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Suparcaps and Math

A project log for 1 Square Inch of Power

An MPPT solar boost converter with 3.3v Output and Supercap energy storage.

OzQubeOzQube 01/18/2016 at 23:110 Comments

Just as predicted last project log, the supercaps arrived!

I soldered one of the 2.7v/5F caps onto the board, then setup the solar panel under the halogen light source ( because night time). I connected the multimeter up to the positive terminal of the supercap, which is also the output of the first LTC3105. The voltage rose to .3v..........but no higher.....

Now I know its a tiny solar panel, so I left it under the light for 5 minutes. I had previously figured out that having the solar cell really close to the light is a good approximation to sunlight.

Or is it?

No increase in voltage.

As I don't have a lab power supply, I can't provide an alternative input that is current limited. I thought I'd connect an AA up directly to the supercap to charge it up to 1.5V. And yes, this worked!

I then put the lamp back onto the solar cell, but no increase in voltage even after 5 minutes.... What's going on?

I checked the solar panel voltage again. It's outputting 1.48V. Exactly what it should be with the MPPT resistor setting. Maybe I have a short? I desoldered the second LTC3105 so I can isolate the first one. Tried again. No increase.

I went back to one of the reference web pages that are similar to this project - http://wiesel.ece.utah.edu/redmine/projects/essc/wiki

I decided to do the math to see how long it should take the solar panel to charge the supercap.

From the website:

Theoretically, solar output is 1.5V @ 44.6mA. So 0.0669W. Using the LTC3105 datasheet, it shows a graph for efficiency and power loss for certain input voltages and output currents. Now I'm just using the basic Ohms law calculations for the output current.

So assuming 100% efficient, output power is .0669W. At 2.5V, the output current is 26.76mA. Efficiency at this current vs the 1.5v input is a smidge over 70%. So output power is .04683W. At 2.5V, that's 18.732mA. (

V = 2.5V C = 6F Ic=.018732

I had to check the math, so (V*C)/Ic works to get their results ( and using the math from another source)

2.5V/.018A/6F = 800 s = 13 minutes

Looks like I'll have to test during the day to make sure the solar panels are actually outputting the required current, as I also found that the halogen lamp easily makes the cell output the right voltage, but output current is questionable.

To be continued....

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