Directed Energy, Inc.

            What's meant by Peak Output Current?

            Because DEI voltage pulsers are designed to drive capacitive loads, they must be capable of high peak output current. Here’s the explanation.

            The current required to charge a capacitor is:

                  I = C (dV/dt)

            where I is current in amperes, C is capacitance in farads, and (dV/dt) is the rate at which the voltage changes in volts per second.

            Because pulsers deliver pulses with very fast rise and fall times, (dV/dt) can be a large number. For example, if a capacitor is charged from 0 V to 1500 V in 25 ns, the voltage rate of change is (1.5 • 103)/(2.5 • 10-8), or 6 • 1010 volts per second.

            DEI's specifications generally assume a load of 50 pF plus an additional amount for the pulser’s output circuit capacitance and the coaxial output cable. If the total load capacitance is 250 pF, the current is:

                  I = C (dV/dt) = (2.5 • 10-10)(6.0 • 1010) =  15 A

            This is the peak current required during the leading edge of a pulse; it’s also the current returned to the pulser during the falling edge. Little current flows to or from the load at other times. In this example, the pulser must have a peak current rating of at least 15 A. If it does not, the pulser may be damaged.

            The pulser’s 50 Ω to 90 Ω output impedance acts to limit the current when the load is a short circuit, which is what happens at the start of the pulse (the voltage across the capacitor is zero and can’t change instantaneously).

            So, to determine the maximum peak output current of a PVX-4150 (a 1500 V pulser with a 93 Ω output impedance), divide the voltage by the output impedance:

                  I = V/R = 1500/93 = 16 A

            This is the absolute maximum output current referred to in the manual, and it is not sustainable for two reasons: (1), the internal capacitor bank discharges into the load, thus the available energy decreases with time; and (2), continuous operation at that current would exceed the instrument’s safe operating area.

            Updated: 07 Feb 2019 05:59 AM
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