Tuesday, July 30, 2019

1:4 Transmission Line Transformers: Coax Impedance versus Length

I have been working on an HF Power Amplifier project that uses a Guanella 1:4 impedance transformer in its output stage to transform a load of 50 ohms down to 12.5 ohms across the drains of the push-pull FETs in the PA.



These sorts of broadband impedance transformers (Guanella, Ruthroff, and others) are created with transmission lines (usually coax, but sometimes twisted pair), and the rule for determining the transmission line's Characteristic Impedance (Zo) is that it should equal the geometric mean of the impedances at the output side and at the input side of the impedance transformer.

In other words...

Zo = √(Zin*Zout)

In the amplifier design, above, the goal is to transform an output load of 50 ohms down to 12.5 ohms, so the coax cable's Zo should be 25 ohms (i.e.  √50*12.5).

A note regarding the impedance transformation being classified as 1:4 rather than 4:1...  Some authors use 4:1 as the impedance transformation of a transformer whose impedance at its output is four times the impedance of its input (e.g. Ruthroff), while others call this a 1:4 transformer (e.g. Davis/Agarwal).  I prefer the latter classification, 1:4, which is the one I will use in this blog post.

Although this equation represents the general rule for selecting the transformer's transmission line Characteristic Impedance, it seemed to me that if the length of coax were very short compared to the wavelength of the highest frequency of the transformer's operation, then transmission line impedance would not be critical and I could use whatever coax was at hand.  (As a rule of thumb I generally consider a circuit to be "lumped element" (i.e. we can ignore transmission line effects) when distances are no more than, say, 1/20th of the shortest wavelength being used).

And so I got to wondering...how does the input impedance of a 1:4 broadband transformer, be it a Ruthroff or a Guanella implementation, vary with transmission line Zo and line length?

Ruthroff''s original article, "Some Broad-Band Transformers" contains loop-equations used to derive the input impedance of his 1:4 impedance transformer as a function of load impedance, line length, and line characteristic impedance.  Assuming a lossless line, his equation for input impedance is:

Zin = Zo*(Zload*cosβl + j*Zo*sinβl) / (2*Zo*(1 + cosβl) + j*Zload*sinβl)

(where β is the phase constant of the line and l is line length; i.e. βl is the phase delay from one end of the line to the other.)

But the source of the cosine and sine terms in the above equation were not obvious to me (although clearly they were related to delay), and, also, I couldn't find a corresponding equation for the Guanella 1:4 balun, so it was time to do some digging and deriving...


Some Fundamental Transmission Line Mathematics:

Consider an electrically long transmission line:


And let us define a voltage V1 and current I1 at its "left" end and voltage V2 and current I2 at its "right" end:


What is the relationship between the voltages and currents at the two ends of this line?

The voltage at any point on this transmission line can be considered to be the sum of a forward moving wave and a backward moving wave at that point.  That is:

V(@point) = V+(@point) + V-(@point)

where V+ is the forward moving voltage wave  (left to right) and V- is the backward moving voltage wave (right to left).

Similarly, the current at any point of this transmission line can be considered to be the difference of the forward moving current wave and the backward moving current wave at that point.

I(@point) = I+(@point)  -  I-(@point)

where I+ is the forward moving current wave and I- is the backward moving current wave.

Because the current at any point on the transmission line equals the voltage at that point divided by the transmission line's characteristic impedance, Zo, we can rewrite this last current equation as:

I(@point) = V+(@point) / Zo  -  V-(@point) / Zo

Let's now consider the voltage and current at the left-hand side of the transmission line (i.e. the side we would normally drive).  Dropping my "(@point)" subscript, I'll use the equations above to define this voltage and current to be:

V1 = V+ + V-

I1 = V+ / Zo  -  V- / Zo

We can rearrange these to become equations for V+ and V-.

V+ = (V1 + I1*Zo) / 2      (equation 1)

V- = (V1 - I1*Zo) / 2      (equation 2)

This left-hand side of the transmission line will be my reference point.

