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

+

-+

+--

+

- dPotentialdifference

(+Q) + (-Q) = 0

Most conductors electricallyneutral because strong forceof attraction between oppositecharges

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 10-3V)/(1 x 10-9m) = 100 x 106 V/m

+

-

++

--

+

-Potentialdifference

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 aconducting path results in charge transfer

Movement of positive charge (+Q) from high to lower potentialMovement of negative charge (-Q) from lower to higher potential

V = E x d

Need to know not just magnitude of voltage difference between two pointsbut which point is at a higher potential

We use consistent sign convention (reference polarity)

Networkelement

+

-

v

v = (voltage at + terminal) - (voltage at - terminal)

For charged plates:

+Q

-Q -Q

+Q

E

+

-

v = + E x d

v v

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

C DA

When circuit elements connected together a network is formed

va

ia

+

-

ib

ic id

1

2 3Network with 4 elements 3 nodes

Because individual elements connected together this imposes constraintson possible values of voltage and current

Kirchoff’s laws:

+

-

vc

+

-

vd

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

Application of Kirchoff’s laws

Ideal resistor Rv

+

-

ii1

i2

v1 v2

Slope = 1/R

v = pressure headR

I = rate of flowVolume/unit time

A real resistor

Ohm’s law

Application of Kirchoff’s laws

v1

v2

R1

R2

i1

i2

Vtotal

Itotal = i1 = i2

series parallel

Vtotal

R1 R2

i2i1

Vtotal = v1 + v2

Vtotal = Itota; (R1 + R2)

Itotal = i1 + i2

Vtotal = v1 = v2

Vt/Rt = V1/R1 + V2/R2

1/Rt = 1/R1 + 1/R2 = (R2 + R1)/R2R1

Application of Kirchoff’s laws

R1

R2vo

+

-

+ -

+

-

i1

i2

Apply KVL: -vo + v1 + v2 = 0

vo = v1 + v2 Ohm’s: v1 = i1R1

v2 = i2R2Vo = i1R1 + i2R2

Vo = i1(R1 + R2)

i1 = i2 = vo(R1+R2)v1 = vo R1(R1+R2)

v2 = vo R2(R1+R2)

Voltage across series resistor gets divided proportional to resistance of each element

Calculate vo and I for vo = 30V and R1 = R2 = 10K ohms

Loading or “output impedance”

R2 = 10Kv = 30V

+

-

vo

Rload = 10K

R1 = 10K

10K

10K

10K 10K 5K

vo = 10V

10K//10K = 100K/20K = 5K

Capacitors

+

-

++

+ +

++

+++

--

--

--

-- -

+Q

-Q

Electric field proportional toboth voltage & total charge

Q = Cv

to find v-i characeristicstake 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

+V Vc

time

large I = big dV/dt

smalI I = small dV/dt

I = CdV/dt

constant V

1K

+10 V

1 µF

Vc

Vc

+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

Vc = Vsource (1-e-t/RC)

Low pass/high cut filter high frequency shunted to ground

VC

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

Vs

+

-

i = CdV/dT

iC

Vs = A

slope = A/t1

t1t2t3

ic(t) is proportional to rate of change of source voltage Vs

t

Sourcewaveform

Response waveformiC

CA/t3

CA/t2

CA/t1

ic is rectangular pulse when dV/dt = constant

As Vs gets faster, response (iC) gets larger in amplitude but shorter duration

What happens when dV/dt ?

R

vs

i

C

+

-

Imax = A/R (Vs/R)

Real source networks have non-zero source resistance and finite rise times:

A

vs

tr

slope (dV/dt of source voltage) = A/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

Capacitor always lags voltage

then RC >> tr

Transient response of RC circuit

vs

A

time

v=0, t<0v=A, t>0

vs vc

+

-

R

+-

R

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 place3) Final time interval long enough after step so vs acts as DC source

Voltage on capacitor: ic = dV/dt, V= constant, i=0

vc initial

vc final?

R C

ic

Transient response for parallel RC circuit

Use KVL: vc + iR = 0 ic = CdV/dt

vc + RCdV/dt = 0

Kirchoff’s laws will give differential equation with networks containing capacitors

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

ir

When t = , t V = V/e ~ .37V

Network model of neuron

Electrical impedance

Measure of opposition of circuit to AC currentComplex ratio of V to I in an AC circuit: magnitude and phase

Impedance = Z = V(t)/I(t)

Z = (X2+R2)

Zc = 1/jC = 1/sC

the impedance of a capacitor (Zc)decreases as frequency (w = 2pf) increases

Filters

Fc = 1/2πRC

low pass

high pass

Vo/Vi = Xc/Z

1/wC

(R2 + 1/(wC2)1/2

=

1((RwC)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+

-vcc

+vcc

i1

i2-+

+

-

vo

f(v+-v-)

Supply voltage

vo volts

+vcc

-vcc

positive saturation

negative saturation

(v+-v-) millivolts

slope A > 10000 to 106

|v+-v-| < vcc/A ~ 10-3 V

Voltage-controlled voltage source

Why do we need op amps?

cellV V

Re = 10 MΩ

10 KΩ

What is measured membrane potential?

Vm = 100 mV x 1

107/104 + 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

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

summing amplifier

unity gain voltage follower

current voltage converter

Vo = - (V1 x Rf/R1 + V2 x Rf/R2 … + Vn x Rf/RN)

Vo = G(0-V-)

if = (V0 - Vi)/Rf

if

ii ia

ii + if + ia = 0

ii + (Vo + Vo/G)/Rf = 0

Vo = GVi or Vi = Vo/G

Find Vo

ii + (Vo - Vi)/Rf = 0

Vo = -ii Rf

1 + 1/G

Vo = G(Vin-Vo)

Vo = G1+G

Vin

Op amp circuits in real devices

Frequency response

Stability

Noise

A realistic patch clamp amplifier