All Posts on one Page


  • 6. Give examples of point defects in a crystal. Explain what kind of defect is a dopant atom. July 29, 2022

    Defects in semiconductor crystals are:

    1. Point defects
    2. Line defects
    3. Surface defects
    4. Volume defects

    Of point defects, there are native defects (happening without the addition of other atoms) and non-native defects due to contaminants. Native defects are vacancies and self-interstitials. Non-native point defects are substitutions or interstitials. Dopants must be substitution defects.

  • 5. What is the behavior of the semiconductor energy gap Eg of direct-bandgap semiconductors vs. the lattice constant a? July 28, 2022

    For direct bandgap semiconductors, the bandgap is inversely proportional to the lattice constant. For ternary semiconductor compounds such as AlGaAs, varying the Al composition will change the bandgap of the compound, as well as the lattice constant. Ternary compounds vary the concentration of the metals, Al and Ga. As such this leads to a variation in Al concentration from AlAs with no Ga concentration to GaAs, with no Al concentration. AlAs and GaAs are lattice-matched and can therefore be grown together using conventional methods. AlGaAs is direct bandgap for Al concentrations below 0.45.

    For a quarternary compound InGaAsP, the concentration varies between InAs, InP, GaAs, and GaP. These concentrations are not lattice matched to each other and vary significantly in bandgap.

    The relationship between bandgap and lattice constant also enables an increased bandgap with compressive lattice strain and a reduced bandgap for tensile strain.

  • 3. In an out-of-equalibrium semiconductor, where Nc=10^18, Nv=10^19 cm^(-3), evaluate the position of the quasi-Fermi levels… July 27, 2022

  • 3. Explain why ferroelectric crystals already have, with zero applied electric field, a net dipole moment. What is the Curie temperature? July 26, 2022

    Ferroelectric materials possess a dielectric polarization without an applied electric field. Examples of ferroelectric materials include potassium dihydrogen phosphate (KH2PO4), barium titanate (BaTiO3) and Lithium niobate (LiNbO3).

    Ferroelectric materials undergo a change based on temperature. Above a temperature called the Curie temperature, the crystal is cubic, but below the Cure temperature they are tetragonal. Undergoing the change to tetragonal, with an electric polarization.

    Ferroelectric materials are also piezoelectric, meaning that they produce an electric charge under a mechanical load.

    4. Quote at least one important ferroelectric material exploited in electrooptic modulators.

    LiNbO3 is a highly attractive ferroelectric crystal used in electrooptic modulators, due to the strong electrooptic effect, or change in refractive index under an electric field.

  • 1. Sketch the planes denoted, according to the Miller notation, as (101), (001), (1¯1¯1¯). July 25, 2022

  • 2. What is the wurtzite crystal structure? Quote at least one important wurtzite semiconductor. July 24, 2022

    Wurtzite crystal structure is a hexagonal pattern found in some semiconductors. This differs from the cubic crystal structure of diamond and zinc-blende latttices.

    While in the cubic semiconductor structures, the distance between points of the Bravais lattice are the same in three directions, which are separated each by 90 degrees. In the hexagonal structure, only the distance between two of the three points are the same. The two points that are the same distance from each other are separated by 90 degrees, while the third point is separated by 120 degrees.

    Materials with a wurtzite crystal structure include GaN, AlN, InN, and ZnO.

  • 1.1 A bullet with a mass of 10 g flies at a velocity of 1000 m/s. Determine the de Broglie wavelength. Why might we use equations from classical physics for description of bullet flight? July 23, 2022

    See response below:

  • 1. Describe the crystal structure of Si and GaAs. Explain the difference between the diamond and zinc-blende lattice cells. July 22, 2022

    Both Silicon and GaAs are cubic semiconductors. Silicon is a diamond crystal, while GaAs is a zinc-blende crystal. Diamond differs from zinc-blende lattice in the composition of the atoms in the lattice.

    Since Si is a pure substance, only Si atoms are present in the crystal lattice. A diamond crystal has 8 atoms, which include the corner atoms, atoms on faces, and internal atoms.

    In the GaAs semiconductor compound, the corner and face atoms will be of a different atom than the internal atoms. For instance, the corner and face atoms can be the metal Gallium while the internal atoms are Arsenic.

  • 1.1 Consider the (001) MBE growth of GaAs by MBE. Assuming that the sticking coefficient of Ga is unity… July 21, 2022

  • Converting from normalized SFDR (dBHz^(2/3)) to real SFDR (dB) January 7, 2022

    SFDR is frequently written in the units of dBHz^(2/3), particularly for fiber optic links. Fiber optic links can often have such high bandwidth, that assuming a bandwidth in SFDR is unhelpful or misleading. Normalizing to 1Hz therefore became a standard practice. The units of SFDR for a real system with a bandwidth are dB.

