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  • Solar charge confusion

    I have an IP camera that I want to place at the edge of my property using a solar panel.

    I already have the camera and it is normally powered by a usb cable. It does NOT have an internal battery..
    I do not want to use an off the shelf camera designed for this, they are every where but I want to use this camera
    It's a WYZE v3 for the curious,

    So I obviously need a panel, a battery, and maybe a charge controller.

    The camera specs state 1A but doubt it actually takes that .. but that is the spec.

    I guess my question is what wattage panel would I need to charge the battery up
    to full charge during the day ( and run the camera ) then power the camera all night.

    I live is San Diego so lots of sun.

    Would something like this work

    John Titor, when are you.

  • #2
    The way to work this kind of calculation is to understand that a solar cell will produce a certain number of watts of power for as long as it has enough sunlight. The watts measurement is the amount of power created at any particular moment. The watt hours is amount of time that it can create it's rated number of watts. Example: a 10 watt solar cell that gets full sun from 9 am to 4 pm produces 10 watts for 7 hours which is 70 watt hours (70 Wh).

    Then we move to batteries... They store a certain number of amp hours, or milliamp hours. That's the amount of current (in amps) that they are designed to produce for a set period of hours. Example: A 15 amp hour battery will produce 1 amp for 15 hours, or it will produce 3 amps for 5 hours, or 15 amps for 1 hour.

    Last, there are the conversions of volts and amps to watts. It's simply volts times amps equals the watts that are being produced or consumed.

    So in your case, if the camera uses 1 amp from a 12 volt source, that's 12 * 1 which equals 12 watts. If you want to run that camera 24 hours a day, you would need a battery that is rated 24 hours times 12 watts, or 288 watt hours. Your solar panels would have to create 288 watts for 1 hour, or 42 watts for 7 hours to recharge the battery. And don't forget that besides that 288 watts to charge the battery the panel also needs to generate 12 watts per hour to power the camera's during that time.

    Did that help?

    At the end of the project, there is a profound difference between spare parts and left over parts.

    Location: SF East Bay.


    • #3
      Dan's write up sounds good.

      The first thing I would do is look for package solutions: solar cells, charging controller, and battery all in one. Here's a start using Dan's number (~100 WH):

      If you go this route, be sure there is an USB output with the charging Voltage and current you want/need.
      Last edited by Paul Alciatore; 05-27-2022, 09:09 PM.
      Paul A.
      SE Texas

      And if you look REAL close at an analog signal,
      You will find that it has discrete steps.


      • #4
        I think you should measure the current drawn by the camera under various conditions. It may have a motion detector so it may draw less current when nothing is moving. At night, it may have infrared LEDs that may take a fair amount of current. And it may draw more current when the voltage is lower, so it will deplete the batteries faster as they lose charge.

        I need to sign up in order to see what you have in your wish list, and I'm not going to do that. I think you will want a MPPT controller, such as these: ($28 and up)
        Paul , P S Technology, Inc. and MrTibbs
        USA Maryland 21030


        • #5
          Originally posted by danlb View Post
          So in your case, if the camera uses 1 amp from a 12 volt source,
          Great description. Umm? Would it need what a USB port supplies? Is that 5vdc? I dont know, just asking. JR


          • #6
            Originally posted by danlb View Post

            Then we move to batteries... They store a certain number of amp hours, or milliamp hours. That's the amount of current (in amps) that they are designed to produce for a set period of hours. Example: A 15 amp hour battery will produce 1 amp for 15 hours, or it will produce 3 amps for 5 hours, or 15 amps for 1 hour.
            Ummm, point of order (so to speak), Dan:

            Discharge rate is an important consideration - capacity is not linear but dependent on the rate at which the battery is discharged. The one hour rate is usually specified by the constant C.

            Capacity of a lead-acid battery in amp-hours is specified at 0.1C - that is, a 15 amp-hour battery will produce 0.1 amp for 150 hours but 1 amp for less than 15 hours. Likewise, at 0.05 amps it'll last longer than 300 hours.

            Geez, I hope I didn't get myself tied up in knots with that explanation.

