Charging multiple devices from 25W <span class="highlight">solar</span> <span class="highlight">panel</span>

If you have a small inverter (100W) plug this into the car lighter connector and use the 240V to run all those chargers. This is what I do to charge camera and phone batteries.

Lachlan,

don't give up just as yet.

All your devices seem to have their own battery which is good.

The only thing you have to find now is a small inverter which provides just enough juice for these chargers.
Get the smallest inverter you can find, it only needs to output 25 Watts max.

As has been mentioned before, inverters are designed to be powered by low impedance sources such as batteries.

But because even small batteries depend on a charge controller to prevent too much charge from the solar panel damaging the battery, they only add complexity to your situation.

Try this: grab your 25W panel and wire it in parallel to a 10,000 mF/35V electrolytic capacitor. Observe the polarity on the capacitor or it'll go bang. The capacitor only weighs a fraction of a battery, and best of all it can't be overcharged so you don't need a charge controller.

Be aware that the capacitor will charge up to the maximum solar panel o/c voltage wich can be as high as 22V.
Your small inverter needs to have an input rating up to and including this voltage which it probably will.
If the input rating is lower than this, report back and I'll give you some advice on how to overcome this.

This capacitor/solar panel combo will most likely offer a low enough impedance to make your inverter power up.
Once powered up, the inverter should easily stay on even under lower light conditions.
Just be aware not to draw more than say 20 Watts in full sun, or your inverter might undervolt because there is not much of an energy buffer on the input side.

20 Watts of power will probably just be enough to supply all 240V battery chargers in your gear. But don't turn the lappy on while it's charging ;)

To answer your question about the speed of charging: I don't think that any batteries in your gear can take a charging rate exceeding the 20 Watts equivalent, so the answer is no. You'll just have to juggle with the charging sequence because you probably won't be able to power all chargers at the same time. So in a way, yes the devices could charge slower in their entirety, if they can't be charged all at at the same time.

Hope that this will work out for you.
Seems to be a pretty extreme trip you're planning!

Best regards, Peter

Peter you truly dont have a bloody clue what you are waffling on about. What you are suggesting is bizarre and completely bloody useless.

Where you going to buy a 25W inverter?
The inverter would take more power to power itself than a 25W panel could supply.

In winter the OP would need min 80w panel, regulator and 40Ah battery and may just make,

OP would be better purchasing 12V power supplies for all the gear. Using an inverter when trying to conserve energy is the last thing you would do.

Faulic_McVitte,

Yes I do.

It doesn't have to be a 25W inverter, it can be larger, like 150W.

They're readily available and are designed to receive 12V power from the ciggy lighter.

You obviously have overlooked the fact that the "OP" is trying to minimize weight and number of items, due to the nature of his trip.

BTW, my own 3kW 24V true sinewave inverter (weighing over 20kg) draws less than 0.5A when a small 240V load of a few Watts is connected to it.

So why on earth do you think a small 150W inverter would draw more juice to power its own electronics, than a 3kW unit?

I think you just haven't grasped the concept of an inverter.
It merely passes on power from a battery/solar panel/capacitor on its input to a load on its output.
An inverter functions as an impedance converter.
And if the load on the output is small (as is the case with this "OP"), then the loading on the input source will also be small.

Best regards, Peter

Peter lets pretend you know what you are talking about. How much USEABLE power is the OP going to get out of a 25W panel a day in winter say in Wagga Wagga, NSW?

How much is a 'few watts' on 240V AC compared to DC?

Unbelievable you try and defend the undefendable incorrect information.

Tragic what people get told here at times.

Faulic_McVitte,


Some people don't have to pretend....
And no need to play the drama queen here Faulic.

You're asking how much power can be expected from a 25W solar panel on a cool winter day in Wagga Wagga?

Make it a touch over 20 Watts in full sun.

If the inverter gobbles up say 5 Watts to power its own circuitry plus any conversion inefficiency of around 10%, then there'll be just under 15 Watts left for any 240VAC loads.

Because we're only concerned with small battery charger loads, this should be sufficient.

I expect the small laptop charger to demand the most power, with a real chance to exceed the offered maximum of 14W.
In this case the "OP" could either buy a slightly larger solar panel, or he'd need to go the small-battery-solar-charge-regulator-way.

