used above is the fudge factor, and is used to account for losses in the cable, CC, etc etc.


I hope this was able to help?!

[powerfulpup]

I found a chart online and the insolation for the nearest city to me is 4.53 averaged over the year. Apr-Sep are 4.5 to nearly 7, winter months drop off considerably. Unfortunately, as I speculated, I don't think there is a way to measure the camera's consumption reliably or easily as every picture likely draws a different amount.

You did give me some helpful math though and it's a good jumping off point for me, and I thank you.

What about having the inverter and charger share terminals... is that okay to do?

[solarIdiot]

What kind of camera is it. Model number, usually the cameras spec sheet will have that information.

I was seeing a lot of the cameras have a 1.3A draw, however I don't think any of these were using AA batteries.

[inetdog]

I found a chart online and the insolation for the nearest city to me is 4.53 averaged over the year. Apr-Sep are 4.5 to nearly 7, winter months drop off considerably. Unfortunately, as I speculated, I don't think there is a way to measure the camera's consumption reliably or easily as every picture likely draws a different amount.

You did give me some helpful math though and it's a good jumping off point for me, and I thank you.

What about having the inverter and charger share terminals... is that okay to do?

One way to get a rough estimate would be to load the camera with rechargeable batteries of a known capacity and fire off photos until the batteries go dead. This will at least give you a ballpark. Also check the power consumption just sitting there waiting for the timer to fire. You do not need to be accurate within 25%, but within a factor of 2 will be a big help.

Inverter and charger should both be connected directly to the battery terminals, with fuses in each connection appropriate to the CC or inverter current.

[daz]

You did give me some helpful math though and it's a good jumping off point for me, and I thank you.

No problem. Glad to be able to help!

What about having the inverter and charger share terminals... is that okay to do?

As a rule of thumb, you always connect your load to the same point as the charging terminals. So where-ever the charge controller connects to, you must also connect the leads for the inverter.

Unfortunately, as I speculated, I don't think there is a way to measure the camera's consumption reliably or easily as every picture likely draws a different amount.

You do not need to know the exact amount of power every photograph takes, just the maximum amount (or even the average amount). It is just a base figure to work with.

You will need to work out an approximiate somehow, as it is impossible to design a system for something in which you do not know how much it needs. This can lead to an under-sized system that will fail, or a very expensive over-sized system. As Inetdog has said, perhaps try using rechargeable batteries, ones which you know the capacity of, and then take photos until the batteries are flat. That way we can work out more or less how much power you need for the camera (from how long the rechargeable batteries last for).
Alternatively you will have to try get hold of the datasheet for your camera, as the datasheet should have the power consumption numbers.

[powerfulpup]

Thanks for your input all, this is very helpful to someone not very knowledgeable in this area.

It is a Canon A2000IS. A fairly simple point and shoot. According to the Canon website, 2 AA alkalines will yeild around 240 pics. I have Eneloops in there and while I said 250+, I was being conservative on purpose. Let's call it 300. For argument's sake I can take 300 pics on a set of fully charged batteries. Here is my source for that:

http://www.usa.canon.com/cusa/support/consumer/digital_cameras/powershot_a_series/powershot_a2000_is#Specifications

Thesen Eneloops are 2,000 mAh batteries, 1.2V each. There are two of them in the camera, does that mean I have 4,000 mAh at 1.2V total availabe? If so: 4,000/300 = 13.3 mAh per shot. Am I going about this the correct way? I found no info on the specs page about the camera's "idle" power consumption. Orrrrrrrrrrr, is it 2,000 mAh at 2.4V available? And then my math would be... halved?

If I took a picture every minute, every minute I'm using 13.3 mAh on top of whatever the idle consumption is. 300 shots, at a shot a minute, would be five hours, correct? (Again, not factoring in idle consumption). I can of course actually drain the batteries, but that'll be a few hours likely. If that's the best route, I'll defer to your expertise and do that and get back to you all with some numbers.

