\documentclass[a4paper,11pt,titlepage]{article}
\usepackage[pdfborder= 0 0 0 1]{hyperref}
\usepackage{graphicx}
\usepackage{pst-circ}
\author{Carlos Gil Soriano\\BE-CO-HT\\
\href{mailto:carlos.gil.soriano@cern.ch}{\textbf{\textit{carlos.gil.soriano@cern.ch}}}}
\title{\textbf{{\LARGE Technical Specifications for a 30 V output level Pulse
Converter}}}
\date{\today}

\begin{document}
\pagenumbering{roman}
\maketitle{}
\pagebreak 

\setcounter{page}{2}
\pagestyle{empty}

\paragraph{System Description and Purpose}
This documents aims to explain different approaches to the design of a fast,
stable and reliable pulse level converter compliant with ANSI/IEEE Std 390-1987.
The ideas covered in this techinical report can be applied to different input
triggers and output pulse shapes.

\pagebreak
\setcounter{tocdepth}{3}
\tableofcontents
\pagebreak

\pagenumbering{arabic}
\pagestyle{plain}
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\section{ANSI/IEEE Std 390-1987}
\subsection{Pulse Shape}
ANSI/IEEE Std 390-1987 \cite{IEEEStandard} defines the following notation for
defining a pulse shape:
\begin{itemize}
  \item Peak working voltage
  \item Pulse amplitude, \textit{A$_{M}$}
  \item Rise time, \textit{t$_{\textnormal{r}}$}
  \item Pulse duration, \textit{t$_{\textnormal{p}}$}
  \item Fall time, \textit{t$_{\textnormal{p}}$}
  \item Trailing edge amplitude,  \textit{A$_{\textnormal{T}}$}
  \item Tilt, \textit{A$_{\textnormal{D}}$}
  \item Overshoot, \textit{A$_{\textnormal{os}}$}
  \item Backswing, \textit{A$_{\textnormal{BS}}$}
  \item Returnswing, \textit{A$_{\textnormal{BS}}$}
  \item Ringing, \textit{A$_{\textnormal{RI}}$}
  \item Leading edge,  \textit{A$_{\textnormal{L}}$}
  \item Quiescent value,  \textit{A$_Q$}
  \item Leading edge
  \item Pulse top
  \item Trailing edge
\end{itemize}
\subsection{Tests}
The Pulse Converter must comply the following tests carried out with test
conditions, as specified in [1]:
\begin{itemize}
  \item Pulse Response
  \item Temperature Rise Test
\end{itemize}
\subsection{Marking}
The Pulse Converter marking shows:
\begin{itemize}
  \item Manufacturer
  \item ID
  \item Brief pulse specifications: top
  \item Source and load impedances
\end{itemize}
\pagebreak
\section{Triggered Monostable Blocking Oscillator}
\subsection{References and tips}
A blocking oscillator is usually employed in synchronization applications due to
its simplicity, the little number of elements required -one BJT, a transformer,
few diodes, resistances and capacitors- and the sharp slope of the rising edge it provides. Because of these reasons, a monostable blocking oscillator is selected for the pulse conversion over other possibilities --such as flyback converters, for instance.\\ An intuitive introduction to pulse converters can be found in
\cite{StantonBook}, where the blocking oscillator is sketched pretty
simplisticly. Millman's book, \cite{MillmanBook}, of{}fers an easy and
straightforward view on the pulse top state of the circuit.\\
The more mathematical insight of the switching state is found in Linvill's
classical literature \cite{LinvillBook} and \cite{LinvillIREPaper}, which warns the reader of the aproximations taken out for
the sake of simplicity. It should be noted that, for a proper switching
analysis, some values of the model -such as base resistance- are usually
difficult to find in the datasheet of the manufacturers, and must be inferred
--a Ning-Tang method for the base resistance, i.e. Apart from this, it is a
good advice to carefully checking these values in the SPICE models
provided by the manufacturers. Some manufacturer's base
resistance parameter model corresponds to the intrinsic value of it and not to the intrinsic plus extrinsic one, as it is requiered for a good matching with
the mathematical analysis. This will produce a misleading simulation which will
turn out into an unexpected outcome for the designer. Thus, we encourage not to
give 100\% confidence to the simulation results due to this innacuracies with
respect to the aproximate model used in the analysis.\\ A comprehensive study of the
switching state based on an extension of Linvill's aproximation is done in \cite{McDonaldIEEEPaper}. However, we have considered it as overcomplicated compared to the more reasonable original Linvill's aproximation.\\
Last but not least, Norman's guide for the design of monostable blocking oscillator \cite{NormanIEEPaper} constituted the invaluable help which serves as
the reference of this design.
\subsection{Topologies evaluated}
Three blocking oscillators circuits were considered for the pulse converter.
The first two are positively collector-emitter feedbacked and the later has a
base-collector feedback.
\begin{description}
  \item \textbf{Circuit 1}\\
  This circuit can be found in \cite{LinvillIREPaper}. The main drawbacks it
  presents lay on the charge it produces the output load and a worse triggering
  option compared to Circuit 2.\\
  
