typos, README

.gitignore

@@ -2,3 +2,5 @@
 *~
 *#
 *.tab
+*.aux
+*.log

Makefile

@@ -14,3 +14,18 @@ SPICE = ngspice
 %.tab: %.dat
 	ngspice2gpt.awk $< > $@
 
+
+PSTOPS=pstops
+PDF2PS=pdf2ps
+PS2PDF=ps2pdf14 -dEPSCrop
+PDFLATEX=pdflatex
+PS2EPS=ps2epsi
+
+%.pdf: %.eps
+	$(PS2PDF) $< $@
+
+%.eps: %.ps
+	$(PS2EPS) $< $@
+
+%.ps: %_a4.pdf
+	$(PDF2PS) $< $@

README.md

@@ -8,8 +8,7 @@ Design guidelines
 - *f*=125 kHz seems to fit quite well with reachable inductances and reasonable current scales.
 - A single asymmetric drive may be sufficient.
 
-This is what I came up with.  Some components are there only to keep `ngspice` happy.
+After a serious [math excersize](hvosc.pdf) this circuit is not understood, hopefully.
 
-![Schematics](hvosc-spice-1_sch.png)
+![Schematics](hvosc-doc_sch.png)
 
-![Plot](hvosc-spice-1.png)

hvosc-doc.sch

@@ -0,0 +1,128 @@
+v 20220529 2
+C 49800 46900 1 90 0 capacitor-1.sym
+{
+T 49100 47100 5 10 0 0 90 0 1
+device=CAPACITOR
+T 49650 47300 5 10 1 1 0 2 1
+refdes=C2
+T 48900 47100 5 10 0 0 90 0 1
+symversion=0.2
+T 49650 47450 5 10 1 1 0 0 1
+value=100pF
+}
+N 49200 48000 50100 48000 4
+N 47900 46900 50100 46900 4
+N 47900 48000 48300 48000 4
+N 46500 48000 47500 48000 4
+N 49600 48000 49600 47800 4
+N 47500 47900 47500 48000 4
+N 47900 47900 47900 48000 4
+C 49200 48000 1 180 0 inductor-dot-1.sym
+{
+T 49000 47600 5 10 0 0 180 0 1
+device=INDUCTOR
+T 48500 48100 5 10 1 1 0 0 1
+refdes=L3
+T 49000 47400 5 10 0 0 180 0 1
+symversion=0.2
+T 48500 47700 5 10 1 1 0 0 1
+value=6.93mH
+T 48500 48800 5 10 1 1 0 0 1
+partname=B64290L0045X830
+T 48500 48400 5 10 1 0 0 0 1
+n=50
+T 48500 48600 5 10 1 0 0 0 1
+AL=2.77µH
+}
+N 47900 47000 47900 46900 4
+N 47500 46900 47500 47000 4
+C 46600 47900 1 270 0 capacitor-1.sym
+{
+T 47300 47700 5 10 0 0 270 0 1
+device=CAPACITOR
+T 46750 47500 5 10 1 1 180 2 1
+refdes=C1
+T 47500 47700 5 10 0 0 270 0 1
+symversion=0.2
+T 46750 47400 5 10 1 1 0 8 1
+value=2.5nF
+}
+N 46800 46900 46800 47000 4
+N 45500 46900 47500 46900 4
+N 46800 47900 46800 48000 4
+C 47900 47900 1 270 0 inductor-dot-1.sym
+{
+T 48300 47700 5 10 0 0 270 0 1
+device=INDUCTOR
+T 48050 47300 5 10 1 1 0 0 1
+refdes=L2
+T 48500 47700 5 10 0 0 270 0 1
+symversion=0.2
+T 48050 47100 5 10 1 1 0 0 1
+value=4.43mH
+T 47950 46500 5 10 1 1 0 0 1
+partname=B64290L0045X830
+T 47950 46100 5 10 1 0 0 0 1
+n=40
+T 47950 46300 5 10 1 0 0 0 1
+AL=2.77µH
+}
+C 47500 47900 1 270 0 inductor-dot-1.sym
+{
+T 47900 47700 5 10 0 0 270 0 1
+device=INDUCTOR
+T 47450 47300 5 10 1 1 0 6 1
+refdes=L1
+T 48100 47700 5 10 0 0 270 0 1
+symversion=0.2
+T 47450 47100 5 10 1 1 0 6 1
+value=277µH
+T 47550 46500 5 10 1 1 0 6 1
+partname=B64290L0045X830
+T 47550 46100 5 10 1 0 0 6 1
+n=10
+T 47550 46300 5 10 1 0 0 6 1
+AL=2.77µH
+}
+L 47700 47100 47700 47800 3 10 1 0 -1 -1
+L 47800 47100 47800 47800 3 10 1 0 -1 -1
+C 45600 47900 1 0 0 resistor-2.sym
+{
+T 46000 48250 5 10 0 0 0 0 1
+device=RESISTOR
+T 46050 48200 5 10 1 1 0 4 1
+refdes=R1
+T 46050 48000 5 10 1 1 0 4 1
+value=1kΩ
+T 45800 48600 5 10 0 0 0 0 1
+symversion=0.1
+}
+N 45600 48000 45500 48000 4
+C 44700 47900 1 0 0 input-1.sym
+{
+T 44700 48200 5 10 0 0 0 1 1
+device=INPUT
+T 44750 48000 5 10 1 1 0 1 1
+value=IN
+}
+C 50100 47900 1 0 0 output-1.sym
+{
+T 50200 48200 5 10 0 0 0 1 1
+device=OUTPUT
+T 50350 48000 5 10 1 1 0 1 1
+value=OUT
+}
+C 44700 46800 1 0 0 input-1.sym
+{
+T 44700 47100 5 10 0 0 0 1 1
+device=INPUT
+T 44750 46900 5 10 1 1 0 1 1
+value=INR
+}
+C 50100 46800 1 0 0 output-1.sym
+{
+T 50200 47100 5 10 0 0 0 1 1
+device=OUTPUT
+T 50350 46900 5 10 1 1 0 1 1
+value=OUTR
+}

