Divider for Recording High-Voltage Nanosecond Pulses
Abstract Generated abstract
This paper analyzes a capacitive voltage divider built into a cable for recording high-voltage pulses with nanosecond rise times. Using an equivalent circuit that includes cable impedance, oscilloscope capacitance, and lead and tube inductance, it derives conditions under which the divider reproduces an exponential pulse front with limited distortion and oscillation. The calculation gives parameter relations for maintaining front-duration error and superposed oscillations below about 10 percent, and expresses the minimum recordable pulse front in terms of the division coefficient, cable impedance, and cathode-ray tube parameters. An example for a 13LO-5A tube and 75 ohm cable indicates that such dividers can, in principle, record pulse fronts of about 10^-9 s without substantial distortion, with practical limits set by tube electron transit time.
Full Text
Divider for Recording High-Voltage Nanosecond Pulses
G. A. Mesyats
A capacitive divider built into a cable is used for recording high-voltage pulses with a rise time of \(10^{-9}\) sec [1], [2], [3]. However, in the literature known to us [4], [5], [6], etc., there is no analysis of the operation of such dividers. A schematic construction of the divider is shown in Fig. 1.
Fig. 1. Divider arrangement: \(i\) — cable insulation, \(zh\) — core, \(o\) — braid.
A metal plate is placed on the polyethylene insulation of the cable; it forms a capacitance \(C_0\) with the cable core. The second capacitance is the intrinsic capacitance of the oscilloscope plates, \(C_p\). The division coefficient of such a divider is determined from the expression
\[ K=\frac{C_p+C_0}{C_0}. \tag{1} \]
Let us calculate under what conditions such a divider will reproduce the minimum rise time with permissible distortions.
The equivalent circuit of the divider is shown in Fig. 2. Here the capacitance
\[ C=\frac{C_p}{K}, \]
\(R\) is the characteristic impedance of the cable, and the inductance
\[ L=L_p+L_v, \]
where \(L_p\) is the inductance of the lead-in conductors 1 and 2
(Fig. 1), \(L_v\) is the intrinsic inductance of the tube leads. Let \(L = k_L \cdot L_v\), where \(k_L > 1\) is the inductance coefficient. We assume that the voltage on the pulse front varies according to the law
\[ U_n = 1 - e^{-at}; \tag{2} \]
Since, to obtain high-voltage pulses with a front of the order of \(10^{-9}\) sec, a spark gap is most often used, and the switching characteristic of the gap is close to an exponential [4]. The duration of the pulse front \(t_{\phi}\), determined between the values 0 and 0.9, will be
\[ t_{\phi} = \frac{2.3}{a}. \tag{3} \]
Fig. 2. Equivalent circuit.
The voltage on the plates of the oscillograph in operator form is written as follows:
\[ U_{\mathrm{out}}(p) = \frac{1}{K_s}\, \frac{2a}{(p+a)(2LCp^2 + RCp + 2)} . \tag{4} \]
Introducing the new operator \(p' = \dfrac{p}{a}\), we reduce the number of independent parameters of equation (4) from 4 to 2. In this case we obtain
\[ U_{\mathrm{out}}(p') = \frac{1}{K}\, \frac{2}{(p' + 1)(2Bbp'^2 + Bp' + 2)} . \tag{5} \]
where \(B = aRC,\quad b = a\dfrac{L}{R}\).
For the case of an oscillatory charge of the capacitance \(C\), we obtain
\[ U'_{\mathrm{out}}(x) = \frac{1}{K_s} \left[ 1 - \frac{1}{Bb - \dfrac{B}{2} + 1}\, e^{-x} - \frac{Bb - \dfrac{B}{2}} {\left(Bb - \dfrac{B}{2} + 1\right)\sin z} \, e^{-\frac{x}{4b}} \sin(\omega x - z) \right]. \tag{6} \]
where
\[ \omega=\sqrt{\frac{1}{Bb}-\frac{1}{16b^{2}}}; \qquad \sin\varphi= \frac{\omega\left(Bb-\dfrac{B}{2}\right)} {\sqrt{Bb-\dfrac{B}{2}-1}}; \qquad x=at. \]
The error \(F\) introduced by the divider is
\[ \begin{aligned} F&=\frac{1}{K}\left(1-e^{-x}\right),\\ U_{\text{out}}&=\frac{1}{k_g}\left[ \frac{Bb-\dfrac{B}{2}} {1-Bb-\dfrac{B}{2}}\,e^{-x} -\right.\\ &\left.\hspace{2.8cm} -\frac{Bb-\dfrac{B}{2}} {1+Bb-\dfrac{B}{2}}\, e^{-\frac{x}{4b}}\sin(\omega x-\varphi) \right]. \end{aligned} \tag{7} \]
The quantity \(F\) varies with time, and at \(t=0\), \(F=0\). It follows from the equation that the distortions introduced by the divider will be small if the quantity
\[ m=Bb-\frac{B}{2}\ll 1 \tag{8} \]
is small, and the amplitude of the sinusoid
\[ A=\frac{Bb-\dfrac{B}{2}} {\left(Bb-\dfrac{B}{2}+1\right)\sin\varphi}\ll 1. \tag{9} \]
If (8) is satisfied, condition (9) is written as
\[ \frac{\sqrt{z}-\sqrt{z-16m}}{\sqrt{16-z}}\ll 1, \tag{10} \]
where
\[ z=\frac{B}{b}. \]
Thus, by selecting appropriate values of \(m\) and \(z\), one can obtain small distortions of the pulse front.
