edfa667479
Signed-off-by: David Rotermund <54365609+davrot@users.noreply.github.com>
92 lines
2 KiB
Markdown
92 lines
2 KiB
Markdown
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## Main equations
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$$c_m \frac{d V}{dt} = -i_m + \frac{I_e}{A}$$
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$$t_m(V) \frac{d m}{dt} = m_{\infty}(V) - m$$
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$$t_h(V) \frac{d h}{dt} = h_{\infty}(V) - h$$
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$$t_n(V) \frac{d n}{dt} = n_{\infty}(V) - n$$
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$$t_a(V) \frac{d a}{dt} = a_{\infty}(V) - a$$
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$$t_b(V) \frac{d b}{dt} = b_{\infty}(V) - b$$
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$$i_m =\bar{g}_L (V - E_L) + \bar{g}_{Na} m^3 h (V - E_{Na}) + \bar{g}_{K} n^4 (V - E_{K}) + \bar{g}_{A} a^3 b (V-E_A)$$
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## alpha-values
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$$\alpha_n(V) = \frac{0.02mV^{-1} (V + 45.7mV)}{1 - \exp(-0.1mV^{-1} (V + 45.7mV))}$$
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$$\alpha_m(V) = \frac{0.38mV^{-1} (V + 29.7mV)}{1 - \exp(-0.1mV^{-1} (V + 29.7mV))}$$
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$$\alpha_h(V) = 0.266 \exp(-0.05mV^{-1} (V + 48.0mV))$$
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## beta-values
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$$\beta_n(V) = 0.25 \exp(-0.0125mV^{-1} (V + 55.7mV))$$
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$$\beta_m(V) = 15.2 \exp(-0.0556mV^{-1} (V + 54.7mV))$$
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$$\beta_h(V) = \frac{3.8}{1 + \exp(-0.1mV^{-1} (V + 18mV))}$$
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## time constants
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$$\tau_n(V) = \frac{1.0ms}{\alpha_n(V) + \beta_n(V)}$$
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$$\tau_m(V) = \frac{1.0ms}{\alpha_m(V) + \beta_m(V)}$$
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$$\tau_h(V) = \frac{1.0ms}{\alpha_h(V) + \beta_h(V)}$$
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$$\tau_a(V) = 0.3632ms + \frac{1.158ms}{1.0 + \exp(0.0497mV^{-1} (V + 55.96mV))}$$
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$$\tau_b(V) = 1.24ms + \frac{2.678ms}{1.0 + \exp(0.0624mV^{-1} (V + 50.0mV))}$$
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## asymptotic values
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$$a_\infty(V) = \left( \frac{0.0761 * \exp(0.0314mV^{-1} (V + 94.22mV))}{1 + \exp(0.0346mV^{-1} (V + 1.17mV))}) \right)^{1 / 3} ms$$
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$$b_\infty(V) = \left(\frac{1}{1 + \exp(0.0688mV^{-1} (V + 53.3mV))} \right)^4 ms$$
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$$n_\infty(V) = \alpha_n(V) \tau_n(V)$$
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$$m_\infty(V) = \alpha_m(V) \tau_m(V)$$
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$$h_\infty(V) = \alpha_h(V) \tau_h(V)$$
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## Suitable initial conditions
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$$m(t=0) = 0.010$$
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$$n(t=0) = 0.156$$
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$$h(t=0) = 0.966$$
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$$a(t=0) = 0.540$$
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$$b(t=0) = 0.289$$
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$$V(t=0) = -68.0 mV$$
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## Parameter
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$$c_m = 0.1 \frac{\mu F}{mm^2}$$
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$$\frac{I_e}{A} = 0.35 \frac{\mu A}{mm^2}$$
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$$\bar{g}_L = 0.003 \frac{mS}{mm^2}$$
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$$\bar{g}_{Na} = 1.2 \frac{mS}{mm^2}$$
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$$\bar{g}_{K} = 0.2 \frac{mS}{mm^2}$$
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$$\bar{g}_{A} = 0.477 \frac{mS}{mm^2}$$
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$$E_L = -17.0 mV$$
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$$E_{Na} = 55.0 mV$$
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$$E_K = -72.0 mV$$
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$$E_A = -75.0 mV$$
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