[inetdog]

...
These are 2,000 mAh batteries, 1.2V each. There are two of them in the camera, does that mean I have 4,000 mAh at 1.2V total availabe? If so: 4,000/300 = 13.3 mAh per shot. Am I going about this the correct way? I found no info on the specs page about the camera's "idle" power consumption. Orrrrrrrrrrr, is it 2,000 mAh at 2.4V available? And then my math would be... halved?

If I took a picture every minute, every minute I'm using 13.3 mAh on top of whatever the idle consumption is. 300 shots, at a shot a minute, would be five hours, correct? (Again, not factoring in idle consumption).

Just for sanity, when you are looking at different voltages, it is safer to talk about watt-hours instead of just amp-hours. Two 1.2 volt 2,000 mAh batteries each hold 2,400 watt-hours. The power will be the same whether the two batteries are in series or in parallel. Then eventually you will need to figure out how the PV system will provide the specific voltage and current needed by the camera.

[Sunking]

Two 1.2 volt 2,000 mAh batteries each hold 2,400 watt-hours. .

Me thinks you misplaced a decimal point and forgot a battery. 2.4 volts x 2 amp hours = 4.8 watt hours. Or a single 1.2 volt @ 2000 mah = 2.4 watt hours. You only missed it by 3 decimal places, or 10 cubed

I wished your math worked, we could run our homes on a few rechargeable flashlight batteries. If primary batteries like Alkaline or Lithium were used would still cost to much for 20 cents worth of electricity.

[daz]

Thanks for your input all, this is very helpful to someone not very knowledgeable in this area.

It is a Canon A2000IS. A fairly simple point and shoot. According to the Canon website, 2 AA alkalines will yeild around 240 pics. I have Eneloops in there and while I said 250+, I was being conservative on purpose. Let's call it 300. For argument's sake I can take 300 pics on a set of fully charged batteries. Here is my source for that:

http://www.usa.canon.com/cusa/support/consumer/digital_cameras/powershot_a_series/powershot_a2000_is#Specifications

Thesen Eneloops are 2,000 mAh batteries, 1.2V each. There are two of them in the camera, does that mean I have 4,000 mAh at 1.2V total availabe? If so: 4,000/300 = 13.3 mAh per shot. Am I going about this the correct way? I found no info on the specs page about the camera's "idle" power consumption. Orrrrrrrrrrr, is it 2,000 mAh at 2.4V available? And then my math would be... halved?

If I took a picture every minute, every minute I'm using 13.3 mAh on top of whatever the idle consumption is. 300 shots, at a shot a minute, would be five hours, correct? (Again, not factoring in idle consumption). I can of course actually drain the batteries, but that'll be a few hours likely. If that's the best route, I'll defer to your expertise and do that and get back to you all with some numbers.

Well, let's assume those figures are correct, and use your 1 photo per minute requirement. I will not be calculating idle power requirements, as we can add a fudge factor afterwards for it (calculated guess!). We will use your average insolation only for this example, 4.5. You need to tell me what your lowest solar insolation figure is, as you always calculate for worst case scenario (ie using winter insolation, not average or summer insolation).


Example:

--calculating battery capacity (in WH)--
1.2V x 2000mAH x 2 = 1.2V x 4000mAH = 4800mWH = 4.8WH

--Calculating power per photo----------
4.8WH/300 photos = 0.016WH per photo

--calculating required photos per day---
1photo per minute x 60minutes = 60 photoes per hour
60photos per hour x 24 hours = 1440 photos per day

--calculating power requirements------
1440 photos x 0.016WH = 23.04WH per day

--calculating battery requirements-----
@50% DOD

23.04WH/12V=1.92AH x 2 = 3.84AH

@30% DOD

23.04WH/12V=(1.92AH/30)100=6.4AH

--calculating required solar array------
If we use a PWM charge controller, then...

