High efficiency energy transfer between two caps - how? - Page 1
Forum section: electronics / beginners

author: Faranight
author_rank: Unknown
timestamp: June 01, 2022, 09:33:52 am
content: Greetings,This topic is something I don't fully understand so perhaps someone can point me in the right direction.Suppose you have a circuit like the one posted below. There are two capacitors -C1andC2.What is a good way (the unknown element) to transfer energy from capacitor C1 to capacitor C2 with low loss (this is an energy harvesting application)?The following conditions apply:C1 is continuously being charged by a high impedance (R1) voltage source (5V)C1 has a parallel load (R2) connected to itThe load (R2) requires a minimum voltage of X volts (suppose X = 2.0V) to operate, so the voltage on C1 must never drop below this valueC2 should start to charge only after the voltage on C1 has reached X voltsThe transfer of energy from C1 to C2 must be highly efficient (no zener diodes !!)C2 is a supercap and is much bigger than C1 that is alu electrolyticC2 is allowed to fully discharge, but it must not affect voltage on C1My idea was to use a buck converter as the unknown element. The problem I face is that all buck's I've found use a feedback loop on the output because they regulate the output voltage. In my case I would need to regulate voltage on C1 which is located on the input (would need an input feedback loop). Are there any bucks with this option? Can standard bucks be adapted to work this way i.e by using a resistor divider on the input and an inverting opamp? Is there a better solution to this?Thanks in advance!

author: dmills
author_rank: Super Contributor
timestamp: June 01, 2022, 12:02:00 pm
content: Hysteretic buck converter?Close the switch whenever C1 exceeds say 3V, open it at 2.1V, catch diode and inductor, job done.Control the switching device with a micropower comparator with a little hysteresis.

author: Marco
author_rank: Super Contributor
timestamp: June 01, 2022, 12:11:12 pm
content: Some buck converters have an adjustable UVLO.

author: tszaboo
author_rank: Super Contributor
timestamp: June 01, 2022, 12:17:19 pm
content: Sure. Energy harvesting chip like ADP5090. There are a few variations for this.

author: eugene
author_rank: Frequent Contributor
timestamp: June 01, 2022, 02:27:37 pm
content: If VC2is always less than X (the minimum allowed voltage on C1) then the contents of the yellow box can be a simple switch.1. Let C1 charge from the external energy source. When VC1= Vhigh, close the switch in the yellow box (open the valve!)2. Let charge flow from C1 to C2. When VC1= Vlow, open the switch (close the valve.)3. GOTO 1Both Vhighand Vlowcan be whatever works well, but must both be greater than X.If the range of VC2is too great to allow the above method to work, then the yellow box could contain a boost converter of some sort. But it seems to me that it would be easier and simpler to put the boost converter after C2.

author: dmills
author_rank: Super Contributor
timestamp: June 01, 2022, 02:59:04 pm
content: That manages (at best) 50% efficiency because charge is conserved between the two caps, and you really want energy to be conserved not charge.

author: golden_labels
author_rank: Super Contributor
timestamp: June 01, 2022, 03:01:11 pm
content: Quote from: eugene on June 01, 2022, 02:27:37 pmFaranight asked about efficient transfer. If the yellow box is a switch, it’s a resistive element. The moment you close the switch, you have voltage between the capacitors and corresponding current. Which means a lot of joules go into air as heat. It doesn’t matter how small the resistance is. The amount of energy to be transferred is exactly the same: lower resistance gives higher power for shorter time, higher resistance gives lower power for longer time.That’s the reason why you can’t simply have PWM-controlled on/off switch as a step-down converter. One needs an inductor between the capacitors. Initially it sees high voltage drop, hence low voltage drop on resistive parts. Most of the energy is stored into the inductor, instead of heating the circuit. Then the inductor gives it back.

author: eugene
author_rank: Frequent Contributor
timestamp: June 01, 2022, 04:52:44 pm
content: Quote from: dmills on June 01, 2022, 02:59:04 pmBoth are conserved according to fundamental laws of physics, which is to say that neither are created or destroyed. I'll assume that what you're trying say is that when a direct connection is made between two capacitors charged to different voltages, all of the charge that leaves one appears at the other, but that this is not the case with energy.So the question is, where does the energy go?Quote from: golden_labels on June 01, 2022, 03:01:11 pmOK, but what alternative is there that doesn't violate any fundamental laws of physics? What method is there to get energy from one capacitor to the other without having current flow through any conductors?

author: Marco
author_rank: Super Contributor
timestamp: June 01, 2022, 05:08:15 pm
content: Quote from: eugene on June 01, 2022, 04:52:44 pmResistance, because of I2R even when resistance gets smaller the power loss stays constant. At a theoretical R=0 you get an infinity*0 situation, even superconduction can't save you.

