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Negligible resistors?

  1. Mar 19, 2009 #1


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    Hi all,

    In the circuit 1, R1 is ever, at least, 100000 times lower than R2.
    Is it possible to assume the second circuit is then valid?

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  2. jcsd
  3. Mar 20, 2009 #2
    Both circuits can be built, so they are "valid" circuits. The circuit on the left is an RC transmission (delay) line with leaky capacitors, while the circuit on the right is a bunch of resistors and capacitors in parallel.
  4. Mar 20, 2009 #3
    what is the second circuit for, DC analysis?
  5. Mar 21, 2009 #4
    Both are a bunch or parts connected together. But is either one a circuit?
  6. Mar 21, 2009 #5


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    Maybe the explanation wasn't good enough.
    Since R1<<R2 is it possible to transform the circuit on the left to the one at the right?
  7. Mar 21, 2009 #6


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    So what you were really asking is whether the second circuit is valid as an approximate equivalent of the first. Even that question is a little vague. Equivalent in terms of input impedance or equivalent in terms of some two-port parameter like vout/vin or whatever (though no vout or vin are labelled on the diagram).

    In general though it will be a good approximation at sufficiently low frequencies but very much a poor approximation at frequencies starting somewhere about [itex]\frac{1}{2 \pi R_1 C_1}[/itex] and higher.
  8. Mar 21, 2009 #7


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    Exactly. It is a huge simplification but it may help to compute the resulting (parallel/serial) resistance and capacitor.
  9. Mar 21, 2009 #8
    It really depends on what you're trying to do with the circuit. In many ways it is the [tex]R_2[/tex] that is irrelevant (essentially an open in many cases).

    It's late, but, it seems like you are modeling some sort of delay or transmission line. The voltage transfer function is just 3 voltage dividers back to back, giving (check my math on this): [tex]\frac{1}{6 \ast \left(s \ast R_1 \ast C_1 +1 +\frac{R_1}{R_2}\right)^3}[/tex] with [tex]R_1<<R_2[/tex], this reduces to [tex]\frac{1}{6 \ast \left(s \ast R_1 \ast C_1 +1 \right)^3}[/tex] giving you 3 poles close to [tex]-\frac{1}{R_1 \ast C_1}[/tex].

    From the transfer function, you know its, ac response, and it's impulse and step responses. You can even do the root-locus if you really wanted.

    This is assuming the load impedance is essantially an open.
    Last edited: Mar 22, 2009
  10. Mar 21, 2009 #9
    Why dont you calculate both, compare results and post us in :)
  11. Mar 21, 2009 #10
    Look up "RC transmission line" on the web. I got several thousand hits. It is a special form of lossy delay line that has dispersion.
  12. Mar 23, 2009 #11


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    That is good idea and I'll simulate these circuits.
    I use a demo version of http://www.spectrum-soft.com/index.shtm" [Broken]
    I'll take a look at these hits.
    Last edited by a moderator: May 4, 2017
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