I’ve often wondered how “ground” works.

Let me explain: ground is of course the zero voltage reference level, which we implicitly think of when talking of a 3.3V or 5V signal, for example. You can only measure voltage between two points – it’s a relative concept.

But we all (well, I do, anyway) loosely talk about 0V and “ground” as if it’s an obvious thing. Not so, if you talk to electricians or engineers involved with power circuits.

First of all, there’s no such thing as a single ground potential. You can push a metal rod into the ground and think you’ve hit a solid reference but you might be surprised what two such metal rods pushed into the ground at some distance will do in a thunderstorm!

So why is grounding and ground level so tricky?

The problem is resistance. Every copper wire has some resistance. And where current flows in a resistance, you get a voltage drop (Ohm’s law, again – it’s everywhere!).

Let’s take a simple circuit, and let’s stay in the domain of JeeNodes and Physical Computing:

A 3.3V power supply feeds a JeeNode, which in turn controls a lamp. In normal low-power cases, things behave exactly as you’d expect. Ground is tied to the power supply, but both the JeeNode and the lamp are also connected on one side to ground, i.e. 0V.

I’ll redraw this to make the wiring resistance explicit:

Those resistance values will be very small, let’s say about 10 milliohm, i.e. one hundredth of one ohm.

Now imagine that you’ve got a super-powerful 33 Watt lamp instead, which draws 10 A. And assume the power supply is a big fat one which can supply those currents, and that the Relay Plug is also a beefy souped-up version.

With the lamp off, and the JeeNode drawing about 10 mA, the voltage drop over R1 and R2 will be around 0.00001 V (10 µV). Virtually undetectable.

Let’s turn the lamp on. Now a 10 A current will flow through R1, R2, R3, and R4. That means the voltage drop over each one will be 0.1V – so the JeeNode, for example, will be running at 3.1V now!

But that’s not all. Since the “ground” level of the JeeNode, i.e. the wire between R2 and R4 is now floating 0.1V above the power supply ground, everything the JeeNode does starts to float. The signal to the relay will switch between 0.1 and 3.2V – it’ll never be 0V (to ground) in fact.

For a relay plug, that doesn’t really matter – it’ll still switch on and off as intended. But for things like analog signals and ADC/DAC conversions, this really messes things up. If you don’t plan for it, all your analogRead() measurements could easily be 0.2V off, that’s a 6% error.

Now imagine driving the lamp with a big fat MOSFET instead and using PWM. It’s not hard to see that the JeeNode is almost literally going to be in a roller-coaster ride as the switching ends up affecting everything!

Remember seeing the lights dim briefly when a big heater is turned on? That’s due to cable resistance (and inductance) and floating grounds, in essence. So forget about an absolute ground – there’s no such thing.

Does that mean we can’t measure properly? Of course we can. Because there are some simple ways to deal with these resistance effects. The first obvious trick is to use fat wires for stuff that draws a lot of current. This lowers the resistance, and hence the voltage drop. Reduced voltage drop is why cables are sometimes much thicker than strictly needed w.r.t. current-carrying capacity.

But that doesn’t mean you need to waste copper all over the place. Have a look at this design:

Same circuit, but the JeeNode’s “ground level” is back in the microvolt range, even though it’s still switching 10 A. That’s because the current drops caused by the lamp currents no longer affect the JeeNode. R1 and R2 carry just 10 mA for the JeeNode again, even when the lamp is on: thin wires from power supply to JeeNode are ok.

Tomorrow, I’ll try to apply some common-sense to AC mains and multi-device grounding…