You don't.How would one go about simply cutting the voltage in half, say to 60V but allow the full current necessary to operate the bulb in the AC realm??
You don't.
Power in Watts = Voltage x Current.
also= Current(squared) x Resistance
and= Voltage(Squared/Resistance)
Not sure what you are referring to with respect to "That Rectifier". The rectification I had in mind was full bridge which does not cut the voltage in half, see figure 1. Figure 2, on the other hand does indeed cut the efective voltage considerably.That rectifier will only pass half the ac waves + its own voltage drop(negligible)= Abt half the voltage out and brightness + More lethal DC volatge
I'm quite sure jj didn't say that, any more than I did.In other words, your comments regarding a single diode is almost a solution to reducing the voltage in half. Adding a smoothing cap would produce a crude sine wave but with only a positive amplitude, a rippled DC voltage at about 80VDC maybe?
by Jake Hertz
by Aaron Carman
by Aaron Carman
by Jake Hertz