L Network Blocking DC

Discussion in 'Homework Help' started by phil3992, Apr 20, 2014.

  1. phil3992

    Thread Starter New Member

    Apr 20, 2014
    Hi, I am currently revising for my finals. With this question all I can find is the answer of what an non blocking L network would be. The information that is given is as follows: f =10MHz, L1=0.318uH rs =50Ω & rl= 250Ω.

    The given design looks like this:


    This is what I have done to find the unblocking L network:

    L = xl/2Pif

    xl = RL/Q

    Q= Squ root of RLarger/ rsmaller - 1

    Q = 2 xl= 125 so L = 1.99uH

    From this i found C 1/2piFXL

    Making non blocking 127.32pF

    I know this is correct for non blocking but for the life of me i cannot find blocking value for C.

    I have been given the answer for this being 132.63pF but I don't understand how to get there. My lecturer explained how to do this by the following:

    1. first calculate the reactance of L1,
    2. use a capacitance C1 with the same value of the reactance of L1 to resonant L1.
    3. add the reactance of C1 calculated to the reactance of C2 calculated for the L matching network (without taking into account L1)
    4. finally calculate the final value of the total capacitance of C1 and C2. ( in farads)

    But I am struggling to understand this^^^^

    can someone please explain and show me the process off gaining the value 132.63pF for the blocking as I am totally lost?

    ( Sorry the question is long winded but I wanted to explain what I understand and what I actually need!)
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    The problem I have with the suggested approach is that it provides no insights as to the underlying analytical basis.

    Granted, once you have done the analysis yourself, there may some time economies to be gained in new problems with solutions of the same type, by applying a step-by-step computational algorithm.

    Presumably you are familiar with frequency domain circuit analysis using complex algebra. Draw the entire schematic and derive the equation for the impedance looking into the LHS of the matching network with the 250 ohm load connected. Clearly this impedance will have two unknowns L_match & C_match. But you would be expected to realize that if the matching is working correctly then the aforementioned impedance would be the complex conjugate of the effective source impedance comprising the series equivalent of 50 Ohms and 0.318uH at 10MHz.