Solar Charge Controller Ideas... Thinking out loud
Beau Schwabe
Posts: 6,568
I’m working on a charge controller for my solar setup.· Right now I have 15 solar panels each capable of producing 15 Watts (225 Watt system)… A conventional solar setup tells you to connect all of your panels in parallel.· This method requires HIGH current cables and much of the power is lost in the wires used to go from your panels to your charge controller.· Other power generation methods also suffer from this.·
As an example, placing all of my panels in parallel would create a current potential of almost 20Amps … The distance that I must go is about 50 feet, and the wire that I am using is household 10 gauge wire rated for 30 Amps·· BUT!!· The “IR drop” (ß Current loss over wire resistance)· is about 1 milli-Ohm per foot (http://www.powerstream.com/Wire_Size.htm) ·… in my case that’s 100milli-Ohms (ß100 because it’s round trip there and back) ··0.1 Ohms does not sound like much, but if you push 20 Amps through it, then you loose 2 Volts (Amps X Resistance = voltage across resistance) …. Again, 2 Volts doesn’t seem like much, but when you only start out with 12 Volts ( <-- under load ), loosing 2V before you get to do anything with it is quite a bit· (about a 17% loss)
What I have done or have an idea for, is to place all of the solar panels in series…. 12Volts times 15 panels = 180 Volts under load·…. In series, the panels only produce about 1.25 Amps·· (225 Watts / 180 Volts = 1.25 Amps)…. Using the same wire rated for 30 Amps, I still have the same amount of IR drop, but because I’m only pushing 1.25 Amps, I’m loosing 1/8th of a volt instead of 2V
The Charge controller that I·have an idea for takes the 180Volts and Chops it up with PWM (Pulse Width Modulation) for direct 12V battery Charging … The PWM duty cycle creates an average 15V supply for charging (about 8% duty cycle)… The technique that I would like to implement·stagers multiple duty cycle PWM’s together and places the output of each of them in parallel… for each PWM output, the current will be·about 1.25 Amps ( <-- I think )… by placing them in parallel, the current effectively multiplies by the number of parallel PWM outputs…. The maximum is 225 Watts / 15V or ·15 Amps (ßdictated by the number of solar panels I currently have) if it were 100% efficient.· I don't expect that of course, but I do expect something much better than 83% simply due to IR drop in initial power availability.
From here,·a parallel battery·bank that uses a distributive inverter method rather than a single inverter method located directly at the panel or junction box and I have minimal IR drop from my solar panels.
A distributive inverter system basically uses an inverter for each breaker circuit that you have on your breaker panel, or each circuit in your home.· This is easier to expand or make repairs. ·A single inverter system uses one inverter for the entire house…· this is usually costly and hard to expand or repair.· With the distributive system, you can also have mixed mode inverters where some are pure sine, and others are modified sine inverters, depending on their particular use.· A distributive system can also combine inverters (if they are phase matched) depending on the load demand of the entire system.· This way energy can be saved during low usage periods by minimizing the number of inverters in use, where as a single inverter system would be running all of the time.
I just want some thoughts or ideas, mainly on the method for folding the PWM duty cycle for current·gain without using a transformer.·
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Beau Schwabe
IC Layout Engineer
Parallax, Inc.
As an example, placing all of my panels in parallel would create a current potential of almost 20Amps … The distance that I must go is about 50 feet, and the wire that I am using is household 10 gauge wire rated for 30 Amps·· BUT!!· The “IR drop” (ß Current loss over wire resistance)· is about 1 milli-Ohm per foot (http://www.powerstream.com/Wire_Size.htm) ·… in my case that’s 100milli-Ohms (ß100 because it’s round trip there and back) ··0.1 Ohms does not sound like much, but if you push 20 Amps through it, then you loose 2 Volts (Amps X Resistance = voltage across resistance) …. Again, 2 Volts doesn’t seem like much, but when you only start out with 12 Volts ( <-- under load ), loosing 2V before you get to do anything with it is quite a bit· (about a 17% loss)
What I have done or have an idea for, is to place all of the solar panels in series…. 12Volts times 15 panels = 180 Volts under load·…. In series, the panels only produce about 1.25 Amps·· (225 Watts / 180 Volts = 1.25 Amps)…. Using the same wire rated for 30 Amps, I still have the same amount of IR drop, but because I’m only pushing 1.25 Amps, I’m loosing 1/8th of a volt instead of 2V
The Charge controller that I·have an idea for takes the 180Volts and Chops it up with PWM (Pulse Width Modulation) for direct 12V battery Charging … The PWM duty cycle creates an average 15V supply for charging (about 8% duty cycle)… The technique that I would like to implement·stagers multiple duty cycle PWM’s together and places the output of each of them in parallel… for each PWM output, the current will be·about 1.25 Amps ( <-- I think )… by placing them in parallel, the current effectively multiplies by the number of parallel PWM outputs…. The maximum is 225 Watts / 15V or ·15 Amps (ßdictated by the number of solar panels I currently have) if it were 100% efficient.· I don't expect that of course, but I do expect something much better than 83% simply due to IR drop in initial power availability.
