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Thread: OT: help with low voltage regulation

  1. #11
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    Comes to mind that with fresh batteries you will have an excess voltage of about .35v or so- but when the batteries discharge, your 'low dropout voltage' will disappear. If you had a regulator with a dropout voltage of say .1v, then the best case you would have is for the battery to be called 'dead' at 1.3 volts. You still won't be using much of its charge, and that's assuming you could find a regulator with that low of a dropout voltage.

    Since this is an application where the current is drawn in short pulses, with no loading down between pulses, I wonder if there's a way to regulate only during the time the current is flowing- or to regulate the duration of the pulse. The duration would be shortest when the battery is fresh, and would lengthen as the voltage drops.

    I'm also wondering about using two cells in series. You start with a higher voltage, but you get to use more of the charge in the cells, so even with 'wasting' some of the voltage (to run the regulator) you might end up with a longer run time for the pair of cells in series rather than in parallel. A bi-stable switch in a circuit with a voltage sensor might allow you to charge a capacitor to the required voltage, then cut off until the cap is discharged by the clock, at which time it triggers on until the cap voltage is restored, at which time it triggers off until the cap is discharged again, etc. No or very little current would be drawn during the off time.

    One thing to be aware of- most if not all regulator chips require a minimum load current to function properly. Between times when the clock is drawing the impulse, the regulator will be totally unloaded, and the output voltage will then rise above the design value. This may be the worst application to use any normal type of regulator, low dropout voltage type or otherwise.

  2. #12
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    Here's an idea for a circuit. I haven't tried to draw it out but the idea is this- use a series connected pair of cells so you have enough voltage to operate the 'controller circuit', then use a series transistor with a fairly high value of base resistor. This would allow just enough transistor 'turn-on' to charge a cap up to the wanted voltage, or just a tad beyond. A second pair of transistors is hooked up base-emitter-base-emitter to act as a voltage sensor. The second transistor in this pair has its collector tied to the base of the series transistor. The collector of the series transistor connects to a cap, which is where the first transistor of the second pair has its base tied to also. No current goes through the transistor pair until the cap voltage comes up to about 1.2v or so, and the turn-on current for the series transistor also goes into charging the cap. That turn-on current is only drawn away, or wasted, during the time between the cap having come up to voltage and the clock drawing the pulse out of the cap.

    This circuit would need a minimum voltage of about 1.8 or so to operate, so that represents each supply cell delivering current until it's voltage has dropped to about .9v. That's pretty good usage of the charge in it, so the pair of series connected cells would probably last as long or even longer than a pair of parallel connected cells which wouldn't be able to discharge to that level and still power the circuit.

    In the above description of the circuit, you would figure out the value of the turn-on resistor so you would get the full 1.2 volts into the cap with the batteries depleted to about 2 volts for the pair, and within the time between the clock asking for the next pulse. The value of the cap would be decided based on the 'power' required to reliably click the clock mechanism over.
    Last edited by darryl; 05-24-2012 at 08:18 PM.

  3. #13
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    Here's a crude drawing of the circuit I was thinking might work. The two cells feed voltage through Q1 and through R. Q1 turns on and feeds current to the capacitor. When the voltage rises to about 1.2 or so, the B-E junction of Q2 begins to conduct and brings Q3 into conduction. In turn, that pulls current from R towards battery negative, making Q1 turn off and cease charging the capacitor. The three transistors can be pretty much any small-signal npn types and can all be the same. 2SC458 would be good, or 2N2222, or 2N3904. Any small npn transistor that's not leaky would work. R could be anywhere from 10k to 100k- the right value would be arrived at by trial and error. The larger the value of the cap, the lower the value of the resistor would be. The cap should be either a low leakage type or a typical electrolytic rated at 16v or more- the value could be 10uf or so, but this again is experimental. The value should be large enough to fully click the mechanism. It might take a 47uf cap to do it. The entire circuit will not need much space.