  \begin{figure}[htb]
    \begin{center}
      \psset{unit=0.8}
      \begin{pspicture}(11,5)
      \pnode(0,0){A}\pnode(1.75,0){B}\pnode(4,0){C}\pnode(6.5,0){D}\pnode(7.5,0){E}\pnode(10.5,0){F}
      \pnode(0,3){G}\pnode(1,3){H}\pnode(3,3){I}\pnode(4,3){J}\pnode(6.5,3){K}\pnode(7.5,3){L}\pnode(10.5,3){M}
      \pnode(0,4.5){N}\pnode(10.5,4.5){O}
      \wire(A)(B)
      \resistor[dipolestyle=zigzag](A)(G){$R_L$}
      \wire(G)(N)
      \capacitor(N)(O){$C$}
      \transformer[primarylabel=$n_1$,secondarylabel=$n_2$](L)(E)(M)(E){$ $}
      \wire(K)(L)\wire(D)(E)
      {\psset{fillstyle=solid,fillcolor=black}
      \diode(C)(J){$D1$}\diode(K)(D){$D2$}}
      \wire(J)(K)\wire(I)(J)
      \battery(C)(B){$V1$}
      \battery(D)(C){$V2$}
      \transistor[transistorinvert,TRot=90, basesep=1.5](B)(I)(H)
      \wire(H)(G)\wire(O)(M)
      \end{pspicture}
    \vspace{25pt}
    \caption{Linvill's circuit, taken from \cite{LinvillIREPaper}}
    \end{center}
  \end{figure}  
 
  \item \textbf{Circuit 2}\\
  It is the circuit chosen for the design. With the inclusion of a resistance in
  the positive feedback loop it is really simple to accomodate the pulse width
  to the designer needs.
  \vspace{20pt}
  \begin{figure}[htb]
    \begin{center}
      \psset{unit=0.8}
      \begin{pspicture}(14.5,8.5)
      \pnode(1.5,1){A}\pnode(6.5,1){B}\pnode(10,1){C}
      \pnode(4.5,3){D}\pnode(6.5,3){E}\pnode(8.5,3){F}\pnode(10,3){G}\pnode(11.5,3){H}\pnode(14.5,3){I}
      \pnode(1.5,5.5){base}\pnode(4.5,5.5){J}\pnode(6.5,5.5){K}
      \pnode(0,7.5){L}\pnode(0.5,7.5){M}\pnode(4.5,7.5){N}\pnode(6.5,7.5){O}\pnode(8.5,7.5){P}\pnode(11.5,7.5){Q}\pnode(14.5,7.5){R}\pnode(10,7.5){Taux}
      \pnode(0.5,8.5){S}\pnode(10,8.5){T}
      \wire(A)(B)\wire(C)(B)\wire(A)(base)
      \ground(B)
      \battery(B)(E){$V1$}
      \battery(E)(K){$V2$}
      \wire(D)(E)\wire(E)(F)
      \resistor[dipolestyle=zigzag, unit=0.75](D)(J){$R$}
      {\psset{fillstyle=solid,fillcolor=black}
      \diode[unit=0.75](J)(N){$D$} \diode[unit=0.75](K)(O){$D$}}
      \coil[dipolestyle=elektorcurved, arrows=-](F)(P){$ $}
      \psdots(8.25,6.25)
      \coil[dipolestyle=elektorcurved, arrows=-](G)(Taux){$ $}
      \psdots(9.75,6.25)
      \coil[dipolestyle=elektorcurved, arrows=-](H)(Q){$ $}
      \psdots(11.25,6.25)
      \wire(T)(Taux)\wire(C)(G)\wire(P)(N)
      \resistor[dipolestyle=zigzag, unit=0.75](S)(T){$R_E$}
      {\psset{fillstyle=solid,fillcolor=black}\diode[unit=0.75](Q)(R){$D$}}
      \resistor[dipolestyle=zigzag, unit=0.75](R)(I){$R_L$}
      \wire(H)(I)
      \transistor[basesep=0.5, transistortype=PNP, TRot=90](base)(N)(M)
      \wire(S)(M)\wire[arrows=o-](L)(M)
      \rput[bl](0,6.75){\text{$IN$}}
      \end{pspicture}
    \caption{Norman's circuit, taken from \cite{NormanIEEPaper}}
    \end{center}
  \end{figure}