But since electrolytic caps aren't expensive, the experiment could be well worth it.

I recommend to get this cap wired up in parllel to the panel, then power up the inverter and then start with the smallest loads first.
Tilt the solar panel away from the sun to get an idea of how much lack-of-solar-leeway each load allows, before the inverter undervolts.

Although not quite the conventional way of doing things, it's not rocket science and might work out allright.

Best regards, Peter

I have a Sharkshield Feedom 7 & the charger is in 2 parts an inverter box which is powered by 12V DC @ 1.5A. This is powered by a 12Vdc cord or a 240v plug pack suplying 12V @ 1.5A. Getting close to the solar panell rating alone. Daryl

Peter to say you don't have a damn clue is been damn polite. I cannot believe you know so little and purport to know it all. To say you are clueless is been mighty kind.

You obviously have ZERO idea how much 20W at 12VDC will give you on 240VAC.

The most you would get in winter at Wagga Wagga from a 25W panel in ideal conditions may be 17.5W and you have no idea how long you would be able to get that power over the day.

Peter, please go and do some basic homework prior to answering posts here and leading people right up the garden path in the wrong direction.

Keep it nice guys,

Ok Faulic,

no need to try and hide behind a curtain of strong but meaningless words.

Let's get down to the nitty gritty of solar cells instead, if you dare.

Why exactly do you think a 25 Watt rated panel would drop a whopping 30% of its rated power on a cool but sunny winter day in Wagga Wagga?

Let me give you some hints:
First find out under what conditions solar panels are rated, watch out for 'standard test conditions'.
Then, compare the STCs with the operating conditions you will encounter on a typical cool winter day in Wagga Wagga.
Concentrate on things like 'temperature derating of solar cells' and how this might affect the output of a 25W panel.

You've got the whole world wide web at your fingertips - just go and find us some links etc which you deem suitable to back up this 30% drop of output power.

Feel free to ask for help if you get stuck.
But if you fail to put some effort into this, I consider any additional discussion with you meaningless and a waste of time.

Best regards, Peter

Peter you cannot even answer the basic questions asked of you. Do you know about Ohms Law and other basic laws of physics.

Forget about STC's and anything else is all irrelevent to this post.

How about you answer these two questions you keep avoiding.

1) How much USEABLE power is the OP going to get out of a 25W panel a day in winter (JUNE) say in Wagga Wagga, NSW?

2) How much is a 'few watts' on 240V AC compared to DC?

3) Tell everybody how much 20W at 12VDC will give you on 240VAC

No doubt you will duck and weave to avoid answering these very basic questions.

Isnt 20W on 12v roughly equal to 20W on 240v AC? If you ignore the losses thru the transforming medium? It was when I did electrical engineering anyways

i'm not real familiar with wagga, only passed through there a few times so i can't help with that one.

but unless the laws of physics have altered in the last few minutes i'd say that a few watts of AC would be about the same amount as a few watts of DC.
And i'd reckon it'd be a pretty good conclusion that 20W at 12V would be about 20W at 240 Ac or at any other voltage.

If we introduced ohms law into the discussion, oh I see you have, then we could discuss this as follows:
P = V * I -> 20W = 12V * I -> I = 20/12 = 1.666666667A
P = V * I -> 20W = 240V * I -> I = 20/240 = 0.08333333A

so yep, the current is different but there seems to be a confusion with any suggestion that the actual power available would alter just becuase the voltage changes.

Now if we were to suggest that an inverter isn't 100% efficient and that some power would indeed be lost in the conversion process then that would be a reasonable observation.
An awful lot of these discussions seem to follow a common theme where numbers and units are thrown into the discussion with no regard or understanding of Mr Ohms law and the basic laws of physics. Quickly followed by a descent into unflattering comments.
Lets all have a nice day and may everyone have the joyous experience of learning that sometimes we don't allways know everything we think we know ;)
Even old dogs can learn to roll over and play dead. :)

Bonz you are correct. A watt is a power of force.