            There are no stupid questions. But there are lots of stupid answers. This is the internet.

            Location: SF Bay Area


            • #7
              I figure that a 33 Ah, 12V battery and a 30W panel should be plenty safe. The true, correct, and proper answer to this question depends on a surprising number of variables; I've gone through this for some of my own equipment, and it's surprising how much capacity you can actually need for a low-power device. Those numbers I quoted above are very dependent on your actual power draw; you may only need half as much storage and panel capacity, but the information we have to go on isn't that specific.

              To make the calculations, the main issues we're concerned with are (1) how much energy does the device actually use over a 24-hour period, and (2) how many hours of direct sun you receive per day. We could take geographical issues into consideration as well, but that's mostly pointless for San Diego. The power draw of the camera is likely the most important piece of information. As mentioned, it's probably (significantly?) less than the 5W (5V/1A) stated on the camera's spec sheet, but we don't know how much. We'll assume that the camera draws the full 5W continuously to be on the safe side.

              As Dan mentioned, energy is measured in watt-hours = volts x amps x hours. Assuming that the camera is running 24 hours per day (otherwise, why bother?), 5W x 24H = 120 Wh. To convert that to amp-hours, we simply divide by the battery voltage. I'll assume we're using a 12V battery. 120 Wh / 12V = 10 Ah. There's an added wrinkle if you're using a lead-acid battery; the conventional wisdom states that they shouldn't be discharged below 70% of their capacity for best lifespan. In that case we would divide our 10 Ah rating by 0.3 (30%, the amount of capacity we will actually use); that gives us 33 Ah.

              For the solar panel, you'll need it to produce 120 Wh (+ maybe another 10% for losses - call it 130 Wh) over the course of a day. Now, we have to determine how much energy a given panel can produce in a day. I looked up how many hours of direct sun you are likely to get in San Diego, and came up with an estimate of 5.7 hours per day. 130 Wh / 5.7 h = 22.8 W. We have to derate a bit because a solar panel will almost never produce its rated power; I would guess at something like 75%. 22.8W / 0.75 = 30.4W, call it a 30W panel.

              See, it's easy!


              • #8
                Jim raised a very important point that I'd chosen to leave out. Thanks Jim.

                I looked up the web page for the camera in question and it states that the power usage is 1 amp at 5 volts which is 5 watts. It also shows that they have it set up to use a USB cord and power supply.

                At the end of the project, there is a profound difference between spare parts and left over parts.

                Location: SF East Bay.


                • #9
                  Sorry about the link .... here it is outside my wishlist


                  The 33ah battery is a little ouchy price wise. Maybe a motorcycle battery.

                  Oh .. and yes ... some really good info.

                  I have a cheap little usb checker, gonna go out to the shop and
                  see if it will tell me the draw of the camera
                  Last edited by Mike Amick; 05-27-2022, 10:40 PM.
                  John Titor, when are you.


                  • #10
                    A 15 A-h battery would be rated at 15 amps for 1C, and the A-h rating for a SLA battery is defined at 1/20 C, so it should provide 0.75A for 20 hours, although for best life it's best to stop discharging after about 15 hours or 75%
                    Paul , P S Technology, Inc. and MrTibbs
                    USA Maryland 21030


                    • #11
                      You won't want a small lead acid battery like that. I'd suggest a used but still pretty good automotive or RV battery. The issue being that none of the lead acid batteries live long lives if they are deep cycled. And a motorcycle battery would pretty well need to give it's all every night. So it'll be rough on it and only have a fairly short life span. Go with a big car, truck or RV battery and the amount you drain each day is a small portion of the total charge so you totally avoid the deep cycle issue.

                      The solar panel outputs are all based on the angle of the sun. I don't see this budget project having a sun following mount. So you'll get peak charge over the 4'ish hours around high noon and a tapering off charge as the angle of the sun to the panel runs to higher angles outside of that mid day spell. You'll likely find that the battery only charges over about an 8 hour period in the winter and 10 hours in the summer due to this. And that the charge rate on each end starts out low and "bell curves" up at noon then down again.