author: DavidAlfa
author_rank: Super Contributor
timestamp: June 01, 2022, 05:32:37 pm
content: A transistor in avalanche mode would do this. Very common for simple blink circuits.

author: eugene
author_rank: Frequent Contributor
timestamp: June 01, 2022, 05:45:35 pm
content: Quote from: Marco on June 01, 2022, 05:08:15 pmRight! But what's the alternative? It's pointless to reject the simple solution on the grounds that it's only 50% efficient unless you can offer an alternative solution that does better. Specifically, a solution that doesn't involve charges moving through conductors resulting in i2r losses.(I admit that I don't understand how DavidAlfa's solution might do that. My belief is that it's impossible, but I respect DavidAlfa enough to give him a chance to explain.)

author: bdunham7
author_rank: Super Contributor
timestamp: June 01, 2022, 06:27:12 pm
content: Quote from: eugene on June 01, 2022, 05:45:35 pm

author: Faranight
author_rank: Unknown
timestamp: June 02, 2022, 05:09:05 am
content: bdunham7:That's a buck converter, an asynchronous one.Quote from: tszaboo on June 01, 2022, 12:17:19 pmADP5090 has maximum input voltage of 3V3. If the C2 charges fully then voltage on C1 will eventually be 5V, so it won't work. I checked this IC before and dismissed it.The closest I found is BQ25570 that also has an integrated buck, but... it's out of stock everywhere!!!

author: woodchips
author_rank: Frequent Contributor
timestamp: June 02, 2022, 09:26:54 am
content: Is this the 'loss of 50% of charge' question?Have two caps, 1F, one discharged, one charged at 10V so energy, Q=1/2CV^2. When connected in parallel you now have one cap of 2F but with 5V across it. Energy was Q=50 Coulombs, now 25 Coulombs.This is one of those obscure corners of electronics and only 1-2% of books even mention it. Even with zero ohm conductors etc you still lose 50% so must be the radiation etc effects from the current flow.What is interesting if you really start looking at this is that winding a spring also loses 50% of the energy, likewise a flywheel. Also seems to be true of any method of storing energy, not batteries though. It is the 1/2CV^2 or similar equation with whatever energy you are using, it is the 1/2 bit that does the damage.Anyone got any comments that aren't just puff, a book reference for example. I have had the 'you are wrong' thrown at me before but no proof or better, references.

author: Doctorandus_P
author_rank: Super Contributor
timestamp: June 02, 2022, 11:11:40 am
content: Quote from: woodchips on June 02, 2022, 09:26:54 amI'll give it a puff.It's not the 1/2 bit that does it, but the quadratic energy stored with linear voltage change.That it's 1/2 is just from your simple example of equally sized capacitors

