Wayneh: The battery and solar panel are 'hooked up together' and the potential diode in the IC is quite far away. In trying to understand stuff I don't understand how a diode in the IC would prevent the battery from back feeding into the solar panel and discharging the battery when it is dark?

Did you try looking at the link I posted? There is a schematic there showing you where the diode goes.

 

Did you try looking at the link I posted? There is a schematic there showing you where the diode goes.

ErnieM: Got the QX5252 schematic. There was just a bit of confusion with regards to Wayneh comments. Wayneh was talking about the QX5252 schematic when I thought that he was referring to my schematic. My wiring is different from the QX5252 and my chip diode, if I have one in my chip, is not used to prevent the battery from discharging back thru the solar panel. I have a separate diode doing that.

 

1) What does CDS stand for?

2) Any idea on what the resistance across the light sensor would be?

3) In bright sun light, 90 degree to sun, the solar panel puts out 100 ma and the voltage of battery goes from 1.3 v to 1.7 volts. Is the solar panel putting out in excess of 2 volts?

4) Since I had the panel 90 degrees to the sun when in actuality the panel would be may be 30 degrees to the sun (panel facing straight up) would the output drop by say half in actuality?

5) Using the hz function on multimeter: When the light is on it reads across the inductor 127 khz. Does this seem right? Can the voltage booster factor be estimated from this?

6) The LED still can turn on when the battery is at 0.9 volts (no load, 0.8 volts when light is on). A white LED has a range of 3.2-4.3V. Since the LED can come on at 0.9v is it reasonable that the inductor is boosting the voltage by a factor of 3.5? At full battery of 1.3 volts a 3.5 factor would bring the LED voltage to 4.5?

7) When the battery is full the draw from the battery is 50ma. With a boosting factor of 3.5 would the average current be 50ma/3.5 =14ma ? Because of the frequency/oscillating would the LED be consuming double the 14ma (28ma)?

 

1) What does CDS stand for?

Cadmium Sulfide (CdS). It's what photocells are made of.

2) Any idea on what the resistance across the light sensor would be?

Nope. Measure it in full sunlight and in the dark.

3) In bright sun light, 90 degree to sun, the solar panel puts out 100 ma and the voltage of battery goes from 1.3 v to 1.7 volts. Is the solar panel putting out in excess of 2 volts?

Probably. What's the voltage before the diode?

4) Since I had the panel 90 degrees to the sun when in actuality the panel would be may be 30 degrees to the sun (panel facing straight up) would the output drop by say half in actuality?

Yep it drops, not quite by half

5) Using the hz function on multimeter: When the light is on it reads across the inductor 127 khz. Does this seem right? Can the voltage booster factor be estimated from this?

AFAIK these things opperate in that range.
Can't estimate from that, but it is boosting the voltage to exactly what the LED needs because the LED is what controls the voltage.

6) The LED still can turn on when the battery is at 0.9 volts (no load, 0.8 volts when light is on). A white LED has a range of 3.2-4.3V. Since the LED can come on at 0.9v is it reasonable that the inductor is boosting the voltage by a factor of 3.5? At full battery of 1.3 volts a 3.5 factor would bring the LED voltage to 4.5?

The min voltage probably just depends on having enough to run the circuit. See #5 for the output voltage.

7) When the battery is full the draw from the battery is 50ma. With a boosting factor of 3.5 would the average current be 50ma/3.5 =14ma ? Because of the frequency/oscillating would the LED be consuming double the 14ma (28ma)?

Power out equals power in (minus the efficiency), but make them equal for now. So voltage in (at the inductor) times current in (thru the inductor) (input power) is the LED voltage times LED current.

As this is a set up converter the voltage in is less than the voltage out, while the current in is less than current out.

 

Cadmium Sulfide (CdS). It's what photocells are made of.
Nope. Measure it in full sunlight and in the dark.

Probably. What's the voltage before the diode?

Yep it drops, not quite by half

AFAIK these things opperate in that range.
Can't estimate from that, but it is boosting the voltage to exactly what the LED needs because the LED is what controls the voltage.

The min voltage probably just depends on having enough to run the circuit. See #5 for the output voltage.

Power out equals power in (minus the efficiency), but make them equal for now. So voltage in (at the inductor) times current in (thru the inductor) (input power) is the LED voltage times LED current.

As this is a set up converter the voltage in is less than the voltage out, while the current in is less than current out.

ErnieM:
1) What is AFAIK?
2) Do you mean the current in is MORE than the current out?

 

As far as I know.
Yes, output current drops. You can't get more power out than in, so current must drop (at least) in inverse proportion to the voltage increase.

 

Cadmium Sulfide (CdS). It's what photocells are made of.
Nope. Measure it in full sunlight and in the dark.

Probably. What's the voltage before the diode?

Yep it drops, not quite by half

AFAIK these things opperate in that range.
Can't estimate from that, but it is boosting the voltage to exactly what the LED needs because the LED is what controls the voltage.

The min voltage probably just depends on having enough to run the circuit. See #5 for the output voltage.

Power out equals power in (minus the efficiency), but make them equal for now. So voltage in (at the inductor) times current in (thru the inductor) (input power) is the LED voltage times LED current.

As this is a set up converter the voltage in is less than the voltage out, while the current in is less than current out.

ErnieM: Can you try to explain to me in laymans' terms how the LED controls the voltage. Is it thru the IC or?

ErnieM said:"Can't estimate from that, but it is boosting the voltage to exactly what the LED needs because the LED is what controls the voltage."

 

This thread is continued on a new thread: "Solar Light: supplementing charging to battery" http://forum.allaboutcircuits.com/posts/813643/