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I have a photodetector with differential outputs that are each 50-ohm terminated. When I connect one output to the oscilloscope to look at the signal, I see a higher signal amplitude with higher channel impedance on the oscilloscope. According to my very basic understanding, maximum power transfer should occur when the impedance is matched. Could you help me understand what is happening here?

The pictures are taken from my scope, and you can see the amplitude of the signal when the termination is 1-Mohm is more than double what I get when the termination is 50-ohm. The cable connection to is as follows:

PD --> 50-ohm SMA --> SMA-BNC adapter --> 50-ohm BNC --> Scope

enter image description here enter image description here

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Do not confuse power transfer with voltage transfer.

If the source is indeed 50 ohms, and your scope is set to 50 ohms, it becomes a voltage divider and what you see is half the voltage that you will see when the scope is set to high impedance.

schematic

simulate this circuit – Schematic created using CircuitLab

Power transfer has to do with the relationship between the currents and the voltage. If the load resistor is zero, the current is max but the voltage across the load is zero so no power is transferred. If the load resistor is infinite, the voltage is max but the current is zero, so again no power is transferred. When Load resistance = Source Resistance V*I is at it's maximum.

enter image description here

But scopes do not work on power, they work purely on voltage. (or arguably current)

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  • \$\begingroup\$ Thanks. Yes, I suspected I was confusing both. Now, I'm digitizing the signal using a digitizer card with 50-ohm inputs, and all the information is encoded in the signal's amplitude, since the photodiode current is first converted by a trans-impedance amp. Dumb question #1: wouldn't the voltage reduction as such result in worse SNR? Dumb question #2: What would be a good solution to get the voltage transferred with minimal loss? A voltage follower maybe? \$\endgroup\$
    – Mohamed Tarek
    Commented Dec 5, 2017 at 17:51
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    \$\begingroup\$ @MohamedTarek You should ask those as new questions. But it depends on the circuit. \$\endgroup\$
    – Trevor_G
    Commented Dec 5, 2017 at 17:55
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    \$\begingroup\$ @MohamedTarek well if you want the voltage transferred with minimal loss presumably you'd use the high-impedance mode? But you'll get a more distorted signal. 50-ohm mode is to match the transmission line to reduce reflections (because all the power goes into the oscilloscope!) \$\endgroup\$
    – Criticizing Israel not allowed
    Commented Dec 6, 2017 at 0:52
  • \$\begingroup\$ What is your base time period signal ? Is it 2us ? Long cables ? \$\endgroup\$
    – user288518
    Commented Jun 22, 2021 at 7:38
  • \$\begingroup\$ @ SamGibson - ok for your remark. But I think that the paragraph "line transmission theory" in my link is the answer. It explain why the voltage is the "double" (or whatever) when you change the load impedance of scope from 1 Meg="infinity" to 50 Ohm. \$\endgroup\$
    – user288518
    Commented Jun 22, 2021 at 14:20

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