Next, let's assume that the transmission line is being driven with a sinusoidal waveform.  Therefore, at any point on the transmission line, both V+ and Vare sinusoids in the form of Vcos(ωt - φ), where φ is the phase delay from the left-hand side of the transmission line to any point on the transmission line to its right.

Given this sinusoidal form, we can simplify our calculations by using phasors (see this reference:  http://www.ee.ic.ac.uk/hp/staff/dmb/courses/ccts1/01800_LinesB.pdf) in lieu of cos(ωt-φ).  (Note:  transmission line is assumed to be lossless!)

So, using the left-hand side of the transmission line as our "reference" point, the voltage and current at the right-hand side of the transmission line (V2 and I2) can be expressed as the left-hand side voltage and current (see the equations for V1 and I1, above) with a phase-delay factor applied to them.

V2 = V+e-jβl + V-ejβl           (equation 3)

I2 = (V+/Zo)e-jβl  - (V-/Zo)ejβl         (equation 4)

Where l is the distance between the left side and the right side of the transmission line, and β is the transmission line's phase constant.

You will see a form similar to the above in the Wikipedia entry for "Telegrapher's Equation":


Note that the exponents are expressed using the term jkx rather than jβl.  The two forms are equivalent, as I will explain:

The units for βl are radians, as are the units for kx.  x and l are equivalent -- they are simply length (or distance), in meters.  And thus k and β are identical, as are their units (radians per meter).

Note also that k equals ω/u, (or, alternately, ω√LC) where ω is frequency of the driving sinusoidal signal (radians/sec) and u is the propagation speed of waves traveling within the transmission line medium (meters/sec).  And therefore u equals 1/√LC.

Note that the negative exponent represents the phase delay for a wave moving from left to right with respect to our left-hand transmission line reference point, while the positive exponent represents the phase delay for a wave moving from right to left, again with respect to our left-hand transmission line reference point.

Continuing...recall Euler's formulas...

ejβl = cos(βl) + jsin(βl)

e-jβl = cos(βl) - jsin(βl)

We can apply these to equations 3 and 4 to replace the complex exponentials with sines and cosines.  The result is:

V2 = V1*cos(βl) - j*I1*Zo*sin(βl)           (equation 5)

I2 = I1*cos(βl) - j*(V1/Zo)*sin(βl)           (equation 6)

It is important to note that we can also derive equations for V1 and I1 in terms of V2 and I2 (i.e. using the right-hand side of the transmission line as our reference point instead of the left-hand side -- note that this will flip the sign of the exponents, and thus the signs of the sine terms (Davis/Agarwal)).  The resulting two equations are:

V1 = V2*cos(βl) + j*I2*Zo*sin(βl)           (equation 7)

I1 = I2*cos(βl) + j*(V2/Zo)*sin(βl)           (equation 8)

If you refer to the Ruthroff article, you will see that these last two equations are two of the four loop equations specified in his Appendix A:


Let's use these equations to derive some input impedances!


Input Impedance of the Guanella 1:4 Impedance Transformer:

Let's first derive the input impedance of a Guanella 1:4 Impedance Transformer as a function of load impedance, transmission line characteristic impedance, and line length.

Here is a schematic representation of a Guanella 1:4 transformer.  It consists of two separate 1:1 transmission-line transformers, each wound with transmission line having a characteristic impedance of Zo and a length l.


Let's annotate this drawing with voltages and currents for both 1:1 transformers:


From inspection, it is clear that:

V1 = V3 = Vin

I2 = I4

But does V4 equal V2, and does I3 equal I1?  It would seem so from symmetry, but it's always a good idea to verify mathematically.  How do we do this?

We can write out the input voltages of the two transformers using equation 7:

V1 = V2*cos(βl) + j*I2*Zo*sin(βl)

V3 = V4*cos(βl) + j*I4*Zo*sin(βl)

Knowing that V1 = V3 and that I2 = I4, we can make these substitutions into these two equations and quickly prove that V4 = V2.

