An antenna will collect charge from charged particles hitting it, snowflakes and raindrops. It also acts as a capacitor to store the charge. A bleeder resistor may be used to to drain the charge before it builds up and makes any damage.

If there is nothing in place to drain the collected charge from the antenna, it may be manifested with high voltage and arcing at the equipment end. This is because the capacitance of the antenna is small. If it were larger, the voltage potential against ground would be lower.

Arcing and associated high voltage spikes caused problems at my linear amplifier. In the amplifier the SWR –measuring circuitry consists of a toroid coil (transformer), picking its primary voltage from the antenna lead, to a pair of Germanium diodes and other components. The spikes were too much for the diodes, which I had to replace - and figure out a system to prevent this of happening again - I came out with simple solution: a bleeder resistor.

**Reader's Question: Using the charge collected by the antenna?**

In theory one could use this collected charge and its stored energy to charge, say, a battery. However there are some obstacles for practical use.

- the charge collected is very, very tiny; It would take a very long time to collect enough to charge a battery

- the voltages associated are very high; it would be difficult if not possible at all to find a practical circuitry to lower them to the battery level and simultaneously maintain the small stored energy

**Back to the school bench and do some analysis**

Charged particles are either negative or positive, so the system is DC, making the analysis easier. Let’s think of charging a small rechargeable AA cell. Typical rechargeable AA battery cells are rated with a charge between 800 to 3000 mAh. Let’s assume we want to charge a 1.2 Volt and 2400mAh battery cell from empty to full. This means we need 2.4Ah or 2.4x3600As or Coulombs of charge moved into it (1 Ampere = 1 Coulomb/second).

**A closer look**

If the antenna capacitance is say 200pF and the potential (voltage) between antenna and the ground when charged, just before arcing, was 6kV, and arcing happened twice per second, we can calculate the amount of charge and energy stored and released in the system. Let’s also assume the charge collected is more or less uniformly distributed across the whole antenna. Also we assume a steady flow of charged particles hitting the antenna from snowing/rainfall.

Now the charge collected by the antenna capacitor raises the voltage to say 6kV before arcing. Electromagnetic theory tells: Q = C x U where Q = charge stored in the capacitor, U = potential across the capacitor, C = capacitance. This implies U = Q / C, or the smaller the capacitance, the larger the voltage for any given amount of charge.

**Collected charge and current to the antenna**

For the antenna in case: Q_{Ant} = C_{Ant} x U_{Ant} = 200x10^{-12} x 6000 Coulombs = 1.2 x 10^{-6} C = 1.2μC (microCoulombs). This charge is collected twice per second, so the charge collected from snowfall/rain in unit time of one second will be 2.4μC.

Assuming the flow of charge to the antenna is steady and 2.4μC/s, then the current I_{Ant} to the antenna wire, which is defined as amount of charge moved per unit time or I = Q / T, would be 2.4μA. This would be across the whole exposed antenna. Note that this implies Q = I x T, so the antenna cumulative charge in a second can also be expressed as 2.4μAs.

**Direct battery charging with the charge collected?**

If we could somehow move this charge flow (=current) hitting the antenna without losses directly to a battery cell, ie. charging it from “empty” to full with this steady 2.4μA current, it would take 2400 x 10^{-3}Ah / (2.4 x 10^{-6}A) = 10^{6} h = 1 million hours, 114 years – or a couple of lifetimes. Really a trickle charging. Obviously a direct connection moving the charge to the battery would not be efficient.

**Usable energy of the stored charge in the antenna capacitor**

However, remember that we have a capacitor in place – and high voltage. Theory tells that the energy which can be extracted from a charged capacitor is E = 1/2 x C x U^{2}.

The energy, in form of stored charge and voltage potential in the antenna capacitor, just before arcing, would be E_{Ant}=1/2 x C_{Ant} x U_{Ant}^{2} = 0.5 x 200x10^{-12} x 6000^{2} Joules = 3.6 x 10^{-3} Joules = 3.6 milliJoules = 3.6 milliWattseconds. We would have this energy available 2 times per second, or total of 7.2 mWs.

The usable energy in a 1.2 Volt 2400mAh AA cell is roughly 1.2V x 2400 mAh = 2.88 VAh = 2.88 x 3600 Ws = 10368 Ws (or Joules). This energy is the same as the potential energy of 1 kg mass at about 1km height, ie. in ideal situation you could lift 1 kg mass up 1km in the air with an AA cell. Pretty impressive!

**Antenna energy vs battery energy**

Now when taking the ratio of the above antenna capacitor’s charged energy to AA battery’s energy, one finds that the AA cell has roughly 1.44 million times more energy stored, than is collected by the antenna capacitor in a second.

This means that the antenna should collect, and after reaching 6kV, store forward the charge twice a second, for over 1.44 million seconds, for the stored energy to reach the level inside of an AA-cell. This is nearly 17 days. A long time still. I would not like to live in a place where there were constant raining/snowing for 17 days or more in a row .

The apparent difference between 1 million hours and 1.44 million seconds comes from the assumption that the charge in the antenna would be at 6kV potential instead of 1.2 Volts and we could convert the energy it contained into energy in the battery.**Any practical circuitry?**

The second obstacle would be how to realize a practical charging circuitry – from 6 kV down to 1.2 Volts, without huge losses.

For example, one might just think of adding a larger capacitor C_{2} to the bottom of the antenna, where the collected charge would be moved. This capacitor would be switched in as soon as the voltage reaches the arcing level. Following the U = Q / C law, we could lower the voltage to closer to the battery charging voltage. This capacitor would get the charge twice per second.

Let's assume we want to lower the charged antenna system potential to 2V, to be able to charge a 1.2 Volt battery cell. Then a charge amount of 2.4μC will be moved from the antenna capacitor to a second larger capacitor

C_{2} = Q_{Ant} / U_{2} = 2.4 x 10^{-6} / 2 = 1.2 x 10^{-6} Farads = 1.2μF

Which is actually quite normal capacitor for 2 volts (but not for 6 kV). The 200pF antenna will be negligible when in parallel with this. However the energy usable from C_{2} will be much less as it follows E = 1/2 x C x U^{2} law.

E_{2 }= 1/2 x C_{2} x U_{2}^{2} = 0.5 x 1.2x10^{-6} x 2^{2} = 2.4x10^{-6} Joules = 2.4 microJoules or 1/3000^{th} of the energy usable from the antenna capacitor. Accordingly charging the battery would take at least 3000 times longer and we come back close to a million hours.

**Conclusions**

Of course this is nothing new. Utilizing the static - or friction electricity has been studied for over 200 years - without many practical results. If there were efficient means of extracting energy from static, they would have been invented a hundred years ago. But for me this was good and delighting exercise to refresh some math/physics learned some 30+ years ago.