To suggest you are going to get anything meaningful from a 25W panel converting to 240V is fanciful in the extreme. The minimum inverter self power consumption is going to be 400mA.
Wagga Wagga in June is going to give you a maximum of 2.22 solar hours a day. Under ideal conditions the 25W panel would output 38.85W per day. Take in the losses charging batteries through a 12 to 240V inverter with a 240V battery charger and you may be lucky to achieve 8 to 12W if powering an inverter directly worked.
To suggest to the OP using an inverter with a capacitor with an inverter and a 25W solar panel is completely daft. The inverter would be unlikely to start as it would first sense over voltage and then low voltage and end up doing nothing.

Faulic,

ad 1) because you've no grasp of the concept of energy and power, you asked this question the wrong way around.
If you would have asked about the expected daily energy output of a 25W panel, instead of this nonsense of 'how much power a day in winter...'

ad 2) Watts stay constant when converting from one form of voltage to another, except for conversion losses which are in the order of 10% to 15% in switch mode circuits.

ad 3) if you look at your previous question, you may notice that you basically asked the same thing twice.

Bad thing is, you didn't do your homework.

Just one final advice:
If all this doesn't quite make sense to you, just study the contributions of other posters in this thread who're certainly clued up better in these things than you are.

Best regards, Peter

Hello Peter,

Re:
Quote :
But since electrolytic caps aren't expensive, the experiment could be well worth it.

I recommend to get this cap wired up in parllel to the panel, then power up the inverter and then start with the smallest loads first.

End quote.

Idea is good but one really needs to do the sums before setting this up.

1. The energy stored in the 10,000 micro-Farad capacitor is very small for this job.

This cap will store 2.42 Joules ie. give 2.4W for 1 sec. Energy (Q) stored in a capacitor is Q = 0.5 * C * V**2. As you can see this capacitor holds a very small amount of energy.

Compare this to a 12.5V 20AH (20 x 12.4 WH) or 3600 x 20 x 12.4 Joules ie. battery that stores 900,000 Joules !!!

2. The resistive load on the capacitor is 6 ohms (24W @ 12V) and this gives a time constant (R x C) of 0.06 sec. ie. ti will hold up the 24 W inverter for only about 0.1 sec !!

Hello Jeremy,

I appreciate your time and effort evaluating my idea.

You're quite right, the energy storage potential of a capacitor is very small compared to a battery.

But keep in mind that the purpose of the capacitor in my proposal is not one of energy storage (over seconds or minutes) but one of impedance reduction of the inverter source.

Because a solar cell is in effect a current source, its internal resistance is higher than the one of a typical voltage source like a battery, or a capacitor.

Now, due to the DC/DC converter inside the inverter drawing current from the input smoothing caps in a pulsed form, their charge gets replenished by a current from the source with high ripple on it. This spiky current would cause overproportionally high power losses (I square R), inside the source.
These losses would result in voltage dips in the input stage of the inverter which would cause the electronics to undervolt.

A capacitor's low ESR in parallel to the higher one of the solar panel will limit the voltage dips on the inverter input.

It'll satisfy the pulsed current demand from the inverter without causing the voltage to drop much, while it's constantly receiving charge from the solar panel.

The cap only needs to smooth out things a bit, not store energy for any prolonged period of time.

Best regards, Peter

sorry Jeremy,

the concept of the additional smoothing cap as described above needs to be amended.

The most demanding current drain from the solar panel will certainly happen on power up of the inverter.
If the input source resistance isn't sufficiently low, the inverter will probably not stay on because the beefy smoothing caps will take too long to charge up.
The additional cap on the input will act as energy buffer for the inrush current on power on.


Best regads, Peter

No problem Peter.

Re:

A capacitor's low ESR in parallel to the higher one of the solar panel will limit the voltage dips on the inverter input.

OK on this.

Re:

It'll satisfy the pulsed current demand from the inverter without causing the voltage to drop much, while it's constantly receiving charge from the solar panel.

OK on this also - with reservations!

Question : How much is "too much" ?

Answer: Inverter operates between 9 and 15V so the voltage must not drop below 9V Nor rise above 15V. OK? How much capacity do I need to maintain this voltage ? I don't know but it sounds like we need a voltage regulator here !

Question: How do we manage the start-up when the capacitor is fully discharged? ie when does the inverter switch in?

I have a strange feeling that. if we continue along this path, we will end up designing a current to voltage converter and regulator in the middle of this thread and bore everybody stupid.

