Charging batteries with solar panels
I want to charge a 14.4V battery pack with some small solar panels. Will it work? The panels only produce small voltages. I don't want to try this unless I know cause I don't want to blow up the panels.
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Don't worry. Be happy
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Don't worry. Be happy
Comments
Anyway, the details depend on the current capacity and chemistry of your batteries and the specs of your solar panels. It's possible to use lower voltage solar panels to charge a higher voltage battery by using a voltage boost regulator. There's some loss of efficiency, maybe 15-20%, but it can be done. You'll usually need a regulator anyway (as a charging controller).
The simplest solar charging circuit requires a panel with a voltage under load about 2V above the battery voltage and you'd have a diode in series with the solar panel output to prevent the battery from discharging through the solar panel in less light (or darkness).
Post Edited (Mike Green) : 7/7/2010 1:03:42 PM GMT
and how fast do you want the batts to charge ?
Peter
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"Carpe Ducktum" "seize the tape!!"
peterthethinker.com/tesla/Venom/Venom.html
Never underestimate the bandwidth of a station wagon full of tapes hurtling down the highway. —Tanenbaum, Andrew S.
LOL
But as long as the voltage is high enough, couldn't even a really small panel charge a battery, just very slowly? I'm thinking of a FLED solar engine where a small high voltage, low current panel charges a capacitor and run a motor. Scale that up a bit to a battery and a micro-controller.
The robot would need an inactive off state, and an active on state. You checkpoint progress in EPROM so the robot would keep state between power cycles. Basically a photo popper with a brain.
If the current is too low, the battery won't ever charge.
My ideal solution is attached, but I suspect I need something more. Do I need to disconnect the panels when the equipment (stamps, motors) is attached?
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# of panels = 17 / "small voltage"
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- Stephen
I've ordered new panels now. They are 9V and 70mA peak. I can chain two of them and get 18V, for example.
Should I chain three of them and get 27V? Is there a limit to high the voltage should be?
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Don't worry. Be happy
2) 18V is reasonable. The issue of more voltage has to do mostly with the minimum light level that will cause charging. When there's insufficient light, the voltage (and current output) of the panels drops. Eventually, the output voltage is insufficient for charging to occur. With more panels in series, you may be able to charge in lower light levels, but the extra voltage will be wasted in bright light. With large panels, you actually have to switch in resistors or some other load to get rid of the extra energy. With your small panels and low currents, it won't matter. The 9V rating is probably open-circuit voltage in full sunlight, so the actual voltage output under charging load in realistic light levels will be much lower. You should get a better idea of the situation when you get the panels and can test them in varying light levels for voltage under load at the kind of load you're talking about.
3) Don't mix NiCd and NiMH batteries and don't mix battery packs with different capacities. You're probably ok if they're close given the low currents involved here, but it's generally a bad idea to mix batteries.
I've ordered new panels now. They are 9V and 70mA peak. I can chain two of them and get 18V, for example.
Should I chain three of them and get 27V? Is there a limit to high the voltage should be?
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Don't worry. Be happy
I'm pretty sure you'll need a diode in there as well, so the batteries don't discharge through the solar cells at night.
(nevermind, you have one
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"puff"...... Smile, there went another one.
You really should characterize your solar panels once you get them. As it has been mentioned, the 9V may be an open circuit voltage. Under a true load, you could see something very different.
With solar, you can chain several panels together for increased voltage and for some applications this might be desirable. However, be aware that one of the things that contribute to solar inefficiencies is the internal resistance of a panel itself. When you chain several cells together in series, this internal resistance is accumulative. It's better to change the aspect ratio (Not overall solar surface area) and place all of your panels in Parallel to minimize the effects of internal resistance. This way with an efficient converter you can step-up the voltage to an adequate level for charging.
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Beau Schwabe
IC Layout Engineer
Parallax, Inc.
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Don't worry. Be happy
First determine what your solar cells are capable of, then determine what your load demand is better suited for.
For example:
For a NiCad battery, the recommended charging volts per cell is about 1.65V, meaning you need at least 9.9V of overhead for a 7.2V pack (6 Cells X 1.65V = 9.9V) ... so a single solar cell wouldn't do it. But the question remains would it be more efficient to bump up the voltage from a single solar panel to 9.9V or to drop it down from multiple solar panels in series. Generally it's more efficient to bump the voltage up than it is to bump it down, but again it depends on the circuitry. Even using a 555, it depends on what your load requirements are.
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Beau Schwabe
IC Layout Engineer
Parallax, Inc.
Recommended Design Practice of Off Grid Solar PV Systems
The design process is a fairly simple and straight forward that doesn’t require a lot of technical knowledge. The initial steps are:
Determine the load in energy for a 24-hour period. Not the watts, but the watt-hours.
