Good quality MPPT CC are expensive. It depends on your purchasing capability and your future plans for upgrade.
Announcement
Collapse
No announcement yet.
New partial off grid system help needed
Collapse
X
-
-
Originally posted by Tobym View PostOk great two panels then is it worth getting a 40a cc then or uprate that in a few years as wellComment
-
Ok on the subject of future proofing how do i calculate the size of the cable( inc voltage drop as its about 20-25 meters from panel to battery) for all 4 panels when I eventually put the others on. Is it the amps ie 4.85x4 so 20amp cable the. Add say another 10 amp for drop. If so 30 amp cable then is it ac or dc cable if ac then some domestic 240v 4mm twin core flex will be heaps wont it ?Comment
-
-
There is a program to calculate the sizes of cables and losses with respect to loads and length. Try to search for it. IIRC " Voltage drop calculator". For future estimates, your panels will put out 4.86 amps @140v. That's about 50 amps @ 12v output from your MPPT.Comment
-
Comment
-
50 amps, thats current output from the CC. That's 50AH in 1 hour, 100AH in two hours and so on. Do some reading on " How MPPT works".Comment
-
Originally posted by Tobym View PostOk on the subject of future proofing how do i calculate the size of the cable( inc voltage drop as its about 20-25 meters from panel to battery) for all 4 panels when I eventually put the others on. Is it the amps ie 4.85x4 so 20amp cable the. Add say another 10 amp for drop. If so 30 amp cable then is it ac or dc cable if ac then some domestic 240v 4mm twin core flex will be heaps wont it ?
V=IR
where:
V=voltage (in Volts)
I=current (in Amps)
R=resistance (in Ohms)
As an example, lets say you have a 140V 4.8A solar array, with 20metres of cable.
The cable's resistance can be found on its datasheet. In this case let's assume it is 1.0 Ohms per 100 metres. So for 20 metres the resistance is (1.0/100)20=0.2 Ohms.
Now.......
V=IR
V=(4.8A)(0.2R)
V=0.96V
So in our example, over a 20metre cable carrying 4.8A, you would have a Volt drop of 0.96V. So our solar array comes into the CC at 140V-0.96V=139.04V
This is why is is more efficient to use a higher voltage over a lower voltage, as a higher voltage equals a lower current, which means less Volt drop!Comment
-
Originally posted by Tobym View PostOk how do you get the 50 amp and is that the amp/h that will be going into the batteries
Please do not write amp/h. That literally means amps per hour, and since an amp is already a rate (proportional to electrons going by per second) the rate divided by time is meaningless.
If you keep a current of one amp going for one hour, you have transferred one amp-hour (AH) worth of electrons. It is AH that measure a batteries capacity. And AH times Volts is watt-hours (Wh) which is unit of energy.SunnyBoy 3000 US, 18 BP Solar 175B panels.Comment
-
Ok thanks again so am I correct then in saying the way you got 50a is 35v x4 panels =140v but the amps stay at 4.86 as they are in series so 140x4.86 = 680.4 watt hours (watts per hour yes) into the cc so when 680.4 is divided 12v the batteries are having 56.7 amps put in to them an hour. Please tell me I've got it. So if I run all 4 panels as series pairs then linked in parallel the amps would be 9.72 and the volts would 70. So 70x9.72=680.4/12=56.7 again but there potentially more loss as the voltage is lower so cables would have to increase in size . Sorry for being a bit thick its been a very long time since I used my brain in this wayComment
-
Originally posted by Tobym View PostOk thanks again so am I correct then in saying the way you got 50a is 35v x4 panels =140v but the amps stay at 4.86 as they are in series so 140x4.86 = 680.4 watt hours (watts per hour yes) into the cc so when 680.4 is divided 12v the batteries are having 56.7 amps put in to them an hour. Please tell me I've got it. So if I run all 4 panels as series pairs then linked in parallel the amps would be 9.72 and the volts would 70. So 70x9.72=680.4/12=56.7 again but there potentially more loss as the voltage is lower so cables would have to increase in size . Sorry for being a bit thick its been a very long time since I used my brain in this way
Originally posted by Tobym View Postam I correct then in saying the way you got 50a is 35v x4 panels =140v but the amps stay at 4.86 as they are in series so 140x4.86 = 680.4 watt hours
A 35W panel will put out about 1.94A (@18V). When you put panels in series, you add up the voltage, but the current stays the same. When you place panels in parallel, you add the current, but the voltage stays the same.
So 4x35W panels in series will give you 140W (1.94A @ 72V). Using a MPPT CC, you will be looking at about 140W/12V=11.66A going into the battery.
Originally posted by Tobym View PostSorry for being a bit thick its been a very long time since I used my brain in this wayComment
-
-
Yep I get that it's 4.86a so with 4 of them it means 50a into battery. Ps say I have 2 12v batteries I assume that's 25a amp in each not 50 into eachComment
Comment