From here,·a parallel battery·bank that uses a distributive inverter method rather than a single inverter method located directly at the panel or junction box and I have minimal IR drop from my solar panels.
A distributive inverter system basically uses an inverter for each breaker circuit that you have on your breaker panel, or each circuit in your home.· This is easier to expand or make repairs. ·A single inverter system uses one inverter for the entire house…· this is usually costly and hard to expand or repair.· With the distributive system, you can also have mixed mode inverters where some are pure sine, and others are modified sine inverters, depending on their particular use.· A distributive system can also combine inverters (if they are phase matched) depending on the load demand of the entire system.· This way energy can be saved during low usage periods by minimizing the number of inverters in use, where as a single inverter system would be running all of the time.
I just want some thoughts or ideas, mainly on the method for folding the PWM duty cycle for current·gain without using a transformer.·
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Beau Schwabe
IC Layout Engineer
Parallax, Inc.
Comments
Of course, if they always all get the same amount of sunlight, I guess it doesn't matter.
-Phil
Addendum: If each panel has its own blocking diode, you will lose another 15 * 0.6V = 9V in series, vs. 0.6V in parallel. Or can you eliminate the blocking diodes in a series configuration?
Another thought: If the cells are neither constant voltage nor constant current, is there a way to operate each panel at its optimum power point before combining their outputs?
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Post Edited (Phil Pilgrim (PhiPi)) : 8/21/2008 4:26:20 AM GMT
"...Are they constant voltage/variable current devices?..."
A little bit of both I believe based on their lighting conditions and load demands
"Another thought: If the cells are neither constant voltage nor constant current, is there a way to operate each panel at its optimum power point before combining their outputs?"
Probably, but I wanted to stay away from possible circuitry that would be exposed to the elements.
My main goal is to reduce the IR drop, I suppose that I could wire the Panels in parallel and dedicate a weather proof DC-DC step-up converter in close proximity to the panels to deliver a higher voltage lower current to the PWM charge controller, but how efficient would the DC-DC converter be at this point?· Would I really be gaining anything other than cheaper (not as thick) electrical wires from the panels to the charge controller?
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Beau Schwabe
IC Layout Engineer
Parallax, Inc.
I'm not quite sure I understand your proposal, though. Would each "staged" PWM have its own inductor and diode in a buck configuration?
-Phil
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Few months ago I did mention about the IR drop problem which I am also facing, but you said you have very thick copper "tubes", so it is not a problem.
You still owe me a measurement of the max charging current at noon.
I had also suggested doing away with the diode, because every 0.1 volt counts.
If you look at the V-I graph of most solar panels, the current drops over a cliff if the voltage is more than the rated voltage.
If you don't use 12v appliances, can I recommend a 24v system for your case.
DC-DC converters can be very difficult to scale and upgrade.
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"I'm not quite sure I understand your proposal, though. Would each "staged" PWM have its own inductor and diode in a buck configuration?"
Sort of yes, but I don't think it needs to be that complex.· I guess what I'm looking for is the·equivalent of a high voltage DC input·switch mode power supply capable of delivering the most available current.
william chan,
"...Few months ago I did mention about the IR drop problem which I am also facing..."
Yes, it's all your fault!·
"...but you said you have very thick copper "tubes", so it is not a problem..."
Yes I did, but it's not a very pretty solution... I'm looking for something that has better eye appeal, and still functional.
"...You still owe me a measurement of the max charging current at noon..."
This is a difficult measurement, because this measurement·will vary depending on the current charge level and health of your battery, and what kind of load demands are in place at that moment in time.· Without the battery bank connected, and 3 of the panels (45Watts) terminals shorted together (something I really don't like doing)· I measure about 3.5 Amps at·noon.
The meter I am using is a 50mV analog panel meter in parallel with a·18.8 inch piece of 18 gauge wire.
18 gauge wire is about 6.385 milli-Ohms per foot.· To get full scale (50mV = 5 Amps) with my meter I need a·10 milli-Ohm resistor.
10/6.385 = 1.566
1.566 * 1 foot =·18.8 inches
·
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Beau Schwabe
IC Layout Engineer
Parallax, Inc.
Post Edited (Beau Schwabe (Parallax)) : 8/21/2008 2:17:55 PM GMT
Discharge the battery about 5%, switch of all loads, and then measure the charging current at noon.
