Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Question about LRC Circuit and Filter

  1. Apr 19, 2005 #1
    In the diagram, Z1 and Z2 are the impedence of some combinations of L, R and C. The voltage input is placed across AB, and the output voltage is measured across CD. Z1 and Z2 can be designed to make the circuit a filter for output voltage V_CD. However, if we really place a load across CD, it will have an impedence Z_CD, the whole circuit will be affected and the output behavior V_CD will change. Then, the circuit cannot act as a filter anymore if we place any load (at least not as the same filter when Z_CD is not there). How can we say in the first place that the circuit act as a filter? The "filter" only "works" in the absent of load. How useful is LRC circuit as a filter then?

    Attached Files:

  2. jcsd
  3. Apr 19, 2005 #2
    The short answer is cicuits are designed with very high Z_in(not always but quite often and for the reason you mentioned) thus when you put Z_2 in parallel with a load of Z_in you end up with:


    so as Z_in appraches infinity it has less and less effect on the filtering capability of Z_1 and Z_2.

    Try it. Just pick some numbers for Z_1, Z_2, and Z_in (these can all be real BTW). Choose a small Z_in and reiterate with increasing Z_in's and watch as Z_tot approaches Z_2.

    Alternetly, you design your filter for a specific circuit and account of the Z_CKT you are connecting your filter to.

    Or, you design your filter for a range of Z_in's.

    There are a few ways around the problem you've mentioned you just have to realize that when designing ckts you try to design value for some characteristic X but you do so knowing that every component you use will be off of its rated value by some percentage and interconnecting systems may have unknown Z's. So, real world design is more about dealing with these little problems than simply saying "I want value X exactly and if I can't have then the circuit is a failure."

    Hope this helped.
  4. Apr 21, 2005 #3
    Wouldn't it significantly limit the power transfer to the output?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook