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I have a 2.7 V 100 F supercap, that would equal 364.50 J of energy based on E = 1/2CV^2. If my load is a series of 4 LEDs with a stack forward voltage of 12 V and forward current of 3 A that would equate to a power of 36 W based on P = VI. We know that P = E/t. Hence maximum time that the Load will be on with steady power is ~2-5 s, I'll be strobing the LEDs at 10 Hz and 30% duty cycle, so might be able to get a bit more time. Well that's the theoretical capacity.

A few more calculations for a more realistic look at this, considering how the voltages and current changes with discharge, lead to:

enter image description here

What I'm struggling to understand is in designing a boost circuit for this. I've looked at commercial and booster ICs to no avail. Some issues, the booster ICs that go down to 0.7 V input only have a max output of 5 V. While booster circuits with a higher voltage cap have like a 3.7 V min input. Since we aren't working with a battery but a supercap with much greater power output. I have no idea how to approach this design because the voltage is dropping by self discharge and the current consequently increases.

NOTE: Only reason I'm looking at such a low energy solution is the fact that I'm looking to power the load for hardly 1.5-5 s. Potentially expanding to larger power loads for a similar run time such as 18 A and 12 V with a slightly larger capacitance of course.

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  • \$\begingroup\$ You might be able to use one of the 5 V max output parts with a flyback transformer in order to get a higher voltage out. The devil is in the details. \$\endgroup\$
    – John D
    Sep 5, 2023 at 16:28
  • \$\begingroup\$ It's all about power tho, something with 5V max output and 3A isn't going to translate to 12V and 3A. How is current from a supercap even determined, we're using time to discharge as a parameter in calculation right? \$\endgroup\$
    – roaibrain
    Sep 5, 2023 at 16:43
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    \$\begingroup\$ Well, at 0.1 V the current would be more than 360 A, so if your cap and boost converter were ideal, yes. Otherwise the ESR and max discharge of your cap, and the max input current capability of your boost (inductor rating etc.) as well as the minimum input voltage of the boost will limit what you can do to a much higher voltage. \$\endgroup\$
    – John D
    Sep 5, 2023 at 18:26
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    \$\begingroup\$ 18650 INR battery is about the same size as the cap, cheaper, and will deliver 36W without trouble. Instead of 4 LEDs in series, consider 2x2 LEDs in series or other arrangement to make DC-DC design simpler. \$\endgroup\$
    – bobflux
    Sep 5, 2023 at 20:58
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    \$\begingroup\$ @bobflux I've actually completed a working design with a 70C lipo battery for a 54W led strip (27v 2A) using a constant current boost driver and it works fine. I'm curious in designing one with supercaps - hence my question. \$\endgroup\$
    – roaibrain
    Sep 5, 2023 at 21:05

1 Answer 1

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Here is how I would do it, using a very low voltage boost regulator to pump up a 5 V rail to allow a bog-standard boost controller and external MOSFET to do the heavy lifting. For generating 5 V, MAX17220 will operate down to 0.4 V and will boost to 5 V. At 5 V, you have tens if not hundreds of boost controllers to choose from. I picked MCP1632 at random which was in stock but does not feature synchronous rectification.

This is just a simplification to push you in the right direction. You need to calculate all values. Not all pins are drawn and several things are omitted. Choose a MOSFET and Schottky which can handle your full peak current and with low enough Vgsth to operate on 5 V.

Ideally, choose a synchronous rectifier boost for your main power stage which can do the current limiting for your LEDs directly.

schematic

simulate this circuit – Schematic created using CircuitLab

I just realized the MCP1632 is a better choice for LED drive than I first thought as it have a current source and allow you to program the Vref. This allows you to set it to a very low value with R_r and set your current sense resistor R4 to a low value to not burn excessive power at 3 A. Set it with 3*R4=50µ*R_r. Here is how it would look with direct LED drive:

schematic

simulate this circuit

Ignoring all DC-DC converter losses, this is the minimum current you are looking at for a 12 V 3 A load. It would take ever increasing inductor saturation current and low Rdson on your MOSFET to extract more energy from your supercap. If you want to play around with it yourself, you can create a constant power load with a resistor fed by a rail named Vcap and set the resistor value to R=limit(1m,V(Vcap)**2/36,1000) in LTspice, should be very similar in other SPICE flavors (get it?). 36 is your load in W here. 1m and 1000 are your lower and upper bound for how far it will go. I would say it looks uneconomical at 0.8 V to go any further at that high power level. At 40 A, you are probably better off with interleaving several boost converters, if not for MOSFET losses, than for inductor size.

enter image description here

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  • \$\begingroup\$ very informative answer, I understand things aren't quite accurate. What would be the best way to determine the voltage drop of the supercapacitor as a function of time with two dc-dc controllers here? \$\endgroup\$
    – roaibrain
    Sep 6, 2023 at 16:09
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    \$\begingroup\$ @Roshan You are most welcome! You can ignore the 5 V pump, it will consume tens of mW. The 12 V boost one will however do its damndest to keep the output current 3 A to your LEDs until it either hits it max duty cycle limit or peak current limit set by CS. You can model it as a constant power load of you want to simulate it. \$\endgroup\$
    – winny
    Sep 6, 2023 at 16:53
  • \$\begingroup\$ I've gone through the datasheet and am still un clear with how the CS pin is used to set the peak current value, I know this is leaning away from the main question. If possible some insight on this, would be much appreciated. \$\endgroup\$
    – roaibrain
    Sep 6, 2023 at 18:23
  • \$\begingroup\$ Do you also mind clarifying what the input current from the supercap to the boost will look like, I still keep going back to thinking 2.7V, 100F so, it'll start at >13A current and keep on increasing. Only reason i'm concerned about that is the input current capabilities of boost controllers. I think it's the intermediary boost that's just messing up with this fundamental understanding between source and load. \$\endgroup\$
    – roaibrain
    Sep 6, 2023 at 18:26
  • \$\begingroup\$ @Roshan See my update. \$\endgroup\$
    – winny
    Sep 7, 2023 at 9:17

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