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Simple power supply questions

Discussion in 'Power Electronics' started by thomaskwscott, Nov 6, 2013.

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  1. thomaskwscott

    thomaskwscott

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    Nov 6, 2013
    Hi Guys,

    I'm new to the forum and new to electronics in general so i may be making a few noddy mistakes but i was hioping someone could help out with general principles for a project i am working on.

    I have one of those Android mini TV boxes that takes a 5V 2A power supply and i've mated it to a monitor and lcd controller board i have that takes 5-12V power. This all works great from a basic 5V 2.1A USB power supply that i have but i want to run it on batteries. I tried a 4.8V NI-MH battery (that was actually reading over 5V) and the screen flickered from what i think was under voltaging.

    Looking for a solution to this i found Low Dropout Regulators and particularly this board:
    http://www.ebay.co.uk/itm/190814696...eName=STRK:MEWAX:IT&_trksid=p3984.m1438.l2649
    This would allow me to use a higher voltage battery and get around my issues if i understand correctly?

    If the above is true then I plan to use a 9.6V battery and charger like the one here to power everything:
    http://www.ebay.co.uk/itm/181245672216?ssPageName=STRK:MEWAX:IT&_trksid=p3984.m1438.l2649
    and this raises another issue with regards to charging, I want to be able to use the device whilst charging the battery. I have attached a small picture of my planned solution where in "charging mode" i introduce 2 voltage from both the battery charger and the original power supply for the device. The 5V power supply is introduced downstream of the regulator and is responsible for keeping things running during the charging process the 9,6V charger is responsible for charging the batteries only. My question is whether i have to take further steps to isolate the batteries in this example? In "charging" mode will my device attempt to pull power from the batteries as well and if so will this effect their charging?

    Any help much appreciated


    Tom
     

    Attached Files:

  2. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    Yes, a low dropout regulator will give you a stable voltage from a higher-voltage battery, but it wastes power. You would be better to use a buck switching regulator. These generally have higher dropout voltages than linear LDO regulators, but you don't actually need low dropout voltage - 4.6V is plenty of headroom for any normal regulator. A switching regulator will be much more efficient, so your battery will last longer.

    Regarding battery backup switching, your circuit is workable if you can ensure that the regulator output voltage is a bit less than the voltage coming in from the AC-DC adapter, because in that case, when the adapter is operating, the regulator will see that its output voltage is higher than it should be, so it will "back off" and stop drawing significant current from the battery. You need to make sure that the regulator will not be damaged if an external voltage is fed onto its output (most regulators, linear and switching, will not have a problem with this), and it's probably wise to connect a diode such as a 1N5401 with its anode to the regulator output and its cathode to the regulator input, so that when the battery is disconnected, the regulator is not in a situation where voltage is applied to its output but there is no voltage at its input; this can damage some regulators.

    An alternative arrangement would be to gate the two power sources (battery, and mains-derived power from an AC-DC adapter) at the input of the regulator, not the output. You would use an AC-DC adapter with a voltage higher than the maximum possible battery voltage, and use two diodes (again, 1N5401 would be suitable, or you could use 3A Schottky diodes for lower loss) with their cathodes tied together, feeding into the regulator, and one anode connected to the battery, and the other to the AC-DC adapter.

    In this case, when the AC-DC adapter is running, its diode will pull the regulator input to a high voltage (say +12V or +24V) and the other diode will be reverse-biased, so there will be no load on the battery, and the charger will be in full control of the battery. When the power fails, the voltage from the AC-DC adapter will drop to zero and the other diode will conduct, passing the battery voltage into the regulator. The diode in series with the AC-DC adapter will prevent current from back-feeding into the adapter.

    This arrangement means that the regulator will see a wider range of input voltages, and that's another good reason to use a switching regulator. The actual voltage coming from the AC-DC adapter is not important as long as it's higher than the maximum battery terminal voltage and not too high for the regulator.

    An extension of this scheme is to use a single DC power source which feeds through the diode into the regulator and also powers a linear charger or switching circuit to charge the battery. In this case the DC voltage needs to be somewhat higher than the maximum battery voltage under charge. This method is good if you want to implement the charger yourself and use a single DC input, instead of using an external mains-powered charger for your battery and an external AC-DC adapter as well.

    Edit: Buck switching regulators are available cheap on eBay. I don't vouch for any of them but others have been successful using them I think. YMMV, caveat emptor, etc.
     
  3. thomaskwscott

    thomaskwscott

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    Nov 6, 2013
    Thankyou so much for all the information you have given me and particularly for explaining things in a way I can understand. You've given me much more confidence in the design I was going to implement.

    Thankyou also for recommending the more efficient design. I have an inbuilt fear of high powered batteries that meant i wanted to rely on others for the charging and protection circuitry, maybe now i will have a go myself...
     
