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I have solar cell with peak volt=9V and I=300mA. I want to charge 6V acid lead battery 4.5Ah. I'm thinking to put L7809 voltage regulator to prevent spike in voltage of solar cell and at night time there will low voltage at solar panel.

1.Does voltage regulator prevent voltage back flow from battery to solar panel?

2.Do i use voltage regulator or not?(this will reduce the life of battery if not use? right)

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  • \$\begingroup\$ Zener diodes have nothing to do with your question. You need a diode to prevent back flow, but not a zener diode. You want a diode with low forward voltage. \$\endgroup\$ – user207421 Dec 8 '15 at 19:42
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You must advise battery type and specification.
eg lead acid or NiCd or MimH or LiIon or ...?
How many mAh capacity?

I have solar cell with peak volt=9V and I=300mA. I want to charge 6V battery. I'm thinking to put L7809 voltage regulator to prevent spike in voltage of solar cell and at night time there will low voltage at solar panel.

You will not harm the panel by charging a lower voltage panel. The battery will "clamp" the panel output to the battery voltage and supply whatever current it cam.

You should add a diode between panel and battery to prevent "backflow" when panel voltage is lower than battery voltage. You do NOT need a regulator for basic charging except if the battery is so small that it will not toilerate a 300 mA charge rate - which is unlikely. You MAY need a regulator to stop the battery from overcharging.

The panel will supply about 300 mA into the battery in full midday sun when pointed at the sun and less or much less when cloudy, not midday or not pointed directly at the sun. If the battery has capacity below about 2000 mAh it MAY be fully charged in less than one day. If 2000 mAh it will not fully charge in one day from "flat".

Where are you located? What use will be made of the battery (daily discharged fully or ...?) These factors affect the answer.

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  • \$\begingroup\$ yes sir , it is lead acid 6v 4.5Ah. I don't have zener diode now. I was thinking if voltage regulator have inbuilt P,N junction which will prevent the back flow? \$\endgroup\$ – editinit Dec 8 '15 at 18:33
  • \$\begingroup\$ As above - you do not need a zener diode -= just an ordinary diode between panel and battery. Silicon will do (ef any of 1N4001 ... 1N4007) or a 1A Schottky diode if desired (iN5817 etc) . | A 4.5 Ah battery with 300 mA charge will take 4500/300 = 15 hours of FULL SUN to fully charge. If you are in say Mumbai you get 6.8 full sun equivalent hours in April and about 3.6 in August on average from 4th table here. So full charge would take 15hrs/6.8 = 2.2 days in mid Summer and 15/3.6 = 4.2 days in mid Winter on average. BUT ... \$\endgroup\$ – Russell McMahon Dec 9 '15 at 8:16
  • \$\begingroup\$ ... a 6V lead acid battery needs up to about 6.9V for boost charging ("topping charge") occasionally. Depending on your application, discharge per day etc you may not need a complex regulator. Can you give more details. Where are you located? \$\endgroup\$ – Russell McMahon Dec 9 '15 at 8:18
  • \$\begingroup\$ yes sir, location is mumbai. now i'm thinking to buy 4v 1.5Ah acid lead battery and use with VR L7805 diode 1N4001. \$\endgroup\$ – editinit Dec 10 '15 at 7:46
  • \$\begingroup\$ You did not read my answer properly. You probably do not need a voltage rgulator and if you do use one the requirements are more complex than you realise. \$\endgroup\$ – Russell McMahon Dec 12 '15 at 4:50
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To charge the battery you need the current, not voltage (if you are not about lead-acid ones, see comment by @Andy-Aka below). Therefore you should be mainly interested in using current stabilizer.

Voltage regulator could be used as current stabilizer anyway (look in the PDF, you simply need to connect proper resistor as a fake load and insert the chip in series with the battery). Meanwhile you do not need to use 9V regulator for this. The regulator will consume all unwanted energy (as it usually does in voltage-controlling schematic also).

To prevent the back current in this solution you will simply add a diode (perhaps some Schottky's) in series.

If you will use L7809 as a voltage regulator, you could not prevent some back current, I'm afraid. Look at the internal schematics in PDF for details. You will lose the current through internal output resistors etc.

P.S. it will be hard to charge 6V battery if the cell gives less than 6V + regulator_dropout. Perhaps you should disassemble the battery into 3V or 1.5V cells if possible.

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    \$\begingroup\$ This is bad information if the battery is a lead-acid type. LA batteries largely use voltage regulation; other chemistries use current. The question doesn't say so it's probably ill-advised to assume. \$\endgroup\$ – Andy aka Dec 8 '15 at 10:31
  • \$\begingroup\$ Thanks for info. However I suspect lead-acid batteries are not extremely popular today. Let us ask the author anyway! \$\endgroup\$ – Alumashka Dec 8 '15 at 10:38

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