  \item \textbf{Circuit 3}\\
  However it is the most intuitive among all the designs, the triggering is not
  as independent from the input as Circuit 2.
  \vspace{20pt}
    \begin{figure}[htb]
    \begin{center}
      \psset{unit=0.8}
      \begin{pspicture}(8,9)
      \pnode(3.5,0){GND}\pnode(3.5,9){VCC}
      \pnode(0,1){A}\pnode(3.5,1.5){B}\pnode(3.5,1){Baux}
      \pnode(3.5,3.5){C}
      \pnode(2,4.5){D}
      \pnode(2,6){E}\pnode(3.5,6){F}\pnode(5,6){G}\pnode(8,6){H}
      \pnode(0,9){I}\pnode(2,9){J}\pnode(5,9){K}\pnode(8,9){L}
      \wire(A)(Baux)\wire(B)(Baux)\wire(A)(I)\wire(I)(J)
      \resistor[dipolestyle=zigzag, unit=0.75](C)(B){$R$}
      \ground(Baux)
      \coil[dipolestyle=elektorcurved](J)(E){$ $}
      \psdots(2.25,6.5)
      \coil[dipolestyle=elektorcurved, arrows=-o](F)(VCC){$ $}
      \psdots(3.75,8.35)
      \coil[dipolestyle=elektorcurved](K)(G){$ $}
      \psdots(5.25,8.35)
      {\psset{fillstyle=solid,fillcolor=black}
      \diode[unit=0.75](K)(L){$D$}}
      \resistor[dipolestyle=zigzag, unit=0.75](L)(H){$R_L$}
      \wire(H)(G)\wire(E)(D)
      \transistor[basesep=1](D)(C)(F)
      \rput[bl](3.5, 9.25){\text{V$_{CC}$}}
      \rput[bl](3.175,8.6){\text{$n:1:n$}}
      \end{pspicture}
    \caption{Millman's circuit, taken from \cite{MillmanBook}}
    \end{center}
  \end{figure}
\end{description}
\\
\subsection{Analysis of the chosen topology}
The analysis follows the flow pointed out in \cite{NormanIEEPaper}. The complete
mathematical resolution of the switching state from \cite{LinvillBook} is
included. Furthermore, the fixed-point methond algorithm is added to clarify
how the normalized natural frequency of the circuit is gotten.
\\
  \begin{figure}[htb]
    \begin{center}
      \includegraphics[scale=0.6, keepaspectratio]{blockingStates}
      \caption{States and equivalent circuits}
    \end{center}
  \end{figure}
\vspace{-20pt} 
\subsubsection{Switching on}
When it is switching both on and of{}f, an equivalent circuit is shown in
the figure below. Emitter resistance and capacitance are omitted for
simplicity.
The critical value while switching lies on obtaining the \textit{natural
frequency} of the circuit. This value is closely related with the
\textit{transformer turns-ratio} and, given the aproximations of the model, an
optimun turns-ratio can be calculated. If this turns-ratio is used, the fastest
rise time will be achived an no ringing will be obtained, ideally.
\\
\begin{figure}[htb]
    \begin{center}
    \begin{pspicture}(8,5.5)
      \pnode(0,0){A}\pnode(5,0){B}\pnode(6.5,0){C}\pnode(8,0){D}
      \pnode(0,2){E}\pnode(1,3){Eexrtra}\pnode(3,2){F}
      \pnode(0,4){G}\pnode(1,4){Gextra}\pnode(3,4){H}\pnode(5,4){I}\pnode(6.5,4){J}\pnode(8,4){K}
      \pnode(0,5.5){L}\pnode(6.5,5.5){M}
      \wire(L)(M)\wire(J)(M)\wire(J)(K)
      \wire(A)(D)\wire(H)(F)\wire(G)(E)
      \wire[intensitylabel=$I_E$, arrows=o-o](L)(G)
      \resistor[dipolestyle=zigzag, unit=0.75](A)(E){$r_b$}
      \resistor[dipolestyle=zigzag, tensionlabeloffset=-1,tensionoffset=-0.5,
      tensionlabel=$E_1$, unit=0.75](D)(K){$R_L$}
      \capacitor[unit=0.75, labeloffset=1](E)(F){$C_c$}
      \wire[intensitylabel=$i_L$, arrows=o-](H)(I)
      \coil[dipolestyle=elektorcurved, tensionlabeloffset=-1,tensionoffset=-0.5,
      tensionlabel=$nE_1$, directconvention=false](I)(B){${}$}
      \psdots(5.25, 2.85)
      \coil[dipolestyle=elektorcurved](J)(C){${}$}
      \psdots(6.75, 2.85)
      \Icc[dipolestyle=twoCircles, labeloffset=1,
      unit=0.75](G)(H){$\frac{I_E}{1+\frac{s}{w_o}}$}
      \rput[bl](5.375,3.5){\text{$n:1$}}
    \end{pspicture}
    \caption{Eq. circuit B: switching}
    \end{center}
\end{figure}
\vspace{-20pt} 
\begin{description}
\item \textbf{Kirchoff's equations}\\
The electrical relationship of the circuit are:
\begin{equation}
\frac{E_1}{R_L} + I_E - ni_L = 0 
