Amp Logic

Discussion in 'The Projects Forum' started by Omega9850, Aug 29, 2011.

  1. Omega9850

    Thread Starter New Member

    Aug 26, 2011
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    Hi, I am working on a logic unit (I know they already exist, I like to do things the hard way). One thing that I am unsure about is ampere logic. A diode only permits electricity to pass in 1 direction, but will not begin conducting until the voltage across it hits a certain value. Is there something that doesn't begin conducting until the current (mA) across it hit a certain value? A transistor is one option, but the problem with that is that it can take very high amounts of energy in the collector, and will short if too much energy is applied.
    Thanks!
     
  2. SgtWookie

    Expert

    Jul 17, 2007
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    Well, you really need to explain better what it is that you're trying to do.

    This:
    does not make sense, because as soon as current starts flowing through it, it's flowing.

    Maybe you're confused about current and voltage.

    Current is a quantity of electrons moving through a conductor; a flow of electrons.
    Voltage is pressure.
    You won't have current flow without something pushing the electrons along; namely voltage.

    You can have very high voltage (pressure) without current flow, if the insulation is good enough.
     
  3. Omega9850

    Thread Starter New Member

    Aug 26, 2011
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    Thank you for replying to this post. Yes, I understand what current and voltage are, and the difference between them. What I am trying to achieve is the following.

    Two 3V 100mAh cells with a common ground. The positive terminal of each cell can be turned on or off via a switch. (each cell has its own switch). The cells are hooked in parallel so that when both switches are closed, there will be a total of 3V 200mAh. The load is a device that allows the current to pass once it has reached 110mAh. This would require both switches to be turned on. I do not want to hook up the cells in series, and I do not want to use resistors (unless it is the only option).

    Another thing to keep in mind is that this isn't what I am actually trying to achieve. It is simply a setup that replicates a small part of a project that I am working on.

    Thanks!
     
  4. wayneh

    Expert

    Sep 9, 2010
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    That's still nonsensical. Where is the current going when it's at 109mA?
     
  5. SgtWookie

    Expert

    Jul 17, 2007
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    The least expensive way to do that is to sense the voltage across a resistor using a comparator. A more expensive but low-power way to do it is to use a Hall-effect sensor to detect the current flowing in a conductor. The Hall-effect sensor detects the magnetic field building around the conductor.

    You're telling us what the mAh rating of the batteries is, but not the current flow.

    Simply connecting a pair of batteries in parallel will not cause current to flow anywhere, unless the batteries have different voltage levels; in that case one battery will try to charge the other battery, resulting in high battery temperatures and possible rupturing of the battery packages.
     
  6. Omega9850

    Thread Starter New Member

    Aug 26, 2011
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    Sorry for the late reply. I guess I am not expressing my question properly, although I appreciate the help. Below is the schematic I am attempting to make work. As you will see in the attached image, it is a simple matrix. There will be a 3V 105mA Power Supply connected to every switch. When 1 switch turns on, the power across its junction will become 3V 105mA. But when the 2nd switch on the same junction turns on, the power across the junction will become 3V 210mA. What I want to do is replace the diodes in the circuit with something. A device that everytime the power exceeds 110mA on its junction, it will offload 110mA onto the next junction. What I mean is shown in the second image. I do not want to use any resistors. I appreciate all the help.[​IMG]
    [​IMG]
     
  7. SgtWookie

    Expert

    Jul 17, 2007
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    I have edited the 1st image you linked to in order to label the connections to the switches:

    [​IMG]

    It's OK with us. Since the board is not interactive, time doesn't matter a whole lot.

    3v in reference to where?
    Do you mean that X1 through X9 have a 3v potential on them, and Y1 through Y9 have a 0v potential?
    Or X1 has 3v, X2 has 0v, X3 has 3v, etc..
    Or something completely different?

    When a switch closes, the voltage across the switch becomes nearly zero. Switches do have some small amount of resistance, so it never actually reaches 0v as long as there is some current flow through the switch - but for simplicity's sake, call it 0v for the moment.

    The current flow through the switch will then be controlled by the resistance/impedance of the load, the resistance of the switch itself, and the resistance/impedance of the supply. If the load impedance AND the source impedance are both very low, the switch will burn up.

    In the 2nd schematic, you have shown X1 and Y1 closed. If X1 and Y1 both had 3v connected to them, then no current would flow between points X1 and Y1. Since no other switches are closed, no current flow would occur in the circuit; as there is no path to a lower voltage potential (ie:ground)

    But for example, were point Y2 connected to ground, and switch Y2 were closed, then current could flow via the diode that is connected to the junction of Y1 & X1 through switch Y2 to ground. Since there is no current limiting shown, the power would be dissipated in the voltage source(s), the diode, and the three switches. The mAh rating of the current source doesn't really matter at that point, as the circuit will look pretty much like a dead short across the terminals.

    Power dissipation is voltage times current flow. If you don't have current flow, you won't have a means of detecting "available" current, or "available" mAh.
     
  8. Omega9850

    Thread Starter New Member

    Aug 26, 2011
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    Thank you for the reply SGT Wookie, but I guess this is the best way I can represent my inquiry. Below is an image...
    [​IMG]
    As you can see, I am working on a Logic Unit. I know that Logic Units already exist, but I like to do things the VERY hard way. I already have a full binary system (10Bit) developed for my project. It is compliant with standard ASCII. The main problem that I am facing is this circuit. Say a 0=0v 0mA and a 1=3V 105mA (I know the volts and mA are high for logic). When I plug a 1000000000 into one of the logic inputs, and 1000000000 into the other (so I am plugging a "1" into both logic inputs), the device that connects each logic junction (in this case, the diode) will realize that there is over 110mA flowing through that junction, and take that 110mA and carry it to the 2nd junction, so there is 100mA on the first junction, and 110mA on the second.

    Now the ouput logic I have designed a bit differently than the input. The output is sensitive so that a 0=0V 0mA and a 1=3V 100mA.
    This way, the output would be a 1100000000 (or a 2). Is there any device that does that. I am thinking of trying a Silicon Controlled Rectifier, shorting the Gate and Anode with a small capacitor. (The SCR's holding current would be 110mA).

    I would really appreciate a REAL answer rather than just suggestions, and I would REALLY appreciate no comments that I do not understand Binary. Thank you for your time!
     
  9. Kermit2

    AAC Fanatic!

    Feb 5, 2010
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    There are no 'simple' devices like diodes or resistors which will transfer a portion of the current flow when it exceeds a set value.

    There ARE simple devices called resistors which will give you a varying voltage drop with increases and decreases of current flow.

    The simplest way to do what you want would require a micro controller and several(many) current sensing transformers, or hall effect devices to translate your current levels into binary form.

    Sensing the voltage changes from a varying current passing through a precision resistance is done for a very good reason. Working with current levels as you describe your project will NOT be a simple task.
     
  10. Omega9850

    Thread Starter New Member

    Aug 26, 2011
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    Thank you for your reply, although it will be static rather than dynamic, I think a thyristor will work for what I want. Just one thing, I am not using 8Bit Binary, I am using a custom engineered version of binary. Thanks though!
     
  11. SgtWookie

    Expert

    Jul 17, 2007
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    Once a thyristor is turned on, it remains on until the current through it falls below a threshold value; usually just a few mA's.

    You might have a read through our E-book on thyristors if you haven't already done so:
    http://www.allaboutcircuits.com/vol_3/chpt_7/1.html
     
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