Are nodes with the same voltage considered to be the same node?

Discussion in 'Homework Help' started by lll, Jan 28, 2013.

  1. lll

    Thread Starter New Member

    Mar 7, 2012
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    From my understanding, all grounded nodes are connected to each other. Can the same thing be said for all the black triangles that are -5.7 volts in the picture?
     
  2. tshuck

    Well-Known Member

    Oct 18, 2012
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    Yes, the downward pointing arrow typically denotes a negative voltage, as this example is showing...
     
  3. WBahn

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    Mar 31, 2012
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    And, similarly, all the upward pointing arrows are at +3.7V.

    If there were two positive supplies, say +5V and +15V, then either a different type of arrow could be used for each, or each could have a label, such as VccD and VccA.
     
  4. bountyhunter

    Well-Known Member

    Sep 7, 2009
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    In basic circuit analysis, equipotential nodes can be strapped together.
     
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  5. crutschow

    Expert

    Mar 14, 2008
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    But two separate nodes at the same voltage does not necessarily mean they are (or should be connected as) the same node. If one was the circuit supply 5V power and the other was a stable 5V reference voltage then you would not want to tie them together as bad things would likely happen.
     
  6. WBahn

    Moderator

    Mar 31, 2012
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    This probably needs to be qualified somewhat. For instance, the inputs of an opamp in the active region are generally considered to be equipotential. But, even from an analysis standpoint, they can't be strapped together because current cannot flow from one to the other.
     
  7. bountyhunter

    Well-Known Member

    Sep 7, 2009
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    If they are actually equal voltage (and will stay the same), nothing bad can possibly happen.... but if they do not stay the same voltage for whatever reason, it might. Two nodes of the same voltage can not cause a current flow across them... if any current flows, one is not the same voltage as the other.
     
  8. bountyhunter

    Well-Known Member

    Sep 7, 2009
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    What I said is absolutely correct. Two equipotential voltage nodes can be analyzed as the same nodes strapped together. Op amp inputs are not because in nearly all operation modes, they will not be the same voltage:

    however, if they are actually the same (held at DC voltage) and are the exact same voltage, they can be shorted together with no change. In reality, no real op amp would qualify because there is some input offset voltage.
     
  9. bountyhunter

    Well-Known Member

    Sep 7, 2009
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    Then they are not equipotential if current flows from one to the other, since any current flow can only occur with some electromotive force (voltage) driving it.
     
  10. MrChips

    Moderator

    Oct 2, 2009
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    Suppose you have a non-inverting unity gain buffer amplifier where the output is connected to the inverting input. By definition the output voltage follows the input voltage. Clearly, you cannot tie the non-inverting input to the output.
     
  11. Georacer

    Moderator

    Nov 25, 2009
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    Especially for OpAmp circuits, I too, disagree with considering the OpAmp inputs as the same node.

    In simple OpAmp feedback circuit analysis we do make the assumption that the input terminals have exactly the same voltage (virtual short). But they must never be connected with each other during the circuit analysis, since then the feedback current would have an extra way towards the voltage reference, which ruins the whole circuit.
     
  12. WBahn

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    Mar 31, 2012
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    Not true. The simplest example is a superconducing magnet in persistence mode.

    But let's forget about that and look at the consequences of this claim. By this reasoning, it would also be the case that you can cut any equipotential node at any arbitrary place since, if no current is flowing at the place you make the cut, then you haven't made any change that can have any effect.

    So consider a battery (on the left) and two resistors (center and right) all connected in parallel by a wire on top and a wire on the bottom. Let's call the top of the battery Node A, the top center resistor Node B, and the top of the right resistor Node C. These are equipotential nodes and so, by your claim, there is no current flowing between Node A and Node B or between Node B and Node C. Now, before you claim that there is always SOME voltage drop from Node A to Node B to Node C, consider that this then, by the same reasoning that you used to say that the opamp inputs can't be connected because they aren't really equipotential (and the voltage drop across a wire can easily exceed the voltage offset between opamp inputs), they are not equipotential nodes and therefore can't be connected. But even more to the point, you can't make that claim unless you are also going to claim that no current could flow between those nodes if we were to find some ideal wire that had no resistance (say, use a superconductor).
     
  13. bountyhunter

    Well-Known Member

    Sep 7, 2009
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    man, you are making something simple really complicated:

    You can always connect two points of equipotential with no change. If there is a change, by definition, that proves they were NOT equipotential or were not equipotential under all ccircumstances.

    The whole construct of nodal analysis is built on the fundamental step of selecting one potential as a reference voltage and expressing all the other nodes in relation to it. Any other node at the same voltage is, by definition, that same reference node and is labeled such. It can be any voltage selected as the reference.

    As for op amp inputs, references vs batteries, etc.... if they are the same voltage (and do not change) then they can be strapped together. That's why op amp inputs (in general) can't be.
     
    Last edited: Jan 31, 2013
  14. WBahn

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    IF no charge is flowing between two points on a conductor THEN those two points are at the same potential.

    Fine. True statement.

    However, the converse fallacy applies, namely "If A then B" does NOT imply "If B then A". You are drawing conclusions based on a claim that IF two points on a conductor are at the same potential THEN no current is flowing between those two points. This is not a true statement.
     
  15. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    I find it easier to think of op amps as producing an output that will make the inputs equal, rather than assuming the inputs are equal.

    For voltages to be considered on the same net, they need to be connected. Universal symbols, such as those in the OP, are perfectly OK to assume as a net (wire). If you have two 12V 20A switching supplies, making two 12V rails (computer PSU), they can't be considered equal/same net, as putting a wire between them will destabilize regulation.

    I'm sure most of us know when it is ok to call two points at the same potential the same net, and when not to. Students just learning may think that is a rule (like op amp inputs always being equal), resulting in possible confusion later on.
     
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