Impedance matching attenuator - RF Cafe Forums

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Brian

Post subject: Impedance matching attenuator Posted: Mon Aug 15, 2005 12:02 pm

Hi all,

I'm trying to impedance match a non-50Ohm DUT to a 50Ohm VNA. I've seen an explanation of how to do this using minimum loss pads (see "Spectrum and Network Measurements" by Robert A. Witte), but any commercial attenuators I see are simply rated in dB attenuation. Do I need a special 'impedance match' attenuator, or can I use any which attenuate more than the theoretical minimum loss? Thanks for your help.

Regards,

Brian

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IR

Post subject: Posted: Mon Aug 15, 2005 1:03 pm

Site Admin

Joined: Mon Jun 27, 2005 2:02 pm

Posts: 373

Location: Germany

Hello Brian,

Any resisive element has an attenuation. Therefore, an attenuator that is based on resistors will have an attenuation. This kind of attenuator will decrease (improve) the return loss by 2dB for each dB of attenuation.

The only type of matching networks without loss (theoretical) are reactive matching networks i.e networka that are based on capacitors and inductors. However, practically, these network are lossy due to the Q of its components -especially the inductors.

Hope this helps.

_________________

Best regards,

- IR

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Joe

Post subject: Min loss PadsPosted: Mon Aug 15, 2005 2:58 pm

Yes, you do need special matching attenuators for this task. Most standard attenuators are 50 ohm in and out.

If you are looking to match from 50 to 75 ohms there are some sources out there. I have used a relatively inexpensive source in the past. The insertion loss is 5.7dB.

https://www.smelectronics.us/coaxialimpe ... ngpads.htm

You can also make your own for any impedance combination using 1 series and 1 shunt resistor.

Min Loss Pad:

0--------R1--|--------0

_________|

Z1_______R2_____Z2

_________|

0-------------|--------0

Z1>Z2

S=SQRT(Z1/Z2)

K=SQRT(Z1/Z2) + SQRT(Z1/Z2 - 1)

R1=(Z1/S)*(K-S)

R2=(Z1/S)*(K/(K*S-1))

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Brian

Post subject: Posted: Tue Aug 16, 2005 7:16 am

IR / Joe,

Thanks for the replies.

Brian

Posted  11/12/2012