One very popular application for inductors is storing energy in the present for release in the future. This energy is in the form of a magnetic field, and current is required to maintain the field. In an analogous situation, the capacitor stores energy in an electric field, and a voltage

Question

One very popular application for inductors is storing energy in the present for release in the future. This energy is in the form of a magnetic field, and current is required to maintain the field. In an analogous situation, the capacitor stores energy in an electric field, and a voltage across the capacitor is required to maintain it. As an application of the inductor’s energy storage capability, let us consider the high-voltage pulse generator circuit shown in Fig. 7.26. This circuit is capable of producing high-voltage pulses from a small dc voltage. Let’s see if this circuit can produce an output voltage peak of 500 V every 2 ms, that is, 500 times per second.

Accepted Answer

At the heart of this circuit is a single-pole, double-throw switch—that is, a single switch (single-pole) with two electrically connected positions (double-throw), 1 and 2. As shown in Fig. 7.27a, when in position 1, the inductor current grows linearly in accordance with the equation

i(t) = [latex]\frac{1}{L}\int_{0}^{T_{1}}{V_{in} dt}[/latex]

Then the switch moves from position 1 to position 2 at time [latex]T_{1}[/latex]. The peak inductor current is

[latex]i_{p}(t) = \frac{V_{in} T_{1}}{L}[/latex]

While at position 1, the resistor is isolated electrically, and therefore its voltage is zero.
At time t > [latex]T_{1}[/latex], when the switch is in position 2, as shown in Fig. 7.27b, the inductor current flows into the resistor producing the voltage

[latex]v_{o}(t - T_{1}) = i(t - T_{1})R[/latex] t > [latex]T_{1}[/latex]

At this point, we know that the form of the voltage [latex]v_{o}[/latex](t), in the time interval t > [latex]T_{1}[/latex], is

[latex]v_{o}(t) = Ke^{-(t-T_{1})/ au}[/latex] t > [latex]T_{1}[/latex]

And [latex] au[/latex] = L/R. The initial value of [latex]v_{o}[/latex](t) in the time interval t > [latex]T_{1}[/latex] is K since at time [latex]T_{1}[/latex] the exponential term is 1. According to the design specifications, this initial value is 500 and therefore K = 500.
Since this voltage is created by the peak inductor current [latex]I_{P}[/latex] flowing in R,

K = 500 = ([latex]V_{in}T_{1} R)/L = 5T_{1}(100)/10^{-3}[/latex]

and thus [latex]T_{1}[/latex] is 1 ms and [latex]I_{P}[/latex] is 5 A.
The equation for the voltage in the time interval t > [latex]T_{1}[/latex], or t > 1 is

[latex]v_{o}(t - 1 ms) = 500e^{-100,000(t-1 ms)}[/latex] V

At the end of the 2-ms period, that is, at t = 2 ms, the voltage is [latex]500e^{-100}[/latex] or essentially zero. The complete waveform for the voltage is shown in Fig. 7.28.
It is instructive at this point to consider the ratings of the various components used in this pulse generator circuit. First, 500 V is a rather high voltage, and thus each component’s voltage rating should be at least 600 V in order to provide some safety margin. Second, the inductor’s peak current rating should be at least 6 A. Finally, at peak current, the power losses in the resistor are 2500 W! This resistor will have to be physically large to handle this power load without getting too hot. Fortunately, the resistor power is pulsed rather than continuous; thus, a lower power rated resistor will work fine, perhaps 500 W. In later chapters we will address the issue of power in much more detail.

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