HELP with solar system bank and charger controller

xsr_Gx

Hello,


Our project involves deploying a WSNs for monitoring purposes. This involves a camera and a microphone. Our daily average Amp-hours and Watt-hours are 2.5 and 12.1, respectively. We are considering using a 6V 15AH Lead Acid battery and a 12V, 20W solar panel with the following specs:

*Short circuit current = 1.3A
*Open circuit voltage = 21.6V
*Max power voltage = 17.2V
*Max power current = 1.17A

If our calculations are right (correct me if i'm wrong), if we start with our battery fully charged, the by the end of the day our battery will have 12.5 AH or about 83%. Considering the worse case, say we get at most 2.5 hours of good sunlight, and if we consider that it provides 1A, then we produce 2.5 A-hrs from the panel in one day, or 12V*2.5 = 30W-hrs/day.

Does it mean that it takes (2.5Amp-Hrs)*(6V) = 15 W-hrs to charge the battery again to 100%? Does this mean that if we get the conditions described above, we can charge the battery back again in one day since our panel is producing 30W-hrs and since we only used 2.5 A-Hrs of the battery, which means it only needs 15W-hrs?

Another question, we are considering using a MPPT charge controller since we are decreasing our voltage from 12V to 6V. Is this a good solution? Any information will be appreciated. The battery has the following charge method at constant voltage:

cycle usage (repeating use) :

-Initial current 4.8A or smaller , -control voltage 7.25V to 7.45V

Trickle use:

-Initial current 1.8A or smaller, -control voltage 6.8V to 6.9V

What is the best way to recharge the battery?

Sr

BB.

Re: HELP with solar system bank and charger controller

Welcome to the Forum!

Lots of questions, so here we go:

xsr_Gx wrote: »

Our project involves deploying a WSNs for monitoring purposes. This involves a camera and a microphone. Our daily average Amp-hours and Watt-hours are 2.5 and 12.1, respectively. We are considering using a 6V 15AH Lead Acid battery and a 12V, 20W solar panel with the following specs:

Sounds like you know what you guys are doing... 2.5 AmpHours at 6 volts is a pretty small amount of power. If you have any questions and are looking for some test gear--I would suggest a totalizing DC Amp*Hour / Watt*Hour meter like one of these (suggestion--I have not used these meters or this vendor myself) so you can measure actual load over 24/XX hours (good for variable current/power loads).

*Short circuit current = 1.3A
*Open circuit voltage = 21.6V
*Max power voltage = 17.2V
*Max power current = 1.17A

Note that solar panels are, more or less, current sources whose output is proportional to the amount of sunlight falling on the panels.

These are Vmp=17.2 volt panels--you are going to get only a little bit more current when running them at a charging voltage of ~8 volts on your 6 volt battery using a strait connection or PWM/Analog battery charger brick/chip:

• 8 volts * 1.17 amps = ~9 watts

Also, I would derate the panel to 0.80 of output (current in this case) to allow for real word use (dirt, temperature, etc.).

If our calculations are right (correct me if i'm wrong), if we start with our battery fully charged, the by the end of the day our battery will have 12.5 AH or about 83%. Considering the worse case, say we get at most 2.5 hours of good sunlight, and if we consider that it provides 1A, then we produce 2.5 A-hrs from the panel in one day, or 12V*2.5 = 30W-hrs/day.

• 15 AH - 2.5 AH = 12.5 AH remaining (assuming 100% charged battery at 25C/77F)

Realistically, you are probably going to fully charge the battery to ~90% or a bit more on a daily basis.

Does it mean that it takes (2.5Amp-Hrs)*(6V) = 15 W-hrs to charge the battery again to 100%? Does this mean that if we get the conditions described above, we can charge the battery back again in one day since our panel is producing 30W-hrs and since we only used 2.5 A-Hrs of the battery, which means it only needs 15W-hrs?

It is going to get a bit more complicated here... Roughly a flooded cell lead acid battery is 80% efficient charging with respect to Power and approximately near 100% efficient in AH use (ignoring equalization for a moment).

• Power = I*V = 2.5 AH * 7.35 volts (charging) = 18.375 Watt*Hours

And in your case, you have a way over voltage rated solar panel that is basically current limited--so its power rating at 17.x volts is more or less meaningless here.

For this setup, assuming PWM or other "simple" type of charge controller (on/off, analog pass transistor) it is the available current from the panel that will determine the charging time:

• 2.5 Amp*Hours * 1/0.80 panel derating * 1/1.17 amps = 2.7 hours of full Sun

Another question, we are considering using a MPPT charge controller since we are decreasing our voltage from 12V to 6V. Is this a good solution? Any information will be appreciated.