  4. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    You're welcome :)

    If you want to do the charging yourself, have a look at charge controller ICs. There are lots of types available, from several manufacturers, and the data sheets usually have lots of applications information. You may also find detailed application notes on the manufacturers' web sites.

    A good starting point is Digikey's battery management ICs page: http://www.digikey.com/product-search/en/integrated-circuits-ics/pmic-battery-management/?stock=1

    In the Function column, select the Charge Management options (use Ctrl or Shift to select multiple lines), choose your battery chemistry, and use the Supply Voltage column to eliminate the low-voltage-only devices.

    Through-hole components are a lot easier to work with, but nearly all of the modern devices are available in surface-mount technology (SMT) only. You can buy SMT adapter ("breakout") boards - Google SMT breakout.

    Good luck!
     
  5. thomaskwscott

    thomaskwscott

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    Nov 6, 2013
    Wow, thankyou so much that's an excellent resource. I have already ordered the parts for my project but thanks to your help i'm already working on a mk2 version,
     
  6. thomaskwscott

    thomaskwscott

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    Nov 6, 2013
    I've done a bit of experimenting and am still having some issues. I've swapped in a buck switching regulator (lm2596), wired up the 9.6V cells and adjusted the output to an easy 5.1V. However, when i connect it to the screen and tv box there is a flicker and a high pitch whining sound coming from screen driver module. I've done a little background reading and wonder if the sound is coming from a poor quality DC construction? From what I understand the switching could lead to power interruptions that the screen driver module would not like. Does this sound plausible and if so is there a way around it?
     
  7. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    Layout is important for switching regulators. That's why I was thinking that you might want to buy a fully assembled switching regulator on its own PCB from eBay. (I meant to suggest this in the Edit: paragraph in post #2, but I didn't explain that I meant a fully assembled regulator board, sorry!)

    You said you have "swapped in" an LM2596... You know a switching regulator needs several external components, including an inductor and a diode and significant smoothing capacitance on the input and output? If you don't have those external components, anything could happen, including damage to your loads!

    Have a look at the LM2596 data sheet to see what you need. You can build a switching regulator on stripboard but a preassembled PCB is a lot easier.
     
  8. thomaskwscott

    thomaskwscott

    16
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    Nov 6, 2013
    Hi Kris,

    I'm still a bit wary of breadboard so when i say LM2596 i meant i ordered the pre assembled board from ebay (particularly this one: http://www.ebay.co.uk/itm/221307319695?ssPageName=STRK:MEWNX:IT&_trksid=p3984.m1439.l2649

    I connected my 8 cells to the input and verified 9.6V then adjusted until i could test a no load 5.1V at the output. After that i hooked everything up and again teasted 5.1V at the output but the screen began to flicker and make noise. I wondered if i'd damaged the screen so tested it with the usual power supply and got no noise and no flicker. I then decided to test the current going through the regulator and got a maximum of around 5A with no load. The regular power supply can only provide 2.1A and the ebay add claims up to 3A should be fine. My cells are 3000mAh 1.2V ni-mh cells so should have the power.

    I'm stumped!!!
     
  9. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    Where did you measure that current? At the input, in series with the battery?

    You know that when you measure current, you have to interrupt a circuit and connect the multimeter across the gap? You don't connect the multimeter across the input of the power supply when you're measuring current. Current flows through a wire; it is not measured between two separate points the way voltage and resistance are.
     
  10. thomaskwscott

    thomaskwscott

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    Nov 6, 2013
    Ha, i feel like such a numpty. To test the current i just stuck the multimeter across the two output pins on the voltage regulator board whilst everything was hooked up. I thought it was a little strange that everything cut out when i did it. I hope i've not done any permanent damage.

    I have just tested in series at the battery connections and got between 0.65A (when screen is off and only tv box is running) and 1.2A at its peak with both screen and tv box running.

    This is much less than i thought it would be so encouraging but it still does the whining noise though...
     
  11. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    LOL :)

    At least you have a good attitude!

    The whining noise MIGHT just occur at certain voltages; it may not be a problem at all. Try disconnecting the direct 5V supply, so there's only the regulator in circuit, and adjust the regulator output voltage steadily from 5.1V down to 4.9V or something like that. If the noise comes and goes, and varies in pitch and quality, it's probably nothing to worry about.
     
  12. thomaskwscott

    thomaskwscott

    16
    0
    Nov 6, 2013
    Hi Kris,

    Changine the voltage does seem to have an effect on the whine but i can't seem to eliminate it completely. It seems the higher the voltage goes the less the whine but i don't want to take it above 5.25V as that should be the tolerance on the android tv box.

    My new plan is to order another on of these voltage regulators and have two voltages supplied. The monitor takes 5-24V so i could take that up until the whine stops.

    Does this sound plausible?
     