This is most important to gauge the efficiency of the charging system and charge controller, as a percentage of the rated power.
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"Short circuit current means nothing."
I'm not sure that that is a valid assessment.· From a short circuit current you can determine the internal resistance effect of your solar panels or other power supply.
Looking at Figure 1 this equates to an internal resistance of 6.1 Ohms.
·
Note: (Using 3 panels (45 Watts) for this exercise)
The unloaded voltage from the solar panels is about 22 Volts
The .1 Ohms is from the 50feet round trip of 10 gauge wire
The 0.01 Ohm resistor is the "shunt" across the 50mV Panel meter
·
·
As for my batteries I won't allow them to fall below 11 Volts, so looking at Figure·2 with the internal resistance implemented·the maximum current is 1.77 Amps
·
In Figure 3 you see the batteries·at 14 Volts with a maximum charge current of 1.29Amps
My measurements·seem to agree closely with this paper I found...
http://www.vicphysics.org/documents/events/stav2008/A7TheoryNotes.doc
Note: Keep in mind, that the batteries will also have an internal resistance about 10 milli-Ohms ... Since I have 4 batteries in parallel I have about 2.5 milli-Ohms of internal battery resistance.· If the batteries are not healthy, this value can be much higher.
·
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Beau Schwabe
IC Layout Engineer
Parallax, Inc.
Post Edited (Beau Schwabe (Parallax)) : 8/21/2008 4:53:28 PM GMT
Many modern switch-mode power supplies accept an input from 90 - 260VAC. But I believe one of the first things the input encounters, before it gets switched at a higher-than-powerline frequency, is a bridge rectifier. (They do, nonetheless, usually specify an input frequency range, so I could be wrong about that.) If so, I wonder if you could simply adapt an off-the-shelf 12- or 15-volt 300W supply and use it with your present charge controller.
Here is one that outputs 13.5V @22A (86% efficiency). It does specify an input frequency range of 47-63Hz, but the block diagram in the datasheet shows "rectifiers" right after the EMI filter and active current limiter.
-Phil
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Post Edited (Phil Pilgrim (PhiPi)) : 8/21/2008 4:04:59 PM GMT
> and the wire that I am using is household 10 gauge wire rated for 30 Amps
Beau,
I'd have used 8 gage in this case, if it's solid core. Note that solid is:
10 ga. = 0.9989 Ohms / ft
8 ga. = 0.6282 Ohms / ft
And I'm sure you know that stranded wire is better as it carries more amps and runs a bit cooler. You might laugh, but consider going to your nearest metals recycling place and see if you can get some scrap power line cable· - most places will sell it to you for the cost of the copper (or aluminum <- watch that stuff though), but you have to get it before they strip off the insulation·
cheers
- H in Fla
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Post Edited (CounterRotatingProps) : 8/21/2008 10:40:54 PM GMT
Interesting, I might just look into that ... Thanks
CounterRotatingProps,
What I have at the moment·is 1/4 inch copper tubing, that I figured at one time based on the cross-sectional area of the copper that it's equivalent to about 4 gauge wire.
What I want to do is use something with a little less beef if you will (just easier to work with)... that's where the·10 gauge wire came into play.· My initial thought was that if I could increase the voltage at the source (the Solar panels), that I would have less IR drop to deal with going to my charge controller.· By increasing the voltage my tradeoff is less current in the wires, and less voltage loss over the wires.
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Beau Schwabe
IC Layout Engineer
Parallax, Inc.
Try shorting the diode during measurement and see how much the current increases.
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Post Edited (william chan) : 8/22/2008 6:54:24 AM GMT
I'm sure you know this and will take the necessary precautions, but others on the forum may not: DC at these voltages is far more lethal than AC. If you get zapped with high voltage AC, you at least have a chance to let go. No so with DC. Under the wrong circumstances, one's muscles can seize, and the term "death grip" takes on a whole new meaning. The old rules for working with tube gear apply here: keep one hand behind your back at all times.
-Phil
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'Still some PropSTICK Kit bare PCBs left!
Some thing power transmission engineers have been dealing with for decades. At least you're not going to put up 220,000 volt (mini) towers! Lossy lines was also, historically, the reason power companies went with AC instead of DC transmission. (And that's why Tesla lost out to Edison and Westinghouse, if I remember correctly). Any who, just a passing blurb here...
- H
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Outside of the "box" it will be AC
CounterRotatingProps,
Yes, I have started thinking along those lines... With that length of wire and gauge that I want to use, the inductance will be about 23uH ... Not sure about the capacitance yet, but I should be able to alternate the DC at the resonate frequency of my particular setup to create an effective "transmission line" with minimal IR drop. That's my hope anyway.
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Beau Schwabe
IC Layout Engineer
Parallax, Inc.