    This is a fixed regulator based on the base/emitter voltage drops of Q2 and Q3 in series. The output voltage might be as low as 1.1 volts, and if this isn't enough you would replace Q2 with two shottky diodes in series. That would bring your regulated voltage up to about 1.4 or a bit more, which is going to be fine anyway since you'll still be able to use each cell down to about 1 or 1.1 volts before the regulated voltage begins to suffer much.

  4. #14
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    1) NiCd are pretty darn consistent in voltage until they hit the "knee" on discharge. I'd try it (which you may have done) before dismissing it.

    2) anything that depends on junction voltages directly for regulation is going to be rather temperature sensitive. This may not be an issue, but then again it may, I do not know how sensitive the load is to voltage variation.

    3) Anything which resembles a shunt regulator, even one in which the base voltage is the shunt regulated parameter, is going to draw excess current unless very strong precautions are taken to reduce current to the absolute minimum. That affects battery life, and if not done carefully, may mean that a NiCd is no worse in battery life.

  5. #15
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    Well, a regulator like this will do quite nicely what you want:
    http://cds.linear.com/docs/Datasheet/3620fa.pdf

    How are your soldering skills?

    edit:
    Well, maybe this is slightly less masochistic to solder:
    http://cds.linear.com/docs/Datasheet/3404fb.pdf
    However you see the efficiency drop at very low currents.

    You could also go the switched capacitor regulator route, that's also not too bad efficiency wise. Something like this:
    http://datasheets.maxim-ic.com/en/ds/MAX868.pdf
    They even send out free samples

    You really need to measure the supply current to the clock to be able to select a sensible regulator.

    Igor
    Last edited by ikdor; 05-25-2012 at 09:29 AM.

  6. #16
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    JT, your points are noted regarding the 'shunt regulator'. In the circuit I drew, the application is very low power, so there's little temperature change to affect the regulation much. This is not a perfect regulator, nor a very good one, but it is much better than the use of a cell directly. The intention was to provide a lesser change in supply voltage for the clock to minimize the inaccuracy of the clock over the lifetime of the batteries.

    As to the wasted current shunted through the transistors, it is very little and only occurs once the capacitor has come up to voltage. If the timing is right, the clock takes the charge soon after, so the time during which the current is 'wasted' is relatively short. The current through the resistor would be higher than that shunting to ground through the Q2/Q3 pair, but that current also flows into the load during the cap charge cycle, so is not wasted until the voltage on the cap comes up. This is not a shunt regulator in the traditional sense, but I do agree that shunt regulators are wasteful.

    I'm not aware of any regulator chips that might be purpose-built for this type of application, but I would think there probably is one. I'll go back and check out those links- maybe I'll learn something.

  7. #17
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    I'm afraid all the voltage regulation circuitry will eat battery life. Can't you just put a diode in series with the battery? I believe an inline diode will give you a .3 volt drop.

  8. #18
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    You can drop the voltage from the battery with a series diode, but that won't improve the regulation at all.

  9. #19
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    Default the NEW rechargables...

    I have had good luck with the "new breed" of rechargables. Eneloop brand from sanyo.

    Very low self discharge. Retains 75% of its charge after 3 years of storage. Comes precharged (mostly) from the factory, by photovoltaics no less. Not stupidly expensive. And you really will get many many charge/discharge cycles out of them. It's what rechargeable batteries were meant to be.

    As my old ni-cad and LiPo batteries poop out, I replace them with eneloops.

    http://www.eneloop.info/home/the-new...d-eneloop.html


    Not complicated. Might work. Might replace all your other rechargeables too. The amp hour rating may be a tad lower than some other brands, so compare that if it's important.

    Finest regards,

    doug

  10. #20
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    Of course, the whole idea here is to be able to feed the clock as constant a voltage as possible over the life of the cell or battery. From some charts I looked at, the nickel metal cell seems to have the flattest voltage curve as it discharges. If you choose one which is optimized for low self-discharge, it might be a viable option.

    Now I'm wondering- is there a non-rechargeable version of the ni-mh cell?

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