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Hi,

I would just like to seek your assistance and idea for the observation below on my experiment:

I have a gadget that is being charge by a 2w 9v solar panel, the problem is I haven't place a blocking diode and I observed that when I tested the charging slot there is about 0.16v output. To clear it that was supposed to be my input voltage.

1. Question is why does it produce an output voltage even though it is imy input voltage?

2. Would the 0.16volts output have a significant effect on my system during night time and or bad weather? Please give facts and computation if possible.

3. If I want to step down my input voltage of 9volts to 4.5 volts can you give me the idea how to do it.

Well, I am asking you this because your insights and helps for me in the past are very helpful for my science project.

Thank you teachers

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The solar panel will give you 9V in bright sunshine. Anything less than that and the voltage will be less.

Are you charging a battery from the solar panel?  If so you need the diode otherwise the battery will discharge through the panel when the panel output is less than the battery voltage. Also consider when the sun goes behind a cloud. Use a schottky diode as they have a lower voltage drop than silicon.  If there's no battery than a good sized capacitor would keep the circuit running for a brief period of less light.

Is the 0.16 volts from the 'gadget' or the panel?  If it's the panel, it will just be what the cells produce in very low light conditions.  Id it's from the circuit, did you only measure it once?   It is possibly from a small charge left in the circuit's capacitors.

A voltage like that will not interfere with the circuit, but depending on what it's connected to, a supply voltage which drops off slowly can have an effect on how the circuit performs.  You may get unexpected behaviour as the voltage drops.

To regulate the voltage you need a low dropout voltage regulator.  These are much better than the normal 7805 type in this application as they can work with an input voltage much closer to the output voltage.

Well Sir, actually I just have the theory when I kept my solar panel connected to my battery charger. After a few weeks without disconnecting the panel to my system I noticed that my battery is losing charge on it. Then I measured my system's charging or input port. It gives me a measurement of .16v and a resistance of 50.6ohms so using ohm's law (I=V/R) I found out that I am losing about .0495Amps during the night.

I am wondering though what if my voltage input from my solar panel say about only 5v during cloudy day and my battery has 6v would it there be a reverse voltage charging and my solar panel would sip in an approximately with my resistance at 50.6 ohms using ohm's law again am i losing an approx .098 Amps per hour?

Thanks again mentor.

If you put in a series diode, this problem will go away.
Your NiMH batteries will normally be about 1.2V each, rising to about 1.35V when fully charged.  That will give you 3.6V to 4.1V.

When the output of the panel is below the battery voltage, no current will flow as the diode will stop the battery discharging through the panel.

When the voltage tries to go above the battery voltage (with an extra 0.3V for the diode), the battery will clamp this to its terminal voltage, so current will flow into and charge the battery.  The brightness of the sun determines how much current flows, but the terminal voltage will not change, except slightly when fully charged.

The best way of checking the charge current is to put your multimeter in the supply to the battery and measure directly.  You can then see if the battery is charging (+ve current) or discharging (-ve current) through the radio in different levels of sunshine.  Bear in mind your solar panel specification is based on brightest lighting conditions, but will quite often be less.  (Although I'm guessing you're in India where there is no shortage of sun #;¬)

If the radio normally runs off 3 x non-rechargeable cells you may want to consider putting another NiMH in series as a non-rechargeable outputs 1.5V as opposed to the NiMH cell 1.2V.  This would give you 4.8V to 5.1V which is closer to what it normally uses.  The extra 1/2 volt won't do any damage.

I've just answered your other question and now I see you're charging batteries, you don't need a regulator.  They will regulate the voltage to your radio.

oh i almost forget. the easiest way would be using a voltage divider since you are using 9v

use a blocking diode.