I am a student researching physics and I am currently fixing some old equipment (RF signal attenuators for our acousto-optical modulators), I found the control voltage passing through a low-pass filter box as shown below, but I am not sure what this topology is called and why it has been used. Could someone enlighten me?

enter image description here

EDIT: Apparently this is a type of circuit suggested by Mini Circuits to use mixers as signal attenuators. We are putting signal into IF port, but usually IF port is where you pick your (RF +/-LO) signal. So if a high frequency noise reflects back from the LO port it will show up in the IF port, and the circuit is supposed to filter out whatever high frequency noise is reflecting back into the IF port.

  • \$\begingroup\$ Looks like something to match a 50ohm transmission line. What does ground attach to? The guard on the cable? what cables attach to the box? Is it BNC? \$\endgroup\$
    – Voltage Spike
    Jul 9 '19 at 17:00
  • \$\begingroup\$ @laptop2d, the ground is the ground of the BNC. The output of the box above connects to a mixer (Minicircuits ZAD-3H+). Specifically, it connects to the IF port of the mixer. The signal is some DC voltage, so that it controls how much power goes from RF to the LO port of the mixer. \$\endgroup\$
    – wcc
    Jul 9 '19 at 19:45

It’s an attenuator. The 5k and 50 ohm work as a voltage divider, the output is at 50 ohm impedance.

  • \$\begingroup\$ I don't think it works as a voltage divider, since in the ideal voltage divider one resistor is directly connected to ground, and your divided voltage is measured halfway between the two resistors in series. And at low frequency (the signal is mostly DC with some time-dependent ramp), there is only 5 kOhm resistor between in and out. \$\endgroup\$
    – wcc
    Jul 9 '19 at 19:47
  • \$\begingroup\$ The cap appears as a low impedance to an AC signal, so effectively it's a 100:1 attenuator. \$\endgroup\$ Jul 9 '19 at 20:34
  • \$\begingroup\$ For the AC signal I agree, but now I am confused why the first capacitor connected to the ground is even there. \$\endgroup\$
    – wcc
    Jul 9 '19 at 20:51
  • \$\begingroup\$ It's not clear to me either. From the description, seems like this block is meant to pass DC and attenuate AC. But the first cap would also attenuate AC even more, depending on its value and source impedance. \$\endgroup\$ Jul 9 '19 at 21:04
  • \$\begingroup\$ It seems the 5 kOhm resistor and the unknown cap form a first order low pass (and it seems the cutoff is around 500 kHz). And then the (50 Ohm + second capacitor) forms a voltage divider for AC signal but still boggles my mind why this is necessary. \$\endgroup\$
    – wcc
    Jul 9 '19 at 21:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.