Charging a supercapacitor

Discussion in 'General Electronics Chat' started by cmartinez, Jun 2, 2018.

  1. cmartinez

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    Jan 17, 2007
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    I'm building a circuit in which a supercapacitor will be used as a battery. So far, all of my tests have been successful. However, the power source is very limited, and I need to build a circuit as efficient as possible.

    A small generator will be producing an AC source of about 7 VAC (that would be 14V peak to peak). I have not yet tested the power source, since I'm in the building phase, but I don't expect it to be able to deliver more than a tenth of a Watt, if I'm lucky. My plan is to rectify that source using schottky diodes (which, although they normally cannot carry too much current, they have a very low forward voltage of about 0.4V) but after that things get a little confusing for me.

    An MCU should be powered from the same rectified source while the supercapacitor is being charged. The MCU can run at voltages of between 2.4V and 5.5V, and I'd like it to start running at the rated minimum 2.4V and keep going until the supercapacitor has been fully charged to 5V. It's important that that voltage is never exceeded. I expect the supercap to take several minutes (up to 15 or 20) while it's being charged by the power source. The supercap I've chosen for this application is a 0.33 Farads one.

    So here are the three conditions I'd like to meet:
    • To run an MCU while the supercap is being charged. The MCU should start running at 2.4V, so maybe I need some sort of "snap on" circuit that will deliver the 2.4V to the MCU once the supercap has reached that level of charge, but not before.
    • To charge the supercap to a maximum voltage of 5V, even while the MCU is still running
    • To let the MCU know when the supercap has reached the desired 5V.

    The MCU I'll be using (an AT89LP4052) already has an internal comparator that I could use for that last condition. But since both the supercap and the MCU will be subject to the same voltage, I'm not sure how to go around it. Perhaps use a zener diode to that effect?

    Also, how do I limit the voltage reaching the supercap, wasting as little energy as possible?

    Here's a small graph of how I'd like things to behave:

    Capture.PNG

    Any ideas on where I should start?
     
  2. OBW0549

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    Mar 2, 2015
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    My approach would be to power the MCU from a LP2951 micropower LDO, set (via an external voltage divider, see data sheet) for a 2.4V output. This will give you regulated MCU power and a stable reference voltage for its ADC. The LP2951 also has an open-collector ERROR- output which goes low whenever the regulator output goes out of regulation and is more than 5% low; this can be connected to the RESET input of your MCU to keep it from running until its supply voltage is up.

    The LP2951 has a maximum output current of 100 mA, but from your description it doesn't sound like you need that much current anyway.
     
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  3. DickCappels

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    This is what I use to charge supercaps.

    [​IMG]

    It is a boost converter with a negative output voltage, taken from figure 13 of the ON Semiconduuctor data sheet MC34063A/D, August, 2010 − Rev. 23, which is attached. The limiting voltage is set with the two resistors connected to pin 5 of the MC34063 and the current is set with the resistor between MC34063 pins 6 and 7. Consult the datasheet for details.

    Below is tens of millivolts out vs seconds while charging a 100F capacitor.The point is that the current is constant (as indicated by the slope) until the 2.5 volt limit is reached.

    [​IMG]

    MC34063 datasheet
     
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  4. cmartinez

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    OB, my MCU is normally reset by a high signal on its reset pin. What would be the most efficient way of inverting the LP2951 active low ERROR output? Also, my device will only be using less than 15 mA during operation, so that chip is more than enough to fill my needs.
     
  5. cmartinez

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    Many thanks, Dick. Would this circuit work if I were to feed it an unregulated, rectified 7V source? The source would look something like this:

    upload_2018-6-3_12-55-27.png

    But those 7V peak would vary widely. Sometimes it would be almost zero, and sometimes it would be a maximum of 25V
     
  6. DickCappels

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    I use an unregulated source of approximately 8 volts, sometimes higher if not loaded :) I think it will work fine, especially since it is basically a boost converter.

    By the way the circuit performs cycle-by-cycle current limiting so the ripple would not create much of an effect on the output.
     
  7. OBW0549

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    Give the LP2951's ERROR output a pullup resistor, feed it to the gate of an N-channel small-signal MOSFET (2N7000 or equiv.) with source terminal grounded, and pull up the drain of the MOSFET with another resistor. Feed that to your MCU's RESET pin.
     
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  8. cmartinez

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    Yeah, I had already thought of that, but I didn't think that the LP2951's output could drive the 2N7000's gate directly. Your suggestion is reassuring.

    I have a few 2N7000's laying around that I could do some testing with, while I order and wait for the LP2951s to arrive.

    Thanks!
     
