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Impedance matching in a circuit

  1. Dec 3, 2011 #1
    i have designed my circuit, but i am confused about impedance matching. i know we can use a resistor for matching, but is it possible to use a capacitor or inductor for matching the impedance in a circuit.
  2. jcsd
  3. Dec 3, 2011 #2


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    You haven't given much detail so the answer will be general.

    Yes, you can use inductors and capacitors for matching. Usually, you add an inductor in series with a load if the load is capacitive and vice versa.
    The reactance of the inductor and capacitor have to be the same, so clearly this can only apply at one frequency.

    For normal voltage amplifiers, "matching" really just means using appropriate impedances.

    You can drive a voltage amplifier that has a 10K ohm input impedance with another amplifier that has an output impedance of 1000 ohms. You don't get maximum power transfer by doing this, but this is a voltage amplifier and the aim is to not lose too much voltage when you drive one amplifier with another.

    The point is that you can't drive this amplifier with something that has an output impedance of 1 megohm, because that would reduce the voltage to about 1% of what it was before, while you can drive it with something that has a considerably lower impedance.
  4. Dec 4, 2011 #3
    Impedance matching is a "complex" subject and a general answer would be too long for this forum. For general information I recommend the book, RF Circuit Design by Chris Bowick. https://www.amazon.com/s/ref=nb_sb_...+chris+bowick&sprefix=RF+Circuit+Design+Chris

    We can help with a solution to a specific problem and would need the complex source impedance, the complex load impedance and the frequency.
  5. Dec 4, 2011 #4
    Yes!!! In RF, we even avoid using resistor to do matching because it create noise and consume power. Matching are all done with inductors, capacitors and transmission lines.
  6. Dec 6, 2011 #5
    BroadWave Technologies might be able to assist you with this application. We manufacture a wide variety of coaxial impedance matching pads also called minimum loss pads. These devices are used to transform between different impedances. Typical insertion loss is 5.7 dB. Custom impedances are available. Follow this link to our website for standard devices:


    or feel free to contact the factory for a custom device at an off the shelf price.
  7. Dec 6, 2011 #6
    This is just 50 to 75Ω. We are talking about more general matching not limited to real match real impedance. If we want real resistance matching with loss, that would be a cake walk.

    Besides, is it ok for vendors/manufacturers to advertize here on Physics Forums?
  8. Dec 7, 2011 #7
    the input impedance value is 50Ω and the output or load is 25Ω typically . so how to match impedance between those two resistor value and the important thing is , use only capacitor and inductor in the circuit. the input signal is RF signal and the output is a laser diode.
  9. Dec 7, 2011 #8
    Is the signal wideband or one frequency?
  10. Dec 7, 2011 #9
    the frequency range is 2.6 GHz .
  11. Dec 7, 2011 #10
    then it should be easy by using either a parallel L with series C or the other way around. I have to pull out my smith chart to get the value.

    You just want to value right?
    Last edited: Dec 7, 2011
  12. Dec 7, 2011 #11
    yes please
  13. Dec 7, 2011 #12
    I'll try to get back to you in like an hour.

    You know if you do it on pcb, easiest is to do it in quarter wave transmission line in between the two side to transform between 50 to 25 or back. Would you like to do that? If so, give me the pcb material you use and your normal stackup. I can work out the width of the line so you just put it on the pcb. No components needed. 2.6G is at the edge where you have to be careful with the parasitic of the discrete components.
  14. Dec 7, 2011 #13
    could you please show me the maths calculation for impedance matching for the idea you gave as connecting L in series and C in parallel or vise versa. i am really sorry that i could not sent the pcb to you, because this project a very big one, which is used by a research guy too...
  15. Dec 7, 2011 #14
    I did not do calculation as you did not ask!!! I use smith chart and verified by simulation:

    zmmuk9.jpg alogph.jpg

    As you can see in the left side, that is the circuit and value from 50Ω port 1 to 25Ω port 2.

    The right graph is the smith chart simulation to verify that this network work in both direction.

    If you want the math, it's a different story all together. I would have to get back to you some times tomorrow. I have not touch this for 6 years since I was not working. I had to revise the material to get to this. I have to review the formulas to give you the equations. But these values will give you the right matching. I did the design on the smith chart and this is the verification.

