How to run my laser diode?

[(*steve*)]

Just beware that the capacitor MUST be non-polarized (foil, ceramic, etc), since AC flows through. And the minimum voltage rating for it should be = RMS supply voltage * sqrt(2) + some few volts extra for added protection.

It would almost certainly be best to select a capacitor that is specifically rated for mains use.

Here is a good tutorial which illustrates why ceramic capacitors are a VERY BAD choice in this situation.

I have seen this method used by someone who did not fully understand the issue, and like climatex, assumed a ceramic capacitor would be OK. The end result was a LED which worked for a time. Eventually a large enough spike (presumably) came along and the capacitor failed. In this case the end result was a very dead LED, fortunately not an expensive failure.

TBennettcc: what is happening here is that the capacitive reactance is acting to limit current. They will have a shocking power factor, but at these low currents is probably not going to be an issue.

The voltage is kept low by virtue of low impedance of the load. Essentially you have a high voltage source in series with a high value resistance, a classic current source.

Use of reactance rather than resistance minimises the power loss by shifting the phase relationship between voltage and current.

 

[(*steve*)]

You can't compute it exactly, however it is fairly trivial to determine the approximate voltage by looking at graphs of Vf vs I and applying ohms law for the series resistance.