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 • 10^{3})/(2.5 • 10^{-8}), or 6 • 10^{10}
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 • 10^{10}) = 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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