Designing driving LED circuit with electricity consumption in mind

Question

I have a question about driving LED in different, inverted way: the LED should light on when the signal/state is LOW (0V) and the same LED should be turned off when the signal is HIGH (5V).

I have attached a schematic which this solution with two npn transistors to do it simply without specific IC's.

But I wonder what about efficiency. If we think about using ONLY transistors (bipolar or MOSFET, doesn't matters), is it the best way (I mean this circuit configuration strictly) to drive the LED with a reasonable low current drawn by the circuit? I ask because driving LED circuit in inverted way requires to be powered the whole time. In other words - I am asking about the most current-efficient LED control path when using only transistors, when we want the LED to shine (the diode current should be sufficient to be able to shine clearly) when the IN pin is low. Or maybe is it possible to do it by other type of transistor than npn?

Answer 1

If MOSFETs are allowed, then this will probably be your best bet. Depending on how you're controlling it, the gate circuit needs to be adapted. This not only eliminates the idle current through your first transistor, but it draws no current from the control signal in steady state:

^{simulate this circuit – Schematic created using CircuitLab}

But, you could still accomplish nearly the same with a single PNP. The control signal will need to sink a tiny bit of current when the LED is on:

^{simulate this circuit}

Answer 2

The original resistor calculations are wrong and waste a lot of current. Calculate the resistors like this:

Answer 3

The 74LVC1G07 is your huckleberry. Power it with the controller's supply (e.g., 3.3V) to ensure the input-high level is compatible. The output will tolerate 5V and will not have a parasitic leak path. No pullup/pulldown needed, no wasted current.