\end{equation}
\begin{equation}
\frac{E_1}{r_b} + \frac{I_E}{1+\frac{s}{w_o}} - I_E - (n-1)E_1C_ss = 0
\end{equation}
\begin{equation}
i_L - \frac{I_E}{1 + \frac{s}{w_o}} + (n-1)E_1C_ss = 0
\end{equation}
\item \textbf{Fundamental equation}\\
Operating the previous equations yields:
\begin{equation}
(\frac{s}{w_0})^2 + \frac{s}{w_0}\cdot\frac{G_L + g_b}{(n-1)^2 C_c w_0} -
\frac{g_b}{C_c w_0 (n-1)} = 0
\end{equation}
That can be greatly simplified by normalizing the natural frequency and
performing the following changes of variable:
\begin{equation}
x=\frac{s}{w_0} \hspace{25pt} n-1 = \Delta \hspace{25pt} \frac{g_b}{C_c w_0} = k
\hspace{25pt} \frac{g_b + G_L}{C_c w_0} = k
\end{equation}
\begin{equation}
x^2+x(1+\frac{l}{\Delta^2})-\frac{k}{\Delta} = 0
\end{equation}
However, as $\Delta$ is a design parameter we can rewrite the equation to:
\begin{equation}
\Delta^2(x^2+x) - k\Delta + xl = 0
\end{equation}
If a value of the normalized natural frequency of the circuit, $x$, is lower
than $x_{max}$ two possible optimal \textit{turns-ratios} will exist. If it is
higher than $x_{max}$ no real \textit{turns-ratios} exist. We can find an
optimal value of $\Delta$ that corresponds to $x_{max}$, by obtaining a double
root of the previous equation. Hence:
\begin{equation}
k^2 - 4(x_{max}^3 + x_{max}^2)l = 0
\end{equation}
We can get the \textit{normalized natural frequency} by applying
iterations by a fixed-point method, given that it converges --because we expect
a value close to 1
\begin{equation}
x_{max_i} = \frac{k}{2\sqrt{l(x_{max_{i-1}} +1 )}} \hspace{25pt} i = 1,2,3\ldots
\end{equation}
\item \textbf{Results}\\
The expected \textit{rise-time} is:
\begin{equation}
t_{rise}=\frac{2.3}{x_{max}w_0}
\end{equation}
for an optimal \textit{turns-ratio} is:
\begin{equation}
\Delta_{opt} = \frac{k}{2(x_{max}^2+x_{max})} \hspace{25pt} n_{opt} =
\Delta_{opt} +1
\end{equation}
Subsequently, getting a good rise-time depends mainly on choosing the optimal
\textit{turns-ratio} value and using a fast switching bipolar transistor in the
design.\\
\textbf{TIP}: as stated in \cite{NormanIEEPaper}, the triggering signal must be
active for $3t_{rise}$, so as to effectively switching the cirucit state.
\end{description}
\subsubsection{On state}
The on-state is related with the value of the \textit{magnetizing inductance} in
the collector and the \textit{positive feedback resistor}. With this two parameters
the designer is able to chose the \textit{pulse width}.
\begin{figure}[htb]
  \begin{center}
    \begin{pspicture}(11,5)
      \pnode(2,0){A}\pnode(6,0){B}\pnode(8.5,0){C}
      \pnode(2,1){D}\pnode(5,1){E}\pnode(6,1){F}\pnode(7,1){G}\pnode(8.5,1){Gaux}\pnode(10,1){H}\pnode(11,1){I}
      \pnode(0,4){J}\pnode(4,4){K}\pnode(5,4){L}\pnode(7,4){M}\pnode(8.5,4){Maux}\pnode(10,4){N}\pnode(11,4){O}
      \pnode(0,5){P}\pnode(8.5,5){Q}
      \wire(A)(C)\wire(H)(I)\wire(N)(O)\wire(A)(D)\wire(C)(Gaux)
      \wire(E)(G)\wire(Maux)(Q)\wire(P)(J)
      \wire[intensitylabel=$i_c$, arrows=-o](K)(L)
      \wire[intensitylabel=$i_t$, arrows=o-o](L)(M)
      \coil[dipolestyle=elektorcurved, tensionlabeloffset=-1,tensionoffset=-0.5,
      tensionlabel=$V_p$, directconvention=false](L)(E){$L$}
      \coil[dipolestyle=elektorcurved](M)(G){${}$}
      \psdots(7.25, 3.375)
      \coil[dipolestyle=elektorcurved](Maux)(Gaux){${}$}
      \psdots(8.75, 3.375)
      \coil[dipolestyle=elektorcurved](N)(H){${}$}
      \psdots(10.25, 3.375)
      \resistor[dipolestyle=zigzag, unit=0.75](O)(I){$R_L$}
      \resistor[dipolestyle=zigzag, unit=0.75](P)(Q){$R_E$}
      \battery[labeloffset=1](B)(F){$V1$}
      \transistor[basesep=2, transistortype=PNP, TRot=90](D)(K)(J)
      \rput[bl](0.5,3){\text{$r_{EB_T}$}}
      \rput[bl](8,4){\text{$n:1:n$}}
    \end{pspicture}
    \caption{Eq. circuit A: on-state}
  \end{center}
\end{figure}