  9. cmartinez

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    Why would I want to limit the current being fed into the supercap?
     
  10. OBW0549

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    The ERROR output is an open-collector output which can sink at least 400 μA:

    Untitled.png

    I usually use a pullup resistor of 10 kΩ, 22 kΩ, or 47 kΩ, any of which will work well.
     
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  11. cmartinez

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    And therein was my misconception. It's not the ERROR pin that will drive the 2N7000, but rather the pull up resistor.
     
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  12. cmartinez

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    OB, what do you think about the LTC3459? Could this be a single chip solution to what I'm trying to do?

    That is, charge a supercap to a total of 5V at the same time that an mcu is monitoring its charging progress.

    Screenshot_20180604-024450.jpg
    I've already done some experimenting, and the supercap I'll be using will be either a 0.50F or 0.33F.
     
  13. OBW0549

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    Might be worth trying, although you'll need a second regulator to provide Vdd for your MCU, as well as Vref for its A/D converter.

    Note the Absolute Maximum Ratings for this chip; do you have means of preventing its input voltage from going higher than 7 volts?

    I noticed this sentence in the description: "The LTC3459 can be powered from a single lithium ion battery, a 2- to 3-cell stack of alkaline or nickel batteries, or any low impedance voltage source between 1.5V and 5.5V." What happens, I wonder, if the voltage source is NOT low impedance? And just how low is "low"? The data sheet doesn't say.
     
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  14. DickCappels

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    The broad answer is that current has to be limited somewhere so it might as well be somewhere it can be controlled.

    In more detail:

    The power supply chip that shuts off the charging current is limited in the amount of current it can supply safely. This also protects the power source, which in my case is an unregulated wall wart.​

    In the case of Ultracaps, in which the impedance is often very low slapping an ultracap across a power supply is akin to putting a dead short on the power supply's output. That is a bad idea unless current limiting is present (or a fuse).

     
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  15. Sensacell

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    And the switching regulator performs the energy conversion far more efficiently than any linear current limit scheme.
     
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  16. -live wire-

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    So, for charging it, what if you designed a constant power buck/SMPS regulator? That way it will charge most efficiently. Let's say your supercap is 5V 25F, about 315 joules or a little less than 1/10 a watt hour. Given that your generator can supply 1/10 a watt, it should charge it in 1 hour, even with inefficiencies. But let's say you use just a resistor. That's 3 times as long as it should be! Towards the beginning, it charges as fast as it can. It is still pretty good, but then it eventually starts to charge MUCH slower, with much less current.

    What about constant current? Here is what I am assuming. You set a constant current of x milliamps and that is around the PS's maximum current. It may draw less current on the input for a lower output voltage, but it will not allow a higher output current. Well, towards the end, you operate at close to maximum power. You are close to the maximum voltage and at the maximum current. But in the beginning, it is very low voltage but decent current, meaning low power.

    So here is the best solution. Design it to be intelligent and aim for maximum POWER. Maybe use a microcontroller or chip specifically designed for it. It should be relatively simple.
     
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  17. cmartinez

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    I wish it were that simple. Thanks for the explanation. The cap I'll be using is 0.5F, and the source will be a small generator whose power will vary quite wildly. I agree that using a simple resistor is a no-no. At this point, I'm looking at options related to boost-buck converter, like the one in post #3 that Dick so generously shared.

    I don't really need to set a current limit reaching the cap, because the power source will be quite weak. But I do need to simultaneously run an MCU, and I'm currently considering weather to either power it from the same supercap, or to power it from a separate circuit with a much smaller cap. So my choice right now is either to charge the supercap and then start the MCU when voltage reaches the minimum threshold, or to accumulate a little energy inside a separate, smaller cap that will feed the MCU, and then to start charging the supercap.
     
  18. -live wire-

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    Sounds tricky with how much it may vary. Maybe first have a capacitor in the uF or mF after the FBR range to stabilize it some. This way it will be easier to do more stuff. But think about it. You can maybe get a maximum of 7ish volts and 10-15mA normally. But maybe it supplies 30 mA to your supercap at a few 100 to a few mV. That is a lot less power. But if you efficiently converted it to a few 100 mV (75%), that is maybe up to 100mA, and this is far faster. Just think about it.
     
  19. cmartinez

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    I'm afraid I don't quite follow. Raising the voltage reaching the cap is simple, using a pump-charger. But how is the circuit supposed to also also increase the available current?
     
  20. -live wire-

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    My understanding was that the motor was 5-10 volts and low current. And essentially shorting it may not be dangerous, but it seems like if you buck it you could get considerably more power. But I guess it could do that with free energy. ;)
     
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