# Impedance matching in a circuit

1. Dec 3, 2011

### rakesh.kumar

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. Dec 3, 2011

### vk6kro

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.

3. Dec 4, 2011

### skeptic2

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.

4. Dec 4, 2011

### yungman

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.

5. Dec 6, 2011

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.

6. Dec 6, 2011

### yungman

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?

7. Dec 7, 2011

### rakesh.kumar

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.

8. Dec 7, 2011

### yungman

Is the signal wideband or one frequency?

9. Dec 7, 2011

### rakesh.kumar

the frequency range is 2.6 GHz .

10. Dec 7, 2011

### yungman

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
11. Dec 7, 2011

### rakesh.kumar

12. Dec 7, 2011

### yungman

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.

13. Dec 7, 2011

### rakesh.kumar

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...

14. Dec 7, 2011

### yungman

I did not do calculation as you did not ask!!! I use smith chart and verified by simulation:

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 $\epsilon_r=4.3$. The stackup is the distance between different layers, which one is the ground plane, your copper weight.

Last edited: Dec 7, 2011
15. Dec 7, 2011

### skeptic2

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
16. Dec 7, 2011

### yungman

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.

17. Dec 8, 2011

### yungman

Ok, The calculation involve quite a bit of math, instead of me typing it out, this is a link that explain most of it:

http://www.millertechinc.com/pdf_files/MTI%20TN113%20Series%20to%20Parallel%20Impedance%20Transformation.htm

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:

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

$$X_P=\frac 1 {\omega C_1}\;\;,\;\; X_S= \omega L\;\;,\;\; R_P=50\;\;,\;\;R_S=25$$

$$Q_S=Q_P=\sqrt{\left(\frac {R_P}{R_S}\right)-1}=\sqrt{\left(\frac {50}{25}\right)-1}=1$$

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

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

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
18. Dec 8, 2011

### skeptic2

You're correct. I mistakenly used 50 as the reactance of the inductor when it should have been 25.

19. Dec 8, 2011

### rakesh.kumar

very much thanks guys ... for helping me .... especially yungman... let me connect it and test it.. and post my results guys....

20. Dec 8, 2011

### vk6kro

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
21. Dec 8, 2011

### yungman

As I said, I would prefer to do it with distributed element like a tx line to avoid parasitic. If you are interest, we can talk more about your stack up of the pcb and we can do a distributed design that you don't need any components, everything on pcb trace. At 2.6GHz, you have to be careful in choosing components where is distributed element has no such problems. Particular in your case, a quarter wave length trace will do the trick.

22. Dec 11, 2011

### rakesh.kumar

this is just a very small part of my project .... still i have to do a lot of work in this project... i have already characterised the laser, but there was a loss in output . that is the main reason i was looking into this impedance matching.... i totally have 3 months yet to go... within that i have to complete the whole idea of me, so i would not able to continue now with this design but anyway thanks a lot for asking this and helping me a lot...

23. Dec 16, 2011

### rakesh.kumar

as i have searched many places, the value of Q value is Frequency / Bandwidth. hence my frequency is 2.6GHz and bandwidth is 20MHz. then the Q value is to be 130. Is this is the way to find out the value of Q or is the above value shown by yungman. The actual design is shown below, the capacitor is used for blocking the dc supply fom another input the laser.

24. Dec 16, 2011

### skeptic2

Bandwidth is used in different ways. Here I think you are using it to mean the bandwidth of the carrier with the modulating signal. When Q is used with bandwidth, it usually means the bandwidth of a bandpass filter and in that context what you said is correct. The important thing in calculating frequency / bandwidth is determining the upper limit of Q for the bandpass filters.

Usually in impedance matching, it is desired for the Q to be as low as possible in order for the match to be useful over as wide a frequency range as possible. Note in Yungman's example, it has been defined as 1. In bandpass filters the bandwidth is usually just wider than the bandwidth of the signal itself in order to exclude everything but the signal.

Will your carrier frequency always be the same frequency or will you want to use a range of carrier frequencies? In other words, is your objective only impedance matching or do you also want to filter the signal. There are ways of doing both.

25. Dec 17, 2011

### rakesh.kumar

the frequency is fixed to 2.6GHz and not varied. i got confused with the q value due to a web link, please clarify this to me.
http://www.raltron.com/cust/tools/network_impedance_matching.asp [Broken]

it states q = frequency / bandwidth.

Last edited by a moderator: May 5, 2017