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  • mbenkerumass 9:00 am on January 19, 2020 Permalink | Reply
    Tags: 100 ADS Examples   

    027/100 Shunt Reactance on Smith Chart 

    Example 3.5-2A: Measure the effect of susceptance on Smith Chart impedance matching.

    First, build the circuit and run the Smith Chart Matching tool.

    026-3

    A shunt capacitor moves in a clockwise direction across the smith chart tool:

    027-2

    Also note that a shunt inductor moves counter-clockwise across the smith chart tool:

    027-1

     
  • mbenkerumass 9:00 am on January 12, 2020 Permalink | Reply
    Tags: 100 ADS Examples   

    026/100 Lumped Element Smith Chart Movements: Series Inductor 

    Example 3.5-1: Measure the amount of movement caused by the reactance added to the circuit below. Measure the change from the starting point to the end point on the Smith Chart.

    026-1

    The circuit simulated gives the following result:

    026-2

    Recall that the circuit without a series inductor had the following result:

    025-3

    Through this simulation, it is shown that adding a series inductor causes the smith chart diagram plot to move in a clockwise direction.

    Note the change using the Smith Chart matching tool:

    026-3026-4

     

     
  • mbenkerumass 10:36 pm on December 29, 2019 Permalink | Reply
    Tags: 100 ADS Examples   

    025/100 Smith Chart Impedance Plotting 

    Example 3.4-1: Plot the impedance Z = 25 + j25 Ohm on the standard Smith Chart.

    025-1

    In order to plot a schematic simulation on a smith chart diagram, run a simulation.

    025-2025-3

     
  • mbenkerumass 9:59 am on November 26, 2019 Permalink | Reply
    Tags: 100 ADS Examples   

    034/100 Loaded Q and External Q 

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

    Example 4.2-3: Analyze the parallel resonator that is attached to a 50 Ohm source and load as shown.

    034.0

    This problem is specifically asking to define the Q factor related to this circuit. The Q factor is a ratio of energy stored (by an inductor or capacitor) to the power dissipated in a resistor. The Q factor varies with frequency since the effect of a capacitor or inductor also vary with frequency. For a series resonant circuit, the “unloaded” Q factor is defined by the following function: Qu = X / R = 1/(wRC) = wL/R

    The unloaded Q factor of a parallel resonant circuit: Qu = R / X = R/(wL) = wRC

    Overall, the Q factor is a measure of loss in the resonant circuit. A higher Q corresponds to lower loss, while a lower Q indicated higher loss. An “unloaded” Q factor means that the resonator is not connected to a source or load. The above circuit can no longer apply the “unloaded” Q factor formulas due to the presence of a source and a load. There are two further Q factor formulas that need to be considered: loaded Q factor and external Q factor. The loaded Q factor includes the source resistance and load resistance with the resistance of the circuit. The external Q factor refers to only the source resistance and load resistance together.

    For the above circuit, the loaded Q factor for the parallel resonator is defined as:

    Loaded Q = (Rs + R + Rl)/(wL) = (Source resistance + R + load resistance) / (wL)

    The external Q factor for the source resistance and load resistance is:

    External Q = (Rs + Rl)/(wL) = (Source resistance + load resistance)/(wL)

    The relationship between the different types of Q factors are:

    1/(Loaded Q) = 1/(External Q) + 1/(Unloaded Q)

     
  • mbenkerumass 11:35 am on November 25, 2019 Permalink | Reply
    Tags: 100 ADS Examples   

    033/100 Parallel Resonant Circuits 

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

    Example 4.2-2: Analyze a rearrangement of the RLC components into a parallel configuration.

    As observable by the following figures, the resonant frequency and impedance value remain the same for the parallel RLC circuit. What may be understood by this is that the reactance of the inductor cancels out the reactance of the capacitor at this frequency of 505 MHz.

    The input admittance of a parallel resonant circuit is: Y = (1/R) + jwC + (1/jwL).

    The angular frequency, w = 2*pi*f = 1 / (sqrt(L*C)).

