# L Matching Network (Help Needed)

Discussion in 'General Electronics Chat' started by phil3992, Apr 23, 2014.

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1. ### phil3992 Thread Starter New Member

Apr 20, 2014
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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!)