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
Filters/frequency response
Impedance
Operational amplifiers
Voltage
Consider 2 charged plates
- +
+
+
+ -
d
Potential
difference
(+Q) + (-Q) = 0
E = V/d
Most conductors electrically
neutral because strong force
of attraction between opposite
charges
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
+
Potential
difference
+
- -
+
+
d
-
Potential difference between plates
V=Exd
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
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
E
v
+Q
v
-Q
-Q
v=+Exd
v=-Exd
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
When circuit elements connected together a network is formed
ib
2
B
3
ic
-
va
A
Network with 4 elements
3 nodes
C
+
id
+
vc
-
+
vd
D
-
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
Application of Kirchoff’s laws
i
i1
+
v
Ideal resistor
-
Slope = 1/R
i2
R
v1
v2
v = pressure head
R
I = rate of flow
Volume/unit time
A real resistor
Ohm’s law
Application of Kirchoff’s laws
series
parallel
Vtotal
i1
v1
i1
i2
R1
R1
Vtotal
R2
i2
v2
R2
Itotal = i1 = i2
Itotal = i1 + i2
Vtotal = v1 + v2
Vtotal = v1 = v2
Vtotal = Itota; (R1 + R2)
Vt/Rt = V1/R1 + V2/R2
1/Rt = 1/R1 + 1/R2 = (R2 + R1)/R2R1
Application of Kirchoff’s laws
i1
+
R1
+
+
R2
vo
Calculate vo and I for vo = 30V and
R1 = R2 = 10K ohms
-
-
Apply KVL:
i2
-vo + v1 + v2 = 0
vo = v1 + v2
Vo = i1R1 + i2R2
Vo = i1(R1 + R2)
i1 = i2 = vo(R1+R2)
Ohm’s:
v1 = i1R1
v2 = i2R2
v1 = vo R1(R1+R2)
v2 = vo R2(R1+R2)
Voltage across series resistor gets divided proportional to resistance of each element
Loading or “output impedance”
R1 = 10K
vo
+
v = 30V
R2 = 10K
Rload = 10K
-
10K
10K
vo = 10V
10K//10K = 100K/20K = 5K
10K
10K
5K
Capacitors
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
I = CdV/dt
constant I
+V
large I = big dV/dt
Vc
smalI I = small dV/dt
VC
time
constant V
Low pass/high cut filter
high frequency shunted to ground
+10 V
Vc
1K
+10 V
@ 5V across R = 1KΩ = 5 mA, so 5 V/ms
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)
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
Vs = A
iC
CA/t3
Source
waveform
slope = A/t1
Response waveform
ic is rectangular pulse when dV/dt = constant
CA/t2
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   ?
Real source networks have non-zero source resistance and finite rise times:
R
i
vs
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
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 initial
vc final
Transient response for parallel RC circuit
ir
ic
R
Use KVL:
vc + iR = 0
C
ic = CdV/dt
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
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 = (X2+R2)
Zc = 1/jC = 1/sC
the impedance of a capacitor (Zc)
decreases as frequency ( = 2pf) increases
Filters
Fc = 1/2πRC
low pass
Vo/Vi = Xc/Z
1/C
(R2 + 1/(C2)1/2
1
((RC)2 + 1)1/2
high pass
=
Operational amplifiers
Supply voltage
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
vo volts
+vcc
vv+
i1
i2
- +
+
-
positive saturation
+vcc
slope A > 10000 to 106
vo
(v+-v-) millivolts
f(v+-v-)
negative saturation
-vcc
Voltage-controlled voltage source
-vcc
|v+-v-| < vcc/A ~ 10-3 V
Why do we need op amps?
Re = 10 MΩ
cell
V

V
100 mV
10 KΩ
What is measured membrane potential?
Vm = 100 mV x
1
107/104 +
= 100 µV
1
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
KCL
if = Vo/Rf
Vin/Ri + Vo/Rf = 0
Vo = -Rf/Ri Vi
Vo = (1 + R1/R2) x Vin
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
if
ii
ia
ii + if + ia = 0
ii + (Vo - Vi)/Rf = 0
Vo = GVi or Vi = Vo/G
Find Vo
ii + (Vo + Vo/G)/Rf = 0
Vo = -ii Rf
1 + 1/G
Op amp circuits in real devices
Frequency response
Stability
Noise
A realistic patch clamp amplifier