    Now consider that the real system has a specific bandwidth. The real SFDR can be calculated using the following formulas:
    SFDR_real = SFDR_1Hz – (2/3)*10*log10(BW)

    Here are a few examples.

  • Where do the units of SFDR “dB·Hz^(2/3)” come from? December 28, 2021

    The units of spurious-free dynamic range (SFDR) are dB·Hz^(2/3). The units can be a source of confusion. The short answer is that it is a product of ratios between power levels (dBm) and noise power spectral density (dBm/Hz). The units of dBHz^(2/3) are for SFDR normalized to a 1Hz bandwidth. For the real SFDR of a system, the units are in dB.

    If we look at a plot of the equivalent input noise (EIN), the fundamental tone, OIP3 (output intercept point of the third order distortion), and IMD3 (intermodulation distortion of the third order), a ratio of 2/3 exists between OIP3 and SFDR. This can be recognized from the basic geometry, given that the slop of the fundamental is 1 and the slope of IMD3 is 3.

    Now, we need to look at the units of both OIP3 and EIN. The units of OIP3 are dBm and the units of the equivalent input noise (a noise power spectral density) are dBm/Hz.

    SFDR = (2/3)*(OIP3 – EIN)

    [SFDR] = (2/3) * ( [dBm] – [dBm/Hz] )

    Now, remember that in logarithmic operations, division is equal to subtracting the denominator from the numerator. and therefore:

    [dBm/Hz] = [dBm] – 10*log_10([Hz])

    Note that the [Hz] term is still in logarithmic scale. We can use dBHz to denote the logarithmic scale in Hertz.

    [dBm/Hz] = [dBm] – [dBHz]

    Substituting this into the SFDR unit calculation:

    [SFDR] = (2/3) * ( [dBm] – ( [dBm] – [dBHz] )

    This simplifies to:

    [SFDR] = (2/3) * ( [dBm] – [dBm] + [dBHz] )

    Remember that the difference between two power levels is [dB].

    [SFDR] = (2/3) * ( [dB] + [dBHz] )_

    The units of [dB] + [dBHz] is [dBHz], as we know from the same logarithmic relation used above for [dBm] and [dB].

    [SFDR] = (2/3) * [dBHz]

    Now, remember that this is a lkogarithmic operation, and a number multiplying a logarithm can be taken as an exponent in the inside of the logarithm.Therefore, we can express Hz again explicitly in logarithm scale, and move the (2/3) into the logarithm.

    (2/3) * [dBHz] = (2/3) * 10*log_10([Hz]) = 10*log_10([Hz]^(2/3))

    We can return our units back to the dB scale now, giving us the true units for SFDR: dBHz^(2/3):

    [SFDR] = [dBHz^(2/3)]

  • Keysight ADS – Microstrip Line Design February 26, 2020

    The goal of the project is to design a 50 ohm microstrip line at an operating frequency of 10 GHz and phase delay of 145 degrees.


    The following ADS simulation will be composed of four major parts:

    a) Designing the microstrip lines using two models (I.J. Bahl and D.K. Trivedi model and E. Hammerstad and Jensen model). The insertion loss (S(2,1)) will be plotted over the range of 10 MHz to 30 GHz.

    b) Assuming reasonable dielectric losses, results should be compared to part a

    c) Creation of ideal transmission lines with same parameters compared to part a and b

    d) Showing dispersion on the lossless microstrip line. This is compared to the ideal line.


    The LineCalc tool (which uses the Hammerstad and Jensen model) within ADS is used to design the second line with the correct specifications. The first circuit is designed using hand calculated values.


    The following shows using the LineCalc tool to get the values for the second schematic.


    The simulation is shown below.


    A new substrate is created with a loss tangent of .0002 for the second schematic. The simulation results in:


    An ideal transmission line circuit is created and compared with both the lossy and lossless lines.


    In order to demonstrate dispersion, the phase velocity must be calculated. As shown by the values compared from 0 GHz to 30 GHz, the phase velocity does not change for the ideal line, but does for the microstrip line.



  • Keysight ADS – Frequency Dependence of Microstrip Lines February 23, 2020

    The following ADS simulation will demonstrate how the characteristic impedance and effective dielectric constant change with frequency.  In the simulation, a quarter wave section of multi-layer microstrip line is used to demonstrate frequency effects. The result are expected to show that the dielectric constant and the characteristic impedance are inversely related. When the frequency of the electric field increases, the permittivity decreases because the electric dipoles cannot react as quickly. The multi-layer component is used in place of an ideal component because frequency dependence must be demonstrated. An “MLSUBSTRATE2” component is used with the updated dielectric constant and Dielectric loss tangent.


    For S parameter analysis, two terminated grounds are required. The effective dielectric constant must be solved for by unwrapping the phase of S(2,1). The results show the characteristic impedance (both real and imaginary parts) increasing with frequency and the dielectric constant decreasing.


