help for solar charger

[sarathssca]

Hello,

Could you please help in the construction of a solar charger. My solar panel is rated as 6V,3W and i am looking to charge a 6V,5AH lead acid battery. If i am directly connecting the solar panel to the battery its working (voltage of about 6V to 6.5V and current of about 175mA) . But if i am connecting a 6.8V zener diode in parallel to the battery for over charge protection , its not working.

I think when using zener diode in parallel to battery, all the power from the solar panel is is drained via zener diode and battery is not getting any power.
Is my understanding correct ?
Could you please help how to charge the battery with over charge protection or detection?

thanks
Sarath

 

[CafeLogic]

Zener voltage varies some with current. You probably have it backwards though. The cathode (side with the line) should be connected to positive on your charging circuit. Normal charge voltage for a 3-cell lead acid would be around 6.9V-7.3V so you probably don't need it anyway.

 

[colin55]

You need a diode in series with the positive lead of the solar panel and battery and you need a few more cells to increase the charging current to 300-400 mA.
You need a simple circuit to "jack up" the output voltage of the solar panel to charge the battery.

 

[sarathssca]

Hi,

Thanks for the reply.
I have connected the zener diode properly with cathode to positive. But the solar panel (6V,3W rated) was not working when zener diode(6.8V) was connected.

1. When not using zener diode the solar panel was supplying 6-6.5V and 175-200mA. So for charging 6V 5AH lead acid battery does I need to directly connect solar panel to battery via a diode, without any other regulation or over charge cutoff protection ?

2. While testing the solar charger I noticed that the output was not coming when ever the sun light was interrupted (clouds or tree shade). For this can I use a capacitor in parallel to the solar panel? So that the capacitors are charged and when ever some small interruption to sunlight occurs capacitors will provide a smooth output to the battery.

thanks
S Sarath

 

[CafeLogic]

6-6.5V at 175-200ma for a 6V 5AH lead acid would barely qualify as a preservation charge. You could probably supply it for the life of battery. Note that the battery will never reach full charge. The diode that colin suggested is a good idea to prevent back flow of current but unless you boost the output voltage, you probably can't afford the voltage drop, make sure to use a Schottsky if you do that.

You would need a capacitor bigger than your head for that to almost work. No worries though, you don't need a smooth output the battery, the battery can be thought of as a really high value chemical capacitor. To give you in example, if you buy a black and decker cordless drill, you get a cheap pile of crap B&D NiCD charger with it. All it is a a transformer and a full-wave diode bridge, the output of which leads straight to the battery. If I remember correctly, they use a 17V transformer for the 14.4V Drill. Hence the charge voltage continuously varies from 0V to 25V. I wouldn't recommend that, and I was shocked when I saw B&D doing it, but it can work. The least of your worries, is the over-voltage, solar panels are naturally limited and 7V is not going to hurt anything.

If you get some more charging current, and higher voltage, you will need over charge protection.

 

[ronv]

It sounds like your panel is not well matched to your battery. Something like this is what you are talking about.
**broken link removed**

 

[colin55]

You haven't done what I suggested. You are wasting your time unless you carry out the instructions.

 

[ronv]

Solar

Attached is a picture of how to hook it up. You need the blocking diode to prevent the battery from discharging thru the solar panel when the battery voltage is higher than the solar panel voltage (solar panel in the dark). You may have a problem when you do this because it looks like your panel doesn't put out enough voltage for a 6 volt battery. But try it and lets see how much current you get. After that lets see if you need the zener to protect the battery from overcharge.

 

[sarathssca]

Hello...
thanks for ur replies.
I will try them this weekend and get back ...
thanks
S Sarath

 

[sarathssca]

Hello,

Today I tried the circuit to charge my 6V 5AH lead acid battery with my 6V 3W solar panel. Today it was bright and bit cloudy, but not fully sunny, The following are my observations:
1. Open circuit voltage on my solar panel = 9.7V.
2. I Connected the battery directly to the panel via a diode. After connection solar panel output voltage was 6.5V, battery voltage was 6.18V and a current of 150-190mA was flowing into the battery. Does it means that the battery was charging and the circuit was working fine?
3. To the above circuit I connected a 6.8V zener diode across the battery (over charge protection) but it didn't made any difference. Is it because the input voltage from the panel was around 6.5V and zener diode was of 6.8V?

Please let me know if this circuit is working?
thanks
S Sarath

 

[colin55]

The circuit is working. You do not need the zener. A 5AH cell can take 50mA to 100mA all day long without overcharging. You are only giving it 190mA for a few hours a day and this will not produce overcharging.
Your battery will take over 28 hours to charge at 190mA so that if the sun shines 6 hours per day, it will take nearly 5 days to charge.

 

[Nigel Goodwin]

No need for anything else, the panel is too small to overcharge the battery - the simple series diode is all that's required.

 

[sarathssca]

Hello,

Thanks for the suggestions.

Could you please advise on using the attached circuit for undervoltage indication for my charged 6V,5AH lead acid battery.
Shall i set the low voltage cutoff be about 4.5V for 6V,5AH battery?
thanks
S Sarath

 

[colin55]

The zener is not needed

 

[CafeLogic]

Could you please advise on using the attached circuit for undervoltage indication for my charged 6V,5AH lead acid battery.
Shall i set the low voltage cutoff be about 4.5V for 6V,5AH battery?

