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  • mbenkerumass 2:23 pm on November 17, 2019 Permalink | Reply
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    019/100 Open-Circuit Transmission Line 

    100 ADS Design Examples Based on the Textbook: RF and Microwave Circuit Design
    November 2019
    Michael Benker

    Example 2.11-2B: Calculate the input impedance of a quarter wave open-circuited microstrip transmission line for a given length of time.

    The reactance of a lossless open circuit transmission line can be modeled as being equal to the characteristic impedance multiplied by the cotangent of the electrical length of the transmission line in degrees.

    X = Z * cot(Θ)

    To construct this circuit, a termination of 1 MOhms is used to simulate an open circuit. As the electrical length in degrees varies with frequency (the wavelength), a static electrical length of a transmission line varied over many frequencies will suffice to demonstrate the reactance of a varying electrical length transmission line. The following circuit was created with a transmission line optimized for 10 GHz, similar to the Short-circuited Transmission Line:

    019.1019.2

    The results above are consistent with the theoretical model of an open circuit transmission line being modeled by a cotangent relationship. At the optimized frequency (at which the transmission line length is quarter-wave) it can be observed that the impedance of the line is measured to be zero. At a half-wave length and other multiples of a half wave length, the transmission line generates high levels of resonance.

    019.3

     
  • mbenkerumass 1:33 pm on November 16, 2019 Permalink | Reply
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    018/100 Short-circuited Transmission Line 

    100 ADS Design Examples Based on the Textbook: RF and Microwave Circuit Design
    November 2019
    Michael Benker

    Example 2.11-2A: Calculate the input impedance of a short-circuited microstrip transmission line for a given electrical length of the line.

    This circuit was built with a quarter-wave microstrip synthesized for 10 GHz with given substrate (conductivity of gold) using the LineCalc tool.

    018.1018.2

    The following results conclude that a short-circuited quarter-wave transmission line has high impedance, similar to an open circuit. A short circuited transmission line that is not a quarter-wave transmission line will not have high impedance as demonstrated by frequencies far outside of the range of optimization (10 GHz). This phenomena is is consistent with electromagnetic theory on transmission lines.

    018.3018.4

    Theoretical relationship between transmission line length (short-circuited) and it’s imaginary impedance component:

    018.5

     

     
  • mbenkerumass 8:26 pm on June 29, 2019 Permalink | Reply
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    016/100 Example 2.9-1 Waveguide Insertion Loss 

    100 ADS Design Examples Based on the Textbook: RF and Microwave Circuit Design
    Michael Benker
    Example 2.9-1: Consider the model of a one inch and a three inch length of the waveguide as used in an X Band satellite transmission system. Display the insertion loss of the waveguides from 4 to 8 GHz.

    377 Ohms simulates free space

    016.1016.2

     
  • mbenkerumass 7:10 pm on June 27, 2019 Permalink | Reply
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    014/100 Example 2.4-1 VSWR Measurement of Series RLC 

    100 ADS Design Examples Based on the Textbook: RF and Microwave Circuit Design
    October 2019
    Michael Benker
    Example 2.4-1: For series RLC elements, measure the reflection coefficients and VSWR from 100 to 1000 MHz in 100 MHz steps.

    014.1014.2

     
  • mbenkerumass 4:45 pm on June 26, 2019 Permalink | Reply
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    013/100 Example 1.5-2B Physical Capacitor Q Factor versus Frequency 

    100 ADS Design Examples Based on the Textbook: RF and Microwave Circuit Design
    October 2019
    Michael Benker
    Example 1.5-2B: Calculate the Q factor versus frequency for the modified physical model of an 8.2 pF multilayer chip capacitor.

    013.1013.2

    013.3

     
  • mbenkerumass 4:30 pm on June 25, 2019 Permalink | Reply
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    012/100 Example 1.5-2A Dissipation Factor in Capacitor 

    100 ADS Design Examples Based on the Textbook: RF and Microwave Circuit Design
    Michael Benker
    Example 1.5-2A: Calculate the Q factor versus frequency for the physical model of an 8.2 pF multilayer chip capacitor.

    012.1012.2

     
  • mbenkerumass 4:12 pm on June 24, 2019 Permalink | Reply
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    011/100 Example 1.5-1 Single Layer Capacitor 

    100 ADS Design Examples Based on the Textbook: RF and Microwave Circuit Design
    Michael Benker
    Example 1.5-1 Consider the design of a single layer capacitor from a dielectric that is 0.010 inches thick and has a dielectric constant of three. Each plate is cut to 0.040 inches square. Calculate the capacitor value and its Q factor.

    Capacitance formed by a dielectric material between two parallel plate conductors:

    C = (N-1)(KAεr/t)(FF) pF

    A: plate area
    εr: relative dielectric constant
    t: separation
    K: unit conversion factor; 0.885 for cm, 0.225 for inches
    FF: fringing factor; 1.2 when mounted on microstrip
    N: number of parallel plates

    011.1011.2

     

     
  • mbenkerumass 3:49 pm on June 23, 2019 Permalink | Reply
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    010/100 Example 1.4-6 Magnetic Core Inductors 

    100 ADS Design Examples Based on the Textbook: RF and Microwave Circuit Design
    Michael Benker
    Example 1.4-6 Design a 550 nH inductor using the Carbonyl W core of size T30/ Determine the number of turns and model the inductor in ADS.

