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For my project, I would like to amplify the output from a photodetector for subsequent analysis with an oscilloscope.

The output of the photodetector is a current. Since we are using a voltage-amplifier, we would like to "convert" this output to a voltage by having a resistor after the photodetector but before the amplifier. Would you recommend using a resistor in series (e.g. a 50 Ohm series resistor) or would you rather use a t-piece with e.g. a 50 Ohm termination?

It would be really interesting to know the difference between those two options.

Thanks a lot for your help!! :)

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    \$\begingroup\$ Do some calculations of how much voltage you will get with what you propose. Also know that amplifiers exist that amplify a current. They're called Trans Impedance Amplifier or TIA. Do a search on "Trans Impedance Amplifier" to see how that is done. Especially if you need a fast response (high bandwidth) from your photodetector then converting the current into a voltage using a resistor is not going to work, you either a voltage that is too small or the bandwidth will be very limited. \$\endgroup\$ Jan 12 at 10:02
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    \$\begingroup\$ It depends on the type/model of "photodetector". \$\endgroup\$
    – Andy aka
    Jan 12 at 10:19
  • \$\begingroup\$ I also considered a trans-impedance amplifier, but currently I just have a voltage amplifier that I can use. How does the resistor limit the bandiwdth and how can I calculate the cut-off frequency? Regarding the type of the photodetector, I am using a InGaAs photodetector with 12 V reverse bias. \$\endgroup\$ Jan 12 at 12:01
  • \$\begingroup\$ @Sandro Camenzind -- The issue is most photodiode applications produce very faint currents. If you work out the current you are looking at, it'll determine how complex an amplifier you need. Feel free to include specific part number / datasheet link etc \$\endgroup\$
    – Pete W
    Jan 12 at 13:35
  • \$\begingroup\$ Do you know how to use a voltage amplifier as a TIA? It's very easy (but requires inverting configuration). \$\endgroup\$ Jan 12 at 14:30
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You could do it like the first diagram. The input current generates a voltage across R1, which is buffered and amplified by the amplifier. Unfortunately, you run into a very nasty compromise if you want speed from this arrangement. The voltage on the input of the amplifier has to move, which means it has to charge Cstray. Cstray and R1 form a lowpass time constant on the response of the system. Choose a big R1 for lots of gain, or a small R1 for lots of bandwidth, but you sacrifice the other parameter.

Cstray is inescapable. Photodiodes tend to have a large self capacitance, even if you can reduce that of connectors and cabling, and the amplifier input.

schematic

simulate this circuit – Schematic created using CircuitLab

In the second diagram, the virtual ground connection of the second amplifier 'short circuits' Cstray to ground. The amplifier input node does not have to move. You can choose R2 for gain alone, the amplifier takes care of the bandwidth. Spend more on the amplifier, you get more bandwidth. This configuration is called a TransImpedance amplifier, as R2 defines the ratio of voltage out to current in.

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  • \$\begingroup\$ That's great, thank you for the explanation! Out of curiosity: If the amplifier has an input impedance of let's say 50 Ohm, then in practice I would be in the situation shown on the left but I would not need to add anything else to the circuit, right? \$\endgroup\$ Jan 12 at 13:39
  • \$\begingroup\$ The other advantage of the right hand version is that the photodiode always has 0V across it; in the LH arrangement, you'll see the current output reduce once voltage approaches 0.5V (and the diode gets closer to conduction). \$\endgroup\$ Jan 12 at 14:34
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If the output of your sensor is current, then you convert to a voltage by connecting a resistor on the output to ground.

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  • \$\begingroup\$ So I would have this connection with resistor and ground in parallel to the signal that goes to the amplifier? \$\endgroup\$ Jan 12 at 11:58
  • \$\begingroup\$ Yes. The non-grounded pin of the resistor is where the voltage signal is. This is connected to the input of the amplifer. \$\endgroup\$ Jan 13 at 12:19

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