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An Explanation of a Classic Semiconductor Riddle

Back in 1996, Bob Pease posed an experiment in an April Fools column. “Take an ordinary NPN transistor, ground the base, pull the emitter up to 12 V with a …read more

Back in 1996, Bob Pease posed an experiment in an April Fools column. “Take an ordinary NPN transistor, ground the base, pull the emitter up to 12 V with a 1 KΩ resistor and measure the collector voltage referenced to ground.” Do the experiment, and you might be surprised to find a small negative voltage present on the collector. [Filip Piorski] has always loved the riddle, and has explained how it works in a Youtube video.
The key to the trick is the breakdown voltage of the transistor; normally somewhere around 7-8 volts for a typical small NPN transistor. At this point, where the base-emitter junction enters the breakdown regime, it begins to emit light. This light actually travels through the silicon lattice, where it reaches the base-collector junction, which acts like a photodiode under the right conditions. This generates the negative voltage seen at the collector under these conditions.
[Filip] goes on to try the experiment with a TO-3 transistor with the top cut off so he could visualise the effect in action. His photos, taken in a dark room, show tiny flecks of light appearing at spots on the silicon die. If you’ve got more insight on the effect in action, drop a comment below.
It might seem like a simple curiosity, however silicon junctions and their light emissions are an area of active research in semiconductor physics. Video after the break.

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