Capacitor charging above solar cell voltage?

  • #1

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This has been confusing me a lot lately. I built a type of battery derived from a solar cell that outputs a constant voltage of 0.5 Volts.

My original plan was to work out a method to boost this voltage to around 1V to power a joule thief and light an LED using a capacitor charged by the cell. However upon testing the cell with some capacitors I found that it would charge a variety of different capacitors all the way up to 1.6V if left connected for long enough (It would greatly slow after about 1.4V).

Now this doesn't make much sense to me as it isn't possible to charge capacitors to a higher voltage than the input. I took all measurements with the same multimeter and have verified that the charged capacitors are indeed 1.6V (they will briefly light up the LED in the joule thief that only works down to 0.8V).
 

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  • #2
CWatters
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What is the open circuit voltage of the solar panel?
 
  • #3
What is the open circuit voltage of the solar panel?
Connecting the cell to the multimeter with no other load gives 0.46V (I rounded up in the post).
 
  • #4
Charles Link
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Your multimeter could have a low internal impedance. (thereby reading too low for the open circuit voltage). Your solar cell could have a higher internal impedance than the capacitor. Capacitor voltage reading might be close to correct but solar cell reading inaccurate. Presumably you measured the solar cell during bright sunlight.
 
  • #5
Your multimeter could have a low internal impedance. (thereby reading too low for the open circuit voltage). Your solar cell could have a higher internal impedance than the capacitor. Capacitor voltage reading might be close to correct but solar cell reading inaccurate. Presumably you measured the solar cell during bright sunlight.
I know that the capacitor reading is relatively correct because it is able to power the joule thief LED circuit that only runs from approximately 0.8V and higher. Would a different multimeter perhaps provide more accurate measurements of the cell or should I just use capacitor readings as the accurate measurement?
 
  • #6
Charles Link
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I know that the capacitor reading is relatively correct because it is able to power the joule thief LED circuit that only runs from approximately 0.8V and higher. Would a different multimeter perhaps provide more accurate measurements of the cell or should I just use capacitor readings as the accurate measurement?
I seem to recall having previously encountered a similar problem reading the DC circuit voltages with a cheaper multimeter. It was reading too low because of a low internal resistance. I think I used a buffer circuit in conjunction with the multimeter to read the voltage more accurately.
 
  • #7
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Why are you using a capacitor to power the LED?
 
  • #8
jim hardy
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Connecting the cell to the multimeter with no other load gives 0.46V (I rounded up in the post).


What is the nature of the light shining on your solar cell? Sunlight or indoor lamps at your workbench ?
 
  • #9
PhysicoRaj
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Why are you using a capacitor to power the LED?
Perhaps a nice idea, caps provide much higher current than the source that charged them.
 
  • #10
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Very true, but half the energy extracted from the solar cell is stored in the electric field of the capacitor. The other half is dissipated.
 
  • #11
What is the nature of the light shining on your solar cell? Sunlight or indoor lamps at your workbench ?
The light source is very very dim - I'm using a tritium phosphor vial as the source with a tube of reflective foil around it to redirect some of light back into the cell. This is also why I have to use the capacitor, because without storing the energy over time and releasing it in a burst the current produced directly from the panel is far too low to light and LED, or really do anything.
 
  • #12
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A betavoltaic device (uses a semiconductor to convert beta decay electrons to electric current) might be more efficient.
 
  • #13
jim hardy
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The light source is very very dim - I'm using a tritium phosphor vial as the source with a tube of reflective foil around it to redirect some of light back into the cell.
Then you can't knock loos a whole lot of electrons, can you ?

Try measuring the current your photocell makes by switching your DMM to lowest current range.

I wager you get about 0.5 volts/10 megohm = 50 nanoamps.
You might need an electrometer to measure such feeble current
and you didnt say how big is your capacitor

Here's what i think is the answer to your original question ----
......
When you measure voltage across your capacitor with such a feeble light source
50 nanoamps will charge 1 microfarad at 50 milllivolts per second,initially
as capacitor voltage increases, the DMM steals current via its 10 megohm input impedance
and at 1/2 volt it's stealing it all so charging ceases.

That's why you only measure 1/2 volt
and when you disconnect the meter charging re-commences

What do you think ? I could be all wet, frequently am.
So, Give it a test.

old jim
 
  • #14
Then you can't knock loos a whole lot of electrons, can you ?

Try measuring the current your photocell makes by switching your DMM to lowest current range.

I wager you get about 0.5 volts/10 megohm = 50 nanoamps.
You might need an electrometer to measure such feeble current
and you didnt say how big is your capacitor

Here's what i think is the answer to your original question ----
......
When you measure voltage across your capacitor with such a feeble light source
50 nanoamps will charge 1 microfarad at 50 milllivolts per second,initially
as capacitor voltage increases, the DMM steals current via its 10 megohm input impedance
and at 1/2 volt it's stealing it all so charging ceases.

That's why you only measure 1/2 volt
and when you disconnect the meter charging re-commences

What do you think ? I could be all wet, frequently am.
So, Give it a test.

old jim
After making the original post I was able to get a much higher quality multimeter and test the device. I measured just about 50-60 nanoamps of current. What was interesting was that the original multimeter measured 0.46V, while the new one measures 0.51V, which seems to back up what you suggested since the new multimeter likely has a different internal impedance.

As for the capacitor types: 1uF tantalum cap is charging to over 1.2V but gets discharged before quickly when I connect the multimeter so I assume this is going to 1.6V because the 100uF aluminum cap is charging to 1.6V (although it takes a long time to get to that, 1.4V is much faster). I have also had the same results with a 470uF one.
 
  • #15
A betavoltaic device (uses a semiconductor to convert beta decay electrons to electric current) might be more efficient.
This would definitely be more efficient but Tritium is gaseous so I wouldn't be able to put it onto a semiconductor without good equipment, and solid beta emitters are really hard to find in large enough quantities.
 
  • #16
jim hardy
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What was interesting was that the original multimeter measured 0.46V, while the new one measures 0.51V, which seems to back up what you suggested since the new multimeter likely has a different internal impedance.
Shine a flashlight on you solar cell and see if it makes more current ?
 

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