# What is actually transformed in quarter wave transformer?

I'm reviewing some stuff on quarter wave transformer for a transmission line but since I have not had any experienced in handling these type of stuff, I am struggling to understand this concept intuitively

http://www.ittc.ku.edu/~jstiles/723/handouts/The%20Quarter%20Wave%20Transformer.pdf

What I don't get is that all the calculation involved in this "quarter wavelength transformer" is to calculate the input impedance Zin at a quarter wave length away from the load. What is the point of that?

What is actually done in the field to change the transmission line network into a matched network using quarter length transformer?

• I don't understand your question - I understand the title of your question but not the last sentence in your question. To answer the title - it is an impedance that is trasformed into a different impedance in a quarter wave transformer. Sep 30, 2014 at 17:53
• but what do you mean by transformed? to me nothing is transformed because Zload is still Zload, and Zin is but what you would measure quarter wave length away, nothing physically happend to the circuit by computing the input impedance else where Sep 30, 2014 at 18:05
• question is what you do after getting the input impedance a quarter away? Sep 30, 2014 at 18:06
• Perhaps what you are missing is that the 1/4 wavelength piece has a different impedance than the input piece of coax. (transmission line) So for instance with 75 ohm input coax and 1/4 wavlenght of 50 ohm you could match to a 33 ohm load. (at one frequency.) This is a nice trick, but I don't know if it's used much in practice. One thing that is done is 1/4 wavelength AR coatings in optics. Same idea. Sep 30, 2014 at 18:07
• So the way this works is due to an interference effect. The impedance mis-matches cause some reflection of power. With just the right length of line the reflections have the right phase to cancel on the input side. I was looking for a good picture on the web but couldn't find one. Sep 30, 2014 at 18:19

This is the formula that is interesting: -

$Z_{IN} = \dfrac{Z_1^2}{R_L}$

As George has "said" in a comment - if you have a 75 ohm system feed and you want to deliver the full power to a 33 ohm load, use a quarter wavelength of 50 ohm cable and the formula numerically becomes: -

$Z_{IN} = \dfrac{50^2}{33}$ = 75.75 ohms. (Near enough)

• so the 75 ohm line is connected to a 50 ohm cable that is connected to a 33 ohm load and this will deliver full load? why not use an appropriate system feed to begin with to avoid this matching business? Sep 30, 2014 at 18:15
• @IllegalImmigrant why not indeed! It's a theoretical mechanism for transforming impedances and, like George said in his comment, it's not really used much but having said that I hope someone reminds us of an application where it is. Sep 30, 2014 at 18:17
• I don't know if it's used much in electronics. The idea is used in all sorts of other applications. In ultra sonics, optics, maybe even acoustics. There are always materials with different "impedances" and you want to couple the most amount of power from one to the other. Sep 30, 2014 at 18:23
• You would think by adding on more line, your circuit will be forced to produce more power. Sep 30, 2014 at 19:18
• When you look into the math, no it doesn't work like that. The power delivered to the line is dictated by the characteristic impedance of said line no matter how long it is, even if infinitely long. Sep 30, 2014 at 22:49

So once place impedance transformers get used a lot is in matching networks for amplifier circuits.

Say you want to amplify a 20 MHz signal on a 75 Ohm transmission line, but the laws of physics dictate that the amplifier you can make has an input impedance of 33 Ohms. In that case you want to amplify the signal without reflecting all of your power you put a section of 50 ohm transmission line that has an electrical length corresponding to 1/4 a 20 MHz mode in between your 75 Ohm transmission line and the amplifier. This guarantees your 20 MHz signal will not be reflected and you'll amplify the signal with good fidelity.