\begin{description}
\item \textbf{Kirchoff's equations}\\
From the collector we can get the following equation:
\begin{equation}
i_c = \frac{V_p}{n^2(R_L+r_{EBt})} + \frac{V_p}{n^2R_L} + \frac{V_pt}{L}
\end{equation}
During the on-state we can define $h_{FB_t} = \frac{i_c}{i_E}$, thus the
previous equations can be rewritten as:
\begin{equation}
\frac{V_ph_{FB_t}}{n(R_L + r_{EBt})} = \frac{V_p}{n^2(R_L + r_{EBt})} +
\frac{V_p}{n^2R_L} +
\frac{V_pt}{L}
\end{equation}

\item \textbf{Results}\\
When the transistor is close to leave its saturation state, we can change all
its parameters for the large signal model ones. This yields to obtaining the
\textit{width} of the pulse:
\begin{equation}
T_p = \frac{L (h_{FB} - 1)}{n^2[R_E + r_E + (1+h_{FB}r_B)]} - \frac{L}{n^2R_L}
\end{equation}
As it is noted in \cite{NormanIEEPaper}, adding $R_E$ helps to desensitizing
the circuit from $r_E$
\end{description}
\subsubsection{Switch of{}f}
The switch-of{}f state is governed by the same equations of the switch-on state.
\subsubsection{Of{}f state: \textit{Recovery time}}
Of{}f-state is reached when the circuit completely removes all the current from
the \textit{magnetizing inductor} through the snubber consisting of the
\textit{diode} and \textit{discharge resistor}. Once the current is completely
removed, a new triggering can be faced by the monostable oscillator.
\begin{figure}[htb]
    \begin{center}
      \begin{pspicture}(16,7)
        \psset{unit=0.8}
        \pnode(2,0){A}\pnode(6,0){B}\pnode(12.5,0){C}
        \pnode(5,2){D}\pnode(6,2){E}\pnode(7,2){F}\pnode(9,2){G}\pnode(11,2){H}\pnode(12.5,2){I}\pnode(14,2){J}\pnode(16,2){K}
        \pnode(2,4){base}\pnode(5,4){L}\pnode(7,4){M}
        \pnode(9,5){N}\pnode(11,5){O}\pnode(12.5,5){P}\pnode(14,5){Q}\pnode(16,5){R}
        \pnode(5,6){S}\pnode(7,6){T}\pnode(0,6){U}\pnode(4,6){V}\pnode(9,6){Y}\pnode(11,6){Z}
        \pnode(0,7){AB}\pnode(12.5,7){BC}
        \wire(A)(C)\wire(A)(base)\wire(C)(I)\wire(J)(K)
        \coil[dipolestyle=elektorcurved, intensitylabel=$i_L$](N)(G){$L$}
         
