0
\$\begingroup\$

I'm currently facing an interesting problem regarding RC low-pass filters. In my experimental setup, I have a reverse-biased photodetector (DET01CFC from Thorlabs). To amplify this signal, I make use of a low-noise amplifier (LNA-1440 from rf-bay).

As the photodetector has an internal capacitor, it functions as an RC low-pass filter with a cut-off frequency given by f = 1/(2piR*C), i.e this limits the bandwidth of our circuit. This is however no problem since with the 50 Ohm source impedance of the photodetector this cut-off frequency is way larger than our signal (which is around 300 MHz).

However, I was thinking that since according to Ohm's law the signal power increases with the increasing resistance, it is tempting to introduce an e.g. 1 kOhm feed-through termination between the photodetector and the amplifier to trade some bandwidth against the output signal (higher resistance increases the signal strength according to Ohm's law but according to the previous paragraph it also decreases the bandwidth)

Did anyone of you try something like this before and do you think it makes sense? I am also a bit concerned about the impedance matching, so maybe you have some thoughts about this as well :)

Thank you very much for your help!

\$\endgroup\$
3
  • 1
    \$\begingroup\$ Where did you read that increasing resistance causes increased signal power? That is not at all what Ohm's Law says. \$\endgroup\$ – Elliot Alderson Jan 18 at 18:59
  • \$\begingroup\$ Ohm's law can help you calculate the power absorbed by a resistor. But if you absorb some power in your 1kohm feedthrough resistor, then that is power that won't be available at the input of your LNA. \$\endgroup\$ – The Photon Jan 18 at 20:02
  • \$\begingroup\$ I read in the manual of the photodetector that for best frequency response, a 50 Ohm termination should be used whereas if bandwidth is not important, one can increase the amount of voltage for a given light level by increasing the resistance (see section 4.7 in thorlabs.com/…) \$\endgroup\$ – Sandro Camenzind Jan 19 at 7:14

Your Answer

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

Browse other questions tagged or ask your own question.