2. Building a Passive Neuron Batteries, Resistors, and Capacitors

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Lawrence McCormick
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pass 2. Building a Passive Neuron brc 2. 1.

1 pass 2. Building a Passive Neuron brc Batteries, Resistors, and Capacitors The signals in the brain arise from the motion of charged particles - typically ions of sodium, Na +, chloride,, potassium, K +, and calcium, Ca ++. Figure 1. The typical neuron pumps chloride out of the cell to the degree that its intracellular concentration is one tenth of its extracellular concentration. This imbalance establishes a nonzero rest potential, V. We impale the cell with an electrode that delivers current I and measures the resulting potential, V. The energy required to move charge is called voltage while the rate at which charge moves is called current. A battery is a source of constant voltage, e.g., V. A resistor is a device that resists (or diminishes) current. For example, if a resistor of resistance R is connected across the terminals of a battery of voltage V then the current, I, through the resistor is I = V/R. (2.1)ohm In V g Out I mcon Figure 2.1 We model the cell membrane as a battery of size V in series with a resistance R. 2

2 A capacitor is a device that stores charges in proportion to the voltage V across it. In particular, the charge Q stored by a capacitor of capacitance C will be Q = C V. (2.2)far When we say that current is the velocity of the charge we mean that current is the slope of the charge graph. That is I(t) = change in charge change in time for some small time step,. For example, consider = Q(t + ) Q(t) (2.3)iq Q t Qt Figure 2.2. For small, please confirm that if we substitute this Q into (2.3) we find I(t) = 0 when 0 < t < 1, I(t) = 1 when 1 < t < 2 and I(t) = 2 when 2 < t < 3. If we now combine equations (2.2) and (2.3) we arrive at a current voltage relationship for capacitors, I C (t) = C m V (t + ) V (t) (2.4)civ I stim V R I C m I C mcap Figure 2.3. We add to the model above a parallel current that mimics the membrane s ability to store charge. 3

3 The fundamental circuit law is a simple balance law that requires current in to balance current out. In our case this reads I stim (t) = I (t) + I C (t) V (t + ) V (t) = S (V (t) V )/R + C m S, and it is very important to make sure that we are balancing consistent quantities. The convential unit of current in the brain is one millionenth of an Ampere, written micro ampere, or µa for short. (2.5)rc1kcl 1. A typical chloride resistance, R, is 10/3 kω/cm 2, i.e., 10/3 thousand Ohms per square centimeter. 2. A typical surface area, S, is 10 5 cm A typical chloride reversal potential, V, is 70mV, i.e., 70 millivolts. 4. A typical membrane capacitance, C m, is 1µF/cm 2, i.e, 1 micro farad per square centimeter. The convential units for brain voltage and time are mv (milli-volts) and ms (milli-seconds) respectively. Please confirm that (2.5) is indeed equating µa to µa. We typically rearrange (2.5) to express the future in terms of the present, i.e., we solve for V (t + ) in terms of quantities at t, V (t + ) = (1 /τ)v (t) + (/τ)v + (/S/C m )I stim (t), (2.6)pmem where τ = R C m is called the membrane time constant. This allows us to deduce V () from the known V (0), and from there to step from V () to V (2) and so on. passsoft Building the Neuron in Software It is not hard to code (2.6) in Octave. V = -70; S = 1e-5; R = 0.3; Cm = 1; tau = R*Cm; = 0.05; a = /tau; Imax = 1e-5; b = *Imax/S/Cm; ton = 2; toff = 40; tfin = 60; N = tfin/; V(1) = V; for n=1:n, t(n+1) = n*; V(n+1) = (1-a)*V(n) + a*v + b*(t(n+1)>ton)*(t(n+1)<toff); end 4

plot(t,v) 66.5 67 67.5 voltage (mv) 68 68.5 69 69.5 70 0 10 20 30 40 50 60 time (ms) rc1modresponse Figure 2.4. The result of the simulation coded above.

Building the Neuron in Hardware It is not a simple matter to deliver a prescribed current, I stim.

4 plot(t,v) voltage (mv) time (ms) rc1modresponse Figure 2.4. The result of the simulation coded above. Please experiment with the various parameters. passhard Building the Neuron in Hardware It is not a simple matter to deliver a prescribed current, I stim. We rather drive a known, 672 Ω, resistor with the NI mydaq Function Generator, and record its response with the associated Oscilloscope. 5

5 NIrc Figure 2.5 The red wire connects the Function Generator (pin AO 0) to the (vertical) 672 Ω resistance. This serves as I stim into the membrane RC circuit, R = 1.17kΩ and C = 9.87µF. On selected a 10 Hertz, 4 Volt peak-to-peak square wave from the Function Generator, the Oscilloscope (black wire to pin AI 0+) captures the charge, plateau and discharge that we expect from our numerical simulation. The green wire is to ground, and the small black jumper connects pin AI 0- to ground. We next switch from square to sine waves and note that peak-to-peak response diminishes as we increase frequency. Please record, on a sheet of paper, the response Vpp for frequencies 10 0, 10 1, 10 2, 10 3 and 10 4, and graph your results in Octave, like that below. We define gain to be the ratio of the Vpp of the response to the Vpp of the stimulus gain frequency NIrcgain Figure 2.6 The membrane circuit behaves like a low pass filter. ptran The Limiting Cell Equation and its Gain Function Please check that if we define, R R /S, C C m S and v(t) = V (t) V then (2.5) takes the simpler form RI stim (t) = v(t) + τ v(t + ) v(t). For those that have been exposed to calculus you might recognize that the difference 6

6 quotient v(t + ) v(t) approaches the derivative v (t) as approaches 0. The limiting passive neuron equation is then RI stim (t) = v(t) + τv (t). (2.7)rc1ode There are many ways to analyze such a differential equation. To begin we will demonstrate that if a sinusoid goes in then a sinusoid comes out. More precisely, we suppose that I stim (t) = I 0 (f) exp(2πift) and v(t) = V 0 (f) exp(2πift) where f denotes frequency. (An Octave digression/appendix on exp(2πift) would be nice, simply typing help plot3 gets you very far). On substituting these into (2.7) we find RI 0 (f) exp(2πift) = V 0 (f) exp(2πift) + (2πifτ)V 0 (f) exp(2πift) On canceling the common exponential we find RI 0 (f) = (1 + 2πifτ)V 0 (f), and rearrange slightly to arrive at the analytical Gain Function G(f) V 0(f) I 0 (f) = 1 R 1 + (2πfτ) 2. As this function decreases as f increases we speak of the passive neuron as a Low Pass Filter. 7

PHYSICS 171 UNIVERSITY PHYSICS LAB II. Experiment 6. Transient Response of An RC Circuit

PHYSICS 171 UNIVERSITY PHYSICS LAB II Experiment 6 Transient Response of An RC Circuit Equipment: Supplies: Function Generator, Dual Trace Oscilloscope.002 Microfarad, 0.1 Microfarad capacitors; 1 Kilohm,