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Quarter Wave Transformer: Understanding Formula 
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#1
Aug2413, 11:20 AM

P: 22

Hi,
The quarter wave transformer, as I understand, is supposed to allow matching an input impedance to an output impedance. The formula pertaining to a quarter wave transformer transmission line is $$Z_\mathit{in} = \frac{Z_0^2}{Z_L}$$ I thought that the condition for a line being matched to the load was [itex] Z_L = Z_0 [/itex], so then wouldn't $$Z_\mathit{in} = \frac{Z_L^2}{Z_L} = Z_L$$ implying that the characteristic impedance has nothing to do with the match? I don't understand what I'm missing here! I don't see how the matching criteria that is in the book ([itex] Z_L = Z_0 [/itex]) can be met with the quarter wave line. Thanks. 


#2
Aug2413, 01:02 PM

P: 1,810

Generally you want to match a specific output impedance to a specific impedance and need to select the correct Z_0 for the match. The formula to find Z_0 is sqrt(Z_in*Z_out).
So for instance if you want to match Z_in = 50 to Z_out = 100, you would plug in the two values into the formula and get a Z_0 of 70.7 ohms. Your misunderstanding is that an input or output will always see the impedance of the line. This is true only if both the input and outputs are equal to the line impedance. When the input and out impedances are different, the impedance along a line that satisfies the formula will vary as a raised sinewave from one value to the other, with the minimum value of the sinewave equal to the lower impedance and the maximum value equal to the higher impedance. So with a 70.7 ohm line, the input will see 50 ohms and the output will see 100 ohms. What will happen if the impedances are swapped and a 100 ohm input and 50 ohm output are connected in place of the 50 and 100 ohm values? The transmission line will convert the 100 ohm input to the 50 ohm output. 


#3
Aug2413, 05:22 PM

Sci Advisor
PF Gold
P: 2,603

A 1/4 wave matching section is usually a 1:1 BALUN .... its purpose is more usually to match the unbalanced coax to the balanced input of an antenna. Being a 1:1 balun there is no impedance transformation. The secondary purpose of this balun is that it stops RF current flowing back down the braid of the coax and it then becoming a radiator.
hope these thoughts help Dave 


#4
Aug2413, 10:29 PM

P: 22

Quarter Wave Transformer: Understanding Formula
Okay, so basically it's a quarter wave because in a quarter wave, the impedance goes from a minimum (at, say, the beginning of the line) to a maximum (at, say, the end of the line) and so basically the sine wave (representing the impedance) can then be scaled to meet the requirements of input and output impedance.
What was confusing me was that they have Z_L in the formula when really, Z_L is not the load impedance but rather, the impedance that is seen when looking from the load into the impedance line. 


#5
Aug2513, 07:55 AM

P: 1,810

Your formula $$Z_\mathit{in} = \frac{Z_0^2}{Z_L}$$ is the same formula as the one I gave you.
This is much easier to understand with a Smith Chart. Smith chart  Wikipedia, the free encyclopedia On a Smith Chart the center point is always the characteristic impedance Z_0. All values are normalized to Z_0 = 1 so they all must be divided by Z_0 for the Smith Chart. That makes Z_in on the Smith Chart equal to 0.707 and Z_L = 1.414 on the Smith Chart. Now draw a circle, centered at 1 with the circumference intersecting Z_in and Z_L. One complete revolution around that circle is equal to 1/2 electrical wavelength of a transmission line. A half revolution equals a 1/4 wavelength of the transmission line. This shows how the impedance transforms from one value to another. Only values on the horizontal centerline are real, all others are complex. This means that a transmission line can also be used to match any two impedances, real or complex. Z_L is the load impedance and also the impedance of the transmission line at the load when properly designed. 


#6
Aug2613, 08:08 PM

Sci Advisor
P: 4,030

This behaviour of quarter wave lines is used a lot in antenna design.
For example, if you want to put two 50 ohm antennas in parallel, you could step the impedances up to 100 ohms with 70 ohm quarter wave lines then put these ends in parallel to get 50 ohms again. The disadvantage is that the relationship only applies exactly at one frequency. Also, there are only a few types of cable readily available, so getting one you calculate can be a problem. 


#7
Aug2813, 08:57 AM

Sci Advisor
Thanks
PF Gold
P: 12,135

You actually want two mismatches  one at each end of your matching line. The effect is to give reflections both at the load and at the input to the matching line. These reflections will cancel so no signal returns to the source. (That's the same effect as you get with Blooming on a lens, to reduce reflections) 


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