Let us determine the magnitude of the error \(F\) for a specified error \(\Delta x\) in the length of the pulse front,
\[ F=0.1-e^{-(x_{\varphi}+\Delta x)}. \tag{11} \]
For an error of \(\pm 10\%\), the quantity \(\Delta x=0.23\) and the error \(F=+0.02;\,-0.026\). In order for the distortions to be small, it is necessary to reduce the amplitude of the superposed oscillations. The influence of the oscillations will be insignificant if the amplitude \(A\) is small. Let
\[ A=\frac{\sqrt{z}-\sqrt{z-16m}} {\sqrt{m-1}\,\sqrt{16-z}}=0.1. \tag{12} \]
the actual amplitude will be smaller by a factor of \(e^{x/b}\).
Since \(m < 1\), \(az < 16\), equation (12) is simplified and takes the form
\[ \sqrt{z}+\sqrt{z+16m}=0.4. \tag{12'} \]
From (12′) we determine \(m=f(z)\)
\[ m=0.01-0.05\sqrt{z}. \tag{13} \]
For values of \(m\) calculated from (13), the first term of expression (7) for the error \(F\) will be much smaller than the second; therefore
\[ F \approx \frac{1}{4}\left(\sqrt{z}+\sqrt{z+16m}\right)e^{-x/b}\sin(\omega t+\varphi). \tag{14} \]
For values \(z>0.04\), the maximum value of \(F\) at \(x=x_{\varphi}\pm\Delta x\) lies within the limits specified for \(\Delta x=\pm 10\%\).
Thus, under the considerations stated above, condition (13) is necessary and sufficient for the error in the recorded pulse front not to exceed \(10\%\), and for the amplitude of the oscillations on the front to be less than \(10\%\).
From equation (8) we determine the front duration \(t_{\varphi}\) as a function of \(m\) and \(z\), taking (3) into account,
\[ t_{\varphi}=\frac{9.2RC_n}{K\left(1+\sqrt{1+16mz}\right)}. \tag{15} \]
Instead of \(m\), substitute its value from (13):
\[ t_{\varphi}=\frac{9.2C_n}{K\left(1+\sqrt{1+0.16z-0.8z\sqrt{z}}\right)}. \tag{15′} \]
Let us express \(z\) through the division coefficient \(K\), the wave impedance of the cable \(R\), and the parameters of the cathode-ray tube.
Since
\[ B=a\cdot\frac{RC_n}{K},\qquad b=a\cdot\frac{k_L-L_t}{R},\qquad \text{and}\qquad L_0=\frac{1}{4\pi^2 f_0^2 C_n}, \]
where \(f_0\) is the resonant frequency of the deflecting system of the tube, then
\[ z=\frac{4\pi^2R^2C_n^2f_0^2}{Kk_L}. \tag{16} \]
Substituting \(z\) into (15′), we obtain the dependence of the minimum pulse front that can be recorded by this divider without substantial distortions on all the parameters of the recording system.
As an example, let us calculate \(t_{\varphi}\) for a 13LO-5A tube and a cable with a wave impedance of \(75\ \Omega\).
For this tube \(f_0=600\cdot10^6\ \text{Hz}\), \(C_n=5\cdot10^{-12}\ \text{F}\). With careful layout of the device one can obtain \(k_L \simeq 1\).
Substituting all these quantities into (16), we obtain
\[ \tau \simeq \frac{2}{K_g}. \]
Thus, for the given tube and cable, \(t_\phi\) depends only on the division coefficient \(K_g\). The dependence of \(t_\phi\) on \(K_g\) is shown in Fig. 3.
Fig. 3. Dependence of the pulse-front duration on the division coefficient.
It follows from the graph in Fig. 3 that, theoretically, at large \(K_g\) it is possible to record pulses with a front shorter than \(10^{-9}\) sec. In practice, the minimum value of \(t_\phi\) will be limited by the transit time of the electron in the field of the deflecting plates of the tube [4].
Thus, as a result of the calculations it has been shown that a capacitive divider built into the cable can reproduce, without substantial distortions, pulses with a front duration of \(10^{-9}\) sec.
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