  13. thomaskwscott

    thomaskwscott

    16
    0
    Nov 6, 2013
    I connected verything up in the same way i have done many times before last night and the screen stopped working. It will power on and show images but then turns off and will not work again until i dosconnect it from power and reconnect. I tested the voltage coming off the back of the regulator and it shows 5.25V as normal until I connect everything up. Then it struggles a bit before diving to less than 4.5V as the screen turns off. I think the batteries may be dead as it all works fine from the mains power but when I test the batteries with no load they show 9.7V (8 x 1.2V cells) is this expected? I have an 8 cell nimh charger completely seperate to the circuit so can charge the batteries there but I'm very worried about overcharging them (batteries scare me in general). The charger claims to be protected and the only difference between a normal rc car battery and mine is that i have the cells in a plastic battery holder. Do you think i am ok to charge them up?
     
  14. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    If the monitor takes 5~24V you could power it directly from the batteries, couldn't you?

    If the regulator output voltage is dropping when the load is connected, then either the regulator can't supply enough current, or the batteries can't supply enough current to the regulator.

    Can you measure the amount of current that the screen draws? Is this more than the regulator is rated to supply?

    Also can you measure the voltage on the battery with the regulator connected with no load, and with the screen connected? How far does the battery voltage drop? If it drops to under 8V then the regulator may be dropping out.

    Eight 1.2V cells should measure about 9.6V, yes. When freshly charged they will be higher, and when under heavy load and when discharged they will be lower.

    If your charger is specified to charge the type (chemistry) of battery that you have, and the capacity you have, then yes you can try charging it, but make those measurements first.
     
  15. thomaskwscott

    thomaskwscott

    16
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    Nov 6, 2013
    Hi Kris,

    I took the measurements as you suggested and got some very strange results:

    Battery voltage:

    No Load: 9.57V
    Load: 3.84 - 5.5V

    The current measurement was just 600mA but that is with the screen failing to power up. I remember checking current before and finding a peak draw of 1.4A. The regulator board i have is theoretically rated to 3A and is not getting hot so I hope i'm within tolerances.

    Do you have any idea what could be causing the voltage drop under load?
     
  16. BobK

    BobK

    7,682
    1,688
    Jan 5, 2010
    Insufficient battery power.

    Bob
     
  17. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    Yes, the battery can't handle that much load current.

    A battery can be (very roughly) modelled as a voltage source with a series resistance. This resistance is called the battery's internal resistance.

    You can think of a resistor as a tension spring. This spring is hooked on a point 9.6 feet high. While there's nothing attached to it, the other end of the spring will also be 9.6 feet high (pretend the spring has zero length).

    Once you start pulling down on the spring (which is equivalent to drawing current out of the battery), the voltage at the bottom end of the spring starts to drop. The more current you draw (the more force you apply), the further it will drop. If it drops down to 4~5V, the regulator can only produce 2~3V (because of its dropout voltage), which is not enough for your load. So it will misbehave and stop working.

    Resistance is measured in ohms. A LOW resistance corresponds to a STRONG spring, which is hard to stretch. This is what you need, because your load needs a lot of current, and you don't want the spring to stretch much.

    A HIGH resistance corresponds to a WEAK spring, which stretches easily. At the moment, your battery's internal resistance is too high. This can be because it's not properly charged (it may still show a reasonable terminal voltage with no load), or because one or more cells are damaged or worn out.

    If you can get access to the individual cells in the battery, you can check to see if one of them is faulty. For each cell, connect the multimeter across the cell, and connect the load to the whole stack as normal. After a short time, look at the cell voltage. If it has dropped drastically, or gone negative, that cell is faulty. Disconnect the load once you've measured the voltage, then try the next cell.

    You may be able to replace an individual cell, or you may be able to charge it by itself to possibly resurrect it. I'm no expert on this; Google some appropriate keywords, including the chemistry of the cells (NiMH).

    Of course, it could just be that the whole battery needs to be charged, so you should try that first.
     
    Last edited: Nov 28, 2013
  18. thomaskwscott

    thomaskwscott

    16
    0
    Nov 6, 2013
    Thanks for the replies, I hopes it's just a case of battery charging and have them on charge right now. They are supplied as seperate cells in a plastic holder so when the charger says they're done i will check each cell individually as recommended above.

    I've put some plastic boxes over the top just incase a cell blows and i'm checking for them getting hot every 30 mins. Is there anything else i should be doing to make sure this is safe? As i say i am petrified of batteries and really don't want to be on the recieving end of a blown cell.
     
  19. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    If you're worried that the charger may not be a proper match for the batteries, put the whole lot outside, or do it in the garage. I had a pack of NiMH AA cells overheat on me while I was developing a charger board and I can tell you that it doesn't take long for them to overheat, and when they go, there's a LOT of smoke and it smells awful!
     
  20. thomaskwscott

    thomaskwscott

    16
    0
    Nov 6, 2013
    All good news tonight! The light on the charger just went green so i hooked the cells back up for my test and everything worked perfectly (including no whine from the lcd controller). Many thanks for the help, without it i wouldn't have had the confidence to try any of this and would probably have blown something up a long time ago.
     
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