I just bought a new 26" fluorescent tube for my old Maytag electric range. It costs $15 instead of the $3 you might pay for a standard-length 30" at Home Depot. Instead of tossing out the old tube, I plan to salvage the electrical connectors and buy a few dozen high-power LEDs and construct a permanent replacement.
It would be nice if I could make it "drop-in" compatable. In other words, I would prefer to leave the ballast in place as well as the starter circuit. This way, if someone were to put the proper model of tube in later on, it would not explode in a shower of poisonous sharp fragments. (Like my wife if I pass away).
The second choice would be to disconnect the ballast and starter circuits so that a proper tube would simply do nothing. I would need to do this while my wife is out of the house, because she gets nervous when she sees me with wire cutters and screwdriver after the light-dimmer fiasco.
In either case, the first thing to do is either use a pair of diodes to split the AC into two separate strings of LEDs in series, or to use a full-wave rectifier feeding a single string of LEDs in series. Which is preferable?
Will the current in a string of LEDs be enough to "fool" the starter circuit into sensing a lamp that has properly started? (When the old light stopped working, the light would repeatedly re-start every few seconds. I assume that the current was not enough to heat up the bimetal strip in the starter.)
I presume that I do not need to provide a load to simulate the action of the filament coils at each end of the tube, unless this would end up interfering with the manner that the starter circuit uses to sense the current through the mercury vapour. Is this likely? I would use only two pins from the *same* side of the tube. (One from each end of the tube, but directly opposite, *not* diagonally opposite.) Does it matter which pair I use? (Before or after the ballast, and how do I tell which side is before or after?)
My next question is if I can assume a uniform voltage drop across each (forward conducting) active LED. I do not have the exact specs because I have not yet gone shopping for LEDs, but to keep the math simple lets assume I'm supposed to limit the current to 100 mA. If the LED is 100 mW, does that mean I can assume a one-volt drop across the LED? If so, do I need to add another resistor to limit total current, or would 120 LEDs in series provide sufficient resistance to handle 120 volts? I see no point in wasting power in a current-limiting resistor.
And how many LEDs in series does it take to remove the need for a rectifier? After all, the LEDs are diodes. Is each one "good" to block a fixed amount of back-voltage? If so, does the voltage depend on the junction chemistry, or it it a rating specified for the particular model of LED?
Also, does anyone have an idea where I can find a plastic tube 25 mm diameter? I'll probably end up breaking the glass. I understand that the phosphors are quite toxic, not to mention the mercury vapour. This will surely be an "outside" job during the summer.
PS - One university-level electronics course, so I probably know enough to be dangerous.