Energy in the Battery

First I will show that the battery isn’t going to contain a lot of energy. Suppose the goal is 1 watt of output from the battery (yes, I’ve shifted to power — which is energy/time — away from energy. I’ll fix that in a minute.) The constraint is that the battery is no more than 10nm on a side. A (smallish) car (gas or electric) can produce 150KW of power. Thus, it takes only 150000 such batteries to power a car — assuming the battery can produce that 1 watt of power continuously for a significant amount of time. How big would a car battery made from “chip batteries” have to be, to power a car? Let V be the volume of the required battery.

$$V=150000*{\left({10}^{-8}\right)}^3=1.5*{10}^{-19}{m}^3$$

How big is that? Assuming it’s a cube, then the length of the side is the cube root of that.

$$L={(1.5*{10}^{-19})}^{\frac{1}{3}}=5.31*{10}^{-7}m$$

That’s half a micron, or .02 inches on a side. Let’s assume that the battery only produces 1 W for one second. There are about 2.4 million seconds in four weeks. After multiple unit conversions, a half-cup of such batteries would power a car continuously for four weeks. If the GWF could make such batteries they could make a lot of money selling them to electric car manufacturers.

So that power (or energy) density just isn’t realistic. What is more realistic?

I’m going to consider a battery, 10nm on a side, with the same power density as an electric car battery. In an electric car, the goals are going to be the same as they will for the chip, namely, as much energy/volume and as little weight as possible. Since it’s easy to track down numbers for Tesla car batteries, that’s what I’ll use. Electric car batteries need to be inexpensive as well, and therefore, it’s unlikely that the batteries in electric cars are truly the bleeding edge battery technology, but it won’t matter. It will turn out that for the chip battery to be useful it has to be absurdly better than the batteries in an electric car.

How big are the Tesla batteries? The smallest of the three main varieties is

$$.685m*.30m*.079m=.0162m^3$$

(There’s a popular website that gets this wrong by a factor of ten, calling it .163; do the math yourself to check. If the larger .163 value is used, the result is even worse for the theory.)

How much energy does that battery hold? It varies, but 100KWh is the high end, so I’ll use that. (A watt is one Joule per second.)

$$100KWh=1.0*{10}^5*3600\frac{sec}{hour}=3.6*{10}^{8}Joule$$

The energy density per unit volume is:

$$\frac{3.6*{10}^8}{.0162}=2.22*{10}^{10}\frac{J}{m^3}$$

Let’s assume that the chip is actually all battery (it won’t matter in the end) and the energy density is the same as the Tesla car battery. How much energy can be stored in a 10nm-sided cube?

$$2.22*{10}^{10}*{\left({10}^{-8}\right)}^3=2.22*{10}^{-14}J$$

That’s 139KeV, 139 thousand electron volts. How long does the battery have to last? As shown elsewhere, the two most likely lengths of time the chip will remain in the body are 4 and 28 days. What’s the average power output for 4 and 28 days? Power is energy over time. For 4 days:

$$P=\frac{E}{t}=\frac{139000eV}{3600sec/hr*24hr/day*4day}=.4\frac{eV}{sec}$$

For 28 days:

$$P=\frac{139000eV}{3600sec/hr*24hr/day*28day}=.06\frac{eV}{sec}$$

To give you an idea of how much energy that is, a carbon-hydrogen molecular bond energy runs around 400 kJ/mol, which works out to be 4 eV. So even in the 4-day case, breaking a C-H bond requires ten times more energy than is available per second. In other words, the battery doesn’t have enough energy to break a typical molecular bond. The battery can’t power even negligible amounts of chemistry. Running a microcontroller, reading or writing to memory, etc., is hopeless.