# Electricity & Circuits: An introduction for neuroscientists.

## Presentation on theme: "Electricity & Circuits: An introduction for neuroscientists."— Presentation transcript:

Electricity & Circuits: An introduction for neuroscientists

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

Voltage Consider 2 charged plates + - + + - - + - d Potential difference (+Q) + (-Q) = 0 Most conductors electrically neutral because strong force of attraction between opposite charges E = V/d E is electric field, vector field (N/coul) Suppose charge -Q moved from upper to lower plate For a cell membrane: (100 x 10 -3 V)/(1 x 10 -9 m) = 100 x 10 6 V/m

+ - + + - - + - Potential difference 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 V = E x d

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 +Q E + - v = + E x d vv v = - E x d

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 positive current negative current + + + + + + + + - - - - - - - - positive mobile charge negative mobile charge

Circuits B CDA When circuit elements connected together a network is formed vava iaia + - ibib icic idid 1 2 3 Network with 4 elements 3 nodes Because individual elements connected together this imposes constraints on possible values of voltage and current Kirchoff’s laws : + - vcvc + - vdvd Voltage law: algebraic sum of voltage drops around any loop must equal zero (v a + v b + v c = 0) Current law: algebraic sum ofcurrents entering any node must equal zero (i b - i c - i d = 0) Laws apply to every electric network and nerve cells

Application of Kirchoff’s laws Ideal resistor R v + - i i1i1 i2i2 v1v1 v2v2 Slope = 1/R v = pressure head R I = rate of flow Volume/unit time A real resistor

Ohm’s law

Application of Kirchoff’s laws v1v1 v2v2 R1R1 R2R2 i1i1 i2i2 V total I total = i 1 = i 2 seriesparallel V total R1R1 R2R2 i2i2 i1i1 V total = v 1 + v 2 V total = I tota; (R 1 + R 2 ) I total = i 1 + i 2 V total = v 1 = v 2 V t /R t = V 1 /R 1 + V 2 /R 2 1/R t = 1/R 1 + 1/R 2 = (R 2 + R 1 )/R 2 R 1

Application of Kirchoff’s laws R1R1 R2R2 vovo + - +- + - i1i1 i2i2 Apply KVL: -v o + v 1 + v 2 = 0 v o = v 1 + v 2 Ohm’s: v 1 = i 1 R 1 v 2 = i 2 R 2 V o = i 1 R 1 + i 2 R 2 V o = i 1 (R 1 + R 2 ) i 1 = i 2 = v o (R 1 +R 2 ) v 1 = v o R 1 (R 1 +R 2 ) v 2 = v o R 2 (R 1 +R 2 ) Voltage across series resistor gets divided proportional to resistance of each element Calculate v o and I for v o = 30V and R 1 = R 2 = 10K ohms

Loading or “output impedance” R 2 = 10Kv = 30V + - vovo R load = 10K R 1 = 10K 10K 5K v o = 10V 10K//10K = 100K/20K = 5K

Capacitors

+ - + + + + + + + + + - - - - - - - - - +Q -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 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 Capacitors

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

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 VsVs + - i = CdV/dT iCiC V s = A slope = A/t 1 t1t1 t2t2 t3t3 i c (t) is proportional to rate of change of source voltage V s t Source waveform Response waveform iCiC CA/t 3 CA/t 2 CA/t 1 i c is rectangular pulse when dV/dt = constant As V s gets faster, response (i C ) gets larger in amplitude but shorter duration What happens when dV/dt   ?

R vsvs i C + - I max = A/R (V s /R) Real source networks have non-zero source resistance and finite rise times: A vsvs rr slope (dV/dt of source voltage) = A/  r If capacitor is to change its voltage as rapidly as source, current needed is CA/  r 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 v s A/R << CA/  r Capacitor always lags voltage then RC >>  r

Transient response of RC circuit vsvs A time v=0, t<0 v=A, t>0 vsvs vcvc + - R + - R Divide total time into 3 intervals: 1) Initial time before step when v s 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 v s acts as DC source Voltage on capacitor: i c = dV/dt, V= constant, i=0 v c initial v c final ?

RC icic Transient response for parallel RC circuit Use KVL: v c + iR = 0 i c = CdV/dt v c + RCdV/dt = 0 Kirchoff’s laws will give differential equation with networks containing capacitors Solve: separate variables: dV/dt = -(1/RC x v c )  dV/dt = ∫-1/RC x v c  dV/v c = ∫-1/RC x dt ln V = -t/RC + C V = -Ae -t/RC irir When t =  V = V/e ~.37V

Network model of neuron

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 =  (X 2 +R 2 ) Z c = 1/j  C = 1/sC the impedance of a capacitor (Z c ) decreases as frequency (  = 2  f) increases

Filters F c = 1/2πRC low pass high pass V o /V i = X c /Z 1/  C (R 2 + 1/(  C 2 ) 1/2 = 1 ((R  C) 2 + 1) 1/2

Operational amplifiers Integrated circuits made up of transistors, resistors, capacitors Behaves as a high gain linear voltage amplifier Special properties: 1) huge input resistance (  ) 2) negligible output resistance (  0) 3) cheap v-v- v+v+ -v cc +v cc i1i1 i2i2 - + + - vovo f(v + -v - ) Supply voltage v o volts +v cc -v cc positive saturation negative saturation (v + -v - ) millivolts slope A > 10000 to 10 6 |v + -v - | < v cc /A ~ 10 -3 V Voltage-controlled voltage source

Why do we need op amps? cell V  V R e = 10 MΩ 10 KΩ What is measured membrane potential? V m = 100 mV x 1 10 7 /10 4 + 1 100 mV = 100 µV Need to make voltage meter internal resistance > 100 mΩ Need high input impedance device

Some basic linear op amp circuits i i = (V in - V - )/R i i f = (V o - V - )/R f Ideal properties of op amp: V - = V + = 0 i i = V in /R i i f = V o /R f KCL V in /R i + V o /R f = 0 V o = -R f /R i V i V o = (1 + R 1 /R 2 ) x V in

summing amplifier unity gain voltage follower current voltage converter V o = - (V 1 x R f /R 1 + V 2 x R f /R 2 … + V n x R f /R N ) V o = G(0-V - ) i f = (V 0 - V i )/R f ifif i iaia i i + i f + i a = 0 i i + (V o + V o /G)/R f = 0 V o = GV i or V i = V o /G Find V o i i + (V o - V i )/R f = 0 V o = -i i R f 1 + 1/G V o = G(V in -V o ) V o = G 1+G V in

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