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 02, 2022, 11:59:32 am
content: Quote from: woodchips on June 02, 2022, 09:26:54 amYou have some misconceptions about work; let me explain this a bit.A lossless (energy storing) component does not lose any energy, obviously. (How it can be "lossless" might perhaps not be obvious at first, but if not, take it for granted, for now.) If you look at the voltage and current at the capacitor or inductor terminals, or torque and RPM at the flywheel shaft, or force and velocity at the spring, you will find nothing is lost in any of these cases. It's all about the source.When the source is resistive, there is a loss of 50% when going from zero to full, or vice versa.There can be even more loss, say when powering the spring with biological muscles (dissipates power to hold force at fixed displacement, i.e. doing zero net work), or say with aerodynamic forces (an airplane generates lift to push against its weight, but requires continuous engine power to do so, or equivalently: a glider cannot have an infinite glide slope), so it really isn't very special that this might be the case.There can also be less than 50% losses. Suppose we charge a capacitor, stepwise, to 1V, then 2, then 3, and so on, always using a resistive source. Say C = 1F, and leaving several time constants between steps for the voltage to stabilize. The energy delivered by the source is charge times voltage or Esrc = dQ V (that's delta for change in charge, referencing this to total charge in the capacitor), and energy stored by the capacitor is 0.5 C V^2, and charge, Q = V C. Well, the first step has Esrc = (1V)(1F)(1V) = 1J delivered, and 0.5 (1F) (1V)^2 = 0.5J stored, for 50% lost, confirming the base case. The second step adds 1 more coulomb of charge, and 1.5 more joules of energy, but draws only 2 joules: only 25% was lost! And so on; in fact, the loss is a constant 0.5J, but the total stored goes as N^2 so the efficiency goes as (1 - (1 / N^2)) x 100%!And this makes sense if we decompose the system by components. We can subtract out the DC voltage at every step, setting aside the capacitor's energy and charge gains between steps, and think about the AC being a 1V step every so often. Which obviously is a fixed amplitude into an RC circuit, so dissipates constant power. Meanwhile the DC is ramping up slowly (as contradictory as that combination of terms is... but it works out in this case), and stored energy tracks with delivered energy, minus that fixed offset.And, needless to say, lossless elements of complementary nature, feed each other just fine. The energy is never converted into an orthogonal form (like an AC component into DC), nor into heat (which requires a dissipating element i.e. resistor). A lossless resonant circuit is an example of perpetual motion. (Which is why the claim from the other thread, i.e. an eventual stable DC level of sqrt(2)/2 times initial, is impossible.)So it is, that flywheels, batteries, capacitors, inductors, and etc., can all be used with arbitrarily low losses, when switched in useful ways.Note that we could apply this even to extremely small switches: consider that a CMOS logic circuit consists of capacitors and resistive switches. Suppose that, instead of CMOS switches pulling those capacitors up or down (to DC VDD/GND), we had a clever system of synchronous switches, to join such nodes onto common rails that are then pushed up/down by external means (say, a resonant inductor). We could eliminate the losses from such a logic circuit! And indeed, this has been tried, although AFAIK it's still quite experimental, but the experiments have shown it's feasible. The main downside is, besides having so many extra supplies (and they're AC now) in the circuit, the sheer amount of logic required (who switches the switches? why, more switches of course!), and the fact that it must be done at a minuscule fraction of the maximum switching frequency (since Fmax is defined by capacitive reactance matching switch resistance -- no matter how carefully you switch the load at that point, itmustbe lossy!).Which, is hilariously impractical for current types of computers -- where we want fast calculations at high densities, and the power consumption is merely an operating cost. It will likely be mandatory for future architectures, with multiple layers of transistors stacked up -- the power density would literally melt itself with present technology, but by stacking enough low-loss logic, we could have a circuit running at some ~MHz with perhaps as much (parallel or combinatorial) logic to fully compensate for the orders-of-magnitude reduction in clock speed. (Probably this would necessitate new ways to design architectures and data flows; a cubic array computer would be effective for vector and matrix ops, especially for neural net sorts of applications, say.)Tim

author: Siwastaja
author_rank: Super Contributor
timestamp: June 02, 2022, 12:13:18 pm
content: Quote from: eugene on June 01, 2022, 05:45:35 pmObviously a DC/DC converter that includes aninductor. Inductor does "lossless"voltage conversion, which is something you don't get by connecting capacitors directly together.Any DC/DC controller which has functional (i.e., non self destructive) current limitation feature and is absent of "error states" does the trick.By connecting two capacitors directly together, loss of energy is 50% because while charge is conserved, voltage is lost in the process. You can try it yourself if you don't believe it.Inductance enables you to store energy in the magnetic field at certain voltage, then output it at another voltage. Look at the analysis of buck or boost converters for more details. Buck is well suited for the job. Boost has a limitation that initially, unlimited current flows through the diode.For the OP, indeed looking at some self-oscillating, hysteretic buck control would be a fairly simple solution, and a learning experience. Basically, just make the MOSFET turn off when exceeding inductor current limit, then back on when below another lower limit. If the output capacitor is of low voltage,synchronous buckoffers significant efficiency improvement by eliminating diode Vf. Input voltage sense could be ORred together by the overcurrent signal to provide the interesting UVLO-like requirement.With a complete controller IC, "UVLO" is likely not the right feature to regulate the input voltage because it triggers a full reset and soft-start sequence, which takes time.

author: Faranight
author_rank: Unknown
timestamp: June 03, 2022, 05:38:03 am
content: Siwastaja:Yes...Which again brings us back to the original question in my first post.Can I use a regular buck converter with an "input feedback loop" i.e. by wiring an inverted voltage from the input to the output feedback pin?Can this reliably be used?

author: Siwastaja
author_rank: Super Contributor
timestamp: June 03, 2022, 05:43:50 am
content: Seems fine on the surface, so why not just give it a try?The problem is, your case is so exotic that not many have done it. Input voltage regulation sometimes is a thing, but then it tends to be tailored for a specific purpose (such as solar panel MPPT).

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 03, 2022, 11:38:41 am
content: Well, only one amp is needed, but you can't invert with respect to GND, you need an offset voltage or reference there.Likely compensation will be wrong too, and depends on input capacitance and other source characteristics. Using a few-pins regulator isn't a good idea. One with external compensation and VREF would handle both; likely a controller rather than a regulator.Tim

author: woodchips
author_rank: Frequent Contributor
timestamp: June 10, 2022, 07:54:29 pm
content: Finally found the books I was looking for.Leander Matsch, Capacitors, Magnetic Circuits, and Transformers and Electromagnetic and Electomechanical Machines.Amazing to have engineering books that a re just interesting to read, might not understand all of them, but worth the effort in any case.