Yes, an MPPT charge controller would be a great help here--Then you could use the 30 watts of the panel (high voltage/low current) and down convert to the battery (low voltage/high current) using (ignoring losses)

• Power = Imp*Vmp of panel = Ibatt*Vbatt of Battery (like a DC version of a transformer).

However, I don't know if a MPPT Solar Charge Controller that is both small enough (to be cost effective) and works with a 6 volt battery (most work 12-48 volt battery banks).

I am certainly not an expert and you may find some industrial 6 volt MPPT charge controllers out there that will meet your needs.

But if you don't a small PWM 6 volt controller (assuming you find one) and a larger solar panel (because of the voltage miss-match) will work (Solar RE MPPT charge controllers tend to be expensive vs a small PWM type).

If this is a one time project--then you will "waste" the money on an over-sized panel vs the issues of finding/designing/building a small 6 volt MPPT controller.

The battery has the following charge method at constant voltage:

cycle usage (repeating use) :
-Initial current 4.8A or smaller , -control voltage 7.25V to 7.45V

Trickle use:
-Initial current 1.8A or smaller, -control voltage 6.8V to 6.9V

What is the best way to recharge the battery?

For larger batteries, we use the rule of thumb of 5% to 13% of the batteries 20 Hour rating...

• 15 AH * 0.05 = 0.75 amps minimum
• 15 AH * 0.13 = 1.95 amps maximum

Generally, less than 5% can cause charging issues and start having problems with keeping up on self discharge as the battery ages.

Over 13% for flooded cell batteries can cause over heating and for large Solar RE systems is an expensive battery charging system.

For your system, it is small and battery may have less of an overheating issue (smaller batteries have more surface area vs internal volume--so better convective cooling).

You can also look at sealed batteries, many can take higher charging currents (AGM for example) -- but they are sensitive to over charge (too high of charging voltage) and should have a temperature compensated controller (ideally with a remote battery temperature sensor--but you will probably not find such an option on a small 6 volt charge controller).

Note that you will have to decide on flooded cell (add distilled water once a month) or sealed (more expensive, sensitive to over charging / high voltage).

You mention 2.5 hours of sun--Is that Full Noontime Sun Equivalent, or really 2.5 hours during the middle of the day.

Also, depending on location, you will have sun angle and weather changes throughout the seasons (perhaps ~2 hours of average full sun in the winter and >5 hours of full sun in the summer).

There is also an issue that storage batteries really don't like to be cycled below ~50% state of charge (deep cycling reduces life)--And never below ~20% state of charge (possible cell reversal). So, you would want something to cutoff battery load at 6.00 volts or so (~50% capacity at 77F).

Also, again depending on where you will deploy--you may have a week of no sun (heavy cloud cover) where you may average much less than 10% of your normal daily production.

In your case, if you are using a "few small batteries" you can afford to replace them once a year for reliability. Not like full sized Solar RE system where a $5,000-$15,000 battery bank is not unusual.

You might want to read up on lead acid storage batteries here:

Deep Cycle Battery FAQ
www.batteryfaq.org

I will stop now--probably I am taking too many guesses here and confusing you (or me).

-Bill

xsr_Gx

Re: HELP with solar system bank and charger controller

Hello BB, Thanks for the quick response.

The min specs for this project, is to be deployable and extandable. This means that certain min specs must be satisfied, such as small in size, light and portable. Battery and solar panel must meet these requirements.

As i mentioned before our average total current is 2.5Amp-hrs and our average total power is 12.2 Watt-hrs. This is our first time working on a project like this, we have so many questions/doubts. For instance, i would like to know exactly how to size/select the solar panel and the battery bank. Most of our devices (microprocessor, camera, sensors, etc) run at 3.3V. This is why we have selected a 6V battery.

Our idea was to use either a WMP or MPPT solar charger controller to bring the voltage down to 6V. However, as you mentioned, the MPPT is expensive and we havent been lucky finding one that meets our needs. For what i have read, WPM
controllers must match the voltage of our solar panel array and battery bank.
What about our case? our solar and battery bank have different voltages. Is this still possible?

The other option was, bring the battery bank up to 12 V and use a switch a WPM charge controller OR leave the bank at 6V and use a switch regulator that would bring the 12 from teh solar panel down to 6V (comparator would be implemented to prevent battery overcharging or undercharging), then use a second switch regulator from the battery back to the power rail (3.3V).

