The system uses a single 9-volt alkaline battery as the
input source and a simple switching circuit feeds an inverter transformer, is
rectified to a DC voltage and charges the output capacitor (.22 uF) to a maximum
of 2,000 Volts DC. Once the output capacitor reaches 2,000 VCD, the energy is
discharged through a spark gap directly into the output coil. The output coil
multiplies the voltage and generates the loaded output discharge as discussed
below.

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__Electrical Output Energy

The AIR TASER® Weapon generates 10-15 pulses per second. Each
pulse is approximately .4 Joule. Calculated by 0.5 * C * V ** 2 or 0.5. * .00000022
F * 2000 V ** 2 = .4 Joule. AT 15 pulses per second and.4 Joules per pulse,
6 Joules per second is the maximum delivered energy per second.

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__Loaded Output Discharge

When the output is connected to a 4,000 Ohm load, each
individual pulse of 3.5 uS (.0000035 second) in duration and measures 23,600
volts DC peak with a peak current of 5.9 Amps. The DC waveform is ½ sinusoidal.
The time-averaged current is .00022 Amps. The maximum average power output of
the AIR TASER is 6 Watts.

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__Electrical Test Setup

The output of the AIR TASER® Weapon is measured by connecting
a 4,000-Ohm wire would 5 watt resistor submerged in mineral oil (for insulation)
across the output leads. A high-speed (150-MHz) digital Oscilloscope (Tektronix
320) and a high voltage probe are also connected across the outputs.

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__Electrical Derivations

The __peak __current is calculated to be 5.9 Amps.

I (Amps) = E (Volts) / R (Ohms) ; 23,600/ 4,000 = 5.9
Amps).

The __peak__ power is calculated to be 139,240 Watts