parallel resistance Rs of a current source

Discussion in 'General Electronics Chat' started by PG1995, Sep 29, 2011.

  1. PG1995

    Thread Starter Active Member

    Apr 15, 2011
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    Hi [​IMG]

    I have always thought of a current source as a very, very high voltage supply with a very, very large resistor, Rs, in series (I mean this large resistor would make series connection with the load resistor, RL). The larger the series resistor, Rs, the less effect the change in load resistor has on the load current. The Rs should be a variable resistor which is adjusted to get the wanted current through the RL. In other words, a current source is also a special type of voltage source.

    In a regular voltage source the Rs is very small which can be adjusted to vary the voltage across the RL. If you want more volts to appear across the RL then you need to decrease the resistance of Rs.

    Please correct the stuff above if you deem something wrong. Thanks

    Now I'm coming to the main question(s).

    Please have a look on the linked diagram (or attached one):
    http://img843.imageshack.us/img843/3518/imgan.jpg

    A current source is represented as shown in Fig. 3 in the linked and attached diagram with the Rs in parallel configuration. I don't get it. Why? As I say above, in my view, a current source is a special kind of voltage source. They say, Rs, is an internal resistance and makes a parallel connection (as shown in Fig. 3). I don't get it. Do you get where I'm having problem? Please help me. Thanks

    Regards
    PG
     
  2. #12

    Expert

    Nov 30, 2010
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    Here's a guess...the symbol for "current source" implies a perfect current source with therefore infinite resistance. Real current sources are not perfect, so the Rs in your third drawing might be a reminder of that imperfection. It certainly is not in series with the load, so the designation as Rs seems improper.
     
  3. oldtech33709

    New Member

    Sep 24, 2011
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    The s is for source.
     
  4. Adjuster

    Well-Known Member

    Dec 26, 2010
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    Quite simply, a series combination of an ideal voltage source VTh and a resistor RTh (Thevenin circuit) can be replaced by an equivalent consisting of a parallel combination of an ideal current source IN and a resistor RN (Norton circuit).

    Given either circuit to start with, you can set up the opposite type of circuit to behave identically simply by setting the resistance to the same value ( RN = RTh ) and then calculate the source value according to IN = VTh/RTh , or VTh = IN*RN.

    Mathematically, these circuits do the same thing: there is no difference in the voltage current relationship at the terminals, so if you had one or other circuit inside a black box you would be unable to distinguish which type it was.

    A number of more complicated circuit models include the current source and parallel resistor combination, notably transistor equivalent circuits, so you had better get used to the idea.
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    For the two forms of the circuit shown in figures (1) & (3) to be equivalent they need to present the same terminal conditions according the the load applied at terminals a-b.

    When a-b is shorted then the current Is (=Vs/Rs) flows in the short. This will be true in either representation whether you look at figure (1) or (3). What about the case when a-b is an open circuit? In figure (1) you would see a voltage Vs with a-b open. In figure (3) you also see the same voltage Vs when a-b is open. Hence the two are equivalent for the extreme load boundary conditions. If you leave out the Rs from figure (3) the two are then not equivalent.

    What you have actually drawn in going from figure (1) to figure (3) is the Thevenin to Norton transformation.

    If you have a truly ideal current source then the open circuit voltage will then notionally be ∞ V.

    You can of course design practical current sources that involve quite small internal voltages that are nowhere near infinite. As long as they behave as close to an ideal source over their nominal [V-I] range of operation then the design goal has been achieved.
     
  6. PG1995

    Thread Starter Active Member

    Apr 15, 2011
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    Thank you, everyone, in particular Adjuster, t_n_k.

    I think these current and voltage sources as the ones in Fig.1 and Fig.3 in the attached diagram is nothing more than a mathematical construct. They have little to do with real voltage and current sources. What do you say? Please let me know. Thanks
     
  7. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    One can generalize even further. A mathematical model of a physical device or system is just that - a model or a mathematical construct. However, without those models we wouldn't get very far in the analysis, synthesis & design / improvement process. Mechanical engineers model mechanical systems then go on to build machines. Electronics engineers model electronic systems ....blah blah. Presumably, in the absence of the design / modelling process, we'd be stuck with a cumbersome trial & error approach as the alternative.

    When you say that figure (1) for instance, has very little to do with a real voltage source, you are correct in one sense. It is however (albeit quite simple) a useful means of representing the behavior of a real voltage source. When we move on to models of something a little more complex, like an inductor, then we have to include more possibilities such as parasitic resistance and self-capacitance. Then we may need to go further to allow for non-linearity or high-frequency phenomena such as skin effect. And so on it goes.

    The goal of any model is to provide one with a basis for the prediction of actual physical behavior. As we improve our models we hopefully improve the quality of our design predictions and the resulting devices & systems we build.
     
    PG1995 likes this.
  8. PG1995

    Thread Starter Active Member

    Apr 15, 2011
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    Many, many thanks, t_n_k. I'm much obliged for the help.

    Actually when I was writing the previous posting I was also thinking about something else. Just take KVL as an example. How did Kirchhoff come up with his law? Did he arrive at the result mathematically? Or, was it using lab apparatus? Or, perhaps the combination of both? What is your opinion? Please let me know. Thanks.
     
  9. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Check out this paper ....
     
  10. PG1995

    Thread Starter Active Member

    Apr 15, 2011
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    Thank you, t_n_k.

    I have been through the attached document. I think the development of mathematical models, theories, etc. such as Ohm's law, KVL, Thevenin equivalent circuit, was driven by experimentation and observation. I mean Kirchhoff didn't just sit there and formulated his laws. He simply translated his observations into mathematical language. Am I thinking along the right lines? Please let me know. Many thanks.

    Regards
    PG
     
  11. #12

    Expert

    Nov 30, 2010
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    Yes. Experimentation goes hand in hand with developing a theory. Electrons are never wrong. They always obey the laws of physics. If you're going to have a theory, you must be able to get the electrons to cooperate. If they don't, it is you that is wrong. If they do as you predicted, then the experiment is repeatable by other people, and that means it is useful.
     
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