Based on historical weather conditions, during winter we can get about 3 hrs of good sun light per day.

Do you recommend to go from a 20W to a 12 W solar panel since pretty much we would be wasting voltage?

According to you, the power needed in order to recharge the battery is (if we stick to the 20W solar panel):

Power = I*V = 2.5 AH * 7.35 volts (charging) = 18.375 Watt*Hours

and since the charging time depends on teh current from the solar panel, then

2.5 Amp*Hours * 1/0.80 panel derating * 1/1.17 amps = 2.7 hours of full Sun


we would need aprox 2.7 hrs of full sun to recharge the battery?

by the rule of thumb, it is recommended to charge teh battery from .75 Amps to
1.95 amps maximum, but since our panel cannot provide 1.95 Amps and can possibly provide .8Amps, what are the chances that it can keep the battery mostly charged?

We would like to have at least 10 days of battery life (before recharging), this takes into account consecutive cloudy days where there could be no sun. Do you recommend to go for a higher Amp-Hr battery? Bigger solar panel? I really appreciate your time to read and respond. THis gets quite confusing at times.

In addition, the 2.5amp-hrs will be used at night, this means that it would get the power directly from the battery. This way, batteries will be recharge during the day. One more last question, when selecting the right equipment, do we care more about the total average amp-hours or the average watt-hrs?? what is more important when it comes to selecting the right solar panel?

I know, there are too many questions . Thanks in advance.

mikeo

Re: HELP with solar system bank and charger controller

The other option was, bring the battery bank up to 12 V

It seems that this choice would use more commonly off the shelf components which would be simpler to acquire as well as cheaper. A simple inexpensive PWM controller and an 12 volt AGM battery would then match common PV panels. Just use inexpensive switching power supplies like these to get down to the 3 volts you need.
http://www.dimensionengineering.com/DE-SWADJ3.htm
They also have a 3.3 volt switching regulator but only at 1 amp.

BB.

Re: HELP with solar system bank and charger controller

I would do what Mikeo suggests--find a small buck mode switching power supply with the output ratings you are looking for... They are reasonably efficient--but you will need to take the down converter's losses into your overall power use.

There are a lot of options out there for these types of power supplies. It can be overwhelming to hunt down something that meets your needs.

• By the way, what is the typical current profile of your loads... 0.25 amps for 10 hours a night then off, or 0.10 amps for 24 hours per day or what. Where are these to be used (to figure out how much sunlight, operating temperature, etc.).
• How long are these to be deployed? A week, month, or year?
• Can you loose a week's worth of data during bad weather or must it have power 24x7?
• If 24x7, how long without sun would you expecxt to power the devices?
• Will they be unattended or will somebody be out once per week?
• Must they be tied to a base station by wireless network, or can you use flash memory (like USB flash drives) to log the data and just come and swap out data collection.
• On the sensor/network side, is there anything you can do to reduce power requirements (such as directional antenna to reduce transmitter power, reduce transmission times, lower power transceivers, etc.).

xsr_Gx wrote: »

The min specs for this project, is to be deployable and expandable. This means that certain min specs must be satisfied, such as small in size, light and portable. Battery and solar panel must meet these requirements.

Generally, solar panels and lead acid batteries are neither small or light weight.

To make a bullet proof system will add cost and complexity to the setup.

Anything you can do to reduce power requirements will reduce size and costs.

As i mentioned before our average total current is 2.5Amp-hrs and our average total power is 12.2 Watt-hrs. This is our first time working on a project like this, we have so many questions/doubts. For instance, i would like to know exactly how to size/select the solar panel and the battery bank. Most of our devices (microprocessor, camera, sensors, etc) run at 3.3V. This is why we have selected a 6V battery.

Assuming costs are an issue--using off-the-shelf parts will speed development and reduce costs. And using a down converter power supply would allow you to choose a more standard 12 volt battery bank.

Our idea was to use either a WMP or MPPT solar charger controller to bring the voltage down to 6V. However, as you mentioned, the MPPT is expensive and we havent been lucky finding one that meets our needs. For what i have read, WPM controllers must match the voltage of our solar panel array and battery bank.

At least from the little I know, the smallest very good quality MPPT Charge Controller is the Morning Star 12/24 volt 15 amp at around $240 each. And these probably only make sense if you have 100-200 watts of solar panels and need to mount the panels remotely from the sensor pack (i.e., panels need to be 100 feet away in the sun while the sensor pack is in a forest).

