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For the following high speed laser diode pinout:

enter image description here

Another example is:

enter image description here

enter image description here

Is my model below correct?:

enter image description here

  1. My first question is what type of bias-Tee is that L and R form and what is the purpose of them? What I can understand is that an inductor can block RF frequencies (since it is 2πfL), and passes the DC. But I failed to progress the logic from here. Can someone explain how this R1 and L cooperate and explain the use of them?

  2. If I want to pulse this diode using a precision function generator (with 50 Ω output impedance) by 5 V pulse train with 1ns pulse width and 50 ns period how should I achieve that? I'm confused about the pinout of RF input and bias input in Figure 3. RF+ is outer connector and RF- is inner connector. So the K connector's outside is +RF and diode anode. But how to bias the diode in that case it seems there is no separate pin for RF and bias input. Can someone explain the difference between these two butterfly pinout and why second one has same node for RF and bias?

EDIT:

enter image description here

enter image description here

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  • \$\begingroup\$ ty_1917 - Hi, Please remember the site rule which requires that when a post includes content (e.g. text, image, photo etc.) copied or adapted from elsewhere, that original content must be correctly referenced. For online sources, the original webpage or PDF etc. should be linked as a minimum (the references for books / articles should include title, author(s), publisher, edition, page numbers etc.). Please can you edit your question to include the required source links for the original images? Thanks. \$\endgroup\$
    – SamGibson ♦
    Commented May 31, 2023 at 16:37
  • \$\begingroup\$ You really have too many questions here to answer in a single reply. I did my best to hit the main points. If I've missed something important to you please consider opening a separate question to address those points. \$\endgroup\$
    – The Photon
    Commented May 31, 2023 at 16:56

1 Answer 1

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My first question is what type of bias-Tee is that L and R form and what is the purpose of them?

At RF, the inductor will look like an open circuit (assuming you choose a high enough RF frequency for the inductor being used). Then looking into the RF port you just see the ~45 ohm resistor in series with the laser diode. Since the laser diode has a low differential resistance (1-5 ohms is not uncommon), you have a pretty good match for a 50 ohm source.

I would not add the 50-ohm external resistor you show in your proposed diagram for two reasons:

  1. You now have a ~100 ohm termination, not a good match.

  2. Your external diode will have to be connected via some wires or traces, and the inductance of these traces could be significant at RF frequencies. If you're using a K connector, presumably you're working in the 20-40 GHz range, and designing clean terminations is not trivial at these frequencies.

If I want to pulse this diode using a precision function generator (with 50 Ω output impedance) by 5 V pulse train with 1ns pulse width and 50 ns period how should I achieve that?

Attach the pulse generator to the RF port of the laser using coaxial cable.

Use an external DC block in the RF path so that the RF generator doesn't disturb the DC bias point of the laser.

But how to bias the diode in that case it seems there is no separate pin for RF and bias input.

Use the laser anode as a common RF and DC ground. Apply the DC bias to the through the inductor. This will be a negative voltage and your source will need to sink current.

Your pulsed source will also need to pulse negative to increase the laser output power, and go positive to reduce the laser output power. So you will set your pulsed source to be high for 49 ns and low for 1 ns out of each 50 ns period.

I want to pulse this diode ... by 5 V pulse train

You should work out what current you need to drive through the laser to get the optical power you want. Then work out the voltage from that.

Don't start by making an arbitrary choice of what the pulse voltage should be.

5 V into a 50 ohm termination will give you 100 mA pulse current through the laser. If that's what you want then you're fine. If you want more or less current than that, you need to adjust your pulse source voltage, not try to change the laser to work with 5 V pulses.

why second one has same node for RF and bias?

Pin 3 is for DC bias. Pin 12 is for RF. Pins 11 and 13 are for the common (ground) voltage.

I'm not sure why you're saying RF and bias are the "same node".

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  • \$\begingroup\$ Thank you for the answer. I tried to draw and simulate what you said. Could you look at my edit? I provided waveforms for bias current, laser current and RF pulse.Do you think it is fine now? \$\endgroup\$
    – ty_1917
    Commented May 31, 2023 at 16:36
  • \$\begingroup\$ I was confused with what is meant by RF(-) and RF(+) here : i.sstatic.net/m3ryy.png Normally inner side is positive but there inner side is shown as RF(-). What does that mean? \$\endgroup\$
    – ty_1917
    Commented May 31, 2023 at 16:37
  • \$\begingroup\$ And lastly if I dont bias this diode and directly apply 5V pulse train between the anode and and cathode by a function generator what would be the issue? Since the function generator has 50Ohm output impedance I can go upto like 60mA I was thinking, What would be an issue if I pulse this laser without biasing? \$\endgroup\$
    – ty_1917
    Commented May 31, 2023 at 16:37
  • \$\begingroup\$ @ty_1917 RF- on the center pin is just to tell you that light output decreases when the RF input voltage is negative and vice versa. Pulsing the current without bias might work if the signal is slow enough, but the laser isn't low impedance when shut off so you wouldn't have a good RF match for part of your signal cycle. The laser is also more likely to ring and chirp (or even mode-hop) strongly when turned on quickly from the off state. That might not matter or it might matter a lot depending on your requirements. \$\endgroup\$
    – The Photon
    Commented May 31, 2023 at 16:53
  • \$\begingroup\$ Thank you, regarding your comment "RF- on the center pin is just to tell you that light output decreases when the RF input voltage is negative" I applied negative RF as you said and in the waveform here i.sstatic.net/0xFEq.png But in my case as you see the light output(LD current) is increasing when the RF input goes to negative. (?) \$\endgroup\$
    – ty_1917
    Commented May 31, 2023 at 17:07

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