Electronic – Need some help designing a mains frequency converter (220v/50hz input, 220v/60hz output) for <$200USD

The main difference is that the passive circuit has an output impedance roughly equal to the resistor value. The active circuit has very low output impedance. That could mean you can get much more gain from the active circuit before the response time gets to be limited by the capacitance of whatever load is receiving the circuit output.

Also, in the passive circuit the photodetector bias will vary as its output current changes. The bias on the photodetector in the active circuit will be constant. This could be important if you care about the capacitance of the detector changing.

Both of these effects might mean that you can use a higher resistance value in the active circuit, which will reduce the Johnson noise contribution from that resistor, which gets to your question about accuracy. In either case you'll want to add a capacitor in parallel with the resistor in your circuit to limit the noise bandwidth of your receiver. Op-amp vendors have app notes describing how to optimize the noise in transimpedance amplifier circuits (here's one from National)

If you go with the active circuit, you'll also want to pick an amplifier with low thermal drift of the offset current and low flicker noise (assuming your main goal is to measure the dc light power).

I'm not much familiar with phototransistors, but I noticed the datasheet you linked shows nearly 2-to-1 change in collector current for temperature varying from 0 to 100 C. This is probably going to be the limiting factor in your sensor accuracy. Using a photodiode instead of the phototransistor, and making up the gain using the transimpedance amplifier circuit, and carefully selecting the op-amp and resistor you use, you ought to be able to achieve much better accuracy (I'm thinking on the order of a few % over 0 - 100 C, but I may be forgetting some critical issue).

Edit - One more thing

One minor difference (that might be important depending how you use the circuit) is that because its an inverting amplifier the active circuit will require both positive and negative power supplies (or some possibly accuracy-limiting hoop-jumping to avoid them), whereas the passive circuit may be easier to work in a single-supply environment. But even with the passive circuit you'll have to be pretty careful with the next amplifier circuit down the line to keep its accurate when the current is small.

I see that you're referring to the light fixture itself. In this case, the rating is the maximum that the fixture construction can withstand. Remember that the fixture is really only wires, switches, and connectors. It only serves to pass the current from the wall to the light bulb.

In your case, the fixture can handle any voltage up to 250V. You can use a 120V or a 240V bulb. The bulb would have to match the mains voltage.

660W is the largest bulb the fixture has been designed for, so that nothing gets too hot and/or melts. You can use a smaller-wattage bulb.

And the frequency doesn't matter to the fixture, although the bulb you choose might be affected (but probably not...)

That was the simple answer, but there is another consideration. Thanks to @david for pointing it out.

The ratings on the fixture should include maximum voltage and maximum current. The maximum voltage is simply the highest voltage that the fixture is rated to withstand, without arcing or breaking down in some fashion.

The maximum current, if given, will be independent of voltage. Often, however, the manufacturer will use a power rating instead of a current rating (660W in this example). To do this, they have to make some assumptions about the incoming voltage. Those assumptions reflect the manufacturer's expectations of where the product will be used.

For example, a 660W bulb draws 5.5A at 120V, or 2.75A at 240V. Were they expecting a 120V lamp? Probably so, if it was sold in an area that uses 120V. So, they chose wire that can handle 5.5A (or higher). Since this will also work with a 240V bulb, there is no problem.

But, say that you are using a 24V bulb (assuming you had a 24VAC power source). The fixture could certainly handle the voltage, and should work fine. However, if the bulb was 660W (the fixture's "maximum"), then the required current would be 27.5A. The wiring in the fixture will almost certainly fail at this current.

So, if in doubt, check to see if there is a current rating on the fixture :)