Similarly, we can express the input current of each transformer using equation 8:

I1 = I2*cos(βl) + j*(V2/Zo)*sin(βl

I3 = I4*cos(βl) + j*(V4/Zo)*sin(βl

We already know that I2 = I4 and V4 = V2, and using this information we can quickly simplify these two equations to show that I1 = I3.

So the Guanella currents and voltages simplify to the following:


Our goal is to derive an equation for Zin.  Inspecting the diagram above, we know that:
  • Zin = Vin / Iin
  • Iin = 2*I1
  • Vin  = V1
  • I2*Zload = 2*V2
Now, recall equations 5 and 6:

V2 = V1*cos(βl) - j*I1*Zo*sin(βl)         (equation 5)

I2 = I1*cos(βl) - j*(V1/Zo)*sin(βl)           (equation 6)

Substituting these two equations (for V2 and I2) into the equation I2*Zload = 2*V2 and then replacing V1 with Vin and I1 with Iin/2, the result is:

Zload*(Iin/2)*cos(βl) - j*Zload*Vin/Zo*sin(βl) = 2*Vin*cos(βl) - j*2*(Iin/2)*Zo*sin(βl)

Rearranging...

Vin*(2*cos(βl) +  j*(Zload/Zo)*sin(βl)) = (Iin/2)*(Zload*cos(βl)  + j*2*Zo*sin(βl))

Given Zin = Vin/Iin, it is a quick step to our solution:

Zin = (Zload*cos(βl)  + j*2*Zo*sin(βl)) / (4*cos(βl) +  j*2*(Zload/Zo)*sin(βl))

Here is the equation again, in a (perhaps) more readable form...


An interesting check is to see what Zin becomes when the line-length is 0 (i.e. cos = 1, sin = 0) and when the line-length is a quarter wavelength (cos = 0, sin = 1).  The results are:

1.  Line-length = 0:

Zin = Zload / 4

And so, if Zload = 200 ohms, Zin equals 50 ohms.

Note that this result does not depend upon the transmission line's characteristic impedance.

2.  Line-length = λ/4  (i.e. βl  = 90 degrees) and  Zo = 50*200):

Zin = Zo*Zo / Zload

Given Zo = 100 ohms and Zload = 200 ohms.  Zin still equals 50 ohms.


An Additional Note:  After I derived the above equation for Zin and wrote this blog post, I found the following equation in Jerry Sevick's book, Transmission Line Transformers, Fourth Edition, for a 1:4 Quanella impedance transformer:


Assuming that the imaginary term in the numerator is j*(Zload/2)*tanβl and not j*Zload/(2*tanβl), Sevick's equation 1-1 is identical to my equation for Zin of a Quanella 1:4 impedance transformer.


Guanella 1:4 Impedance Transformer implemented with transmission lines with various characteristic impedances and lengths:

So, with either a transmission line-length of 0 or a line-length of a quarter-wave, the Guanella 1:4 impedance transformer acts as a 1:4 impedance transformer, assuming the transmission line's characteristic impedance is Zo = √(Zin*Zout) .

What about transmission line-lengths between 0 and a quarter wave-length?  And what about other coax impedances?


1.  Input impedance for transmission lines of characteristic impedance that are half or double the ideal characteristic impedance:

If I define Zload to be 200 ohms, the ideal Zo is 100 ohms.  If I then use MATLAB to plot the SWR of the Guanella 1:4 Impedance Transformer's input (50 ohm reference for SWR), for coax lengths from 0 to a quarter wavelength, and for three coax characteristic impedances: Zo = 50, 100, and 200 ohms, the plots look like this:


Clearly, for Zo = 100 ohms (i.e. Zo = √(Zin*Zout)  = 50*200 ), Zin is 50 ohms irrespective of line length -- a great result!

But if I were to use, for example, a 50 ohms transmission line rather than a 100 ohm one, SWR can quickly rise.  SWR becomes 1.58:1 at a length of only 1/20th of a wavelength, and 4:1 if line length is a quarter of a wavelength!