Small good quality PWM (I assume you are talking about Pulse Width Modulation--I am not sure what WPM stands for in this context) will cost less than $50 each. For a little more, you can get ones with Low Voltage Disconnects (LVD is not perfect, but it will somewhat protect the battery from being badly killed).

You can read about charge controllers here:

All About Charge Controllers
Read this page about power tracking controllers

What about our case? our solar and battery bank have different voltages. Is this still possible?

The reason to match the solar array's Vmp to Vbatt+2volts (controller voltage drop) is to best use the panels watts (Power=Voltage*Current) which is what you are being sold for good money. You can use higher voltage panels on PWM systems--it is just the extra voltage is wasted (of course, you have to match the voltage/current limits of the charge controller).

You can use higher voltage panels and low voltage battery banks (and there may be good reasons to do this--such as a long distance from the solar panel to battery bank to allow for wiring voltage drop). Otherwise, if Vmp is >>> then Vbatt--it is a waste of panel money (you can sometimes find 6 volt and other Vmp industrial solar panels that may save costs on low Vbatt applications).

--end of part 1--

BB.

Re: HELP with solar system bank and charger controller

--part 2--

The other option was, bring the battery bank up to 12 V and use a switch a WPM charge controller OR leave the bank at 6V and use a switch regulator that would bring the 12 from teh solar panel down to 6V (comparator would be implemented to prevent battery overcharging or undercharging), then use a second switch regulator from the battery back to the power rail (3.3V).

Without knowing the details of your load requirements--this is my guess the direction you will wish to follow (charge controller to 12 volts, switching regulator down to 3.3 volts for your loads).

Based on historical weather conditions, during winter we can get about 3 hrs of good sun light per day.

Average is one thing--actual is something else. I life near San Francisco CA... I have pretty good average collection, but I can go days (almost a week) where my system generates less than 10% of its "normal" sunny winter production because of weather. And I am on the "Sunny Side" of the coastal mountain range (generally, no marine layer effects). On the coast side, during the summer you could almost set your watch for 4 day of marine layer and 2 days of relatively sunny weather. In San Francisco, on the western side of the city, you could go the entire summer without seeing any sun.

If you have the luxury of doing a site survey months/year in advance, you might want to log the solar irradiation with a small battery powered logging sensor like this.

Will you always have the ability to put the sensing device in full sun? Will there be times it is at the edge of a forest or on the backside of a cliff, etc.? Will you need a non-rechargeable or long term, no sun, battery pack for those installations?

Do you recommend to go from a 20W to a 12 W solar panel since pretty much we would be wasting voltage?

It will be a cost/benefit decision... A Vmp 17 volt panel may be 2x a 8.5volt solar panel -- but you might find 17 volt panels are less expensive because more of them are made because 12 volt battery banks are everywhere.

According to you, the power needed in order to recharge the battery is (if we stick to the 20W solar panel):

Power = I*V = 2.5 AH * 7.35 volts (charging) = 18.375 Watt*Hours

That was the power required by the battery--If you are using a PWM controller--you pretty much will define your usage by the panel charging current--then find the Vmp rating that will meet your minimum voltage requirements (say, 7.35 volt battery charging + 2 volt controller drop + 2 volts for wiring drop and hot panel derating):

• Vmp minimum = 7.35 volt battery charging + 2 volt controller drop + 2 volts for wiring drop = 11.35 volts minimum
• Panel Wattage = 11.35 volts * 1.17 amps = 13.3 watts (Ideal panel)
• Panel Wattage = 17.5 volts * 1.17 amps = 20.5 watts (typical "12 volt" panel)

So, the panel wattage rating is "somewhat" flexible depending on your needs versus what you can purchase.

and since the charging time depends on the current from the solar panel, then

• 2.5 Amp*Hours * 1/0.80 panel derating * 1/1.17 amps = 2.7 hours of full Sun

we would need aprox 2.7 hrs of full sun to recharge the battery?

Note, these are "ideal" numbers... It would not be out of the question to double the numbers for safety factor/higher than expected average loads.

by the rule of thumb, it is recommended to charge teh battery from .75 Amps to 1.95 amps maximum, but since our panel cannot provide 1.95 Amps and can possibly provide .8Amps, what are the chances that it can keep the battery mostly charged?

The closer you are to the exact numbers, the more likely that your system will not keep up with the load requirements.

Remember that Lead Acid batteries are not perfect. There are some losses and they have limits which, if passed, shorten life. If a battery, for example, sits for days-weeks at below ~75% state of charge, it will slowly sulfate and lose capacity (and eventually die).

