# Computing stuff tied to the physical world

## Watts, Amps, Coulombs

In Hardware on Nov 26, 2012 at 00:01

Sometimes I see some confusion on the web regarding the units to measure power with.

Here’s a little summary, in case you ever find yourself scratching your head with this stuff:

• Electric potential is sort of a “pressure level” when using the water analogy, expressed in Volts (V)
• Current is the flow of electrons, and is expressed in Amperes (A)
• Charge is the “amount of electricity”, and is expressed in Coulombs (C)
• Power is the product of volts and amperes, and is expressed in Watts (W)

Another measure of power is Volt-Amperes, this is not the same as Watts in the case of alternating current with reactive loads, but let’s not go there for now…

To summarise with the water analogy:

• Volts = how high has the water been pumped up
• Amps = how much water is flowing
• Coulombs = the amount of water
• Watts = how much energy is being used (or generated)

You can probably guess from this list that pumping water up twice as high (V) takes twice as much energy, and that pumping up twice as much (A) also takes twice as much energy. Hence the formula:

``````    Watt = Volt * Ampere
``````

Other equations can also help clarify things. They all add time into the mix (in seconds).

Current is “charge per second”:

``````    Ampere = Coulomb / second
``````

This is also the way I estimate average current consumption when diving into ultra-low power JeeNode stuff: using the oscilloscope to integrate (sum up) all the instantaneous current consumptions over time, I get a certain Coulomb (or micro-coulomb) value. If that’s a periodic peak and the system is powered-down the rest of the time, then the estimate becomes: X µC used per Y sec, hence the average current consumption is X / Y µA. The advantage of working with Coulombs in this way, is that you can add up all the estimates for the different states the system is in and still arrive at an average current level.

Another one: power consumption is the amount of energy consumed over time. This is often expressed in Watt-hour (Wh) or kilowatt-hour (kWh):

• two 100 W lightbulbs running for 5 hours = 2 x 100 x 5 = 1000 Wh = 1 kWh
• one LED strip drawing 2 A at 12 V for 3 hours = 2 x 12 x 3 = 72 Wh

And then there’s the “mAh” unit used with batteries, i.e. milli-ampere-hour. Quite simple again, once you get used to this metric system, and realise that you also need the voltage:

• 2 AA batteries of 1.5V @ 2000 mAh each provide 2 x 1.5 x 2000 = 6000 mWh = 6 Wh
• a 5 mA load on batteries of 2000 mAh will run for 2000 / 5 = 400 hours

Battery capacities are roughly as follows for the most common types:

• an AA cell has 2500 mAh @ 1.5V = 3.75 Wh
• an AA rechargeable cell has 2000 mAh @ 1.2V = 2.4 Wh
• an AAA cell has 1000 mAh @ 1.5V = 1.5 Wh
• an AAA rechargeable cell has 800 mAh @ 1.2V = 0.96 Wh
• a CR2032 coin cell has 200 mAh @ 3V = 0.6 Wh

Wanna be able to run for a week on a coin cell? Better make sure your circuit draws no more than 200 / (24 x 7) = 1.2 mA on average under optimal conditions.

Wanna make it run a year on that same coin cell? Stay under 22 µA average, and it will.

With 2 or 3 AA batteries, you get an order of magnitude more to consume, so if you can get the average under 200..220 µA, those batteries should also last a year (ignoring the fact that batteries always have some self-discharge, that is).

The difference between 2, 3, or 4 AA batteries in series only affects the voltage you get out of them. Chips do not run more efficiently on more voltage – on the contrary, in fact!

For low-power use: run your circuit on as low a voltage as possible, but no lower (wink).

1. Aaaarrr! You have fallen into the word-trap confusion that many do when trying to explain or understand these terms.

Your statement “pumping water up twice as high (V) takes twice as much energy” is correct, by itself (because you do not mention the rate).

However, because (as you state) an ampere is a unit of current flow (i.e., coulombs per second), then the continuation of your statement “and that pumping up twice as much (A) also takes twice as much energy.” is non-sensical. Pumping “twice as much” amperes is pumping at twice the rate, so requires twice the energy per unit time, i.e., twice the power (not twice the energy).

2. Another question one can ask himself: “Does temperature affect battery life?”… You’d be surprised to see that at lower temperatures battery life can be so far from expected. Certainly something to keep in mind as well when using battery powered wireless sensors at temperatures around freezing point or below.