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Trigger Circuit

exa2
Variables and parameters:

yi: voltages, i=1,...,6

$\lambda$: input voltage (bifurcation parameter)

y6: output voltage


System of equations resulting from Kirchhoff's law:

\begin{displaymath}f_1 & = (y_1 -y_3)/10000 + (y_1 -y_2 )/39 + (y_1 +\lambda )/51=
0\cr \end{displaymath}


\begin{displaymath}f_2 & = (y_2 -y_6)/10 + I(y_2) + (y_2 -y_1)/39= 0 \cr \end{displaymath}


\begin{displaymath}f_3 & = (y_3 -y_1)/10000 + (y_3 -y_4)/25.5= 0\cr \end{displaymath}


\begin{displaymath}f_4 & = (y_4 -y_3)/25.5 + y_4/0.62 + y_4 -y_5= 0 \cr \end{displaymath}


\begin{displaymath}f_5 & = (y_5 -y_6)/13 + y_5 -y_4 + I(y_5)= 0 \cr \end{displaymath}


\begin{displaymath}f_6 & = (y_6 -y_2)/10 - [U_{\rm A} (y_3 -y_1)-y_6]/0.201
+ (y_6 -y_5)/13 = 0\cr \end{displaymath}

with amplifier and diodes modelled as

\begin{displaymath}U_{\rm A} (U) = 7.65 \ \arctan \ (1962 U), \quad
I(U) = 5.6\cdot 10^{-8} (e^{25U} -1).\end{displaymath}



Comments


Figure 1
Trigger Circuit with operational amplifier and two diodes.

Figure 2
Trigger, output voltage y6 versus input voltage $\lambda$.


Postscript-Files for better printing results:

This example
Figure 1
Figure 2