Flooded cell batteries should have their electrolyte levels checked at least once per month (until you see what your setup uses).

Sealed lead acid (gel, VRLA, AGM) will not need "watering" but will need a good quality/accurate charge controller (temperature compensated) to prevent overcharging (if you "vent" the battery from overcharging, it will kill the battery). Also AGM and other sealed batteries are slightly more efficient (~90% power efficient vs ~80% for flooded cell).

Small Lead Acid Batteries (like the 15 AH one you are looking at) will probably need to be replaced every year for reliable operation.

You could go with larger batteries--but then you would need more solar panels and have additional costs if you "kill" the larger batteries early (you should get 3 years or more from a properly serviced deep cycle name brand battery).

Also, remember that the battery bank does not usually charge at 100% rated current for 2.7 hours... As the battery approaches 90% state of charge--the current will drop over the next 1-3 hours to near 100% full charge.

Now, in reality, if you get the battery to at least 90% state of charge every few days, and don't spend much time below 75% state of charge--that is about the best you can do for the battery (and keep electrolyte above the plates).

We would like to have at least 10 days of battery life (before recharging), this takes into account consecutive cloudy days where there could be no sun. Do you recommend to go for a higher Amp-Hr battery? Bigger solar panel? I really appreciate your time to read and respond. This gets quite confusing at times.

In general, if you choose a larger battery, then follow the 5% minimum rate of charge. Obviously, if you use a "large" battery--this requirement will swamp the power usage of the device. An expensive solution--but may be worth it for you...

Say you want to use large battery for 10 days of no sun and you will replace the battery once a year. And a maximum of 80% discharge (20% state of charge). Now, we are talking about 12 volts instead of 6 volts, so AH use is cut by 1/2 to 1.25 AH per day:

• 1.25 Amp*Hour * 10 days * 1/0.80 discharge = 16 Amp*Hour minimum at 12 volts

If you do 50% maximum discharge (better life):

• 1.25 Amp*Hour * 10 days * 1/0.50 discharge = 25 Amp*Hour minimum at 12 volts

These are still pretty small batteries (a typical car battery is probably around 80 AH).

In addition, the 2.5amp-hrs will be used at night, this means that it would get the power directly from the battery. This way, batteries will be recharge during the day. One more last question, when selecting the right equipment, do we care more about the total average amp-hours or the average watt-hrs?? what is more important when it comes to selecting the right solar panel?

Really need understand the power usage of your loads--sizing of the down converter is important to know the details. Also, can you turn off the down converter during the day (save standby losses, etc.). Also peak current load can be important for battery sizing (say it uses all 2.5 AH in 10 minutes vs over 10 hours).

You will need to look at your power needs, both voltage and current--then "translate" that into the Mfg data sheet.

Hope we are not frustrating you here... Look at the battery as the "heart" of your system... And each thing you hang off of it (the "Load", the "Charge Controller", etc.) all are sort of independent of each other (size battery to load, size charger to battery, etc.).

-Bill

xsr_Gx

Re: HELP with solar system bank and charger controller

• By the way, what is the typical current profile of your loads... 0.25 amps for 10 hours a night then off, or 0.10 amps for 24 hours per day or what. Where are these to be used (to figure out how much sunlight, operating temperature, etc.)

Here is a more detailed explanation of our current profile:

Microprocessor (5V):
Active time (230mA): 240m/1440m = 16.7% , Av. Daily A-Hrs = 230mA*(240/60) = 920mAHrs , Daily W-Hrs = (5V)*(230mA)*(240/60) = 4.6W-hrs

StandBy (50mA): = 1200m/1440m = 83.3%, Av. Daily A-Hrs = 50mA*(1200/60) = 1A-Hr, Daily W-Hrs = (5V)*(50mA)*(1200/60) = 5W-hrs

Radio Transceiver (3.3V)
Tx ~ 0%
RX (40mA) = 240m/1440m = 16.7%, Av Daily A-Hrs = 40mA*(240/60) = 160mA-hrs, Daily W-Hrs = (3.3V)*(40mA) *(240/60) = .528W-Hrs
StandBy (4uA) = 1220/1440m = 83.3%, Av Daily A-Hrs = 4uA*(1440/60) = .1mA-hrs, Daily W-hrs = (3.3V)*(4uA)*(1440/60) = 260uW-hrs

This give us:

Total average A-hrs /day = 2.08 A-hrs + 20% inefficiencies = 2.5 A-hrs
Total average W-hrs/day = 10.128 W-hrs + 20% inefficiencies = 12.5 W-hrs

as you can see, we well be running these two devices continuously for 240min or 4 hrs. So 625mA for 4 hrs. These will be deployed near Santa Cruz Ca,


Average Temperatures
September max: 24°C (76°F)
January minimum: 4°C (39°F)
Annual mean: 13°C (56°F)


• How long are these to be deployed? A week, month, or year?
We are aiming for at least a year, meaning our system needs to meet a maintenance-free for at least this amount of time.

