- #1

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"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...

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- Thread starter ENGRstudent
- Start date

- #1

- 26

- 0

"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

- 1,497

- 4

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

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Thank you for your help.

- #4

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- #5

- 1,497

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- #6

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- #7

- 1,497

- 4

- #8

- 191

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

- #9

- 1,497

- 4

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

Yikes!!! this is RF Engineer's nightmare. :surprised

- #10

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

- #11

- 1,497

- 4

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

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yes it did

- #13

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