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Charging one battery stack from another

Discussion in 'Power Electronics' started by windinmysails, Jun 20, 2012.

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  1. windinmysails

    windinmysails

    26
    0
    Mar 12, 2012
    Hi team!

    Thanks for all your help in the past! I've got another question (what a pain I am, I know!)

    OK, so here's the situation. I'm building my own windmill which requires active control. I have a 24V lead-acid battery stack at the bottom of the windmill (in the garage). At the top of the windmill I have a 24V motor which is steering the device and which requires power the power. I've got some electronics controlling the direction the windmill is facing being controlled from a very low power PC on the ground with the battery stack. I therefore have 2 wires going up to my windmill to provide power and a USB cable to do the driving with. The power cables are quite thick (2.5mm^s) as I need to be able to drain a few Amps at 24V when I do the steering and the cable is 25M long (to get it away from my house). Steering only happens a couple of times a minute at worse.

    This works ok, but I want to put in place a situation where if the cable bringing the power breaks or the data communication cable breaks, or coms stops for some reason, the windmill returns to a parked position. This should be pretty easy - my plan was to put a relay and another pair of smaller 24V batteries at the top of the windmill and when the main power cable is disconnected (or lower batteries are dead), the relay falls and turns the steering motor to the park position using the power supplied by the top pair of batteries.

    So, I have two questions! 1) how can I charge these batteries and 2) how can I use the batteries to augment the supply from the bottom batteries?

    The charging point is a difficult one. To charge the upper stack I need to have a voltage that is at least a few volts higher than the voltage on the upper batteries. Even if the lower batteries are fully charged, if the upper ones are not pretty empty the voltage loss on the cable means I won't have enough volts to start charging. I'm guessing I need some kind of DC:DC charger or step-up device?

    The augmented supply is also difficult - how can I make it so that supply is being fed to the motor from both batteries rather than the lower stack but also have a cross battery charging system (top to bottom?)

    My initial thought was that I could have DC:DC converter that stepped up and charge the top stack, and that then both the bottom and top stack could be connected through reasonably sized diodes onto the load. I've drawn this in the attached (simplified) diagram.

    Do you think this will work? Or is there something else I need to do?

    Many thanks,

    Neil
     

    Attached Files:

  2. Harald Kapp

    Harald Kapp Moderator Moderator

    10,609
    2,370
    Nov 17, 2011
    Since the upper batteries are only backups (and hopefully almost never used), I think you won't need a charger solution as described by you. Of course, doing it your way is the "correct" idea, don't get me wrong.

    A simpler scheme would use a small series resistor !~1 1 Ohm, possibly the resistance of the cable is enougt) in series from the lower battery to the upper battery. If both batteries are lead acid then the charging characteristic should be similar (excluding different Ah ratings). The series resistance will ensure that most of the charge current goes to the lower battery,

    Harald
     
  3. windinmysails

    windinmysails

    26
    0
    Mar 12, 2012
    Hi Harald,

    Thanks for this - but what I'm not sure is that any charging will actually take place.

    Given that the cable is 25m long, there will be a drop on the cable even at a quite low current drain from the upper batteries as they charge. I guess it's not so much of a problem because the bottom set of batteries are to be charged with a normal battery charger and so their voltage will generally be more than 24 or 25V, so even with a little drop there'll be enough head to charge the top ones, I just worried that no charging would occur.

    Also, since the top ones will be sealed lead acid batteries, do I have to be more careful about how I charge them or is just putting a 1Ohm resistor in ok?

    Thanks again for your help!

    Neil
     
  4. Harald Kapp

    Harald Kapp Moderator Moderator

    10,609
    2,370
    Nov 17, 2011
    A lead acid battery is charged with constant voltage. The only difference the battery on top and at the bottom will see is the voltage drop across the series resistor. As you will use the top batteries only in an emergency (that is ideally never), these batteries will be almost always fully charged. Thus the voltage is equal to the voltage across the lower batteries (or even higher if you drain the lower batteries). Thus the voltage difference across the series resistor is very small which in turn means a small (trickle) charge current (or small current drain drain). In either case the top batteries will be charged or discharged more slowly than the bottom batteries.
    In a situation where the voltage at the bottom is soignificantly higher than at the top, this means a higher voltage drop across the series resistor and thus more charge current for the top batteries. On the other hand will a charged top battery be drained to charge an empty bottom battery. But according to your description this is no use case.

