The heat loss (required to limit current flow to the diode) is shifted to the transistor instead of the current limiting resistor.it will waste considerably more energy as heat
Sorry I didn't make that properly clear - you're wasting heat in the transistor, which is best avoided.If you just have a mechanical switch in series with the base resistor, then either circuit will work fine. You just have to be aware that the transistor in your "positive switch" circuit will drop 0.8v or so from c-e vs 0.1 to 0.2v in the other circuit.
since you MUST drop some voltage and limit current anyhow, NOTHING is wasted. The heat is the same whether it is at the transistor or at the resistor. In fact, if you use the same components on both circuits, and you are satisfied with the 20mA of LEd current brightness vs the 30mA of brightness for the other, hour battery will last longer and you would be using less power with the Positive Switch and, unlike what Nigel said, you would actually be wasting LESS power.
again, this statement from Nigel is FALSE.
The heat loss (required to limit current flow to the diode) is shifted to the transistor instead of the current limiting resistor.