Why are smaller electronic devices more resistant to EMPs?

[dEdt]

According to an Oak Ridge National Laboratory technical report (http://web.ornl.gov/sci/ees/etsd/pes/pubs/ferc_Meta-R-320.pdf ), smaller electronic devices are better able to resist the damage caused by an EMP.

I'm at a loss to see why this is true. According to Ohm's law, the current density in a conductor is proportional to the electrical field inside it. If we agree that it is the large and anomalous E field created by an EMP which causes the damage, then it seems like it shouldn't matter how large or small the device is. What am I missing?

 

[jedishrfu]

I guess you could think of the device like an antenna and so smaller antennas or devices generate smaller currents in the same EM field with smaller currents less likely to burn out a circuit.

 

[dEdt]

Why would a smaller device have a smaller current?

 

[nsaspook]

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/efromv.html

The resulting current depends on the voltage differential across the device. For a given pulse step, smaller dimensions results in smaller voltage potentials and smaller currents for the same resistance. If the step voltage from an EMP is extremely fast the voltage gradient across even small electronic devices (wiring, circuit boards) could be in the tens of volts. This could cause much larger normal supply voltages for a short period of time and cause permanent damage to the junctions in the semiconductors.

 

[dEdt]

nsaspook, although the potential difference across a larger device is greater, so too is its resistance, so the current should be the same in a large device as in a small one.

 

[jedishrfu]

Why would a smaller device have a smaller current?

A smaller device is like a smaller antenna hence a smaller current in the given EM field. Just like a smaller solar cell absorbs less photons and so generates less current.

 

[willem2]

nsaspook, although the potential difference across a larger device is greater, so too is its resistance, so the current should be the same in a large device as in a small one.

The resistances in a larger device don't have to be larger at all. And they won't be relevant in protecting semiconductor devices, which will be the first to get damaged.

The potential difference is proportional to the length of cables or traces on a printed circuit board. There likely will be some reverse biased pn junction between two connectors on a chip, and it will break down when there's too much voltage across it. The resistance of the device isn't relevant at all in this kind of scenario.
Electronic chips often have input protection diodes that clamp the input between 0.7 below ground and 0.7 above the power supply. These diodes will be small however, and they can easily burn out. The resistance of the device isn't relevant here either.

Of course, if the device is connected to external cable, such as the powerlines, telephone cable or a network, the size of the device won't matter, but only the length of the external cables.

 

[nsaspook]

nsaspook, although the potential difference across a larger device is greater, so too is its resistance, so the current should be the same in a large device as in a small one.

As others have said simple resistance and current flow are not the primary factors. Large semiconductor devices with robust PN junctions in a typical power supply are unlikely to be damaged simply from the induced voltage directly on their structures but smaller sensitive digital, RF or audio FET devices could easily be destroyed by smaller levels of energy. Charger cables, ear phones and other external wiring could act as antennas to channel destructive energy into them.

Want to protect your toys from an EMP? Put them inside a metal trash can with newspaper insulation on the bottom and sides so nothing touches inside with the lid sealed with metallic tape.