        \coil[dipolestyle=elektorcurved](O)(H){${}$}
        \psdots(11.25, 4.5)
        \coil[dipolestyle=elektorcurved](P)(I){${}$}
        \psdots(12.75, 4.5)
        \coil[dipolestyle=elektorcurved](Q)(J){${}$}
        \psdots(14.25, 4.5)
        \resistor[dipolestyle=zigzag, unit=0.75](R)(K){$R_L$}
        \resistor[dipolestyle=zigzag, labeloffset = 1.25,
        unit=0.75](D)(L){$R_{snubber}$} \battery[labeloffset=0.8](B)(E){$V1$}
        \battery[labeloffset=0.8](F)(M){$V2$} 
        \wire(D)(H)\wire(BC)(P)\wire(V)(Z)\wire(O)(Z)\wire(N)(Y)\wire(AB)(U)
        \resistor[dipolestyle=zigzag, unit=0.75](AB)(BC){$R_E$}
        {\psset{fillstyle=solid,fillcolor=black}
        \diode[unit=0.75,intensitylabel=$i_2$](L)(S){$D2$}\diode[unit=0.75,
        intensitylabel=$i_4$](M)(T){$D1$}\diode[unit=0.75](Q)(R){$D$}} \transistor[basesep=0, transistortype=PNP, TRot=90](base)(V)(U)
        \rput[bl](12,5){\text{$n:1:n$}}
      \end{pspicture}
      \caption{Eq. circuit C: of{}f-state}
    \end{center}
\end{figure}

\begin{description}
\item \textbf{Kirchoff's equations}\\
At the end of the on state, the magnetizing current is:

\begin{equation}
I_L = \frac{V_PT_P}{L}
\end{equation}
The magnetizing inductance current must equal the current flowing out of the two
multipoles formed by the diode and the resistor and the diode and the zenner diode. The multipole consisting of the diode and the resistor helps to avoid
overdamping while switching off. In the second case, the current decreases
linearly until the diode acts as an open circuit. From this time on, the only
multipole draining current from the inductor is the one with the resistor. The
discharge equation is:
\begin{equation}
i_L = \frac{V_2}{R_c}[1+ \frac{R_c}{L}(t-t_3)]e^{-\frac{t-t_3}{L/(2R_c)}}
\hspace{25pt} t\geq{t_3}
\end{equation}

 The critical damping happens with a snubber resistor value of:
\begin{equation}
R_{c} = \frac{1}{2} \sqrt{\frac{L}{C_s}} 
\end{equation}
where $C_s$ is the \textbf{shunt capacitance} formed by the addition of the
\textit{collector capacitance of the bipolar transistor}, the
\textit{transformer self capacitance} and the \textit{wiring capacitance}.\\The
resistance of $R_{snubber}$ should be less than $R_c$, so as to avoid overdamping:
\begin{equation}
R_{snubber} < R_c
\end{equation}
\item \textbf{Implicit equation}\\
The residual current in the inductance, $I_R$, will be:
\begin{equation}
I_R = \frac{V_2}{R_c}[1+ \frac{R_c}{L}(t_4-t_3)]e^{-\frac{t_4-t_3}{L/(2R_c)}}
\hspace{25pt} t=t_4
\end{equation}
that cannot be resolved, but it is bounded by:
\begin{equation}
I_R < \frac{V_2}{R_{snubber}}e^{-\frac{t_4-t_3}{L/R_{snubber}}}
\end{equation}
Thus, operating along the two recovery stages as in \cite{NormanIEEPaper}:
\begin{equation}
T_R = \frac{V_1T_p}{V_2} + \frac{L}{R_{snubber}}(log{\frac{V_2}{R_{snubber}I_R}-1)}
\end{equation}
\item \textbf{Results}\\
First, we must set a threshold for $I_R$.\\
Then, the resistance of $R_{snubber}$ that makes $T_{R_{min}}$ is:
\begin{equation}
R_{snubber} = \frac{V_2}{I_R}
\end{equation}
but must comply with:
\begin{equation}
R_{snubber} < R_{c}
\end{equation}
for avoid overdamping
\end{description}

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