    Number of turns calculation: N = sqrt(L/A) = sqrt(55nH/2.5) = 14.8

    010.1010.2

    010.3

     

     
  • mbenkerumass 1:09 pm on June 21, 2019 Permalink | Reply
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    008/100 Example 1.4-4 Q Factor of Air Core Inductor 

    100 ADS Design Examples Based on the Textbook: RF and Microwave Circuit Design
    Michael Benker
    Example 1.4-4 Calculate the Q factor of the air core inductor used in previous example 1.4-2.

    008.1008.2

     
  • mbenkerumass 12:57 pm on June 20, 2019 Permalink | Reply
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    007/100 Example 1.4-3 Air Core Inductor Equivalent Network 

    100 ADS Design Examples Based on the Textbook: RF and Microwave Circuit Design
    Michael Benker
    Example 1.4-3 Create a simple RLC network that gives an equivalent impedance response similar to previous example 1.4-2.

    007.1007.2

     
  • mbenkerumass 12:33 pm on June 19, 2019 Permalink | Reply
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    006/100 Example 1.4-2 Air Core Inductor 

    100 ADS Design Examples Based on the Textbook: RF and Microwave Circuit Design
    Michael Benker
    Example 1.4-2 Calculate and plot the input impedance of an air core inductor.

    006.1006.2

     

     
  • mbenkerumass 12:05 pm on June 17, 2019 Permalink | Reply
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    004/100 Example 1.3-1B Parasitic Elements of a Physical Resistor vs. Frequency 

    100 ADS Design Examples Based on the Textbook: RF and Microwave Circuit Design
    Michael Benker
    Example 1.3-1B: Plot the impedance of a 5 Ω leaded resistor in ADS over a frequency range of 0 to 2 GHz.

    004.1004.2

    This indicates a resonance at 500 MHz. This is due to the parasitic iductance and capacitance that exists on a real resistor. The resistor behaves as a combination of series parasitic inductance and resistance, in parallel with a parasitic capacitance.

    The impedance of an inductor is reduced as the frequency increases, while the impedance of a capacitor increases as the frequency increases. The intersection frequency of these two patters meet is the resonant frequency.

    The resonance frequency can be found from equating XL and XC. The formula is:

    Resonant frequency fR = 1/(2*pi*sqrt(LC))

     
  • mbenkerumass 11:47 am on June 16, 2019 Permalink | Reply
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    003/100 Example 1.3-1A Ideal Resistors 

    100 ADS Design Examples Based on the Textbook: RF and Microwave Circuit Design
    Michael Benker
    Example 1.3-1A Plot the impedance of a 50 Ω ideal resistor in ADS over a frequency range of 0 to 2 GHz.

    003.1003.2

    Thereby noting that an ideal resistor maintains constant impedance with respect to frequency.

    You were here and you read it, so don’t forget it.

     

     
  • mbenkerumass 3:30 pm on June 15, 2019 Permalink | Reply
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    002/100 Example 1.2-4 Skin Effect and Flat Ribbons 

    100 ADS Design Examples Based on the Textbook: RF and Microwave Circuit Design
    Michael Benker
    Example 1.2-4 Calculate the inductance of the 3 inch Ribbon at 60 Hz, 500 MHz, and 1 GHz. Make the ribbon 100 mils wide and 2 mils thick.

    002.1002.2

    The flat ribbon inductance is calculated with the following equation:

    L = K*l*[ ln((2*l)/(W+T))+0.223*(W+T)/l + 0.5 ] nH

    l: length of the wire
    K: 2 for dimensions in cm and K=5.08 for dimensions in inches
    W: the width of the conductor
    T: the thickness of conductor

     

     
  • mbenkerumass 3:15 pm on June 14, 2019 Permalink | Reply
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    001/100 Example 1.2-1 Reactance and Inductance with respect to Frequency 

    100 ADS Design Examples Based on the Textbook: RF and Microwave Circuit Design
    Michael Benker
    Example 1.2-1: Calculate the reactance and inductance of a three inch length of AWG #28 copper wire in free space at 60 Hz, 500 MHz, and 1 GHz.

     

    001.1001.2

    > The increase in reactance with respect to frequency represents the skin effect property, in which, as the frequency increases, the current density begins to be concentrated on the surface of a conductor.

     
  • mbenkerumass 11:12 pm on June 13, 2019 Permalink | Reply
    Tags: 100 ADS Examples   

    100 ADS Design Examples, RF and Microwave Circuit Design 

    I found this book has a number of interesting problems that I would like to go through by myself to get some experience with ADS. I may change my mind, however I intend on posting my solutions to my blog (here) as I go through them, if I do. Stay tuned.

    41IqblbUzRL

     
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