It's OK, so long as you understand it works backwards. Meaning the LED will be on when the battery voltage is above whatever set point is. It is also customary to avoid wasting battery life by using a blink, instead of constant, maybe 1 second on 5 seconds off. You could accomplish both of those things (reversing the indicator and blinking) with a comparator and a 555 timer.

As for your chosen set point of 4.5V, that is not really low, it more like deader than a bag of rocks. Somewhere between 5.2V and 5.5V would probably be more appropriate. You should really do some research on lead acid batteries before completing the charger circuit.

 

[colin55]

This is the type of circuit you need:

**broken link removed**
It can be simplified.

 

[sarathssca]

Hello,

Few days ago we had a bright sunlight and I tested the charger. The following were the observations:
1. When connected 6V,3W solar panel directly to the 6V battery via only a diode, the voltage from solar panel was 6.7V and was giving a current of 300mA.
2. When, to the above circuit on connecting a 2ohm resistor in series with the blocking diode and a 6.8V zener diode (for over charge protection) in parallel with the battery, the solar panel was giving a voltage of 7.3V and current of 260mA. And some current of about 80-100mA was also getting wasted through the zener diode.

Does a zener doide required for the battery for over charge protection given the voltage and current from solar panel in bright sunlight? I feel using a zener diode directly in parallel is wasting considerable amount of current. Should I need over charge protection, can I use a transistor shunt parallel to battery with the zener diode driving the base of the transistor, so that current wasted through zener be minimised?
Please advise
Thanks
S Sarath

 

[colin55]

For a 5AH battery you can charge it each day without any need for over-charge protection. If you draw from the battery each day the battery will not be overcharged.

 

[yandoit.tech]

In order for a zener diode (reversed) to work, the input voltage must be higher than the rated voltage of the diode. I use zener diodes to convert high volts low amps to low volts high amps. So if you hook up a 12v zener to a 17.2 volt 20 watt panel, the panel which puts out 17.2 volt and 1.16 amps (V*A=W) is going to convert down to 12v in the zener, and the joules of voltage convert to joules of amperage, so you then get 12v 1.6 amps. Zeners only work if the voltage going into them is higher than the diodes rated voltage. It should also be noted that the avalanche breakdown (the phenomenon utilized by zeners) cannot occur if your input voltage is lower than the zener's volt rating. If the voltage is lower than the zener's rating, the zener acts as a regular blocking diode (if placed in reverse, in series). You will want to include a regular schottky diode coming off the panel, as a zener placed in reverse will leak joules back into the panel overnight.


It is called a dc-dc step-up converter. It will convert your 6 volts off the panel to another voltage you specify. Just remember v*a=w and a=w/v. When you convert from one voltage to another, your amperage changes (It should also be noted, that most voltage converters are not electrically efficient, and will dissapate some of your watts in the form of heat, so you will loose amperage from that.

Also, when charging 12 volt deep cycle batteries, a 14.2 volt charge is applied. SO for a 6 volt battery, you would apply 7.1 volts. i.e., you need to find a way to increase the system voltage to at least 7.1 volts to charge your 6 volt battery.
Batteries are not entirely efficient when they charge. They actually dissapate roughly 1/3 of the amps going into them as heat. Therefore, you will need to divide your amps coming out of your panel by 1.3 and then use that value to find how many hours it takes to charge your batteries. I like to call this value T, even though I'm pretty sure that is not what it is called, i think T sounds nice. So your 0.5 amps is going to look more like 0.38 amps to the battery, and you are going to need to adjust charging times accordingly.
First, you need to know how many hours of rated output you are going to be getting from your panels. You can find this out by hooking up the panel directly to a multimeter in the early morning and wait till the volts and amps read similar to what your panel is rated (Take note of the time of day). Leave it hooked up to the meter, and watch it thoughout the day. There will be a point where your volts/amps will start to drop (Usually around 4 o clock, 5 o clock) Take note of the time of day. Then figure out how many hours of full sunlight you are getting that your panel is able to produce its rated output. (I usually estimate this as six hours)
Take the T value for your amps, and multiply it by your hours of full sunlight. This will give you Amp/hours OR Ah (E.G. 0.38a*6h=2.28Ah). This is the amount of energy available for you to charge the battery in a given day.

So your small panel (Ignoring the incorrect charging voltage) would take 2.1 days so charge your 5Ah batterry, given you were to provide it with .38 amps for 6 hours a day. That is totally fine, lead acid can be charged over a period of time. It is important however, that you _do not_ use a lead acid battery below 50% of its charge. If you run a lead acid out below this value, sulfation of the electrodes within the batteries will occur, and this ruins lead acid cells. So make sure when you get to using the thing, you don't run it down more than 50%. I wouldn't run it down more that 60% to stay on the safe side.

A cheap charge controller is also a good idea, however, I think they are mostly made for 12v pv systems. The sunforce 7A solar charge controller does a nice job, fteches a price for about $35 dollars on ebay, and I think is even a MPPT carger (Multiple point power tracking) It will charge your batteries until the cell (batt.) voltage increases beyond a certain value, and then shut off the charger to prevent over charging.

You can also run a load off the panel while charging the battery, given the panel puts off an excess wattage not being used by the charger, and your load does not exceed this excess wattage.

Hope this was a fun lesson in solar electric basics!
Until next time