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 Top
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 Top 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)) Top
Brian Post subject: Posted: Tue Aug 16, 2005 7:16 am
IR / Joe, Thanks for the replies. Brian
Posted 11/12/2012
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