So, if I wanted to use 50 ohm coax to wind a Guanella transformer to match a 200 ohm load to 50 ohms, I would need to make sure that the length of coax was very short!  At a length of only 3 percent of a wavelength, SWR is already about 1.3:1

Here's a plot of showing how the Zin impedance changes as the transmission line length is increased from 0 to a quarter wavelength, for non-ideal transmission lines of 50 ohm and 200 ohm characteristic impedance and a load of 200 ohms.  Note that the 50 and 200 ohm plots start in the center of the circle (Gamma = 0) when length is 0 and move away from that ideal match as the transmission line is lengthened):


In the case of my PA (see start of this blog post), in which the Guanella transformer is transforming a load impedance of 50 ohms to 12.5 ohms, if I were to wind this transformer with 50 ohms coax in lieu of the preferred 25 ohm coax, I would get the same result as the above plot for 200 ohm coax, as you can see below:



2.  Characteristic impedance with an error 10 percent or 20 percent from the ideal characteristic impedance:

Suppose the transmission line characteristic impedance is only a little off, say, for example, 10% or 20%, from our ideal characteristic impedance.  What happens?

Here are the SWR plots for transmission lines with characteristic impedances of 80, 90, 110, and 120 ohms, as well as our "ideal" characteristic impedance of 100 ohms, for transmission line lengths from 0 to a quarter wavelength:


A 10 percent error in Zo results in impedances that are still probably acceptable even if the line were a quarter wavelength long (i.e. SWR between 1.2:1 and 1.3:1).

A 20 percent error becomes more of a problem, as SWR could rise to almost 1.6:1 (although that, too, might still be acceptable, depending upon application).

Below is a Smith Chart containing plots of the 1:4 transformer's Reflection Coefficient for transmission lines with characteristic impedances of 80, 90, 110, and 120 ohms, plotted as the transmission line length increases from 0 (in which Gamma = 0 in all cases) to a length equal to a quarter wavelength.



3.  Transmission line lengths greater than a quarter wavelength:

Transmission line length was limited to 1/4 of a wavelength in the above discussion.  Suppose line length extends further?

Again, assuming lossless line, here are some plots for line lengths up to one wavelength long...

First, given a 200 ohms load, here is a plot of the magnitude of the 1:4 transformer's input impedance for transmission line characteristic impedances of 50, 100, and 200 ohms:


And SWR versus line length for transmission lines of various characteristic impedances:


And finally, the Reflection Coefficient of the transformer's Zin versus line length:


Clearly, if Zo = √(Zin*Zout), then Zin does not vary with transmission line length.


4.  Guanella 1:4 Impedance Transformer Conclusions:
  • The ideal transmission line impedance is Zo = √(Zin*Zout)
  • A 10 percent error in transmission line Zo can result in an input SWR of about 1.2:1 when the line length approaches a quarter wave.
  • A 20 percent error in Zo can result in an input SWR of about 1.6:1 when the line length approaches a quarter wave.
  • Don't use 50 ohm coax to wind a transformer whose ideal transmission line Zo is 25 ohms, unless the coax length no more than about 3 to 5 percent of a wavelength long at the highest frequency of operation.  (At 3% the SWR is about 1.3:1, while at 5% of a wavelength, SWR has risen to about 1.6:1).


Input Impedance of the Ruthroff 1:4 Impedance Transformer:

Let's look at the topology of a Ruthroff 1:4 impedance transformer:


I'll add in voltages and currents (using the same V and I labels and current directions as were used by Ruthroff):


Ruthroff, in his paper, gives an equation for Zin (his equation 7).  I rederived it myself, using the equations I introduced above, but the intermediary equations become very long and the math, although straight-forward, becomes tedious.  So rather than list all the steps in detail, I'll summarize my method:

First, from inspection of the transformer circuit above, recognize the following:
  • Zin = Vin / Iin
  • Iin = I1 + I2
  • Vin = V1
  • V1 + V2 = I2*Zload
And let's also recall the following equations, from earlier in this blog post:

V2 = V1*cos(βl) - j*I1*Zo*sin(βl)         (equation 5)

I2 = I1*cos(βl) - j*(V1/Zo)*sin(βl)           (equation 6)

Because Zin equals Vin / Iin, I want to end up with an equation in terms of Iin and V1 (because V1 = Vin).  To do this, I'm first going to create an equation expressing I1 in terms of Iin and V1 by using the equation Iin = I1 + I2 and substituting equation 6 into the I2 term.  This will give me:

I1 = Iin / (1+cos(βl)) + j*(V1/Zo)*sin(βl) / (1+cos(βl))

Next, substitute this new equation for I1 into the I1 terms of both equation 5 and equation 6.  Equation 5 becomes an equation expressing V2 in terms of Iin and V1 (let's call this equation "A"), and equation 6 becomes an equation expressing I2 in terms of Iin and V1 (let's call this equation "B").

Now take the equation "V1 + V2 = I2*Zload".  Into its V2 term substitute our new equation "A", and into its I2 term substitute our new equation "B".

We now have a very long equation that is solely in terms of Iin and V1 (the latter being equivalent to Vin).  Gather the Iin terms on one side of the equation and the V1 terms on the other side, divide to create the relationship V1/Iin (i.e. Vin/Iin), and then reduce.  (After much pencil pushing), the final equation will be:  

Zin = Zo*(Zload*cos(βl) + j*Zo*sin(βl)) / (2*Zo*(1+cos(βl)) + j*Zload*sin(βl))

And this equation is exactly Ruthroff's "equation 7" from his paper!


Again, as a check of performance, let's look at the input impedance when the length of the transmission is either 0 or a quarter wavelength.

Line-length  = 0:

The cosine terms are 1, the sine terms are 0, and the equation reduces to:

Zin = Zload / 4

Excellent!  It is a 1:4 transformer!

Line-length = λ / 4:

Phase shift is 90 degrees, so cosine terms are 0 and sine terms are 1.  The equation for Zin becomes:

Zin = Zo*(j*Zo*sin(βl) / (2*Zo + j*Zload*sin(βl))

Hmmm...Zin is now a complex impedance with real and imaginary parts no longer simply 1/4 of Zload!

So, unlike the Guanella transformer, which maintained its 1:4 impedance relationship irrespective of line length (assuming Zo was correctly chosen), the Ruthroff transformer only has an ideal 1:4 impedance transformation when the transmission line is very short, irrespective of Zo!

Let's examine this relationship more closely...


Ruthroff 1:4 Impedance Transformer implemented with transmission lines with various characteristic impedances and lengths:

Let's take a look at how the Ruthroff 1:4 impedance transformer's Zin various with transmission line length for transmission lines of various characteristic impedances.

Zload will be 200 ohms, and so the transformer should ideally present an impedance of 50 ohms at its input.

Given the 200 ohm load and the resulting 50 ohms at the transformer's input, the ideal Zo is 100 ohms.  I'll look at SWR and I will also plot Gamma on a Smith Chart for this Zo, as well as for characteristic impedances of 50 and 200 ohms (i.e. halving or doubling the ideal 100 ohm Zo), as well as characteristic impedances that differ by 10 percent and 20 percent from 100 ohms.

Here's a plot of SWR as the lengths of these seven different transmission lines are increased from 0 to a quarter wavelength:


And here's a plot of the Reflection Coefficient (i.e. Gamma), again as the transmission line lengths are increased from 0 to a quarter wavelength.  (Note that when any transmission line's length equals 0, its Gamma is 0, representing a perfect match to 50 ohms).



Transmission line lengths greater than a quarter wavelength:

Transmission line length was limited to 1/4 of a wavelength in the above discussion.  Suppose line length extends further?

Again, assuming lossless line, here are some plots for line lengths up to one wavelength long...

First, SWR for transmission line lengths from 0 to one wavelength long, for various transmission line characteristic impedances:


Note that at 1/2 of a wavelength SWR goes to infinity!