  • Keysight ADS – Transient Propagation February 21, 2020

    The following ADS simulation will demonstrate the effects of transients on a transmission line. A rectangular pulse of duration .5 microseconds will be generated and a net voltage vs time will be plotted for a period of .7 microseconds. The circuit has a mismatched load, producing reflections. A time domain reflectometry analysis will prove that the propagating signal voltage steadily increases after the initial time and as time increases, the reflections will eventually die out and leave a steady state response. This is shown with transient analysis.


    The schematic above contains two circuits for the two parts of the rectangular pulse (one with and one without a time delay). The simulated results are shown below.


    A bounce diagram can also be used to convey Time domain reflectometry analysis, as shown below. This diagram is a plot of the voltage/current at the source or load side after each reflection. This is a general diagram and does not apply to the problem.


  • Keysight ADS – Conjugate Matching February 18, 2020

    This project will use conjugate matching to match a capacitive load of 50-j40 to a generator of impedance 25+j30. Since the generator impedance is complex, conjugate matching is required to match the network, as opposed to in situations of low frequency where the reactive components are negligible. In the example, an L network is used to match the generator to the load. Theoretically, differentiating the power and setting this equal to zero proves that maximum power is transferred when the resistance of the source and load are equal and the reactive portions are equal and opposite phase shift/sign.

    The first step is to use the impedances given to calculate the actual lumped inductor and capacitor values to use for the network to work at 2 GHz. 25+j30 corresponds to a 25 ohm resistor in series with a 2.387732 nH inductor and 50-j40 corresponds to a 50 ohm resistor in series with a 1.98944 pF capacitor.

    The following shows the schematic with the source, matching network and load.


    Running the simulation with Data Display equations yields….


    This shows maximum power transfer at the correct frequency of 2 GHz. The next step is to use the Smith Chart tool. A shunt inductor and series capacitor is used to form the L Network. Exact values can be typed in for these to get the impedance value Z = 0.5 +j0.6 which is the normalized equivalent source impedance (divided by 50).


    With the capacitor and inductor values recorded, these values can be loaded into a separate schematic and compared to the original schematic results.

    Conjugate matching is not achieved with this Smith Chart configuration so there is no peak at 2 GHz.


    Alternatively, the Smith Chart tool can be used from the palette. From this point with the chart icon selected, the network can be created by selecting “Update Smart Component” from the Smith Chart tool window. These results show that it is important to select the proper design network for the specifications for optimal results.



  • Keysight ADS – Quarter Wave Transformer Matching February 17, 2020

    In ADS, a batch simulation can be implemented to run different load impedance simulations. This function will be used to simulate a quarter wave transformer matching system for various loads (25, 50, 75, 100, 125 and 150 ohms),  The system is used to match a 50 ohm line with an electrical length of 60 degrees at 1 GHz.

    The simulation will demonstrate that an unmatched load will generate a constant VSWR at all frequencies. With the implementation of the matching network, the VSWR varies because it is only designed to match the network at a specific frequency. A previous post derived the relationship to find the impedance of a quarter wave matching transformer.


    The VSWR can be plotted by adding equations into the data display window and manually adding equations into the plot window to plot VSWR against frequency for both the matched circuit and the unmatched circuit. The mismatched circuits appears constant over frequency with a very high SWR, as it does not have the matching transformer. The quarter wave transformer is shown to provide excellent matching at specific frequencies.


    For batch simulations, a slider tool can be implemented to show only specific impedances. Clicking on the axes and changing the names to include the index will update the plot with the specific impedances one at a time. The plot is updated to match the slider value for the load impedance.


    With the axes correctly updated, sliding the slider tool will change the plot automatically. Also in the data display window, tables can be added to view specific values at different frequencies.



  • Keysight ADS – Open Circuit Analysis February 9, 2020

    Expanding upon the previous project, open circuit analysis can be used to find equivalent per unit length capacitance and conductance values for the dielectric part of the transmission line.


    The same process is used for the open circuit analysis with new equations for capacitance and conductance. The calculated values from the simulation window are compared to the simulated values from the AC analysis.


  • Microstrip Stub Low-Pass Filter (10 GHz) February 2, 2020

    This is a 10 GHz Stub Low-pass filter, made using ADS.


    First, build the component using the ADS DesignGuide/Smart Component Passive Circuit tool.



    This is the original, equation-based simulation.stublp2


    This is the substrate used for the Low-pass filter.



    This is the Momentum simulation of the layout component.stublp4


    This is the layout component for the 10 GHz Stub Low-pass filter component.


  • Microstrip Lange Coupler (5 GHz) January 26, 2020

    The Lange Microstrip (quadrature) coupler is known for it’s low loss, wide bandwidth and compact layout. Similar to other couplers, it consists of an isolated port, through port and coupled port.

