Technical Article

# Analyzing and Improving the Ruthroff Transformer

April 17, 2024 by Dr. Steve Arar

## We analyze a Ruthroff 1:4 transformer's performance at high frequencies, then learn how to improve its bandwidth using equal-delay networks and how to redesign the circuit for higher impedance transformation ratios.

The previous article in this series introduced two versions of the Ruthroff 1:4 transformer—one unbalanced-to-unbalanced, the other a balun—with simplified analyses that explained their low frequency operation. At high frequencies, however, the method of analysis we used isn’t applicable. Instead, we need to use the transmission line equations to gain a thorough understanding of the circuit’s behavior.

In this article, we’ll examine a more rigorous analysis of the Ruthroff 1:4 transformer. We’ll then use the results of the analysis to address a major shortcoming of these circuits when compared to Guanella transformers—namely, their lower bandwidth. After we learn how to build Ruthroff transformers with improved bandwidth, we’ll learn how to build configurations with higher transformation ratios.

### Analyzing the Ruthroff 1:4 Transformer

Figure 1 shows a Ruthroff 1:4 impedance transformer.

##### Figure 1. A Ruthroff 1:4 transformer.

At low frequencies of operation, we can assume that the transmission line transformer behaves like a magnetically coupled transformer and use the lumped inductor method to analyze it. In this way, we can quickly determine the impedance transformation ratio of the circuit. This analysis doesn’t, however, reveal any limitations that might be imposed by non-idealities such as the line’s non-zero delay.

To address these non-idealities, we need to conduct a transmission line analysis. Assuming that the transmission line is lossless, its ABCD representation describes the line response by the following equations:

$$V_1 ~=~ \cos(\beta l) V_2 ~+~jZ_0 \sin(\beta l) I_2$$

##### Equation 1.

$$I_1 ~=~j\frac{\sin(\beta l)}{Z_0} V_2 ~+~ \cos(\beta l) I_2$$

##### Equation 2.

where Z0, β, and l respectively denote the characteristic impedance, phase constant, and length of the line.

Kirchhoff’s voltage law gives us two additional equations:

$$V_S ~=~ R_S(I_1 ~+~ I_2) ~+~ V_1$$

##### Equation 3.

$$V_{out} ~=~ R_L I_2 ~=~ V_1 ~+~ V_2$$

##### Equation 4.

If you can’t immediately derive Equation 4, note that V1 appears across the lower winding of the transformer in Figure 1.

The above four equations contain four unknowns: V1, V2, I1, and I2. After solving for the output current, we can calculate the power delivered to the load:

$$P_{out} ~=~ \frac{1}{2}R_L |I_2|^2$$

##### Equation 5.

We can now examine how Pout changes with different parameters to find the optimum operating conditions. The results of this mathematical analysis are provided below. For the sake of brevity, we won’t walk through the intermediate mathematical steps.

### The Results of the Analysis

The analysis leads to three key insights. First, if we take the derivative of Pout with respect to Z0 and set the result equal to zero, we obtain:

$$Z_0 ~=~ \sqrt{R_S R_L}$$

##### Equation 6.

which specifies the characteristic impedance for maximum power transmission. This result is independent of the line’s length.

Second, the length of the line is constrained—an important limitation that wasn’t revealed by the simpler analysis we performed in the previous article. By taking the derivative of Pout with respect to RL, we find that a load impedance of RL = 4RS maximizes the output power. To obtain this, the length of the line must be small enough that βl ≪ 1.

Instead of the product of the phase constant (β) and length (l), we can consider βl as a single term that represents the electrical length of the line. This brings us to the third important result of the analysis—namely, that Pout changes with the line’s electrical length. The output power normalized to the power available from the source (PAVS) is given by:

$$\frac{P_{out}}{P_{AVS}}~=~\frac{4 \big (1~+~ \cos(\beta l) \big )^2} {5 \Big ( 1~+~1.2 \cos(\beta l)~+~ \cos^2(\beta l) \Big )}$$

##### Equation 7.

This function is plotted in Figure 2.

##### Figure 2. Variation of Pout with the line’s electrical length.

The output power decreases as βl gets larger—but why? The answer lies in Equation 4, which tells us that the total output voltage is the sum of the voltages at the transmission line’s input and output ports (Vout = V1 + V2).

The voltage applied to the input of the transmission line (V1) travels along the line to produce the line’s output voltage (V2). This transmission delay is represented by a phase shift between V1 and V2. The phase shift depends on the electrical length of the line.

As the phase shift approaches 180 degrees, the two voltages interfere destructively, reducing the magnitude of the output voltage. This is consistent with the curve in Figure 2, which shows zero output at βl = 180 degrees.

Provided that βl ≪ 180 degrees or, equivalently, that the physical length of the line is considerably less than half the wavelength (l ≪ λ/2), the transformer operates as expected. As a rule of thumb, we can avoid this destructive interference by keeping l < λ/10. If more precision is needed, we can use Equation 7 to calculate the exact value of the circuit's insertion loss for a given design.

### Example: Calculating the Line Length Based on Maximum Frequency

Let’s say that a coaxial cable is used to build a Ruthroff 1:4 transformer. Assume the following:

• The maximum frequency of operation is 100 MHz.
• The velocity factor of the coax cable is 0.7.

What’s the maximum cable length we can use if we want to keep the destructive interference below 0.5 dB?

