I understand that the energy extraction from an ultra capacitor drops the voltage to the point where the load can't function. But, is it possible to create a converter that can continuously present a consistent voltage to the load using a dynamic dc-to-dc circuit to allow all the energy in the ultra capacitor to be utilized, even if the current drops? What are the issues?
[...] to allow all the energy in the Ultra Capacitor to be utilised [...]
No. The closest that you can get, to that, is by using a synchronous boost converter (if your cap voltage is always lower than your target output voltage), or a synchronous buck/boost converter (if your cap voltage may be above and below the target output voltage).
A synchronous converter uses only controlled switches, instead of controlled switches plus diodes. The benefit is reduced losses, because the main current does not cause a (relatively big) voltage drop across any diode. Only across Rds_on of the controlled switches, which may be in the milliohm range, and will cause much lower voltage drops (and therefore power losses).
As Vin approaches zero, Iin will have to grow substantially, so as to keep Vin·Iin above Pout (which usually you will want to keep constant). Those high input currents will eventually cause that your Vin voltage will just drop across the first devices of your converter. So, below a certain threshold for Vin, you will not be able to deliver any more power to the output. The converter will still drain your cap, but that energy will serve no purpose any more, at the output.
But, is it possible to create a converter that can continuously present a consistent voltage to the load using a dynamic dc to dc circuit to allow all the energy in the Ultra Capacitor to be utilised, even if the current drops?
All the energy? No. But you should be able to extract most of it with the approach you described.
Let's say you have a 5V output and the capacitor is a 0.5F 5V abs max so you start with 4.5V on the capacitor, and can use a boost converter that will operate from 0.8V - 6V, so the capacitor is the constraint for the upper voltage, and the boost converter is the constraint for the lower voltage.
Energy extracted from the capacitor is 1/2 0.5F * (4.5V^2 - 0.8V^2) = 4.9J; energy remaining in the capacitor is 1/2 * 0.5F * 0.8V^2 = 0.16J.
The amount of energy presented to your circuit is 4.9J * the efficiency of the converter; you can get DC/DC converters that operate in the 90%+ range fairly easily these days.
You can "asymptote towards zero" but you can never get there.
In practice getting 95% + of available energy is 'easy enough'.
Energy in cap is proportional to V^2.
Say Vstart = Vs, Vfinish = Vf. So
End energy as a fraction of start energy is (Vf/Vs)^2.
5V -> 1V you have (1/5)^2 = 1/25 = 4% left
5V -> 0.5V = (0.5/5)^2 = 0.25/25 = 1% left
If you really really really MUST get all possible energy out you can use "energy harvesting" circuits that will accept perhaps 10 mV in and boost it to say 5V BUT the efficincy is unlikely to be more than sau 50% and the energy available is very small.
If you start at eg 5V then stopping at 0.5V = 1% is probably very ample in most cases, and 1V / 4% is probably often acceptable.
My company makes such a thing for holding up the DC bus of a variable-frequency drive during an outage. Exactly what you're describing: use a boost converter to present the load with a fixed voltage, even as the cap voltage drops. I understand that this is done in many other lower-power applications, as well.
As others point out, the problem is that using all the energy isn't practical. The first problem is efficiency; there will always be some loss, so you can't get all the energy into the load. But beyond that, the boost ratios involved become impractical beyond a certain low input voltage.
Some of the components in a boost converter are sized based on the input current to the converter, including the choke and the boost switch. If you assume a fixed load current, that means that a boost converter doing a 10:1 boost has to be much beefier than a boost converter doing a 2:1 boost. Doing a 10:1 boost means your input current is 10x your output current!
And with capacitors, the further down you go in voltage, the less energy you're actually getting out. For example, our boost converters are rated for ~2:1 boost, so we typically only drain the caps to half voltage, meaning we've gotten 3/4 of the available energy (1/2 C V^2) out. We could get more energy, say by draining the caps to 1/4 voltage. But we'd have to rate the converter for a 4:1 boost, which would mean all our components would have to be up-sized for higher peak currents. Control of a boost converter at a high boost ratio also becomes problematic. And all we'd get for our additional expense would be another 12.5% of the available energy. Diminishing returns.
Now, that was just for going from 2:1 to 4:1. Imagine doing that at a 10:1 boost. You get even less energy, but you've had to scale your components up by a factor of ten now. And no matter how low you go, you've still got some energy left in the caps, because there comes a point where your boost converter simply won't pull any more because of the non-ideality of the switch.
So can you get all the energy out? No.
Should you even try to get out as much energy as theoretically possible? No.
Is it still useful to use a boost converter to get a fixed voltage output off an ultracap? Absolutely.