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Regarding my previous question, I made a test with this photodiode with an attenuated(attenuation is unknown) 1550nm pulsed laser where it has 1ns pulse width and 1kHz rep rate. The circuit I used is as follows coupled to 39pF/10MegOhm terminated oscilloscope probe:

enter image description here

And in oscilloscope I obtain the oscilloscope waveform:

enter image description here

The peak in the above pulse waveform is around 235mV.

Now by using this data I want to estimate the optical power seen by the photodetector.

I wanted to find the current and convert it to power by using the 0.85A/W parameter. I basically was dividing the peak voltage to resistor 235mV by (Rload//Rscope). So first I though the photo diode current is 0.235mA.

But then when I made simulation considering the photodiode model I get much more different result(due to Cj).

Here is using the diode data from the specs:

enter image description here

In this case, I match the experimental data only if I set the pulse current peak to 800mA:

enter image description here

So I'm very much confused since there's huge difference between these two currents. And the max reverse current in datasheet is given 10mA and still the diode works. Which of these two more realistic to estimate the input optical power seen by the photodiode?

Extra info:

I also made the same measurement with 50 Ohm scope load: enter image description here

And the scope shows the following waveform:

enter image description here

where the peak is around 396mV

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  • \$\begingroup\$ Or should I integrate the current waveform to relate it to the total power delivered per pulse? But I was looking for peak power not the exceed the datasheet max. \$\endgroup\$
    – floppy380
    Commented Dec 21, 2023 at 18:40
  • \$\begingroup\$ You won't be able to measure the peak power of a 1 ns pulse using a 600 ns fall time detector, at least not directly. The fall time is far too slow, so it will give you an average. You could however calculate it easily enough once you know the pulse energy. \$\endgroup\$ Commented Dec 21, 2023 at 18:44
  • \$\begingroup\$ 2nd 'scope measurement @ 50 ohm load has leading-edge ringing - without ringing, it peaks around 250 mV. Now compare that with 1st measurement across 980 ohm load, which also peaks @ 250 mV. These two measurements don't jive. Take a close look at that leading edge...I suspect it is a linear ramp, up to that 250mV peak. A possible cause is photodiode capacitance effectively integrating that short optical pulse. \$\endgroup\$
    – glen_geek
    Commented Dec 21, 2023 at 21:29
  • \$\begingroup\$ @glen_geek Thats true theres ringing in 50 ohm case which i used bnc with crocodile terminals to a lab scope lecroy. Ringing might be due to not using proobe. And the scope was set to 50Ohm. In the first case however I used this probe farnell.com/datasheets/2238502.pdf Tpp1000 and the moment I plugged in to lecroy scope 50 Ohm choice disappeared. So I took it as 10Meg which is probe impedance. Can that be its 10X and the first one I plot is actually 10 times larger voltage? \$\endgroup\$
    – floppy380
    Commented Dec 21, 2023 at 22:19
  • \$\begingroup\$ But it if was 2V, the current would be around 2mA that means there would be 2V voltage drop across the Rs and the voltage at cathode would be 0.6V and anode 2V which doesnt make sense :(( \$\endgroup\$
    – floppy380
    Commented Dec 21, 2023 at 22:29

2 Answers 2

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How to estimate the optical power given the scope reading for this scenario

You don't have to estimate anything, you're directly measuring it. You are measuring the voltage produced by a photocurrent across a 980 Ohm resistor. Using Ohm's law, your current is I = V/R = V/980 ohms. Since you get 0.85A/W, your power is 0.85I = 0.85/980V or 867 microwatts per volt.

You can integrate those measurements to get energy per pulse or average it over time to get average power. You can't get peak power since your photodiode is too slow, although if you know the pulse duration and you measure the pulse energy, you can can calculate the approximate peak power by dividing.

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The peak doesn't measure the peak current. The photocurrent charges the junction capacitance, changing the detector bias. That charge then bleeds away through your load resistor.

The charge is just CV, (5 nF)(235 mV), 1.2 nC. Divide by 0.85 A/W, get 1.4 nJ/pulse.

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  • \$\begingroup\$ How can we estimate the optical pulse energy or average power by using the scope data in my question? Also knowing the repetition rate and optical pulse width. \$\endgroup\$
    – floppy380
    Commented Dec 22, 2023 at 2:16
  • \$\begingroup\$ Read the answer: it gives the optical pulse energy, 1.4 nJ. \$\endgroup\$
    – John Doty
    Commented Dec 22, 2023 at 2:19
  • \$\begingroup\$ Oh so 1.4nJ per pulse and if the rep rate is 1kHz and pulse width is 1ns, so the optical average power is 1.4 micro Watt and the peak power estimate is 1.4W? Sorry I ask just to be sure I understand you. Going from total charge is interesting(I think it is because the diode is slow so we can do it) \$\endgroup\$
    – floppy380
    Commented Dec 22, 2023 at 2:25
  • \$\begingroup\$ Yes, Power is energy/time. \$\endgroup\$
    – John Doty
    Commented Dec 22, 2023 at 2:28
  • \$\begingroup\$ Ok very interesting; if these are correct the thing I need to know from the vendor if the detector can handle 1.4W peak power for 1ns duration. I also will run this in 3MHz rep rate so I need to know what is the max optical peak power for this diode which doesn't exceed the damage threshold. \$\endgroup\$
    – floppy380
    Commented Dec 22, 2023 at 2:38

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