Determine the size of the solar array to be used.
Determine the battery size.
DESIGN EXAMPLE
The following example is a rough estimate to take to a system designer to discuss cost and objectives. He/she will then fine tune the system based on actual components, cable distance, etc… Our basic objective in this simple example is to provide power to a 250 load (light bulb) for 24 hours per day in two different cities (San Diego and Seattle) with 90 % availability. FWIW this is a system I have designed many times using remote cell radio sites. The transmitter is continuous 250 load, and the system will be a typical sized system that can be used in a home application. The only difference is radio sites are designed for 99.99% availability and this one will only be 90%. Getting from 90 to 99.99% greatly increases the cost with a larger solar array, larger batteries, and a standby generator set.
FIRST DESIGN FOR WORST CASE
In this example the worst case is simple to determine because the load is continuous 24 x 7 x 365 of a 250 watt light bulb. So the worst case is the month of December and January when the Solar Insolation is at its lowest point. In some instance, the worst case for the load is the summer and worst case for the resource is the winter, requiring you to perform two designs and then to select the one system that will carry the load through both summer and winter.
So in this example we need to determine the energy needed in a 24 hour period. This is done with watt-hours. To determine the watt-hours is straight forward of Watts x Time (in hours). So 250 watts x 24 hours = 6000 watt-hours or 6 Kwh in a day or 24 hours. Make note of this number as it will be needed latter.
SECOND IS THROW IN A FUDGE FACTOR
You multiply the total 24 hour load energy by 1.5 to account for several factors that would be handled individually in a detailed design. Some of the factors accounted for by this method are all the system efficiencies, including wiring and interconnection losses as well as the efficiency of the battery charging and discharging cycle, and allowing extra capacity for the PV system to recharge the batteries after they have been drained to keep the load going in bad weather. So 6000 x 1.5 = 9000 watts or 9 Kwh. Now take note of this figure.
THIRD DETERMINE SOLAR INSOLATION IN HOURS
Most solar map data are given in terms of energy per surface area per day. No matter the original unit used, it can be converted into kWh/m2/day. Because of a few convenient factors, this can be read directly as "Sun Hour Day” The number you want to use in this example is for December since December days are the shortest. San Diego is shown to receive 4.6 kWh/m2/day in December.. For Seattle, the number is 1.2 Kwh/m2/day. So we need to note 4.6 and 1.2 for our Sun Hour Day as it will be used to determine the solar panel array wattage.
FOURTH DETERMINE THE SIZE OF THE SOLAR PANEL ARRAY.
The size of the array is determined by the adjusted daily energy requirement using the Fudge factor number divided by the sun-hours per day. So for San Diego 9000 / 4.6 = 1956 watts, round up to 2000 watts. For Seattle 9000 / 1.2 = 7500 watts. Note the huge difference; it is because of the Solar Insolation. Location matters and will greatly affect system cost.
FIFTH DETERMINE BATTERY SIZE
All batteries will last substantially longer if they are shallow cycled. That means discharged only by about 20% of their capacity in a given day. Where as deep discharge or cycling means that a battery is discharged by as much as 80% of its capacity. A conservative design of 90% availability will save the deep cycling for occasional duty like several cloudy days in a row. This implies that the capacity of the battery should be about five times the daily load. So that means the capacity should be 5 times the daily load.
To figure the daily load, go back to the original load number before the fudge factor—that is, 6000 watt hours. Add to this a battery fudge factor of about 50% to account for the efficiency of the battery discharge, the fact that only 80% of the battery's capacity is available, and the loss in efficiency because PV systems rarely operate at the battery design temperature.
The end result is that the battery design load is 6000 times 1.5 or 9000 watt hours, which is coincidentally the same as the array's design load, but for different reasons. This is the daily energy taken from the battery, which is now multiplied by five to ensure 20% daily discharge: 9000 x 5 = 45,000 watt hours. This is the battery capacity, which is usually given in ampere-hours so it must be divided by the system voltage; 45,000 / 12 = 3750 Amp Hours. Take note this example is a 12 volt system. If you were using a 24 volt system the battery capacity would be 45,000 / 24 = 1875 Amp Hours. To put this into perspective a 3750 AH battery bank would weight a few tone and the size of a china cabinet or a very large office desk. The floor of this would need to be reinforced concrete built to 3000 psi test.
That’s about it to get you in the ball park. The system designer will then determine the inverter and charge controller requirements and fine tune the system numbers. In this example a 500 watt inverter (2 x Max Load = 2 x 250 = 500) and an MPPT controller of 175 amps for the San Diego system, and 650 amps for Seattle.
Post Edited (alexisrq) : 7/11/2010 4:50:22 PM GMT