Author Topic: Is this calculation correct?  (Read 2609 times)

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AbyssUnderground

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Is this calculation correct?
« on: July 30, 2006, 02:37:07 PM »
Ive just tested my solar array of 6x 1.5w panels. It came out as 0.6A short circuit. I made the following calculation and I was wondering if it was correct:


0.6A * 17.5v = 10.5w


(nominal panel voltage is 17.5v but these are rated as 12v panels and 21v open circuit)


As 6x 1.5w is only 9w, is this feasable? If it is, and Im correct, then Im going to be one very happy person!


All help is greatly appriciated.


Andy.

« Last Edit: July 30, 2006, 02:37:07 PM by (unknown) »

willib

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Re: Is this calculation correct?
« Reply #1 on: July 30, 2006, 08:53:00 AM »
nope cant do that

the current should be measured when hooked to a battery ,because shorting the solar panels through an ammeter is a way to test max current , not the current when charging at 12V , which will be less
« Last Edit: July 30, 2006, 08:53:00 AM by willib »
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AbyssUnderground

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Re: Is this calculation correct?
« Reply #2 on: July 30, 2006, 08:59:41 AM »
Ah right. So I am not getting 10.5w then? I thought the way to get the max current was to short circuit and assume the nominal voltage (17.5v) was correct.


What would be a rough guess as to what the current into a battery would be as I have no low battery to test with at present (an no sun unfortunately, its becomming overcast)?

« Last Edit: July 30, 2006, 08:59:41 AM by AbyssUnderground »

finnsawyer

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Re: Is this calculation correct?
« Reply #3 on: July 30, 2006, 09:13:51 AM »
Try putting a ten watt twenty ohm resistor across the output.  This should simulate the battery.  Then measure the resistor voltage.  If it is less than 12 volts, add more resistance (in series) in say one ohm increments.  A rough way to do it is to divide the open circuit voltage by the short circuit current.  This gives the resistance of the panel.  You can use that to calculate the current into a twelve volt battery.  The problem is that this assumes the solar panel is a linear device, and that its resistance doesn't change with load, a dubious prospect.  The first procedure is accurate and actually can be used to find the internal resistance of the panel as a function of output current.
« Last Edit: July 30, 2006, 09:13:51 AM by finnsawyer »

ghurd

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Re: Is this calculation correct?
« Reply #4 on: July 30, 2006, 09:27:58 AM »
Also keep in mind honestly rated thin-film PVs will make 10 to 20% more power when new.

After a month or 2 the power drops down to the 'rated' power.


The $10 red 1.5W HF PVs make over 110ma (into a battery) when new.

G-

« Last Edit: July 30, 2006, 09:27:58 AM by ghurd »
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Nando

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Re: Is this calculation correct?
« Reply #5 on: July 30, 2006, 09:29:49 AM »
Let me correct a misconception with the panels


Insolation = Solar energy on the panel.


A panel has 2 standard voltages and a peak current defined by the output voltage.

Specs referenced to a certain insolation level

Voc = Volts open circuit

Vmp = Volts maximum power

Imp = Current maximum power


Voc for a 12 volts panel run around 21 volts

Vmp are set around 17 volts ( Vmp (of the diode) times the # of diodes)

Imp when the output is 17 volts or less.


Which says that if the output voltage, controlled by a load is 17 volts or less the current will be Imp ( it is a constant current)


So to clarify : if the Imp = 12 amps and the Voltage,DUE TO THE LOAD, varies from a short = 0 to 17 volts, the Imp will be 12 amps and constant( remember the panel insolation is fixed to a reference).


For this reason it is best to have a charge controller with MPPT ( Maximum Point Power Tracking) to extract the power from the Battery voltage to the 17 volts Vmp.


The Solar Panels if fixed solar insolation will show a constant current up to the Vmp value of it, then starts to drop with the increasing output voltage until it reaches around 21 volts for 0 current -- being the output voltage controlled by the load.