    033.1033.2

     
  • mbenkerumass 9:30 pm on November 24, 2019 Permalink | Reply
    Tags: 100 ADS Examples   

    032/100 Series Resonant Circuits 

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

    Example 4.2-1: Analyze a one port series RLC circuit with R = 10 Ohms, L = 10 nH and C = 10 pF.

    According to the following results, the input impedance at resonance is 10 Ohms, which is the value of the resistor.

    The input impedance of an RLC series circuit is modeled by the following formula, a rather basic expression: Z = R + jwL + 1/(jwC)

    The power delivered to the resonator is: P = |I|^2 * Z / 2.

    032.1032.2

     
  • mbenkerumass 9:13 pm on November 21, 2019 Permalink | Reply
    Tags: 100 ADS Examples   

    023/100 Distributed Bias Feed Design 

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

    Example 2.11-3: Calculate the physical line length of the λ/4 sections of 80 Ω and 20 Ω microstrip lines at a frequency of 2 GHz. Create a schematic of a distributed bias feed network.

    A high impedance microstrip line of λ/4 can be used to replace the lumped inductor from problem 022/100 Example 2.11-2E. Likewise a quarter wave impedance line of a low impedance can replace the lumped shunt capacitor. The 80 Ohm and 20 Ohm transmission lines can be made using LineCalc at 2 GHz. The taper, tee and end-effect element are used to simulate the circuit most correctly and to remove discontinuities between the models.

    023.1023.2

    The return loss null occurs at 1.84 GHz, indicating that the system could be optimized better to adjust center frequency. The high impedance line length is now adjusted to center the frequency to 2 GHz:

    023.3023.4

     

     

     
  • mbenkerumass 8:36 pm on November 20, 2019 Permalink | Reply
    Tags: 100 ADS Examples   

    022/100 Microstrip Bias Feed Networks 

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

    Example 2.11-2E: Design a lumped element biased feed network.

    Bias feed networks are an important application of high impedance and low impedance microstrip transmission lines. The voltage bias may be needed for a device that is connected to the microstrip line, such as a transistor, MMIC amplifier or diode. The inductor in the circuit below is used as an “RF Choke”, which is used in tandem with a shunt or bypass capacitor for a “bias decoupling network.” Lumped elements are typically used for frequencies below 200 MHz.

    The following is a typical bias feed network, followed by a simulation:

    022.1022.2

     

     
  • mbenkerumass 7:45 pm on November 19, 2019 Permalink | Reply
    Tags: 100 ADS Examples   

    021/100 Distributed Inductance and Capacitance 

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

    Example 2.11-2D: Convert the lumped element capacitors and inductors to distributed elements.

    This is the schematic that needs to be changed into distributed element microstrip lines:

    021.1

    The following formulas are needed to calculate the inductive and capacitive line lengths to simulate this schematic using microstrip lines.

    Inductive line length: (frequency)*(wavelength)*(Inductance)/(impedance of line)

    Capacitive line length: (frequency)*(wavelength)*(Capacitance)*(impedance of line)

    In order to know what at which frequency the inductance or capacitance are calculated, let’s run the simulation of the above circuit:

    021.2

    This circuit is centered at 10 GHz, since the circuit behaves as a terminated open-circuited transmission line with an open-parallel resonance at 180 degrees, or twice the length of a quarter wave line.

    The above circuit is then modeled as follows:

    021.3

    021.4

     
  • mbenkerumass 4:10 pm on November 18, 2019 Permalink | Reply
    Tags: 100 ADS Examples   

    020/100 Open-Circuited Transmission Line with Termination 

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

    Example 2.11-2C: Calculate the input impedance of a quarter wave open-circuited microstrip transmission line using termination with end effects.

    An open circuit microstrip line generates a capacitive end effect due to radiation. This radiation is observable in the results from the following simulation. Note that the impedance at 180 degrees is more capacitive than was the open circuit transmission line with out any termination.

    020.1020.2019.3

     
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