• Can you loose a week's worth of day during bad weather or must it have power 24x7?

Assuming we need about 4 hrs to transfer data, then the system will be off for about 20 hrs, The power then must come from the battery itself since this will take place during night time. We want to aim a battery life that can at least power this system from 10-15 days if considering there is no sun and/or cloudy days.

• Will they be unattended or will somebody be out once per week?

This is to be deployed on remote location. The primary goal is to monitor endangered species, so the goal is to avoid human presence as it disturbs these animals. The ultimate goal is as already mentioned, to have a deployment life of min 1 year w/o human intervention.

• Must they be tied to a base station by wireless network, or can you use flash memory (like USB flash drives) to log the data and just come and swap out data collection.

I just explained this better. The power budget above reflects the what we would be using at the base node only. This requires more power since it has to receive data from sensor nodes (5). Data will be stored on a SD card and then sent back to a server via a microwave link.

• On the sensor/network side, is there anything you can do to reduce power requirements (such as directional antenna to reduce transmitter power, reduce transmission times, lower power transceivers, etc.).

The total average current in the sensor node is 356mA-hrs and
the total average power is 1.2W-hrs.

The is less since we only sent ~ 350MB of data to the base node, the base node has to receive this data x 5, since there are 5 sensors.

Thanks a lot guys, your information is really useful!

BB.

Re: HELP with solar system bank and charger controller

How much time will you have to set this up and test?

Can you setup a test system right now and do some logging on power collection right now--see how much useful power you can collect during bad weather.

Also, it sounds like these will be in areas with lots of marine layer influences...

Personally, I would use on of the totalizing DC Amp*Hour / Watt*Hour meter (or many several) to start logging load data and energy collection data right now.

Power supply wise, I would be looking for something like these (PDF download)... They are small switching power supplies that can take ~6-28 VDC to 2 amp of 3.3 or 5.0 VDC out. They are around 85+% efficient and cost $25 each from www.DigiKey.com (note, just a first pass looking for something useful...). There are lots of these type of devices out there. Get a 3.3 volt and 5.5 volt mounted on a perf-board with some external capacitors for better filtering (should have a ground plane too).

If this is a one-off project (for now) and it has to work... I would size the battery and array to be 4x your measured power requirements and the estimated worst case average sun. And I would get a good quality solar charge controller with remote battery temperature sensor and a good quality AGM battery.

Whether you use a 15 amp MPPT controller or a 15 amp PWM controller with LVD and Remote Battery Temp Sensor or the cheaper PWM that does not have an LCD Meter or Data Port--is sort of up to you (prices range fronm $240 to $101)... If you have smaller panels (less than 100 watts) and short cable lengths, you would probably be just as happy with the PWM controller. You could get one "high end" controller and use the meter/computer interface to monitor your test rig--and get the cheaper units for field use.

Also, for the first test system, I would recommend a good quality Battery Monitor. It will give you the battery state of charge at a glance (percentage or AH). Almost manditory for AGM / Sealed battery systems where you cannot use a hydrometer to check battery state of charge (measuring battery state of charge with a DVM is only accurate after ~3 hours of no charge/no load to allow electrolyte to mix). This would probably be overkill for a field installation where nobody is checking.

Besides the normal weather/'marine layer issues, you may also have bird droppings on solar array issues too. Larger will give you a better safety factor for the project.

-Bill

Solar Guppy

Re: HELP with solar system bank and charger controller

The MorningStar Sunsaver Mppt ( 15 amp ) has built in data logging ... with the absorb and float timers you will know if your battery is being fully charged, no need for an external battery monitor

xsr_Gx

Re: HELP with solar system bank and charger controller

Thanks a lot guyz,

I guess i will come back to this later as i need to run some tests. Now, i m going to bug you with a few more questions.

