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Electricity & Circuits: An introduction for neuroscientists

Electricity & Circuits: An introduction for neuroscientists. Topics. Voltage/current: conventions. Circuits/ Circuit elements:. Resistors. Capacitors. Kirchoff’s laws (current & voltage). Voltage sources & voltage dividers. RC circuits:. Response to voltage source.

kamal-monroe
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Electricity & Circuits: An introduction for neuroscientists

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  1. Electricity & Circuits: An introduction for neuroscientists

  2. Topics Voltage/current: conventions Circuits/ Circuit elements: Resistors Capacitors Kirchoff’s laws (current & voltage) Voltage sources & voltage dividers RC circuits: Response to voltage source Filters/frequency response Impedance Operational amplifiers

  3. Voltage Consider 2 charged plates - + + - + + - d - Potential difference Most conductors electrically neutral because strong force of attraction between opposite charges (+Q) + (-Q) = 0 E = V/d E is electric field, vector field (N/coul) For a cell membrane: (100 x 10-3V)/(1 x 10-9m) = 100 x 106 V/m Suppose charge -Q moved from upper to lower plate

  4. + + + + d Potential difference - - - - V = E x d Potential difference between plates If conducting path placed between plates, current would flow because of attractive forces In general, voltage difference exists between two pints if introducing a conducting path results in charge transfer Movement of positive charge (+Q) from high to lower potential Movement of negative charge (-Q) from lower to higher potential

  5. Need to know not just magnitude of voltage difference between two points but which point is at a higher potential We use consistent sign convention (reference polarity) + Network element v - v = (voltage at + terminal) - (voltage at - terminal) For charged plates: + +Q +Q E v v -Q -Q - v = + E x d v = - E x d

  6. Current convention When charge (q) moves in a conductor, we say current flows Positive charge moves from region of high potential to low potential Need to consider algebraic sign of charge in relation to reference direction to determine sign of current Reference direction - + negative current positive current + + + + + positive mobile charge + + + - - - - negative mobile charge - - - -

  7. Circuits When circuit elements connected together a network is formed ib Network with 4 elements 3 nodes 3 2 B id ic - + + A C D va vc vd + - - ia 1 Because individual elements connected together this imposes constraints on possible values of voltage and current Kirchoff’s laws: Voltage law: algebraic sum of voltage drops around any loop must equal zero (va + vb + vc = 0) Current law: algebraic sum ofcurrents entering any node must equal zero (ib - ic - id = 0) Laws apply to every electric network and nerve cells

  8. Application of Kirchoff’s laws i i1 + Slope = 1/R i2 v Ideal resistor R - v1 v2 v = pressure head R A real resistor I = rate of flow Volume/unit time

  9. Ohm’s law

  10. Application of Kirchoff’s laws series parallel Vtotal i1 i1 i2 v1 R1 R1 R2 Vtotal i2 v2 R2 Itotal = i1 + i2 Itotal = i1 = i2 Vtotal = v1 = v2 Vtotal = v1 + v2 Vt/Rt = V1/R1 + V2/R2 Vtotal = Itota; (R1 + R2) 1/Rt = 1/R1 + 1/R2 = (R2 + R1)/R2R1

  11. Application of Kirchoff’s laws i1 + R1 - + i2 + R2 vo Calculate vo and I for vo = 30V and R1 = R2 = 10K ohms - - Apply KVL: -vo + v1 + v2 = 0 vo = v1 + v2 Ohm’s: v1 = i1R1 v2 = i2R2 Vo = i1R1 + i2R2 Vo = i1(R1 + R2) v1 = vo R1(R1+R2) i1 = i2 = vo(R1+R2) v2 = vo R2(R1+R2) Voltage across series resistor gets divided proportional to resistance of each element

  12. Loading or “output impedance” R1 = 10K vo + v = 30V R2 = 10K Rload = 10K - 10K 10K vo = 10V 10K//10K = 100K/20K = 5K 5K 10K 10K

  13. Capacitors

  14. Capacitors + +Q Electric field proportional to both voltage & total charge + + + + + + Q = Cv + + + to find v-i characeristics take derivative with respect to time - - - - - - - - - dQ/dt = CdV/dt -Q i = CdV/dt - Current flows only when voltage changing 1) An ideal capacitor is an open circuit for DC voltages 2) For a rapidly changing voltage capacitor looks like a short circuit