                      If the worse case here is 8 hours of charging with enough to run just the camera for an hour each side of that it means potentially 14 hours of the battery running the camera. And a big enough solar panel that it fully and then some charges during the 8 hour day period.14 hours of 5 watts (camera operating at 5V and 1A current equals 5 watts). is 70 watt-hrs. If you are putting that much back into the battery in an 8 hour period this means you need the panel to supply an overall average over the 8 hours of 70W-hr/8hr= 8.75 watts. Let's round that up to 9 watts. 9 watts at 12 volts is 9W/12v= 0.75 amps. Mind you this is an average. Due to the way the panel's output will follow a bell curve you'll likely want something that can run the camera and still push at least 2 amps into the battery on a sunny day at high noon so you have enough reserve.

                      On top of this the camera is still pulling 5W which at 12 volts and allowing for the converter is going to be pretty close to 0.5 amps. So at least 2.5 amps off the panel at high noon. And likely 3 amps would be safer.

                      The good news is that at 12 volts 3 amps is only 36 watts. So if you opt for a 100w panel it'll easily run the camera and fully charge the battery each day even on fairly bad overcast or stormy days. And if the panel doesn't quite do that then the many amp-hrs in the battery will carry you through a stormy day or two and come back up after a couple of sunny days.

                      So all in all I'd say a 70 to 100 watt panel that comes with charge manager (typical RV kit), a used but still useable deep cycle RV battery and a high efficiency 12v to 5 volt converter .

                      The panel and charge manager kit looks to be from $80 to $100. Check with the local battery places for a battery option but don't go with low amp-hr options to avoid the deep cycle life span shortening.

                      All in all you're probably looking at from $150 to $200 to run your camera. And that's assuming the bad types you're hoping to monitor don't end up stealing the solar panel....
                      Chilliwack BC, Canada


                      • #12
                        Before making my previous post I did look up the USB specs. What I found was that while 5 VDC is more or less standard, there are some USB standards that allow for 20 VDC. While 99.999% of USB ports do supply 5VDC and you are probably safe in assuming the 5 VDC number, it would only take a minute or two with a VOM to check. But do so carefully: do not short those pins in the USB connectors to each other with the meter probes. A breakout board with a USB socket and broken out connections to the pins would be best. And do not expect an exact number. For instance a nominal 5 VDC USB supply may measure several Volts more on a VOM. The next step above that nominal 5 V would be 12 V and numbers between 4 and 10 V can be assumed to be a nominal 5 V.

                        A good general read on USB specs:


                        More on USB. This one mentions the 20 V and even 48 V for Power Delivery via USB:


                        USB breakout boards:


                        I think I am going to order a few of those breakout boards myself, just to have them on hand.

                        And yes, Yes, YES, it is DC as supplied by a computer's USB port. Computers do not have AC power running around inside them. At least most do not. The AC line cord goes to the computer's power supply and all the outputs from there are DC.

                        Originally posted by JRouche View Post

                        Great description. Umm? Would it need what a USB port supplies? Is that 5vdc? I dont know, just asking. JR
                        Last edited by Paul Alciatore; 05-27-2022, 11:38 PM.
                        Paul A.
                        SE Texas

                        And if you look REAL close at an analog signal,
                        You will find that it has discrete steps.


                        • #13
                          So far, so good. Even Sandy Eggo has rain or off-days, so you want more capacity. I prefer two day minimum. The 33 AH battery was mentioned as "ouchy" price-wise, but I cannpt see the item in question, it is still encased n the "wishlist" stuff.

                          BUT, before getting to the battery, there is an important factor..... You have 12V, how does it get to be 5V?

                          One way is with an ordinary regulator, which takes the 1A at 12V and puts out 1 A at 5V. You will not see those anymore, most are going to be SMPS types. SMPS are often best viewed as wattage, not amperes, because they work that way.

                          The 5V 1A is 5W. Low voltage SMPS are often not very efficient, so figure 80%. 5W / 0.8 = 6.25W. The thing will draw 6.25W from the battery, which at a nominal 12V is approximately 0.5A, not the 1A. (more so because "12V" batteries are fully charged at 13.2V, not 12V).

