# Designing driving LED circuit with electricity consumption in mind

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?

• "bipolar or MOSFET, doesn't matter" - well, yes it does, as a MOSFET doesn't require any gate current other than when you're turning it on and off Commented Apr 11, 2022 at 19:23
• Why not a single PNP, or a single-gate inverter? Commented Apr 11, 2022 at 19:33
• that is not driving LED in different, inverted way ... it is controlling LED in inverted way Commented Apr 11, 2022 at 19:33
• Please add reference designators to your schematic. The left two resistors can be increased to 10K. this will reduce significantly the circuit current in both the on and off states. Commented Apr 11, 2022 at 19:41
• Thanks to all for helping:)! I like the option with MOSFET. Commented Apr 19, 2022 at 15:46

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

• Be aware Vin must swing all the way to +5V to turn Q1 off - a 3.3V control signal won't do (unless V1 changed to 3.3V). Commented Apr 11, 2022 at 21:01
• Thanks for the reply! MOSFET seems to be the best in terms of an efficiency cause there's no current on the gate. "Depending on how you're controlling it, the gate circuit needs to be adapted"- what do you mean? For example, when we want to use this circuit to get the signal from the source which cannot give a lot of current, isn't enough to connect the MOSFET gate exactly to the signal source and control the LED? (even without the gate resistor as the MOSFET has very high gate impedance). And 2. question - I see 100 ms; about speed-can we expect that the MOSFET won't be as fast as the PNP? Commented Apr 11, 2022 at 21:13
• @Karlsson What I mean is that you gave no indication of what your IN signal is. For example, it could be just a switch between gate and ground, in which case you need a pull-up resistor. If it's a 5V logic gate, you don't need anything. The 100ms is an arbitrary time I chose for the simulation when the step generator switches from 0V to 5V. If you run the time-domain simulation, you'll see the current through the LED drop to 0 at that point in time. You could replace that step generator with a square wave generator to simulate pulsed switching. Commented Apr 11, 2022 at 21:45
• Thank you for the reply:) In these two versions does it matter significantly if we connect the diode to the collector vs emiter, or drain vs. source? Commented Apr 12, 2022 at 5:56
• @Karlsson Absolutely, it matters! Commented Apr 12, 2022 at 13:02

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

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.

• Not sure where you get the idea that a 3.3V logic high is needed. The question says logic high is 5V. Commented Apr 12, 2022 at 13:05
• Fortunately it doesn’t matter so long as the gate shares its supply with the host. Commented Apr 12, 2022 at 14:25