Re:

The cap only needs to smooth out things a bit

OK on this

Re:

not store energy for any prolonged period of time

I have a problem here.

This is how a capacitor works it stores energy. The question how quickly will it charge and discharge - this is what you refer to as smoothing and this is determined by the circuit time constant - we want a long time constant perhaps 30 seconds (guess) and the figures we are talking about show a very short time constant ie very little smoothing.

Question : What is this "prolonged time" ?

Firstly the start-up transient (load or no load) can be upto 5 x the running current so one will need to have to carry this energy - I don't know how these inverters take to starting with a very slowly increasing input voltage.

The capacitor / load combination make for a relatively fast response so relative to the time variation of the input so the input regulation will be very poor. Remember that a capacitor charged via a constant current source will have its voltage rise in a linear fashion - so one would expect ripples with long straight lines!!

Lucky there's only crap on TV tonight.

Cheers and Beers,

Jeremy.

great info guys, I have the textbooks out and am going thru it

Hello Jeremy,

did a test this morning:
3kW 24V true sinewave inverter powered by a +/-15V 3A regulated power supply.
Load is a 240V fan, causing an inverter input current of 1.4A.

Tried to power up the inverter without the 10,000uF cap first. After a couple of undervolts on power up, it stayed on and supplied the fan with 240V no probs.
Then, wired this cap in parallel to the input, powers up like from a battery, first time every time.

So the crucial time when the cap is really needed is during inrush on power up, as expected.
A 10,000uF cap stores enough energy to feed the inrush, even for this monster of an inverter (quick calc, 70A for 1ms, during a suggested max voltage dip from 16 to 9V).

What's left to fix up, is this high o/c voltage from the panel.
I checked some specs sheets for small inverters and they seem to be somewhat limited in their input voltage rating, 15 to 16V max for the majority of them.

So if it was me, I'd grab 4x4V5W zeners wired in a series string, and then in parallel to the solar panel/cap/inverter input.
Once the inverter load starts loading the solar panel down to below 16V, the zeners will stop bypassing, thus won't gobble up precious solar power during operation.

All up we're looking at around 15 bucks cost of components, combined with minimum weight and complexity.

A couple of photos showing the input voltage/current, 25V10,000uF electrolytic cap, fan load in the back, and the true sinewave.

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Cheers and beers :)

Peter

Definitely a good idea to reg the voltage from the solar panell. Itried a 20W panell with a 22000Uf capacitor & a 150W inverter, The high input voltage destroyed the 240V switching Mosfets IRF740 four ofthem @ $5.74 ea trade price, Ohwell . It would probably be safer to use a small 12V 7Ah sealed battery to even out the load & they are only a bout $20. Daryl

Pretty good Peter.
Re:

3kW 24V true sinewave inverter powered by a +/-15V 3A regulated power supply.
Load is a 240V fan, causing an inverter input current of 1.4A.

what does the input current with no load?

Try the same experiment using a solar panel instead of the regulated power supply unit (PSU) that would be a better test as you would be using a constant current source instead of a well regulated PSU (constant voltage).

What you can do is put a 3.3 ohm (or larger) resistor in series with the PSU output and see what happens

What would be nice to have is a variable (say 0,1A to 10A) constant current power supply to simulate a solar panel. Easy enough to design but this is a fairly hefty PSU as it will be delivering 10A at 22V at the upper end and dissipating about 200W at the lower current (huge heat sinks!!) as I think it sounds expensive to build also.

Yeah zener diodes? - getting to make a regulator OK - good thought.

Try a light load on the 240V output - say an 11W fluoro and see how it goes.

Now to help smooth out variations in the solar current. what about a 1 Farad capacitor that will hold up for about 4 - 6 secs. Audio buffs seem to use them in their car stereo sets. But I think that they cost as much as a small battery.


Hi Daryl,


Re:

The high input voltage destroyed the 240V switching Mosfets IRF740 four ofthem @ $5.74 ea trade price, Ohwell

I'd say Oh! bugger.

Re:

It would probably be safer to use a small 12V 7Ah sealed battery to even out the load & they are only a bout $20. Daryl

Yes. Less messing around.


