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Simple Impedance Matching? 
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#1
Apr2807, 12:33 PM

P: 26

Our professor gave us some homework and one of the problem statements reads:
"Design an impedance matching network used to connect a 300 Ohm transmission line to a 50 Ohm transmission line." Is this as simple as Zmatch = Sqrt(Z1*Z2) ? I'm wondering if there's something I'm missing, because this looks too simple... 


#2
Apr2807, 12:52 PM

P: 1,636

Z = Sqrt(Z1*Z2) is the impedance of a quarter wave trasmission line you could connect between the two impedances Z1 and Z2 to match them.
Another way, is to use an L, Pi or a T network between the impedances Z1 and Z2. These networks are just a combination of L's and C's. The simplest one is the L network with a single capacitor and an inductor. 


#3
Apr2807, 01:14 PM

P: 26

How would I select the size for the L's and C's if I were to use those? What would these networks look like?
Thank you for your help. 


#4
Apr2807, 02:10 PM

P: 321

Simple Impedance Matching?
You can also use shunt stubs. This is covered in most entry level EM or RF books. All you need is a smith chart.



#5
Apr2807, 06:53 PM

P: 1,636

Do you what is a Q of a series and a parallel resonant circuit? If so you could transform a series resoant circuit into parallel and viceversa.



#6
Apr2907, 02:00 PM

P: 26

I'm afraid our professor didn't cover how to use Smith Charts, but I think I remember what a resonant circuit is. However, we weren't given a frequency, so I don't know how I would go about choosing either L or C components.



#7
Apr3007, 02:36 PM

P: 1,636

If you don't know the frequency you can't really design a proper matching circuit with transmission lines nor L match. You would need design a transformer with a proper turns ratio. If I remember correctly, Zp = Zs (Np/Ns)^2 where is Zp is primary winding impedance, Zs secondary, and N is turn ratio primary and secondary. If you choose a proper transormer core, usually a toroid that would match up your impedance more broadly in frequency. Where as the L match only works at a specific frequency.



#8
Apr3007, 03:03 PM

P: 198

i got a 250 ohm resistor in series with the load, what did everyone else get?



#9
Apr3007, 03:18 PM

P: 1,636




#10
Apr3007, 04:05 PM

P: 198

explain, i'm all ears. btw i didn't take reactance into account.



#11
Apr3007, 04:37 PM

P: 1,636

Maximum power transfer from a source to load only occurs when the source and load impedances are equal or more precicily are complex conjugates.
So if you hook up in series that 250 ohm resistor you should have a maximum power transfer because 50 + 250 = 300 ohms. But, resistors dissipate power, so by adding that extra resistor you are dissipating some of that precious power as heat. But impedance matching using L's and C's can achieve the same feat but with pretty much no loss at all. What if you have to match 50 ohms source to a 1500 ohms of an IC receiver chip. If you add a 1450 ohms you are dead since radio signal are very very weak in the first place. Hope that helps. 


#12
Apr3007, 05:08 PM

P: 198

yes it did



#13
Apr3007, 05:10 PM

P: 26

Thank you. I went with the 1/4 wave transmission line since there doesn't seem to be a cut and dry method to do it otherwise without knowing a frequency.



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