                          You need to supply the charge current, and the device current during the charge time, with some extra to take care of charging back after an "off day" weather-wise.

                          If you just wanted the battery to run the camera for 24 hours, it would need to supply 24h x 0.5A, or 12 ampere hours for a full day. Using the figure of approximately 6 hours of good sunlight, the panel should then supply no less than 12 ampere-hours (12 a-h) in the 6 hour charge time, plus supply the camera and losses.

                          To do that requires 18a-h supplied in the 6 hour charge time, or 3A charging, for the direct recharge. The battery is 80% efficient, so you need actually another half amp for losses, for a total panel and controller charge current of 3.5A.

                          That is for one day with replacement charge the next day.

                          The battery would be "OK" at about 36a-h capacity, which means a discharge to 66% daily. Better less percent discharge, so a 48 a-h battery is better.

                          If you want to be able to recharge after a two day drain with minimal to no recharge on day 2, you have choices. Best is to recharge in one day. Acceptable is to take 2 full days to recharge after a 2 day drain. So, that would mean recharging about 18 a-h per day, plus running the camera, plus battery losses.

                          With that, you need 3A to replace the charge, +25% for losses, plus 0.5A for the camera. In that case, the panel should be able to supply 3.75A plus 0.5A, or about 4.25A charge current to the batteries for the 6 hours each day, to take a battery back up to full charge after two days drain with no charging.(due to bad weather).

                          The battery needs to supply 24 a-h without going below say 66% of charge.

                          If I have done all the calculations right, that ends up being a battery of about 75 a-h, and a panel plus controller, which will charge the battery at a max of bout 4.25A. That ensures the battery will not be drawn down too far every day, and that it will be recharged fully in two days, after a period of two days where weather prevents charging.

                          Yes, it seems high.... but, "that's what the numbers say". Special known conditions might allow less capacity of battery and panel, or might need more.

                          Other things.... Panels need to be at 90 deg to the sun , so that the entire panel gets direct sun. If the panel is not at 90 deg, then less light hits it, and it is less efficient, prodcing less outut. Also, panels collect dust. Dust blocks light and lowers panel output.

                          Controllers vary. Some just control the current, others use a maximum power point system, and an SMPS type arrangement which steps down the volts while increasing the current in proportion.

                          A place called "WindyNation" has usable controllers of both the normal and the maximum power point types. They are pretty low cost as such things go.
                          CNC machines only go through the motions


                          • #14
                            My head is spinning reading these posts, I strain to keep up with it, but thankfully I get it.

                            I just measured the draw and got 300ma. which has got to be on my side.

                            as far as the link goes, this is the only way I can do it with some screen captures ... sorry.

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                            Attached Files
                            Last edited by Mike Amick; 05-28-2022, 12:14 AM.
                            John Titor, when are you.


                            • #15
                              If you can get a decent deal on a 100 watt solar panel, then you can be ahead of the game with intermittent charge rates. At the same time, a 90 a/h deep cycle battery might cost little more than a smaller battery, so you would be more able to operate on battery power for days without dropping the battery voltage too much in the discharge phase. That alone will make the battery cheaper when you find that it is lasting a good length of time. It seems to me that you won't really be saving money in the long run by trying to optimize the panel and battery to lower values from the start.

                              A 100 watt solar panel is pretty common, so it should be a fair price. The only drawback with it is it's larger physical size if you compare it to say a 30 watt panel. If size is an issue, then by all means follow the guidelines given by others here.

                              An MPPT controller is ideal, but a PWM controller will also work. I would opt for the former for better efficiency. When it comes to the output, there are many modules available online to convert voltages either up or down, and they are cheap. You need a 12 to 20 volt input, 5 volt output at minimum 1 amp. I would go for one rated for at least 2 amp output, just to help ensure that parts don't run too hot and shorten the life of it.
                              I seldom do anything within the scope of logical reason and calculated cost/benefit, etc- I'm following my passion-