Electronic – Scattering matrix Transmission Matrix lasers

1) Is it really as simple as grabbing a laser drive chip and a three-terminal diode from digikey and hooking them up according to the datasheet? Should the laser drive chip be able to handle all protection mechanisms necessary, or is there typically another device that's needed to handle some other form of protection?

The laser drive chips I'm familiar with are more about applying rapid modulation to the laser than providing DC power. Usually there's an additional power circuit required; and that power circuit is where the protection is normally implemented.

If you have a different type of drive chip in mind, please link the datasheet in your question.

2) Is there a central regulatory body that does any testing to determine what class of laser you have, and whether your product follows all the necessary regulations?

In the US, it's up to the laser manufacturer to self-certify their product. You may be able to find a consultant to assist you with that process if you don't have the expertise.

3) Are there any known issues using lasers with 1mm core plastic optical fiber? I know that POF has very different transmission windows vs. glass fiber, and I know that one of these optimal windows is 650nm.

Would the beam stay narrow inside the fiber, or would it begin to disperse?

The fiber is a waveguide, and the laser power will remain confined within the fiber core. It will attenuate (lose power over distance). There is also a process called dispersion which means different components of the laser power taking different amounts of time to traverse the fiber---but if you're not switching the signal quickly that's not likely to affect you.

Edit: A major difference between POF and glass fiber is that even in its transmission window, POF has much higher attenuation than glass. Attenuation in glass fiber is measured in tenths of dB per km. Attenuation in POF (last time I worked with it, several years ago) is measured in tenths or whole dB per meter.

Would it still be coherent and collimated after going through, say, 15 meters > of POF?

The signal will still be coherent, but the dispersion effect I mentioned above may reduce the coherence length if you've gone through a very long fiber.

The output beam will diverge at a substantial angle (not strictly collimated) when it exits the fiber. The divergence is a diffraction effect and the angle is inversely related to the fiber core diameter --- meaning POF will have a lower divergence angle than smaller-core fibers. In multi-mode fiber like POF the output divergence angle also depends on details of the fiber construction. In general the output divergence angle will be similar to the input acceptance angle.

I am investigating the laser approach, because it seems like most LEDs aren't even capable of 500 uW.

It doesn't matter much what most LEDs can do --- if you can find one LED that meets your needs, that is enough. And I think you should be able to find an LED to produce 1 mW and couple into POF, if you look long enough. But a laser should be able to do it more efficiently (but maybe more expensively).

Edit: Be aware that using an LED does not reduce your safety concerns. 1 mW is still 1 mW and can still be dangerous. You will want the same safety precautions (you mentioned open-fiber control) whether you use a laser or LED. Regulations have not all kept up with the improved capabilities of LEDs in recent years, but that doesn't mean you shouldn't protect yourself and your users.

The problem with using the method you suggest is that if you move the impedance first, you will indeed have 50\$\Omega\$ in the real part plus some other term as the imaginary part, but remember that the stub you are placing is going to be in parallel with the load. If you were to place component such as a capacitor in series with this new impedance that you found, it should be fine.

Think of it this way. Your original load impedance is \$Z_L=X+JY\$, your new impedance after moving toward the generator is \$Z_{L,new}=50+jK\$ (not normalized). If you had the option to place a series capacitor or inductor (which depends on the sign of \$K\$), you would choose it so that this series component has an impedance of \$-jK\$, you add them up and you have a matched system.

Back to the beginning. The problem with the stub is that you place it in parallel with the load. If you try to find the admittance of the load with the series transmission line at this point, you get something like:

$$Y_L=\dfrac{1}{50+jK} = \dfrac{50}{K^2+2500}-\dfrac{jK}{K^2+2500}$$

And all the stub can do for you is to cancel out the imaginary part, it doesn't do anything to the real part. Recall the admittance of the stub has the form \$ Y_{stub}=jP\$ where P could be negative or positive depending on the stub type (short or open).

You can eliminate the imaginary part, but when you try to find the inverse of the admittance (to get the 50\$\Omega\$ equivalent impedance you have been looking for), you get

$$Y_L=\dfrac{50}{K^2+2500}$$

and $$Z_{matched}=\dfrac{K^2+2500}{50} $$

And as you can see you would have successfully matched the impedances if \$K=0\$. But this isn't a possible scenario which you can verify, \$K\neq0\$ . So \$Z_{matched}\neq50\Omega \$.

Why use admittance instead of impedance from the beginning? Because it is algebraically easier to add rather than using the inverse of the sum of the inverse to find the equivalent impedance. The stub is placed in parallel.

Think about it, if you have the impedances \$Z_1\$ and \$Z_2\$ in parallel, in order to find the equivalent impedance you would need to do

$$Z_{eq}=\dfrac{1}{\dfrac{1}{Z_1}+\dfrac{1}{Z_2}} $$

But a simpler way would be to find what you have in the denominator, which are the admittances. Call \$Y_1=\dfrac{1}{Z_1}\$ and \$Y_2=\dfrac{1}{Z_2}\$. If \$Y_2\$ were the admittance of the stub, you can choose it so that the imaginary part of it cancels out the imaginary part of \$Y_1\$, that is, if

$$Y_1=A+jB $$

You would wisely make the stub be \$Y_2=-jB\$ (Remember that \$Y_1\$ and \$Y_2\$ are added in the denominator of \$Z_{eq}\$).

This will make the denominator in \$Z_{eq}\$ purely real. And so \$Z_{eq}\$ will be purely real.

You just have to make sure that \$A=1\$, which you get by rotating to the \$Y=1\$ circle. And since impedances are normalized in the Smith chart(50\$\Omega\$), you would have something like:

$$Z_{eq,actual}=50\dfrac{1}{Y_1+Y_2}=50\dfrac{1}{A+jB-jB}= 50\dfrac{1}{A}$$