hvosc.maxima

@@ -36,4 +36,4 @@ assume(a>0, b>0, c>0);
 SC: solve(LA, [C1, C2]), Uq=a*U1, L2=b**2*L1, L3=c**2*L1, ratsimp;
 SL: solve(LA, [C1, L1]), Uq=a*U1, L2=b**2*L1, L3=c**2*L1, ratsimp;
 SL,a=5,b=3,c=4,C2=100e-12,oo=(2*%pi * 125e3)**2, numer;
-SC,a=5,b=3,c=4,L1=440e-6,oo=(2*%pi * 125e3)**2, numer;
+SC,a=5,b=3,c=4,L1=407e-6,oo=(2*%pi * 125e3)**2, numer;

hvosc.tex

@@ -0,0 +1,156 @@
+\documentclass[12pt,a4paper]{article}
+\usepackage[utf8]{inputenc}
+\usepackage{graphics}
+\usepackage{verbatim}
+
+\title{               Resonant PMT HV cascade driver
+}
+\author{                     Stephan I. Böttcher
+}
+\date{                              \today
+}
+
+\def\units{\,\mathrm}
+\def\MOhm{\units{M\Omega}}
+\def\nF{\units{nF}}
+
+\begin{document}
+\maketitle
+
+\section{Design goals}
+
+For the dynodes supply of a \emph{Photoelectron Multiplier Tube} we
+need a sinus function voltage with an amplitude of about
+$U_{\mathrm{OUT}}=50\units s$ to drive a Cockcroft–Walton cascade.
+
+In the output terminal shall be a capacitor $C_2 \ge 100\units{pF}$ to
+account for capacitive loads.
+
+A regulator supplies a reference voltage to set the resulting tube
+voltage. The input power shall be resonably low.
+
+The Cockcroft–Walton cascade will be located close to the tube inside
+the sensor head of an instrument where interference with detector
+signals must be avoided.
+
+The frequency shall be $f=125\units{kHz}$ to comply with potential
+contraints of the EMI environment.
+
+\section{Circuit}
+
+The circuit shown in Fig.~\ref{fig:doc-sch} with properly chosen
+component values resonates at the desired frequency and thus consumes
+very little input power.
+
+\clearpage
+
+\begin{figure}[htb!]
+  \vspace{-2cm}
+  \resizebox{\textwidth}{!}{\includegraphics{hvosc-doc_sch_a4.pdf}}
+  \vspace{-2cm}
+  \caption{Resonator schematics}
+  \label{fig:doc-sch}
+\end{figure}
+
+\section{Resonator solution}
+
+\def\Uout{U_{\mathrm{OUT}}}
+
+The circuit description is
+\begin{eqnarray}
+  \label{eq:GS}
+  \Phi &=& n_1 I_1 + n_2 I_2, \\
+  U_1 &=& A_L n_1 \dot\Phi, \\
+  U_2 &=& A_L n_2 \dot\Phi, \\
+  U_3 &=& A_L n_3^2 \dot I_2, \\
+  I_1 &=& -C_1 \dot U_1, \\
+  I_2 &=& -C_2 \dot Uout, \\
+  Uout &=& U_2 + U_3, \\
+  L_i &=& A_L n_i^2.
+\end{eqnarray}
+Solving this system for $\ddot U_1$ and~$\ddot \Uout$ provides the