23.04x2=46.08W

46.08W/4.5 solar hours = 10.24W solar panel required --->round this up to next common wattage

So, just using your active camera figures, you will require;
1 x 15W panel,
1 x 7.2AH 12V battery (@30% DOD)
1 x 6A PWM charge controller (or whatever is the closest you can match to the 15W panel; +- 1A)



Let's do a 'guestimate' for your idle power, perhaps 50% of active power requirements? In which case we just add 50% to all the above. That will mean that you will require;
1 x 15W/20W panel,
1 x 12AH 12V battery (@30% DOD) or 1 x 7.2AH 12V battery @ 50% DOD,
1 x 6A PWM charge controller (or whatever is the closest you can match to the 15W/20W panel; +- 0.9A/1.2A)


Will you be using an inverter to step the power up to the line AC voltage, and then back down to the required voltage by the camera's charger?
Or will you be using a DC/DC charger (ie car charge)?

[powerfulpup]

Alright, I started the camera and discovered it did quite a bit better than I anticipated as far as number of pictures. I got about 700 (rounding for ease) without a flash. This was pictures in rapid succession without much down time for the camera display to chew up power. I'd feel comfortable calling it 500 at a longer interval on two topped off Eneloops.

Stealing some of your math:
4.8WH/500 photos = 0.0096 WH per phto

Then you kind of lost me. Why, in the first line of math, did you multiply 1.92AH by two? And why, below that, did you divide by 30?

--calculating battery requirements-----
@50% DOD

23.04WH/12V=1.92AH x 2 = 3.84AH

@30% DOD

23.04WH/12V=(1.92AH/30)100=6.4AH

--calculating required solar array------
If we use a PWM charge controller, then...

23.04x2=46.08W

46.08W/4.5 solar hours = 10.24W solar panel required --->round this up to next common wattage

So, just using your active camera figures, you will require;
1 x 15W panel,
1 x 7.2AH 12V battery (@30% DOD)
1 x 6A PWM charge controller (or whatever is the closest you can match to the 15W panel; +- 1A)



Let's do a 'guestimate' for your idle power, perhaps 50% of active power requirements? In which case we just add 50% to all the above. That will mean that you will require;
1 x 15W/20W panel,
1 x 12AH 12V battery (@30% DOD) or 1 x 7.2AH 12V battery @ 50% DOD,
1 x 6A PWM charge controller (or whatever is the closest you can match to the 15W/20W panel; +- 0.9A/1.2A)


Will you be using an inverter to step the power up to the line AC voltage, and then back down to the required voltage by the camera's charger?
Or will you be using a DC/DC charger (ie car charge)?

As for the inverter question, yes, that's what I intend to do. The camera's power block plugs into a standard 110 receptacle to provide the camera constant power. I've considered the possibility about adding a resistor from the battery and forgoing the inverter and transformer block. At first I didn't like that idea but it's starting to build steam in my though process. I could wire something to step the voltage down to 3V to the battery terminals and then get the correct plug in for the DC input of my camera and avoid the inverter entirely, correct? I suppose I could hook up a cigarette lighter and use one of these?

http://www.parts-express.com/pe/showdetl.cfm?partnumber=265-010&utm_source=google&utm_medium=cpc&utm_campaign=pla

[daz]

Then you kind of lost me. Why, in the first line of math, did you multiply 1.92AH by two? And why, below that, did you divide by 30? :confused

50% is half, correct? So if 1.92AH is 50% DOD, the full capacity of the battery will be 1.92AH x 2 = 3.84AH. 100/50=2, that is why I just multiplied by 2 instead of working it out long-hand. We can work it out the same way as for the 30% DOD as well; 23.04WH/12V=(1.92AH/50)100=3.84AH


In the below calculations, the number in bold is the % figure for DOD. You can change the DOD by just changing that figure. EG 30 below=30% DOD, 50 below=50% DOD.
You could change that to whatever you require, just leave the equation the same, and just replace the BOLD numbers with your required % DOD.

--calculating battery requirements-----
@50% DOD
23.04WH/12V=(1.92AH/50)100 = 3.84AH

@30% DOD
23.04WH/12V=(1.92AH/30)100 = 6.4AH


I hope that explains it a bit better? If not, feel free to let me know!

