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BeO and AlN Attenuators

Tue, 01/29/2002 - 5:03am

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By Patrick Biebersmith, Florida RF Labs

The design of microwave attenuators for high power applications are developed using two preferred ceramics; Beryllium Oxide (BeO) and Aluminum Nitride (AlN). Florida RF Labs has been providing these high power resistive products in both BeO and AlN for nearly 25 years and has found that each material has its own benefits and applications. Microwave attenuators are used to reduce the signal power to a known amount and set power levels for other components following the device. This device is used with high power amplifiers, isolators, and combiners. High power attenuators are manufactured using thick and thin film technologies to maximize their effectiveness. A step-by-step design process will indicate how this is accomplished.


Attenuator Design
The first step in the design process is determining the footprint required to achieve the power and attenuation specified for the attenuator. The resistor values need to be calculated for the specified attenuation, which depends on the type of resistor network being used. The two resistor networks in use are the Pi and Tee structures.

The Pi structure incorporates a three-resistor network in the shape of the Greek letter Π (P). The Pi network is readily producible for attenuation values of 2 to 30 dB. Attenuation values outside this range become more difficult to manufacture due to the limitations of the printing technology and size of the device. The Pi structure attenuation is stable through a large frequency range. The calculations for a 15 dB Pi network follow.

Zo = Zin =Z = 50 ohms
Attenuation = 15 dB
R1 = R2

(1)     10dB/10 = N
     1015/10 = .032

(2)     R1 = Z(sqrt (N)+1)/(sqrt(N) -1)
     70.48 ohms = 50(.17+1)/(.17-1)

(3)     R3 = Z (N-1)/Z(sqrt(N))
     142.35 ohm = 50(.032 - 1)/2(.17)

The Tee structure uses a three-resistor network arranged in the shape of a "T". This structure is easily producible for attenuation values between 2 and 30 dB. The structure provides excellent attenuation stability and is designed to maintain the input and output impedance. Above 30 dB it becomes increasingly more difficult to realize the structure due to limitations in the printing technology and device size. The calculations for a 15 dB Tee network follow. The value from equation 1 is the same for the Tee configuration.

(4)     R1 = (sqrt(N) - 1)Z/(sqrt(N) +1)
          34.90 ohms = (.17-1)50/(.17 + 1)

(5)     R2 = 2Z sqrt(N)/N-1
          18.36 ohms = 2(50)(.17)/.032-1

Material Comparison
High power attenuators are primarily designed on two substrate materials Beryllium Oxide (BeO) and Aluminum Nitride (AlN). Table 1 outlines the material properties of Beo and AlN. A size limitation on these devices is 1.0 × 1.0 inches as the substrate size gets larger there becomes a concern for fractures in the substrate during manufacture of the device. BeO attenuators have a maximum power rating of up to 500 watts using a 1.0 × 1.0 inch substrate. AlN has a maximum power rating of up to 300 watts using a 1.0 × 1.0 inch substrate. AlN if fast becoming the preferred substrate because of cost, availability and material properties are comparable to BeO. Another reason AlN is preferred is the majority of high power commercial applications are using devices up to 200 watts while military applications can be up to 500 watts. Table 1 compares the material properties of BeO and AlN.


Following the calculation of the resistor values the minimum size of the printed resistor can be found. The size of the resistor is determined by the amount of power the resistor will experience, location of the resistor in the network, and the power dissipation of the substrate material used. After the size of each of the resistors has been calculated and the minimum footprint size of the device determined, the next step is to choose the correct resistor network for the device specification. The calculations for determining the minimum footprint of a 150 W AlN Pi and Tee attenuator follow.

The first step is to determine the power in each resistor by calculating the Input and Output voltage.

(6)     Vin = P2Z
          86.6 V = sqrt(150)(50)

(7)     Vout = Vin(10dB/20)
          15 V = 86.6(.177)

     The next step is to use ohms law to calculate the power in each of the three resistors.

(8)     PRin = Vin2/R1
          Pi 106 W = 86.6^2/70.48

(9)     PR2 = Vout2/R2
          Pi 3.2 W = 152/70.48

(10)     PRs = (PRin + PRout) -150
          Pi 40.8 W = (106 + 3.2) -150

Tee the power levels for the associated Tee resistors are the same as Pi resistors. The last step is to calculate the size of the associated resistors.

PD = AlN power dissipation

(11)     ARin = PRin/PD
          ARin = 1.5 e-5 m2

(12)     ARs = PRs/PD
          ARs = 5 e-7 m2

For a symmetric device both Rin and Rout are the same size.

Adding the resistor areas together and two tab pads of .0127 3 .0127 meters can find the minimum footprint. All the information found can be entered into an electromagnetic simulation software package such as Microwave Office or Sonnet to view the device characteristics. This simulation will show that as the frequency increases the attenuation varies widely due to the capacitance of the device over frequency.


Figure 1

To compensate for the capacitance an inductive tuning structure is added to reduce the capacitance and therefore improve the device characteristics. Using simulation software makes this task quite easy. The trade-off that will be encountered is as tuning structures are added, the attenuation varies less from the nominal value, but the return loss gets worse. A middle ground needs to be found to get an acceptable return loss and attenuation stability.

Another method of compensating for the capacitance is to make the device asymmetric. Each resistor network mentioned above has an input and output resistor. In a symmetric device (see Figure 1) the input and output resistor are each sized to handle the maximum power seen by the device. In an asymmetric device, the output resistor is only designed for the maximum amount of power it would receive. The output resistor ends up seeing a very small amount of power and therefore can be reduced in size to compensate for the capacitance. The draw back from this approach is the signal through the device may only be used in one direction whereas the symmetrical devices can use signal in both directions.

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