voltage regulator choice help

Discussion in 'General Electronics Chat' started by s_mack, Jun 12, 2012.

  1. s_mack

    Thread Starter Member

    Dec 17, 2011
    187
    5
    Battery powered device where end users select from a variety of voltages and chemistries as per their preference. Typical selections are:

    8 x AA NiMH
    8 x AA alkaline
    2 cell (7.2v) LiPo
    3 cell (11.1v) LiPo
    3 cell (9.9v) LiFePo4

    The device runs on 5v.

    Due to customizations, the current draw could vary from ~100mA to ~400mA.

    I'd like to select a voltage regulator that's appropriate for all situations, and also efficient.

    Currently, the device has a linear VR with a drop out voltage of about 2V, which makes the 7.2v battery non-ideal when it may otherwise be the best choice.

    So my first thought is to just use an LDO VR with a significantly lower drop out. But then the issue I have with that, is the higher voltage battery options are converting even more voltage to heat.

    Then I think a switching reg would be better... I have no experience at all with switching regs and I see plenty of discussions how they aren't universally more efficient - that it "depends", but I find nothing about what it depends on. And in the data sheets of most of the less expensive ones, the "typical applications" list almost never seems to include battery powered devices, which leads me to believe its likely not the way to go.

    Cost is very much a consideration. @ volume of 500, I'd like the price to be well under $1 - ideally under $0.50.

    Thoughts?

    - Steven
     
  2. panic mode

    Senior Member

    Oct 10, 2011
    1,319
    304
    switching regulators are generally more efficient specially where you have bigger voltage drop. if you draw 400mA from 11V battery, and have linear regulator, you are going to draw same 400mA. If you use switching regulator, your current draw may be some 200-250mA.

    an example of 5V 1.5A regulator that is 7805 drop in replacement is:
    http://www.mouser.com/ProductDetail...RT2lXVNWg5/3Kq428yVOxYAdOa2G8Z2a%2bF%2b/5ymI=

    but this is not going to meet your price even at 500 units.

    you can also make your own, Roman Black made very simple one and it may be just what you are after.
    or maybe http://www.ti.com/lit/ds/symlink/lm5576.pdf

    next option is LDO regulators
    http://www.mouser.com/Semiconductor...ut-LDO-Regulators/_/N-6j76t?P=1z0wa29Z1z0w877

    such as
    http://www.mouser.com/ProductDetail...S4yu56pZvM63%2bVG4ZQhn%2bjSRfnf9MRVcId4a%2bs=
    or
    http://www.mouser.com/ProductDetail...K%2boHS4yu56gPciVN9yJK8CZiltesPflG78KwQ2pxUs=
     
    Last edited: Jun 12, 2012
  3. s_mack

    Thread Starter Member

    Dec 17, 2011
    187
    5
    Thanks. I'm not concerned about drop-in replacement-ability. Its going in a new circuit anyway, so physical compatibility isn't an issue.

    I'll google Roman Black and see if I can find what you're talking about there... thanks.

    - Steven
     
  4. bountyhunter

    Well-Known Member

    Sep 7, 2009
    2,498
    507
    At that price you can forget about a switcher. I guess you will have to live with the linear and the worst case disspation of about 2.5W which will require a TO-220 case regulator and about one square inch of PC board copper for a heatsink..
     
  5. s_mack

    Thread Starter Member

    Dec 17, 2011
    187
    5
  6. Markd77

    Senior Member

    Sep 7, 2009
    2,803
    594
    What it depends on is the percentage voltage change and the efficiency.
    If you have a 80% efficient switching regulator and you need to drop 10V to 9V then a linear regulator is better, at 8V output they are the same, and under 8V the switching one is better.
    Not sure where you will find a switching regulator, inductor and diode for under $0.50 though.
     
  7. panic mode

    Senior Member

    Oct 10, 2011
    1,319
    304
    last two links in my post are low droput linear devices. normal dropout is 0.25V or worst case 0.5V. lower side of 5V for any digital circuit is 4.75V, so even in worst case scenario this should work with battery level down to 5.25V. if the fresh battery is 7.2V and max current draw is 400mA, then 2.2V*0.4A=0.88W, to get 2.4W we would have to use 11V battery.

    MC34063 is low cost and includes switching transistor. but there are few more components including inductor. cost of inductors goes up with current rating and inductivity. this alone can be a dollar or more when rated for 15-200uH and 400mA or so.
     
  8. s_mack

    Thread Starter Member

    Dec 17, 2011
    187
    5
    When you guys are considering costs... are you looking at a breadboard one-off scenario? Because switching regs < $0.50 seem common. The other components are merely pennies. For example, the inductor... here's one rated to 100uH and a full amp and its only 16 cents @ 500 and its probably overkill.



    Anyway, I think I will go with an LDO. I've been reading about switching interference and I don't want to deal with additional noise problems and related testing that could incur. Most people use 11.1 or 8x1.5=12v with this device, so there's considerable waste with linear. However, by using a very low-dropout model, perhaps people will be persuaded to move to a 7.2v and call it a day.
     
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