Small DC/DC Converters

I'm surprised that the input voltage would increase with increasing output current. I would expect the opposite.

Here is a circuit you might want to try. It is designed to scavange energy even at low voltages:

Joule Thief

Example of a joule thief circuit driving an LED. The coil consists of a standard ferritetoroid core with two windings of 20 turns each using 0.15 mm (0.006 inch) diameter wire (38 swg) (34-35 AWG). The circuit can utilize an input voltage down to about 0.35 Vand can run for weeks using a 1.5 VLR6/AA. The battery voltage is usually 1.5 V. The resistor is ~1 kΩ, 1/4 W. The transistorcould be a 2N3904, BC547B, 2SC2500, BC337, 2N2222, 2N4401 or other NPN. V_ceo= 30 V, P= 0.625 W. A white light-emitting diode with V_f= 3.2 V might be used.^[8]

https://en.wikipedia.org/wiki/Joule_thief

I think you may have misunderstood my comment:

"I measured the V_in at which the I_out & V_out starts to collapse and found it varies according to I_out."

The Vin is deceasing (of course) during the discharge. I'm measuring it at the point that the 'boost' no longer functions.

Yes, have looked at Joule Thief - seems appropriate for very low powers.

OK, that makes more sense.

So, the question is: Why?

Whenever you have a voltage regulator or voltage boost circuit, there's a voltage overhead (like a tax) that the circuit takes off the top and the rest you have to pass on to the load. For example, a 5-volt regulator requires about 7-8 volts on the input. In your case, it looks like a bit over 3 volts "overhead". The overhead increases as you draw more current, IR loss.

The capacitance of 2 x 50F capacitors in series is 25F. The energy you can store in a capacitor is E = 1/2 C V². At 6 volts, this comes out to be E = .5 x 25 x 36 = 450 Joules. At 3 volts (where it cuts out) the energy left is E = .5 x 25 x 9 = 112.5 Joules. So you have 75% of your energy available.

The only way around this inefficiency, other than to use a circuit with a lower overhead, is to operate at a higher voltage. For example, 4 x 25 F capacitors would store the same amount of energy (450 J) as 2 x 50 F, but in series, charged to 12 v the unavailable energy would be 56.25 and 87.5% of the energy would be available.

Yes, thanks. I've gone through all the permutations & combinations of series & parallel supercaps and how much I can recover at a fixed output voltage. My intended V_out is 7.4v. to match the output of 2-cell LiPo battery. My next experiment is to be 3 X 50F/3v supercaps in series, charged to 9v. and using a buck/boost converter to get 7.4v out. Unfortunately all the ones I see, and the one I have, have a lower cut-off point at 5v. (worse than the boost converters), so I can only convert from 9v to 5v., which represents 82.6% of the energy - slightly better than the 75%. So, as you say, 4 X 25F would be even better, but still subject to the 5v cut-off.

Note: When the cut-off point has been passed, the output still exists and merely follows V_in.

You need to look at the datasheet for the LTC1871 along with the circuit diagram for the Converter Module #61105

You may need to reverse engineer the relevant part of the circuit diagram if it's not supplied with the board.

You're not using these devices at the most efficient levels...Your using a bus with 5 people on it and complaining about the gas mileage...

Good analogy.

Explain.

It's mostly in the magnetics but switch utilization plays a part as well. See the chart below for a comparison on one company's buck converter against their competition.

It doesn't matter if it is buck, boost, or buck-boost or whatever. They all look similar. Some may also roll off at the high end too.

https://www.maximintegrated.com/en/app-notes/index.mvp/id/4266

https://www.electronicdesign.com/power/fundamentals-buck-converter-efficiency

https://www.analog.com/media/en/technical-documentation/application-notes/an46.pdf

http://www.engr.colostate.edu/ECE562/lecture_supplements/chapter6_switch_utilization.pdf

and so on . . . .

At extremely light load the efficiency is low...the more current you draw the efficiency goes up then levels off and usually declines slowly...Look at the performance curve for your boost component....

...maybe something like this...

Thanks - my intended output is about 30W - 7.4v and 4 amps. The 7.4v is to simulate the output of a 2-cell LiPo battery, which is commonly used to power small model motors, so we are in the 'efficient' area according to you graph. The next voltage of interest would be 3-cell LiPo at 11.1v.

As regards the performance curves - these devices are made in China, have little information and sell for about the price of a cheeseburger on eBay & Amazon. On the other hand they are very well made and I'm staggered that anyone makes a profit on them. Perhaps that's part of the problem.

I wouldn't buy a cheeseburger on eBay

Yes: the units are made in china, but the heart of it is a good old Linear Tech. LTC1871, and the relevant part of the circuitry should be easy to reverse engineer.

You could even then use LTspice to eveluate its performance.

You might be able to, but I am but a humble chemist.

Your V_{in is} 5.5v and V_{out is at} 6v so you are only stepping up half a volt? I don't know that particular device but in my experience with DC to DC converters with that small of a difference they either didn't work or are very inefficient and unpredictable at that small difference. For all the cheap ebay devices I tested when V_in=V_out the converter didn't work at all. You might want to try stepping up to a higher voltage for your testing.

For my use I needed a stable 5v out while the input would vary from 4v to 12v and while I could find buck boost converters capable of this they were more expensive. I ended up using a step up converter to go to about 20v (could have been less) and then from the 20v a step down to go to 5v. A round about way to me indeed but it worked fine and given the cost of the converters was the least expensive solution.