Beginner q: two batteries, virtual ground and open circuits

Discussion in 'General Electronics Chat' started by ogg, Oct 8, 2009.

  1. ogg

    Thread Starter New Member

    Sep 23, 2009
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    Hello,

    I'm using two batteries (actually 2 sets of 2 AA's) to get a +3V, -3V and virtual ground. They are in the configuration shown in the attached image. I have a few beginner's questions about this setup, could you help me think through them?

    If I had one battery, I'd know that if either terminal wasn't connected, the circuit would be open with no power flowing and therefore no drain on the battery. But what happens with my two sets of batteries and three wires?

    If I disconnect the +3V wire, I would imagine there would still be a drain on the lower battery, and the outputs would change to +3V and -3V (where they were virtual ground and -3V). Likewise if I disconnect the -3V wire, I expect +3V and -3V would now come off the top battery. I'm assuming here that the rest of the circuit connected to the wires will drain a current if it can. If I disconnect the virtual ground, I assume that the top and bottom wires now give +3V and -3V.

    Unless I've got anything wrong there (please say!), this leads to my first question: how do you switch the power off with a setup like this? My reasoning is that it requires a two way switch that disconnects both the virtual ground and another terminal. I'm reaching this conclusion because if you disconnect the one terminal, there's a current across the other two, and if you disconnect the 3V and -3V without disconnecting the virtual ground, you have a + and - terminal connected, which although they are on different batteries could make them discharge each other.

    Is this right, and as importantly, am I thinking about this correctly? I'd be grateful for your comments as I'm keen to understand this properly.

    Secondly, I see circuits that use a few capacitors and sometimes diodes around the virtual ground. What are they for? My best guess is that they aim to keep the vg at 0V and counteract the effects of either battery discharging quicker than the other. Is that correct? How should such a circuit be set up, which what values for the caps and when does it need diodes?

    Finally, is it possible to get a virtual ground with a single battery? I tried doing this with a 9V PP3, running wires off the +4.5V, -4.5V and another wire connected to both, shown in the second picture. This didn't work at all - the battery got hot quickly and I disconnected it, I guess I was short circuiting it. Is it possible to get a virtual ground of a single battery?

    Thanks,

    Ogg
     
  2. SgtWookie

    Expert

    Jul 17, 2007
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    The left schematic is fine (except you have the battery symbols reversed; the longer horizontal line represents the more positive side of the cell)
    Same thing; as long as the wires aren't connected to anything, no current flows, so no drain occurs.

    Yes, if you have a load connected between your wires, you would need to break the circuit. It is customary to leave the ground connected, but break the connection from the "rails", eg: +3v and -3v are the rails in your case.

    Leave the virtual ground connected, but switch the +3v and -3v.
    Here's an example:
    [​IMG]

    That could take quite a while to explain. But basically, the capacitors connected between the "rails" and ground are to bypass transient peaks/dips in voltage. Capacitors pass the effects of AC, but block DC. For very simple circuits, caps may not be necessary.

    The diodes you're seeing might be Zener diodes, which are a type of voltage regulator.

    Yes, you did create a dead short across the battery. This is very hard on it.

    Here are a couple of ways to create a virtual ground:

    [​IMG]

    The circuit on the left while simple, is very inefficient, and gives poor regulation.

    The circuit on the right uses an operational amplifier as a voltage follower. The two 4.7k resistors constitute a voltage divider, splitting the battery voltage in half. The opamp actively tries to keep it's output at that same voltage..
     
  3. ogg

    Thread Starter New Member

    Sep 23, 2009
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    Thanks, it's good to get these basics straight. I like the op-amp circuit, I've seen them used to like that to maintain a constant output before.

    In this case, the virtual ground leaves the +3V of one cell connected to the -3V of the other. I don't see why they're not discharged when they're connected like this, it doesn't make intuitive sense to me. Is there something about the chemical or physical properties of the batteries that means both terminals must be connected for them to output anything?
     
  4. SgtWookie

    Expert

    Jul 17, 2007
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    As long as there is not a complete path for current across the battery connections, there will be virtually no current draw, thus virtually no drain on the battery.

    There needs to be a complete path from a negative terminal to a positive terminal for current to flow. If there is not a complete path, there will be no current flow.
     
  5. Wendy

    Moderator

    Mar 24, 2008
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  6. ogg

    Thread Starter New Member

    Sep 23, 2009
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    I do believe you(!), but I'm finding it hard to grasp what the difference is between connecting the + of one battery to it's -, which creates a short circuit, and connecting the same + to the - on another battery, resulting in no circuit.

    I guess there is an underlying point about what makes a connection between a + and - terminal a circuit, or not a circuit, that I'm not seeing here. I have the impression that all + and - terminals are equal, regardless of which battery they are on, but what you're saying implies that this isn't actually the case?

    Thanks for your patience and for the links.
     
  7. ogg

    Thread Starter New Member

    Sep 23, 2009
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    After my last post I realised I could test this experimentally. Here are my results for anyone interested.

    2xAA batteries in series, connected to a 4.7k resistor and a LED in series: the LED is lit and my multimeter reads a voltage drop of 1.8V across the LED.

    The + terminal of one AA battery connected to the resistor, LED and - terminal of a different AA battery in series: LED doesn't light, voltage drop of 0.03V across the LED.

    Testing the voltage drop across the 2 batteries in series shows 3.16V.
    Testing across the + terminal of one and the - terminal of another shows 0.03V.

    This may be self-evident to you but it was news to me! Still don't know why it's not a circuit when you use two different batteries but happy to accept that it isn't.

    Cheers all,

    Ogg
     
  8. jerrytom

    New Member

    Jan 20, 2010
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    @sgtwoki, could u tell me which software u used to stimulate the circuit ? Is any best softwares available to stimulate the circuit ?
     
  9. kingdano

    Member

    Apr 14, 2010
    377
    19

    LTSpice is the best freebie available.

    you can download it from www.linear.com/ltspice
     
  10. Nik

    Well-Known Member

    May 20, 2006
    55
    3
    "finding it hard to grasp what the difference is between connecting the + of one battery to it's -, which creates a short circuit, and connecting the same + to the - on another battery, resulting in no circuit."

    Uh, think about putting a stack of 'C' or 'D' cells into a torch: Barring self-discharge, they only supply power when you thumb the switch and make the circuit. Same applies to 6V, 9V and 12V batteries. IIRC, the term 'battery' comes from having a row of individual Voltaic cells lined up like artillery...
     
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