So, the sensor node consists of the following:

Microprocessor(3.3V):
1)active (16mA) = 242min/day or 16.8%

average current/day = 16mA*(242/60) = 65mA-hrs
average power/day = (3.3V)*(16mA)*(242/60) = 213mW-hrs

2) stanby (1mA) = 1198m/day or 83.2%

average current/day = 1mA*(1198/60) = 20mA-hrs
average power/day = (3.3V)*(1mA)*(1198/60) = 66mW-hrs


Camera (3.3V):
1)active (60mA) = 48min/day or 3.3%

average current/day = 60mA*(48/60) = 48mA-hrs
average power/day = (3.3V)*(60mA)*(48/60) = 158mW-hrs

2) stanby (.1mA) = 1392m/day or 96.7%

average current/day = 60mA*(1392/60) = 1mA-hrs
average power/day = (3.3V)*(60mA)*(1392/60) = 8mW-hrs

Microphone (3.3V):
1)active (.5mA) = 144min/day or 10%

average current/day = .5mA*(144/60) = 1mA-hrs
average power/day = (3.3V)*(.5mA)*(144/60) = 4mW-hrs

Radio Transceiver (3.3V):B]
1)Tx(210mA) = 45min/day or 3.1%

average current/day = 210mA*(45/60) = 157mA-hrs
average power/day = (3.3V)*(210mA)*(45/60) = 519mW-hrs

2)Rx(40mA) = 5m/day or .3%

average current/day = 40mA*(5/60) = 3mA-hrs
average power/day = (3.3V)*(40mA)*(5/60) = 11mW-hrs

3)StandBy(400uA) = 1390m/day or 96.5%

average current/day = 400uA*(1390/60) = .1mA-hrs
average power/day = (3.3V)*(400uA)*(1390/60) = .3mW-hrs



Total average current /day = 300mA-hrs
Total average power/day = 980mW-hrs

Considering a battery bank of 6V, then we should be using 980W-hrs/6V ~ 164mA-hrs per day at 6V, but from above i get ~300mA-hrs;Which number is correct?

We have considered using the following:

-6V, 850mA solar panel
-3 1.2 NiMH, 10AH batteries
-LM2651 Switching Regulator (Vin = 4V - 14V, Vout =1.2V, 3.3V or adjustable)

we will place the 3 batteries in series obtaining a nominal voltage of 3.6V at 10AH. Based on our power calculations (and assuming we get 2 full hours of good sun per day):

(980mW-hrs)*(1/2 sun hours)*(1/.75 panel + charge contr eff)*(1/.8 battery eff)
= 816mW that we need from the solar panel per day, correct??

Now, assuming we have a good day (~2 hrs of GOOD sun) and getting at wors case .5A from the panel, then we get :

(.5Amps)*2hrs = 1A-hr so (6V)*(1A-hr) = 6W-hrs

If during that same day, we use 980mW-hrs +20% (ineff.) ~ 1.2Whrs of power, then we have a surplus of:
6W-hrs - 1.2Whrs = 4.8W-hrs ,correct?

Does this mean that our solar panel is enought to power up the sensor node?

tHANKS A LOT GUYZ!

xsr_Gx

Re: HELP with solar system bank and charger controller

ignore the switching regulator for now, i will explain later when i know these calculations are right.

BB.

Re: HELP with solar system bank and charger controller

xsr_Gx wrote: »

Total average current /day = 300mA-hrs
Total average power/day = 980mW-hrs

First, the relationship between Amp*Hours and Watt*Hours is:

Watt*Hours = Amp*Hours*Volts (which I know you know)

Considering a battery bank of 6V, then we should be using 980W-hrs/6V ~ 164mA-hrs per day at 6V, but from above i get ~300mA-hrs;Which number is correct?

Next, you have to look at the type of down converter that you will be using... A simple Analog 3-T regulator just is a "variable" resistance that drops the voltage from 6 volts to 3.3 volts (in your system example).

Note, that this would then require the use of Amps and Amp*Hours to figure out the sizing of your system. The Amps going into the 3-T analog regulator is the same as what is coming out (less small losses to run the regulator electronics and bias circuits)...

So in that case you would use the 300 mAmp*Hours per day (current in = current out).

