A big question about Z match

I wonder why all the authors of the books give examples only for impedance matching a small source impedance (Rs) with a large load impedance (RL)
Never find the contrary, i.e matching a big source impedance with small load impedance which the case that we face when designing a radio transmitter where a large source impedance and small load impedance (50 ohm antenna) are connected to a resonance LC tank.

One of the most famous examples of this is the book: RF circuit design by Chris Bowick
In Chapter 2 page 39: Example 2-4





I try it using multisim10 and got excellent results, but when replaced RS and RL like this:

I got the following

Is there a way to correct it
Thank you

Hello,

The -6 dB is exactly what it should be.

Your "in" measures the EMF (electromotive force) of the generator. You have a 2 kOhms output impedance (your R1). The network + 50 Ohms load shows also 2 kOhms (as you calculated) so at the "out" of your bode plotter instrument appears half the EMF, and that equals -6 dB. See it as having a voltage divider network with same resistors.

WimRFP said:

Hello,

The -6 dB is exactly what it should be.

Your "in" measures the EMF (electromotive force) of the generator. You have a 2 kOhms output impedance (your R1). The network + 50 Ohms load shows also 2 kOhms (as you calculated) so at the "out" of your bode plotter instrument appears half the EMF, and that equals -6 dB. See it as having a voltage divider network with same resistors.

Click to expand...

Thank you for fast reply,,, I understand you
But why I got a +8dB @ 100MHz when Rs = 50 ohm and RL = 2K ohm?
thanks

That is because of the impedance transform. 10mW into 50 Ohms is 0.7Vrms, but 10mW into 2 kOhms is 4.5 V (16 dB more voltage). I assume that you measure across the 2 kOhms load resistor.

Because of you lose 6 dB (the EMF issue but now when R1= 50 Ohms), the result should be +10 dB.

If you measure at both input and output of the matching network you should measure + 16 dB.

I can't explain your + 8 dB.

yes, I repeated the simulation with R1 (R source) = 50 ohm and R2 (R load) = 2Kohm, I got the +10dB gain
When make R1 = 2K and R2=50 I got -6dB loss
I understand that imedance matching networks did not amplify signals
I think they only allow some frequency to pass attenuating others to some degree

can you please explain from where come that amplification of 10dB?
thanks

I think we have some confusion about the measuring setup. The setups for both situations aren't equal when looking to your numbers.

Your bode plotter just shows 20*log(Vout/Vin), so no power is involved, just voltage.

yes I think so too
Let's start again,, in post #1 examle 2-4, the outher explain that the LC resonance circuit see's the Rs and RL in parallel.
when one of them is small, then the equivalent resistance is smaller than the smallest one and this will load the tank circuit.
He suggested the use a capacitive transformer to match a low Rs to a big RL as a solution for the mismatch.
I experimented the results of Example 2-4 and it was was excellent (+10 dB)
My question is how the solution will be if Rs =2K and RL = 50 ohm

As the author of the book is always used in his illustrative examples, small Rs and large RL.
I had a sense that he can not match impedance except in the case of small Rs with large RL and not vice versa.
thank you WimRFP

When you look into the capacitors from the left side, you can only transform to a higher impedance. When you look from the right side towards the inductor, you can only transform to a lower impedance. So when using the design formulas, the impedance on the left side must always be lower then the impedance on the right side. It doesn't matter on which side the source is (so you can put it on the right side also).

Similar situation you will encounter with the "L matching netwok" containing only one L and one C. When you look into the series component, you can only transform to a higher impedance. So when you need to transform to a lower impedance, the source needs to look into the parallel component (component that goes to ground).

In the past (when using vacuum tubes) you mostly needed to transform a high anode impedance (up to kOhm range) to 50 Ohms. Now with low voltage RF power applications with BJT and MOSFET, you need to transform some Ohms to 50 Ohms.