  15. I = CdV/dt constant I large I = big dV/dt +V Vc smalI I = small dV/dt VC time Low pass/high cut filter high frequency shunted to ground constant V +10 V Vc 1K @ 5V across R = 1KΩ = 5 mA, so 5 V/ms +10 V Vc 1 µF large I  large dV/dt 10V across R = 1KΩ = 10/1K = 10 mA so dV/dt = I/C = 10 mA/1µF = 10 V/ms Voltage across capacitor approaches applied voltage with rate that diminishes towards zero Vc = Vsource (1-e-t/RC)

  16. How fast can the voltage change across capacitor? Important for understanding integration of electrical potentials by membrane capacitance and operation of a voltage clamp circuit Consider an idealized circuit with no source resistance iC + i = CdV/dT Vs - Response waveform Vs = A iC ic is rectangular pulse when dV/dt = constant CA/t3 Source waveform CA/t2 slope = A/t1 CA/t1 t3 t2 t1 t ic(t) is proportional to rate of change of source voltage Vs As Vs gets faster, response (iC) gets larger in amplitude but shorter duration What happens when dV/dt   ?

  17. Real source networks have non-zero source resistance and finite rise times: vs R i Imax = A/R (Vs/R) A + C vs slope (dV/dt of source voltage) = A/tr - tr If capacitor is to change its voltage as rapidly as source, current needed is CA/tr Source network can only supply maximum current of A/R and only when V=0 It is impossible for capacitor to charge up quickly enough to follow vs A/R << CA/tr then RC >> tr Capacitor always lags voltage

  18. Transient response of RC circuit vs R R + A vc vs + time - - v=0, t<0 v=A, t>0 Divide total time into 3 intervals: 1) Initial time before step when vs is DC source or zero 2) Time interval just after step during which transient response takes place 3) Final time interval long enough after step so vs acts as DC source Voltage on capacitor: ic = dV/dt, V= constant, i=0 ? vc final vc initial

  19. Transient response for parallel RC circuit ir ic R C Use KVL: ic = CdV/dt vc + iR = 0 Kirchoff’s laws will give differential equation with networks containing capacitors vc + RCdV/dt = 0 Solve: separate variables: dV/dt = -(1/RC x vc) dV/dt = ∫-1/RC x vc dV/vc = ∫-1/RC x dt ln V = -t/RC + C V = -Ae-t/RC When t = t, V = V/e ~ .37V

  20. Network model of neuron

  21. Electrical impedance Measure of opposition of circuit to AC current Complex ratio of V to I in an AC circuit: magnitude and phase Impedance = Z = V(t)/I(t) Z = (X2+R2) Zc = 1/jC = 1/sC the impedance of a capacitor (Zc) decreases as frequency (w = 2pf) increases

  22. Filters Fc = 1/2πRC Vo/Vi = Xc/Z low pass 1/wC = (R2 + 1/(wC2)1/2 1 ((RwC)2 + 1)1/2 high pass

  23. Operational amplifiers Supply voltage Integrated circuits made up of transistors, resistors, capacitors Behaves as a high gain linear voltage amplifier 1) huge input resistance () Special properties: 2) negligible output resistance (0) 3) cheap vo volts +vcc positive saturation +vcc i1 v- slope A > 10000 to 106 + - vo i2 + (v+-v-)millivolts v+ - -vcc f(v+-v-) negative saturation |v+-v-| < vcc/A ~ 10-3 V -vcc Voltage-controlled voltage source

  24. Why do we need op amps? Re = 10 MΩ cell V  V 100 mV 10 KΩ What is measured membrane potential? 1 = 100 µV Vm = 100 mV x 107/104 + 1 Need to make voltage meter internal resistance > 100 mΩ Need high input impedance device

  25. Some basic linear op amp circuits ii = (Vin - V-)/Ri if = (Vo - V-)/Rf Ideal properties of op amp: V- = V+ = 0 ii = Vin/Ri if = Vo/Rf KCL Vin/Ri + Vo/Rf = 0 Vo = -Rf/Ri Vi Vo = (1 + R1/R2) x Vin

  26. summing amplifier Vo = - (V1 x Rf/R1 + V2 x Rf/R2 … + Vn x Rf/RN) unity gain voltage follower Vo = G(Vin-Vo) Vo = G Vin 1+G current voltage converter Vo = G(0-V-) if = (V0 - Vi)/Rf ii + if + ia = 0 if Vo = GVi or Vi = Vo/G ii + (Vo - Vi)/Rf = 0 ii ia Find Vo ii + (Vo + Vo/G)/Rf = 0 Vo = -ii Rf 1 + 1/G

  27. Op amp circuits in real devices Frequency response Stability Noise A realistic patch clamp amplifier

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