To help us solve this problem more easily, Figure 3 reproduces the power curve in Figure 2 using a logarithmic scale for the y-axis.

##### Figure 3. Variation of Pout with the line’s electrical length. The y-axis uses a logarithmic scale.

If we want the insertion loss to be less than 0.5 dB, the electrical length shouldn’t exceed 80 degrees. Noting that the phase constant (β) is given by:

$$\beta ~=~ \frac{2 \pi}{ \lambda }$$

##### Equation 8.

The length of the line should satisfy:

$$l ~<~ 0.22 \lambda_{min}$$

##### Equation 9.

where λmin is the minimum wavelength inside the cable.

We find λmin by multiplying the cable’s velocity factor by the wavelength in free space. The velocity factor is defined as the ratio of the speed of the electromagnetic wave in the cable to the speed of light in vacuum. In our example, the velocity factor is given as 0.7, producing the following equation:

$$\lambda_{min} ~=~ 0.7 \lambda_{0}~=~ 0.7 ~\times~ \frac{c}{f} ~=~0.7 ~\times~ \frac{3~\times~ 10^8}{100 ~\times~ 10^6}~=~2.1 \; \text{m}$$

##### Equation 10.

where:

λ0 is the wavelength in free space, equal to $$\frac{c}{f}$$

f is the frequency of the signal

c is the speed of light in vacuum.

Note that the velocity factor of a coax cable depends on the dielectric material that separates the cable’s inner and outer conductors.

Using Equations 9 and 10, we find that the cable length shouldn’t exceed 46 cm to keep the insertion loss below 0.5 dB at 100 MHz. A tighter insertion loss requirement would further constrain the cable length.

The low-frequency response of the transformer is determined by the magnetizing inductance of the windings. If we make the line too short, it can prevent us from having the required number of turns, degrading the transformer’s low-frequency performance.

Next, let’s discuss how we can improve the Ruthroff transformer’s frequency response.

### Equal-Delay Transformers

Figure 4 shows a coaxial implementation of the Ruthroff 1:4 transformer.

##### Figure 4. Coaxial realization of a Ruthroff transformer.

To improve the high-frequency performance of this transformer, we can implement the connection between points A and B as an additional phase-compensating line with the same length as the main line. This makes the delay of the A-to-B connection equal to that of the main line as well. The result, shown in Figure 5, is known as an equal-delay transformer.

##### Figure 5. A Ruthroff transformer modified to create an equal-delay transformer.

The circuit can also be viewed as two coaxial lines with their input terminals in parallel and output terminals connected in series. Note that only the upper line of the transformer shown in Figure 5 uses a ferrite core. This is because no current flows in the outer conductor of the lower line, as both ends are grounded.

Depending on the impedance levels involved, equal-delay transmission line transformers can support frequencies ranging from 1 MHz to at least 500 MHz.

### Higher Transformation Ratios

We can also modify the Ruthroff transformer to achieve 1:9 and 1:16 impedance transformation ratios. Figure 6 shows how incorporating two transmission lines can produce a 1:9 impedance transformation ratio.

##### Figure 6. A Ruthroff 1:9 impedance transformer.

Let’s briefly confirm this circuit’s impedance transformation ratio by running through a lumped inductor analysis. In this analysis method, we assume the following:

• The same voltage appears across the primary and secondary windings of each transmission line in the transformer.
• The same current flows through the primary and secondary windings of a line (I1 = I2).

Just like with a magnetically coupled transformer, we use the dot convention to determine the polarity of the voltages and the direction of the currents flowing through the windings.

Since I1 = I2, we can simplify the analysis by expressing all currents in terms of I1, as shown below in Figure 7.

##### Figure 7. Expressing all currents in terms of I1.

Next, applying Kirchhoff’s voltage law, we observe that V2 = Vout and V1 = V2 + Vout = 2V2. This produces the simplified schematic in Figure 8, where all voltages are expressed in terms of V2.

##### Figure 8. Simplified schematic of a 1:9 transmission line transformer with currents and voltages expressed in terms of I1 and V2, respectively.

We see above that the input voltage is Vin = 3V2 and the output voltage is Vout = V2. This shows that the circuit scales the voltage by a factor of 3, producing a 1:9 impedance transformation ratio:

$$R_{in} ~=~ \frac{V_{in}}{I_1}~=~\frac{3V_{2}}{I_1}~=~9 ~\times~ \frac{V_{2}}{3I_1} ~=~ 9 R_L$$

##### Equation 11.

Figure 9 illustrates how we can produce a 1:16 impedance transformation ratio by adding one more transmission line.

##### Figure 9. Adding a third transmission line creates a 1:16 impedance transformer.

Again, the same current flows through all windings. The output voltage is equal to the voltage across Line 3 (Vout = V3). The voltage across each transmission line increases by V3 as we move from one line to the adjacent one higher in the network. As a result, the input voltage in this configuration is four times the output voltage (Vin = 4V3), corresponding to a 1:16 impedance transformation ratio (Rin = 16RL).

### Wrapping Up

The basic Ruthroff configurations provide a lower bandwidth than Guanella transformers. We’ve now seen that we can address this limitation by adding another line to the Ruthroff circuit, thus creating an equal-delay transformer. Additional transmission lines also allow us to build Ruthroff transformers with impedance transformation ratios higher than 1:4.

All images used courtesy of Steve Arar