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\title{Temporary Title}
\author{Steve Van Hooser}
\date{Time and Date}
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{\bf Figure 6: Synaptic Currents}
The spike-mediated synaptic current is modeled as an alpha function with
experimentally-determined (Simon et al. 1994) time constants:
\begin{displaymath}I_{\mathrm{SynS}} = \overline{g}_{\mathrm{SynS}} {\sum_{\mathrm{spikes}}}{f_{\mathrm{SynS}}(t-t_{\mathrm{spike}}) (V-E_{\mathrm{Syn}}),}\end{displaymath} with
\begin{displaymath}f_{\mathrm{SynS}}(t) = \frac{a}{\tau_{\mathrm{fall}}-\tau_{\mathrm{rise}}}(e^{-t/\tau_{\mathrm{fall}}}-e^{-t/\tau_{\mathrm{rise}}})(V-E_{\mathrm{Syn}}).\end{displaymath}
For all synapses except the G3,4 to G1,2 connections, $\tau_{\mathrm{rise}} = 0.0025\mathrm{sec}$ and $\tau_{\mathrm{fall}} = 0.011\mathrm{sec}$. For the G3,4 to G1,2
connections, $\tau_{\mathrm{rise}} = 0.010\mathrm{sec}$, and $\tau_{\mathrm{fall}} = 0.055\mathrm{sec}$. $a$ was chosen so the function $f_{\mathrm{SynS}}(t)$ has a maximum of 1.
To more accurately model the spike-mediated synaptic conductance, this alpha function
needs to be corrected for a dependence on the presynaptic voltage. The correction factor $m_{\mathrm{spike}}$ has Hodgkin-Huxley kinetics with a time constant of 0.2 sec
and the following steady-state activation function:
\begin{displaymath}m_{\mathrm{spike}} = 0.1 + \frac{0.9}{1+e^{-1000(V+0.4)}}
\end{displaymath}
The equation
for the graded synapse is more complicated--the current is approximately proportional to a factor $[\mathrm{P}]$, which itself is proportional to the
presynaptic calcium concentration. At each time step, the factor $[\mathrm{P}]$ increases by the amount of calcium ions that flow into the cell via ion
channels (``$I_{Ca}$'' factor), and decreases by an amount that is both voltage-dependent and proportional to the calcium already present
(``$B$''
and ``$A$'' factors). These equations result from a fit to data obtained from simultaneous measurements of presynaptic calcium current and
postsynaptic synaptic current (see Angstadt and Calabrese 1991 and Nadim et al. 1995 for more details).
\begin{displaymath}
I_{\mathrm{SynG}} = \overline{g}_{\mathrm{SynG}}\frac{[\mathrm{P}]^3}{10^5+[\mathrm{P}]^3}(V-E_{\mathrm{SynS}})\end{displaymath}
\begin{displaymath}\frac{d[\mathrm{P}]}{dt} = I_{\mathrm{Ca}}(t) - B[\mathrm{P}]\end{displaymath}
\begin{displaymath}I_{Ca}(t) = \mathrm{max}(0,-(I_{\mathrm{CaF}}(t)+I_{\mathrm{CaS}}(t))-A(t,V)), B = 10 {\mathrm{s}^{-1}}\end{displaymath}
\begin{displaymath}\frac{dA(t,V)}{dt} = \frac{A_\infty(V) - A(t,V)}{0.2}\end{displaymath}
\begin{displaymath}A_\infty(V) = 0.1/(1+\mathrm{exp}(-100(V+0.020)))\end{displaymath}
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