Here is a plot of the associated Reflection Coefficients:


At a transmission line length equal to 1/2 of a wavelength, the Reflection Coefficient is 1 + j0 (i.e. Zin = infinity).  You can verify this by plugging 180 degrees into the sine and cosine terms of Ruthroff's equation for Zin (i.e. sin() = 0 and cos() = -1) -- the equation's denominator goes to 0 and therefore Zin goes to infinity.


Ruthroff 1:4 Impedance Transformer Conclusions:
  • The ideal transmission line impedance is Zo = √(Zin*Zout)
  • The transformation ratio of 1:4 is only perfectly 1:4 when the transmission line length is 0!

Final Notes...

1.  Modeling versus Testing...

This post uses derived mathematical equations to show how the input impedance of both Guanella and Ruthroff 1:4 impedance transformers vary with transmission line Zo and line length.  The next step would be to verify these simulations by building actual transformers using lines of different characteristic impedances and lengths, but I have not done this.

Do not assume my conclusions are correct.

Verify for yourself!


2.  Use of cos(βl) and jsin(βl) versus cosh(γl) and sinh(γl):

In the discussion above, I followed Ruthroff's method of defining voltages and currents along a transmission line in terms of cos(βl) and jsin(βl).

Other references use the terms cosh(γl) and sinh(γl) in lieu of cos(βl) and jsin(βl).  For example, the Wikipedia entry for Telegrapher's Equations expresses the voltage and current at the input of a Transmission Line, given the voltage and current at its output, as:

V1 = V2*cosh(γl) + I2*Zo*sinh(γl)

I1 = (V2/Zo)*sinh(γl) + I2*cosh(γl)

Compare these hyperbolic trigonometric forms to equations 7 and 8 defined earlier in this blog post:

V1 = V2*cos(βl) + j*I2*Zo*sin(βl)

I1 = j*(V2/Zo)*sin(βl) + I2*cos(βl)

These two sets of equations are actually identical.

To prove this to yourself, all you need to do is prove that cosh(γl) = cos(βl) and sinh(γl) = jsin(βl), which is pretty straightforward.

First, recognize that for a lossless line, γ = jω√LC. (You can verify this by setting the loss terms in the Wikepedia post's equation for γ(s) (see the very last diagram in the Wikipedia post -- it's of the equivalent circuit -- and you will be left with the equation γ = jω√LC).

And because β = ω√LC (per earlier discussion, above), therefore γ = jβ.

Next, there are two important Hyperbolic Trigonometric Identities:

cosh(x) = cos(jx)

sinh(x) = -jsin(jx)

Using the identity for the cosh term and given γl = jβl, then cosh(γl) = cosh(jβl) = cos(j*jβl) = cos(-βl) = cos(βl).

The sinh term is similar:  sinh(γl) = sinh(jβl) = -j*sin(j*jβl) = -j*sin(-βl) = j*sin(βl).

And thus the two forms of expressing transmission line voltages and currents are equivalent.


Resources:

For further reading...

"Some Broad-band Transformers", C.L. Ruthroff, Proceedings of the IRE, August, 1959

"New Method of Impedance Matching in Radio-Frequency Circuits", G. Guanella, The Brown Boveri Review, September, 1944

"Transmission Line Transformers," Chapter Six of Radio Frequency Circuit Design, W. Alan Davis, Krishna Agarwal, John Wiley & Sonse, Inc. 2001

"A novel topology of Broad-band Coaxial impedance transformer", Centurelli, Piatella, Tommasino, Trifiletti, Proceedings of the 40th European Microwave Conference"

Phasors and Transmission Lines


And viewing...

Transmission Line YouTube Series, Professor Gregory D. Durgin, Georgia Tech.  Starting with this video:  https://www.youtube.com/watch?v=7Oz1sazpekM


Standard Caveat:

As always, I might have made a mistake in my equations, assumptions, drawings, or interpretations.  If you see anything you believe to be in error or if anything is confusing, please feel free to contact me or comment below.

And so I should add -- this information is distributed in the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.