It would be interesting to see how the inverter reacts to changes in the input voltage. Does the output frequency and voltage change much?


Jeremy.

allrighty Jeremy,

will get to it in a few minutes, but first let me find this XXXX in the fridge - summer seems to have made a comeback here....

Will let you know soon.

Peter

Beers eh! Have one for me.

Trivia stuff:

Haven't heard about Zeners for along time But did you know that True Zener diodes only exist at low voltages - Zener effect. Above 4 V the break down mechanism is not due to the Zener effect but is an Avalanche break down. The temperature coefficient of the voltages is negative and positive respectively.

Jeremy,

the beer's empty, and finally got a chance to try this:

Tried powering up the inverter with the 6R8 first, but it would undervolt frequently on power up.

Decided to raise the supply voltage to 31V, consequently had to replace the 25V cap with a 4700uF/35 unit (couldn't find 10,000uF/35 in my tool box).
The increase to 31V helped somewhat: inverter powers up, but would undervolt when switching on the 240V fan.

Decreased the series resistor to 4R7 by wiring a 15R||6R8.

Now the inverter powers up and runs the fan like a charm.

The inverter draws about 0.5A with a tiny load (240VAC timer switch) connected.

I think it's safe to say that an inverter could be powered by a solar panel without buffering by a battery.
Remember that this is one big lump of a 3kW unit. I expect the inrush requirement of a small 180W inverter to be a lot lower.
In my 24V setup, the supply voltage was 31V, which is the equivalent to 15.5V in a 12V system (12V solar panel with a 15.5V zener shunt).

As has been shown, even a 4700uF electrolytic capacitor stores enough energy to supply the high inrush current on inverter power up.

Yeah I remember using those 5V6 zeners when designing opamp circuits. At this voltage the neg/pos temp coefficients cancel out, no pesky drift due to reference instability...

Have again included some photos for better illustration.

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Best regards, Peter

Hi Peter,

Like the photos. You must have a stock of resistors!! Every time I need one I have to go to Dick Smiths (32KM) or JCAR (45 KM). That clamp meter: is it an AC only meter or does it measure DC also ? I was considering buying a clamp meter attachment for my DMM.

Yes inclusion of the series resistor does make the circuit behave differently. and that 0.5A standing current is very good isn't it? Under full load the efficiency will be close to 98% and that is really excellent.

Well this experiment shows that this simple system will work but it will certainly need more work for a "mature system". This was a good experiment.

Jeremy,

you're lucky, in our neck of the woods DSE stores don't do 'components' any more....

Does my clamp meter measure DC??
Jeremy, it must be Sat nite hey...of course it does if I use it for measuring the input current of an inverter....

The 0.5A current is only the supply current for the internal electronics of the inverter. On top of this come the switching losses which creep up with higher currents as the drivers for the power mosfets need considerable pumping power to spill out the gate source charge at every switching cycle. So under full load this inverter will just have the same conversion losses as any other similar design, around 10%.

cheers, Peter

Re: Does my clamp meter measure DC??

Jeremy, it must be Sat nite hey...of course it does if I use it for measuring the input current of an inverter....

Need to look closely before engaging the mouth ... silly me!!

Thanks for the explanation Peter - still 90% efficient.

Jeremy.

Re: I would look at the battery option myself it seems more practical,

Yes, of course - no issue!!

We were just investigating Peters' idea of pressing a capacitor into service and seeing how far the idea could be taken.

The main problem I see with putting 240V inverter on a Kayak is one of safety - you know water and all that.

Jeremy

Arhhh. That was great. Digging away in the old *bits box now* been a while but I like this - cheers

Hi Jeremy, For me Adelaide & the stores mentioned are 3Hrs, Agood source for 5W & 10W low value resistors is old TVs & photo copiers & monitors Have wecked heaps for components also old computer power suplies good source for high current shotkey diodes (very low voltage drop through them). For me to buy electronic components from here its cheaper & quick to order from Wescomponents in Sydney parts are here in 24Hrs feight $8. Daryl

Hi Daryl,

I only construct real simple things these days - simple controllers, power supplies etc. the most complicated thing I'm doing these days is wiring the brake controller in the car!! Just got a few power fets for a project and am waiting to get the high side drivers for these from Farnell.

Jeremy