    It is my opinion that the charge should still suffice to park the windmill. Personally I would try that simple scheme before resorting to a more elaborated system (admittedly a possibly more reliable one).
    If you want to play it safe, use a SPDT relay intead of the two diodes. Use a relay with a nominal coil voltage of 24 V as this is the battery voltage. Connect the relay coil to the power from the lower battery. If this power fails for whatever reason, the relay will be de-energized and switch to the upper battery. If you use a DPDT relay, you could also use the second contact to give an alarm signal (for a siren, a light or just for the park automatic of the windmill). The relay will also have a negilgible voltage drop compared with the diodes. Observe the contact rating of the relay. An automotive relay froma a truck might be suitable.


    Harald

    Note that I'm not a battery expert. Anything you do is at your own risk. Even if 24 V is not dangerous per se, you will be handling possibly very large currents which is dangerous. A few strategically placed fuses may be in order.
     
  5. windinmysails

    windinmysails

    26
    0
    Mar 12, 2012
    Hi Harald,

    Thanks for the reply. I'm still a little confused as to how this will work in practice because of the voltage drop, and also because I do want to use these batteries to help with the additional current drain when the steering motor is energised.

    There are some other little bits up on the turbine which drain about 1A. At 4mm^2 cable, over 25m will loose about 10.5V. So, if my batteries at the bottom were full, the ones at the top wouldn't really get any charge given that voltage loss and the loss that would occur across the diode (I'd have to have a diode really, wouldn't I?)

    When the motor fires up and takes a few Amps, the voltage drop on the cable will be more. Let's say it takes 5A, my drop will be about 1.5V. This is the reason I was thinking of using the top batteries to help with the power - the top batteries are right next to the motor so if I had both batteries in parallel to the motor then there would be much less power drawn from the bottom batteries.

    Do you think this is possible or am I trying to do the impossible?

    Sorry to be confusing, I'm just trying to work out the best way to do it!

    Many thanks again,

    Neil
     
  6. Harald Kapp

    Harald Kapp Moderator Moderator

    10,609
    2,370
    Nov 17, 2011
    How do you arrive at a voltage drop of 10.5 V?
    If I take this data http://www.sengpielaudio.com/calculator-cross-section.htm the table in the last quarter of the page gives a resistance of 0.16 Ohm fpr 4mm² cable at 20 m length. Now scale this up (wire resistance is linear with length) to 25 m and get 0.2 Ohm / 25 m. Since you need two wires (+ and -), this makes a total of 0.4 Ohm for the wiring from bottom to top. At 1 A this means 0.4 V voltage drop. Nothing to worry about.

    Harald
     
  7. windinmysails

    windinmysails

    26
    0
    Mar 12, 2012
    Hi Harald,

    The 10.5 was a typo - sorry! I'd originally written 1/2V, but then I was worried it'd get converted to one of the strange symbols like my DC:DC does! So I decimalised to 0.5 but failed to delete the 1! Sorry!

    Absolutely agree, 0.5V (or a bit less) is the drop - but for charging, I need a higher voltage than the other battery set, that's my worry.

    Thanks again, and sorry for the typo!

    Neil
     
  8. Harald Kapp

    Harald Kapp Moderator Moderator

    10,609
    2,370
    Nov 17, 2011
    I still think my setup should work. If not, you can still switch to a more expensive solution.

    Harald
     
  9. gorgon

    gorgon

    603
    24
    Jun 6, 2011
    As long as you are using lead/acid batteries, you can't use higher voltage on one of the batteries, even if you have a drop in the cable.
    The mechanics around this is that the only difference you'll have between the two batteries, is the fact the the one with a small drop will top up slower, but eventually it will reach the rated voltage. If this voltage is higher than the cells can reach, the rest of the voltage will add heat to the battery, the rated voltage will drop further due to the increased temperature and more heat will be made inside the battery. Eventually the battery will heat up or dry out, both alternatives is fatal for the battery, and your backup will be gone.

    I've seen big batteries getting round as balloons, and too big to get out of the cabinet, due to overvoltage and heating.

    TOK ;)
     
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