    As I said, I would do it in transmission line matching if I were you. If you can wait for my slow response, we can work it out sooner or later. I am busy with shopping and other things so it will take a day or so to get all the design for you. Let me know if you are interested.

    I don't need your pcb file. I just need you stack up. My guess is you are using FR4 which [itex]\epsilon_r=4.3[/itex]. The stackup is the distance between different layers, which one is the ground plane, your copper weight.
    Last edited: Dec 7, 2011
  16. Dec 7, 2011 #15
    The match can be either high pass or low pass meaning either an inductor or capacitor can go in either location. The side with the lower resistance takes the series component and the side with the higher resistance takes the parallel component. For a simple resistive to resistive match the math goes like this:

    R1 = 50 ohms
    R2 = 25 ohms
    jX2 = sqrt(R1*R2 - R2^2) = sqrt(50*25 -25^2) = 25
    jX1 = (R2^2 + jX2^2)/jX = (25^2 + j25^2)/j25 = 50

    If jX1 is positive (inductive) then jX2 is negative.
    The plus and minus signs for jX50 and jX25 may be reversed.
    All that remains is to calculate L and C from the reactances.

    For the low pass network in Yungman's example I get C1 = 1.22 pF and L1 = 3.06 nH.
    Last edited: Dec 7, 2011
  17. Dec 7, 2011 #16
    In my calculation inductor is 1.53nH. I double checked already. And the simulation show that the value got to be correct. As you can see the normalized impedance go between 0.5 to 1 and also the other way back. No if and buts about this.
  18. Dec 8, 2011 #17
    Ok, The calculation involve quite a bit of math, instead of me typing it out, this is a link that explain most of it:


    The one I used is low pass where the cap C1 is on the 50Ω side. then from the junction, it connects to L1 which then connect to 25Ω as shown in the previous diagram on the left.

    This is using series to parallel transformation and vise versa. So we know:

    [tex]R_p=(Q^2+1)R_S\;\;,\;\; Q_S=Q_P\;\;,\;\; Q_S=Q_P=\sqrt{\left(\frac {R_P}{R_S}\right)-1}\;\;,\;\;Q_S=\frac {X_S}{R_S}\;\;,\;\;Q_P=\frac {R_P}{X_P}[/tex]

    [tex] X_P=\frac 1 {\omega C_1}\;\;,\;\; X_S= \omega L\;\;,\;\; R_P=50\;\;,\;\;R_S=25[/tex]

    [tex] Q_S=Q_P=\sqrt{\left(\frac {R_P}{R_S}\right)-1}=\sqrt{\left(\frac {50}{25}\right)-1}=1[/tex]

    [tex]Q_P=1=\frac {R_P}{X_P}\Rightarrow \; X_P=\frac 1 {\omega C_1}=50\;\Rightarrow\; C=\frac 1 {50\times 2\pi\times 2.6\times 10^9}\;=\; 1.224\times 10^{-12} F[/tex]

    [tex] Q_S=1=\frac {X_S}{R_S}=\frac {25}{25}\Rightarrow \;X_S=25Ω \;\Rightarrow \omega L=25\;\Rightarrow L=\frac {25}{\omega}=\frac {25}{ 2\pi \times 2.6\times 10^9}= 1.53\times10^{-9}H[/tex]

    You can use this same formulas to get the value if you decided to go with high pass network. But it is not advisable to use high pass as you want to roll off the high frequency. Let me know if you have further question.
    Last edited: Dec 8, 2011
  19. Dec 8, 2011 #18
    You're correct. I mistakenly used 50 as the reactance of the inductor when it should have been 25.
  20. Dec 8, 2011 #19
    very much thanks guys ... for helping me .... especially yungman... let me connect it and test it.. and post my results guys....
  21. Dec 8, 2011 #20


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    Science Advisor

    If you are going to build this, here is a web site that calculates very small inductors like this:

    http://www.consultrsr.com/resources/eis/induct5.htm [Broken]

    A round wire 4 mm (0.157 inches) in diameter and 5.8 mm (0.228 inches) long will have an inductance of 1.53 nH.

    A capacitor with plates of 1 sq cm and spaced 0.072 cm will have a capacitance of 1.224 pF approximately.

    You will appreciate that these are very small components.

    Which raises the question. Do you have the right frequency?
    2.6 GHz is 2600 MHz
    Last edited by a moderator: May 5, 2017
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