# Impedance Matching (Smith Chart)

Discussion in 'Homework Help' started by 60remember, Dec 10, 2013.

1. ### 60remember Thread Starter New Member

Jul 6, 2013
4
0
Hello,

I'm really struggling to understand this one simple thing, would be great if someone could help me:

Given is a simple circuit that consists of a generator and a load. The generator's internal resistance is $Z_G = (20 + j30)\Omega$, the load $Z_L = (40 - j30)\Omega$.

I understand that I have to transform one of them to achieve $Z_L = Z_G*$. According to the sample solution, I am supposed to add an LC circuit to achieve that. Why can't I use a simple parallel resistor? According to the solution, it is impossible to reduce an impedance's real part by adding passive elements. I don't know if I'm missing something really simple here, but why it that? Isn't that exactly what you do when adding a parallel resistor?

60remember

2. ### jegues Well-Known Member

Sep 13, 2010
735
45
If you parallel your load with a resistor you'll drop both the real and imaginary parts.

3. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
Presumably you want a match with the lowest achievable loss. Adding a resistor unnecessarily adds to the losses. Plus how could you obtain the necessary conjugate reactance by just adding a parallel resistance to the load?

At its simplest solution - you can use either a combination two element shunt C plus series L network or a series C plus shunt L (in that order for the signal path from source to load) network between the source and load. You need to specify a frequency if you want the actual L & C values rather than the reactances.

If you do this on the Smith Chart you will at least need to know how to plot the load and generator impedances on the chart. I'd suggest using a normalized base for your chart Zo and then normalize the actual impedances for plotting. Then it's a matter of knowing how suitably placed inductive and capacitive reactances between the source and load allow you to "move" between the load impedance and the conjugate of the source impedance.
Having found the necessary reactances you then transform these back to actual values using your base Zo. From there you need to know the frequency to find the actual values of inductance and capacitance.

As a hint you will probably find the shunt L plus series C option is the easiest to resolve using the Smith chart.

Last edited: Dec 11, 2013
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4. ### 60remember Thread Starter New Member

Jul 6, 2013
4
0
Thank you, that really helps!