    You can build a microstrip Lange coupler using the DesignGuide tool in ADS:



    These are the results for the equation-based simulation. These results admittedly look considerably better.



    This is the substrate used:



    These are the results for the momentum simulation. Admittedly, some tuning would improve this considerable.


    And here is the layout component:


  • 027/100 Shunt Reactance on Smith Chart January 19, 2020

    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.


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


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


  • Microstrip Double-Stub Load Matching January 15, 2020

    The following matches a 50 Ohm line to a 100 Ohm load at 10 GHz using a double-stub component. This was designed using the ADS passive circuit DesignGuide tool. This method is a great alternative to using the Smith Chart matching tool for lumped elements if you need a microstrip line for matching.






    Momentum simulation result (can be tuned to center better at 10 GHz):


    Layout component:




  • 026/100 Lumped Element Smith Chart Movements: Series Inductor January 12, 2020

    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.


    The circuit simulated gives the following result:


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


    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:



  • 025/100 Smith Chart Impedance Plotting December 29, 2019

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


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


  • 034/100 Loaded Q and External Q November 26, 2019

    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.


    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)

  • 033/100 Parallel Resonant Circuits November 25, 2019

    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)).


  • 032/100 Series Resonant Circuits November 24, 2019

    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.


  • 023/100 Distributed Bias Feed Design November 21, 2019

    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.


    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:




  • 022/100 Microstrip Bias Feed Networks November 20, 2019

    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:



  • 021/100 Distributed Inductance and Capacitance November 19, 2019

    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:


    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:


    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:



  • 020/100 Open-Circuited Transmission Line with Termination November 18, 2019

    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.


  • 019/100 Open-Circuit Transmission Line November 17, 2019

    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:


    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.


  • 018/100 Short-circuited Transmission Line November 16, 2019

    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.


    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.


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



  • Directional Coupler ADS Simulation October 30, 2019

    ECE435 – RF/Microwave Engineering, Professor Dr. Yifei Li
    October 2019
    Michael Benker
    Directional Coupler ADS Simulation




  • ADS Coupler Momentum Simulation October 18, 2019

    ECE435 – RF/Microwave Engineering, Professor Dr. Yifei Li
    October 2019
    Michael Benker
    ADS Coupler Momentum Simulation

    Build the ADS circuit.


    Run the momentum simulation and set parameters such as substrate.


    This is a momentum simulation. Let’s see if we can optimize this.


    Export the part to be used as a component in the workspace library in ADS.


    Now run an ADS simulation using the exported component, which uses a database of simulated results.


    If you step into the component, you will see component features.


    Now, tune the parameters to begin optimization.


  • Branchline Coupler – EM Simulation October 10, 2019

    ECE435 – RF/Microwave Engineering, Professor Dr. Yifei Li
    October 2019
    Michael Benker
    ADS Momentum Simulation



  • Quadrature Hybrid Coupler October 4, 2019

    ECE435 – RF/Microwave Engineering, Professor Dr. Yifei Li
    October 2019
    Michael Benker
    Project 4 – Quadrature Hybrid Coupler

    Presentation: Project4_presentation



  • Smith Chart Impedance Matching September 17, 2019

    ECE435 – RF/Microwave Engineering, Professor Dr. Yifei Li
    September 2019
    Michael Benker
    Project 1 – Smith Chart Impedance Matching

    Presentation: proj1_presentationproj1_schematic



  • 016/100 Example 2.9-1 Waveguide Insertion Loss June 29, 2019

    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


  • 014/100 Example 2.4-1 VSWR Measurement of Series RLC June 27, 2019

    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.


  • 013/100 Example 1.5-2B Physical Capacitor Q Factor versus Frequency June 26, 2019

    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.



  • 012/100 Example 1.5-2A Dissipation Factor in Capacitor June 25, 2019

    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.


  • 011/100 Example 1.5-1 Single Layer Capacitor June 24, 2019

    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



  • 010/100 Example 1.4-6 Magnetic Core Inductors June 23, 2019

    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




  • 008/100 Example 1.4-4 Q Factor of Air Core Inductor June 21, 2019

    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.


  • 007/100 Example 1.4-3 Air Core Inductor Equivalent Network June 20, 2019

    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.


  • 006/100 Example 1.4-2 Air Core Inductor June 19, 2019

    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.



  • 004/100 Example 1.3-1B Parasitic Elements of a Physical Resistor vs. Frequency June 17, 2019

    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.


    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))

  • 003/100 Example 1.3-1A Ideal Resistors June 16, 2019

    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.


    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.


  • 002/100 Example 1.2-4 Skin Effect and Flat Ribbons June 15, 2019

    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.


    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


  • 001/100 Example 1.2-1 Reactance and Inductance with respect to Frequency June 14, 2019

    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.



    > 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.