I was just reviewing a circuit and the author was concerned about a voltage loss resulting from using diodes that had a 3v breakdown voltage to be forward biased. It has been a very long time since I was in basic electronics and I'm just not sure my rememberer is working correctly. But, as I recall that is just the voltage required to acheive forward bias and those three volts are not lost. Ie 10v in is 10v out, not 10v in 7v out. Am I remembering correctly?

 

All PN junctions need a forward bias to go into conduction. Germanium doides need on the order of 200 mv, while silicon needs from 6 - 700 mv. LED's need more - that might have been an LED that needed 3 volts - http://www.allaboutcircuits.com/vol_3/chpt_2/6.html

 

The diode forward voltage drop must result in a loss of potential, because the sum of voltage drops and emfs around any loop is zero.

Generally the diode drop is a nuisance because it results in wasted energy. The power loss may be substantial when diodes operate at high currents, e.g. when they are used for power rectification. When calculating the output voltage of a rectifier you do indeed have to subtract the diode forward voltage(s) from the AC input. The voltage drop results in heat being generated, which limits the amount of current that a given diode can safely handle.

Sometimes diodes are utilised to produce deliberate voltage drops: here the requirement may simply be for an approximately fixed voltage, or alternatively the temperature dependency and / or the logarithmic variation of the voltage with current may be of use.