[inetdog]

As for the inverter question, yes, that's what I intend to do. The camera's power block plugs into a standard 110 receptacle to provide the camera constant power. I've considered the possibility about adding a resistor from the battery and forgoing the inverter and transformer block. At first I didn't like that idea but it's starting to build steam in my though process. I could wire something to step the voltage down to 3V to the battery terminals and then get the correct plug in for the DC input of my camera and avoid the inverter entirely, correct? I suppose I could hook up a cigarette lighter and use one of these?
http://www.parts-express.com/pe/show...m_campaign=pla

Before you try to substitute for the camera's associated power block, find out for sure whether the circuitry to control the battery charging is in the camera itself or in the power block. If it is not in the camera, then just using a resistor or just using a voltage convertor will not be safe.

[daz]

As for the inverter question, yes, that's what I intend to do. The camera's power block plugs into a standard 110 receptacle to provide the camera constant power. I've considered the possibility about adding a resistor from the battery and forgoing the inverter and transformer block. At first I didn't like that idea but it's starting to build steam in my though process. I could wire something to step the voltage down to 3V to the battery terminals and then get the correct plug in for the DC input of my camera and avoid the inverter entirely, correct? I suppose I could hook up a cigarette lighter and use one of these?
http://www.parts-express.com/pe/show...m_campaign=pla

Sorry, forgot to answer your other question!

I've considered the possibility about adding a resistor from the battery and forgoing the inverter and transformer block.

This is definitely not a good idea! Using that method, you will lose a lot of power, which in turn means you need a lot more solar panels (ie...going to cost ALOT more!).

You said the camera takes 2x1.2V batteries, I am just going to assume they are in series for 2.4V, so to work out power lost due to the resistor, we will use Ohm's LAW.
Ohm's Law states that voltage is directly proportional to current times resistance;
V=IR

V= in Volts
I = current in Amps
R = resistance in Ohms

--to calculate V-------
V=(V1 - V2) ; where V1 is your source voltage, and V2 is the load voltage
V=(12V - 2.4V) ; not entirely 100% accurate as that is the full charge of the batteries, you should use the discharged value. For this example though, I do not want to confuse you!
V=9.6V

--to calculate I-------
We said that each photo took 0.016WH of power. So....
0.016WH / 2.4V = 0.007A (I just rounded up, easier to work with!)
0.007A = 7mA per photo


So.........to work out the resistor we need to use....
--to calculate resistance required----
V=IR
9.6V=(0.007A)R
9.6V/0.007A=R
1371.43=R

So for your idea to work, you would require a resistor of value 1372 Ohms (which is no-where near a common resistance!).


--to calculate power lost through the resistor------
Power lost = (9.6V)(0.007A)
Power lost = 0.067WH

As you can see....using a resistor causes you to lose more power than what the actual camera requires for the photo!!! (0.016WH per photo compared to 0.067WH lost "reducing" the voltage to feed the camera from 12V)


A far more economical method would be to use an inverter or DC/DC converter. The one you linked to is a DC/DC converter, and even if it was cheap chinese, would be far more efficient than using a resistor!
Most DC/DC converters use switchmode power supplies, so they could be any-where between 80-95% efficient, which is way better than the resistor!! The resistor uses about 400% more power than the required power per photo!

[powerfulpup]

Before you try to substitute for the camera's associated power block, find out for sure whether the circuitry to control the battery charging is in the camera itself or in the power block. If it is not in the camera, then just using a resistor or just using a voltage convertor will not be safe.

As I understand it, the DC in bypasses the AA batteries in the camera. I can't charge them while they are in the camera, I have to take them out and put them in my wall charger. The DC in exists (it would seem) so you can download photos to your computer w/o using up battery power, or so you can show movies/slide shows without using up your battery.

So if I get the proper battery, will something like this guy do the job?

http://www.amazon.com/Instapark%C2%AE-Black-Mono-crystalline-Solar-Panel/dp/B004FX10ES/ref=sr_1_1?ie=UTF8&qid=1369291051&sr=8-1&keywords=15w+solar+panel

And a DC to DC converter, forget the resistor. That'd be better in terms of efficiency than the inverter would be, correct?