Noah Weiss & Susan Koons
Neuroscience: 3ed
Neuroscience: 3ed
Neuroscience: 3ed
Neuroscience: 3ed
• Resistors: Linear or non-linear
F(V,I)=0
V=IR
I=f(V)
V = h(I)
dVC
 IC
• Capacitors: C
dt
dA  VA
• Pumps: dt
• Kirchhoff’s Current Law:
The principle of
conservation of electric
charge implies that:
The sum of currents
flowing towards a point
is equal to the sum of
currents flowing away
from that point.
i2
i3
i1
i1 = i2 + i3
• Kirchhoff’s Voltage Law
The directed sum of the electrical potential
differences around any closed circuit must be
zero. (Conservation of Energy)
R2
R1
VR1 + VR2 + VR3 + VC =0
R3
• Neurons can be modeled with a circuit model
– Each circuit element has an IV characteristic
– The IV characteristics lead to differential
equation(s)
• Use Kirchhoff’s laws and IV characteristics to get the
differential equations
dV
• Solve for I and use C C  I
C
C
dt
• To find I use the current law: A  A  I
A
1
2
A
– Additionally, define the absolute current IS  A1  A2
– Assume a linear resistor with (small) resistance γ in series
with the pumps
• Use Kirchhoff’s laws to get:
 dV
C
C I
 I Na  f K VC  EK   I A
ext



dt


   I S VC   I A 
I A




 I S   I A VC   I A 




  VC  ENa  hNa  I Na 
 I Na

• Assume the “N” curve
doesn’t interact with
the “S” curve
– All three parts of “N” are
within primary branch of
“S”
– Also, let ε = 0:
 dV
C
C I
 I Na  f K VC
ext


dt


   I S VC   I A 
I A




 I S   I A VC   I A 




 I Na  g Na VC  ENa 



 EK   I A

I
V
K
Na
• Substitute the 4th equation into the 1st
 dV
C
C I
 g Na VC
ext


dt




   I S VC   I A 
I A




 I S   I A VC   I A 




 ENa   f K VC  EK   I A



• Nullclines: Set the derivatives equal to zero
– Nontrivial nullcline in the 2nd and 3rd equations are
same
– Re-arrange and obtain the following:
Nullcline equations


 I  Iext  g
Na VC
A

V   I
A
 C
 ENa   f K VC  EK 



• Let Iext  0
– Analyze the nullclines: vector field directions
– Assume C<<1: singular perturbation
– IA nullcline intersects VC nullcline in primary
IA
branch
IA nullcline
VC nullcline
Vc
• Increase Iext to shift the VC nullcline upward
• To get an action potential:
• The “N” curve has 2 “knee” points at V1 and V2
Nullcline equations


 I  Iext  g
Na VC
A

V   I
A
 C
 ENa   f K VC  EK 




0,
if VC  V1

f K VC   G1VC   G2  G1  VC V1  if V1  VC








 G2  G1 V2  V1  if V2  VC



• The “S” curve is merely linear by assumption
(i.e. gNa is constant)
• Some algebra shows that Iext must satisfy:


Iext >= VC  1  gNa  G2   g Na ENa V1  G2  G1  with VC  V1,V2 


 V2
 dV
C
V  E   I
C I

I

f
 C

ext

Na
K
K
A


dt


V   I 
I


I

 A
S  C
A 


V   I 
I


I
 S
A  C
A 


I


I

V

E

h

 Na
Na
Na  Na 
C

Inside the cell
Outside the cell
• Recall the equations for
one node:
– There is no outgoing
current
• Consider a second node
that is not coupled to
the first node
– It should have the same
equation (but with
different currents)
 dV
C
C I
 I Na  f K VC  EK   I A
ext



dt

 dI
 A   I V   I 
S C
A
 dt

 dI S
  I A VC   I A 



 dt

 dI Na
 VC  ENa  hNa  I Na 



dt


2

C



2
 dI A

 dt

 dI 2
 S
 dt

 dI 2
Na


dt

C
dV
2  I 2  f 2 V 2  E 2   I 2
 Iext
Na
K  C
K 
A
dt
  I S2 VC2   I A2 


  I A2 VC2   I A2 



2  h2  I 2 
 VC2  ENa
Na  Na 


• Couple the nodes by
adding a linear
resistor between them

1
VC2 VC1
 dVC
 1

1
1
1
 Iext  I Na  f K VC  EK   I A 
C
R1


dt

 1
 dI


 A   I 1 V 1   I 1 
S C
A
 dt
 1
 dI S
  I 1A VC1   I 1A 



 dt

1
 dI Na
 VC1  E1Na  hNa  I 1Na 




dt

 dV 2
VC2 VC1
 2

2
2
2
C
C
  I Na  f K VC  EK   I A 

R1


dt

 dI 2
 A   I 2 V 2   I 2 
S  C
A 
 dt

 dI 2
 S   I 2 V 2   I 2 
A  C
A 
 dt

 dI 2
Na  V 2  E 2  h  I 2 

Na
Na  Na 
C

dt
• This is the general
equation for the nth
node
• In and out currents
are derived in a
similar manner:

















dVCn
n1  I n  f V n  E n   I n  I n
C
 Iout
in
Na
K  C
K 
A
dt
dI An
  I Sn VCn   I An 


dt
dI Sn
  I An VCn   I An 


dt
n
dI
n h In 
 Na  VCn  ENa
Na  Na 
dt
 I
if n  1
 ext
n1   n
Iout
V  V n1
C
 C
if n  1
n

1

R

 n1
n
VC  VC
if 1  n  N
Iinn   Rn

if n  N
0
C=.1 pF
Forcing current
C=.1 pF
C=.1 pF
C=.1 pF
C=.01 pF
C=.01 pF
C=.7 pF
C=.7 pF
(x10 pF)
(ms)
(x10 pF)
(ms)
(x100 mV)
(x100 mV)
The Importance of Myelination- Myelinated Axon
(ms)
• Myelination matters! Myelination decreases
capacitance and increases conductance
velocity
• If capacitance is too high, the pulse will not
transmit
• First model that shows a pulse that travels
down the entire axon without dying out