Nando

« Last Edit: July 30, 2006, 09:29:49 AM by Nando »

Nando

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Re: Is this calculation correct?
« Reply #6 on: July 30, 2006, 09:34:39 AM »
Heck I forgot a point for further clarification


There is another current = Isc = Short circuit current


This current is defined at Vout = less than 0.1 volts


Isc is about 6 to 7 % higher than the Imp


Nando

« Last Edit: July 30, 2006, 09:34:39 AM by Nando »

AbyssUnderground

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Re: Is this calculation correct?
« Reply #7 on: July 30, 2006, 09:44:08 AM »
Thanks for the info. Unfortunately Im having a hard time understanding it all. All I wanted to know is what is a rough estimate as to what I can put into the batteries. I dont have a 10watt resistor on hand at the moment.
« Last Edit: July 30, 2006, 09:44:08 AM by AbyssUnderground »

ghurd

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Re: Is this calculation correct?
« Reply #8 on: July 30, 2006, 09:45:25 AM »
Then 85ma per PV
« Last Edit: July 30, 2006, 09:45:25 AM by ghurd »
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dinges

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Re: Is this calculation correct?
« Reply #9 on: July 30, 2006, 09:49:29 AM »
So, his 'expected' power would be:


Isc *.94 * 12V = .6 * .94 * 12 = 6.7 W ?


Thanks for the explanation; good to know that Isc is about 6-7% more than Imp.

« Last Edit: July 30, 2006, 09:49:29 AM by dinges »
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RogerAS

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Re: Is this calculation correct?
« Reply #10 on: July 30, 2006, 12:42:42 PM »
Abyss,


A much simpler way to figure this out, without a crash course in algebra and electrical engineering, is to place an accurate amp meter between the output of the panels and the battery. Output from the panel goes in one side of the meter out the other to the battery. Most little digital volt meters (less than $20 at Wally World) will give you a pretty good idea as to what's actually going into the battery, which is all that really counts. Just don't push more than 5 or 6 amps through one without a shunt and then reading that as milliamps. TomW has a posting somewhere here that covers how to do that. Do a search for shunt and see if it doesn't pop up.


Here's how it works:


Battery voltage, 12.5V for example. PV output to the battery, 1.75 amps for example. The solar panel output is 21.875 watts. Simple math,  12.5V X 1.75A = 21.875watts.


All those abbreviations are technical mumbo jumbo that don't ammount to a hill of beans and mean next to nothing in the real world. What matters is how much power the PV pushes into your battery, and nothing else. It's easy to measure and works every time.


RogerAS

« Last Edit: July 30, 2006, 12:42:42 PM by RogerAS »

AbyssUnderground

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Re: Is this calculation correct?
« Reply #11 on: July 30, 2006, 01:33:12 PM »
That was a very easy to understand post, thanks! :-)


I just flattened my battery to charge it with the PV as a test but it was only low enough to put on charge an hour ago and now the sun has gone and its gone overcast :-(


Ive put the battery on its mains charger as I don't particularly want it to sulphate. Ill retry the test tomorrow if I get chance.

« Last Edit: July 30, 2006, 01:33:12 PM by AbyssUnderground »

AbyssUnderground

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Re: Is this calculation correct?
« Reply #12 on: August 02, 2006, 12:53:33 PM »
Just managed to test the panels in full sun:


13.07v * 0.51A = 6.53watts ... correct?


P.S: The battery was at 12.5v when I started the test but quickly rose. It levelled at 12.85v after a 10 minute rest (no charge).

« Last Edit: August 02, 2006, 12:53:33 PM by AbyssUnderground »

ghurd

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Re: Is this calculation correct?
« Reply #13 on: August 02, 2006, 06:52:15 PM »
You would be better off thinking in Amps, instead of Watts.

The amps will stay quite stabile while the voltage (watts) rises.


If the battery is dead, say 10.5V, X 0.51A = 5.355W.

If the battery is near regulation, say 14.0V, X 0.51A = 7.14W.

The amp-hours is the same. There is NOT a 33% increase.


G-

« Last Edit: August 02, 2006, 06:52:15 PM by ghurd »
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AbyssUnderground

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Re: Is this calculation correct?
« Reply #14 on: August 03, 2006, 01:46:31 AM »
Right, I see. So in "estimation" to my calculations I have 6w at 12v? Thats fine with me when you consider its rated at 9w.


So 0.51A final current. :-) Im happy with that.

« Last Edit: August 03, 2006, 01:46:31 AM by AbyssUnderground »