Driving an Arduino with 12 V

Question

I have an Arduino that I want to drive using a 12 V source. Of course, this can be achieved using a voltage regulator or similar solution. Notably, there is a loss in the conversion (at best, this is roughly 5 % with the best switching regulators), which is acceptable. That is all fine.

However, I also know that I need to drive a few LEDs. Let us for now ignore the fact that the LEDs work with 3 - 3.6 V and that the source, being a lead-acid battery actually differs in voltage depending on charge. Assume that I have a LED with exactly 3.5 V drop and place two in series with the Arduino, producing a voltage distribution as 3.5 V, 3.5 V and 5 V.

In the below circuit diagram, the Arduino is modelled as a resistance (commonly called an R-duino), but I suspect it is not very accurate.

Exactly, what will happen? My guess is that since LEDs are passive and the Arduino consists of active components, the current draw will be decided by the Arduino, causing the LEDs to shine more brightly when the Arduino draws more current and shine less when the Arduino is in sleep/idle. If we disregard the losses in cables and such, could this solution be considered having 100 % efficiency? If not, then where are the losses (note that light emission is desired and not a considered loss)?

Answer 1

In the below circuit diagram, the Arduino is modelled as a resistance (commonly called an R-duino), but I suspect it is not very accurate.

You're right - it's not very accurate. The Arduino runs on a steady 5 V supply but its current varies depending on its load and what it is doing internally.

Assume that I have a LED with exactly 3.5 V drop ...

Nope, we can't assume that as no LEDs behave this way. There are two problems:

• V_F, the forward voltage, is not a constant but varies with current.
• V_F varies from device to device.

Figure 1. I_f vs V_f for an LED nominal (blue) and spread (shaded area). Source: Variations in Vf and “binning”.

My guess is that since LEDs are passive and the Arduino consists of active components, the current draw will be decided by the Arduino, causing the LEDs to shine more brightly when the Arduino draws more current and shine less when the Arduino is in sleep/idle.

Correct, but note that when the Arduino is in idle that the If/Vf curve shows us that the voltage drop across the LEDs will decrease. This will increase the voltage to the Arduino and the end result will not be good.

If ... could this solution be considered having 100 % efficiency?

Nothing is 100% efficient, but you are right in that if your load didn't require a particular voltage - or could at least tolerate some variation in voltage - then your circuit makes use of power that would otherwise be wasted in the regulator.

^{simulate this circuit – Schematic created using CircuitLab}

Figure 2. Both circuits use 12 V x 10 mA = 120 mW. The second gives four times as much light for the same energy.

Note that very often we use this idea. If we require an LED indicator on a 12 V supply then 10 V will be lost in the resistor. We can add in additional LEDs without affecting the power budget.

If not, then where are the losses (note that light emission is desired and not a considered loss)?

LEDs dissipate heat as well as light.

Answer 2

Assume that I have a LED with exactly 3.5 V drop and place two in series with the Arduino, producing a voltage distribution as 3.5 V, 3.5 V and 5 V.

But that assumption is plain wrong.

Diodes have a flat V/I curve the higher you get in current, but for low current draws, the voltage across a diode changes drastically with the current you draw.

Don't use a diode for voltage regulation. Full stop.