+differential equations
+\begin{equation}
+  \label{eq:DEE}
+  \begin{array}{ccccccc}
+    L_1C_1 & \ddot U_1
+    & = & - & \left(1 + \frac{L_2}{L_3}\right) \,U_1
+    & + & \frac{\sqrt{L_1L_2}}{L_3} \,\Uout, \\[2ex]
+    L_3C_2 & \ddot \Uout
+    & = & + & \sqrt{\frac{L_2}{L_1}} \,U_1 & - & \,\Uout.
+  \end{array}
+\end{equation}
+Solving that equation yields real solutions for the voltages, $U_1$
+and~$\Uout$ are in phase.  These solutions do not look pretty, please
+refer to \texttt{hvosc.maxima} if you want to look at them.
+
+\goodbreak
+
+Treating the solution as the next system of equations to solve, yields
+usefull formulas, with the parameters
+\begin{eqnarray}
+  \label{eq:PARS}
+  a &=& \frac\Uout{U_1}, \\
+  b &=& \frac{n_2}{n_1}, \\
+  c &=& \frac{n_3}{n_1}, \\
+  \omega &=& 2\pi f,
+\end{eqnarray}
+and solving for $C_1$ and~$L_1$:
+\begin{eqnarray}
+  \label{eq:SL}
+  C_1 &=& C_2 \,a \,\left(\frac{c^2}{a - b} - b\right), \\
+  L_1 &=& \left(1 - \frac ba\right)\,\frac{1}{c^2\omega^2 C_2}, \\
+  L_2 &=& b^2 L_1, \\
+  L_3 &=& c^2 L_1.
+\end{eqnarray}
+
+
+\clearpage
+
+\section{Implementations}
+
+The $10\units{mm}$ core in our inventory has a single turn inductance
+of $A_L = 4.09\units{\mu H}$.  With $C_2=100\units{pF}$ and
+$f=125\units{kHz}$, $a=5$, $b=3$, $c=4$ this yields
+\begin{eqnarray}
+  \label{eq:T38}
+  n_1 &=& 10, \\
+  n_2 &=& 30, \\
+  n_3 &=& 40, \\
+  L_1 &=& 409\units{\mu H}, \\
+  L_2 &=& 3.65\units{mH}, \\
+  L_3 &=& 6.48\units{mH}, \\
+  C_1 &=& 2.5\units{nF}.
+\end{eqnarray}
+These values were simulated with spice.  Please refer to \texttt{hvosc-spice-r.ckt}.
+
+
+Unfortunately, that core of T38 material is not fit for frequencies
+above 100kHz.  A bigger core from N30 material, type
+\texttt{B64290L0045X830} with more gain suggests this set of
+parameters
+\begin{eqnarray}
+  \label{eq:N30}
+  A_L &=& 2.77 \units{\mu H}, \\
+  a &=& 7, \\
+  b &=& 4, \\
+  c &=& 5, \\
+  C_2 &=& 100\units{pF}, \\
+  C_1 &=& 3.03\units{nF}, \\
+  L_1 &=& 278\units{\mu H}, \\
+  n_1 &=& 10, \\
+  n_2 &=& 40, \\
+  n_3 &=& 50.
+\end{eqnarray}
+Or
+\begin{eqnarray}
+  \label{eq:270pF}
+  C_2 &=& 270\units{pF},\\
+  C_1 &=& 8.2\units{nF}, \\
+  L_1 &=& 103\units{\mu H}, \\
+  n_1 &=& 6, \\
+  n_2 &=& 24, \\
+  n_3 &=& 30.
+\end{eqnarray}
+
+\end{document}