But, the power required is different... This will be the sum of the power used to power the electronics at 300 mAmpHours and the drop from 6 volts to 3.3 volts across the regulator:

• 300 mAH * 3.3 volts = 990 mWH to power your electronics
• 300 mAH * (6v-3.3v) = 810 mWH lost as heat on the 3-T regulator
• 990 mWH + 810mWH = 1,800 mWH total

A switch mode regulator however runs basically at Power In - Losses = Power Out. So in this case, we will use the Watt value instead of the Amps (note there is a bit of discrepancy in the Power... 980 mWH given, vs 990 mWH calculated from mAH--I just used your numbers so you can see where the came from and where they go):

• Power Out * 1/efficiency = Power In
• 980m mWH * 1/0.85 = 1,153 mWH for 85% efficient switching regulator

Of course, you can plug mAH in instead and figure out the current ratio from 6 volts to 3.3 volts:

• Load AH * (output voltage/input voltage) * 1/efficiency = Batt AH
• 300 mAH * (3.3v/6v) * 1/0.85 = 194 mAH (at 6 volts)
• 194 mAH * 6 volts = 1,165 mWH (bascially same as above with 98/99 round off error)

We have considered using the following:

-6V, 850mA solar panel
-3 1.2 NiMH, 10AH batteries
-LM2651 Switching Regulator (Vin = 4V - 14V, Vout =1.2V, 3.3V or adjustable)

we will place the 3 batteries in series obtaining a nominal voltage of 3.6V at 10AH. Based on our power calculations (and assuming we get 2 full hours of good sun per day):

(980mW-hrs)*(1/2 sun hours)*(1/.75 panel + charge contr eff)*(1/.8 battery eff)
= 816mW that we need from the solar panel per day, correct??

You are going to hate me...

Yes, you are correct in your equations but missing some information and making some other wrong assumptions.

1. The solar panel voltage Vmp (voltage maximum power) needs to be (roughly) > 7.25v+2v controller drop (for lead acid) = 9.25 volts, not the 6 volts you have listed.
2. Charge Controller efficiency. If PWM, keeping Vmp close to Vbatt is more efficient, but 6v-out/8v-in=0.75 losses right there. ~0.80panelderating*0.75charge-controller-eff=0.60 or 60% overall efficiency--so more losses than you have planned for.
3. You used 0.80 or 80% for battery efficiency--NiMH are around 66% efficient (and I believe that varies with charging current and temperature and can be as low as 40% or less in some cases).
4. You have two ways to recharge NiMH batteries... One is at charge rates at higher than C/10 (C/6 or C/2 or even faster than C/1) but you need a pretty sophisticated charge controller (not Lead Acid simple) to do that correctly or the batteries will be damaged (and battery temperature needs to be monitored). Or you can recharge at C/10 rates or slower and not even need a charge controller.
5. There will be roughly 4-6 hours per day to recharge. If you used a small capacity battery, you would need to charge at C/4 to C/6 rates to fully recharge every day--and you would need a Charge Controller. If you did a C/10 rate (a much larger battery pack) with just a solar panel and blocking diode, you can "overcharge" the batteries without any immediate damage (have to replace pack once a year?).
6. A problem with batteries if you run them to "dead" and have 3 or more batteries in series (to get your 6 volts)--a problem is that there is always one cell with less capacity so it "goes dead" first. When you have your 6 cells (for 6 volts) in series, the other 5 are not yet dead and will begin to reverse charge the battery--which will pretty much kill it immediately unless you have some sort of shutdown circuitry. Once way to minimize the possible damage is to test cells for capacity (some battery chargers have a capacity test function)--you would "match" cell capacities for each pack--then the "all go dead" about the same time.
7. Which is why you see may "cheap" devices using only one or two rechargeable batteries in series. With on or two batteries, you can avoid the "dead cell" detector--there is not enough voltage to reverse bias the "weak cell". However, this forces you to use a "boost" type switch mode power supply which is a bit more complex and a bit less efficient than the "buck" type you are looking at right now.
8. Another issue--A 6 volt lead acid battery is ~5.5 volts when dead and 7.25 volts when charging. A 5 cell NiMH pack will be about 5 volts when dead and about 9 volts when charging--that means your power supply needs a wider range and the solar panel Vmp needs to account for the much higher peak charging voltages of the NiMH (9volts+2volt drop = Vmp of 11 VDC)

Now, assuming we have a good day (~2 hrs of GOOD sun) and getting at worst case .5A from the panel, then we get :

(.5Amps)*2hrs = 1A-hr so (6V)*(1A-hr) = 6W-hrs

If during that same day, we use 980mW-hrs +20% (ineff.) ~ 1.2Whrs of power, then we have a surplus of:
6W-hrs - 1.2Whrs = 4.8W-hrs ,correct?

Does this mean that our solar panel is enough to power up the sensor node?

Not really, see previous book length manuscript... :cool:

-Bill