Biomedical Engineering Reference
In-Depth Information
EXAMPLE 3.6
In a current step experiment, a 100-ms-long pulse of 1 nA of current is applied
into a
cell using a microelectrode and thereby drive the membrane potential of the cell from its
resting level of
−
70 mV to a steady state level of +30 mV. The capacitance of the cell mem-
brane is 0.5 nF. What is the membrane time constant? What is the membrane potential 50
ms after the end of the current pulse?
ΔΦ
0
= −
70 mV,
ΔΦ
SS
= +
30 mV
C
=
0.5 nF
=
5
×
10
−
10
F
I
=
1 nA
=
1
×
10
−
9
[A]
[
]
30
( 70) 10
−
3
−−
ΔΦ
− ΔΦ
SS
rest
8
R
=
=
=
110
×
Ω
m
−
9
I
110
×
inj
τ
=
RC = (1
×
10
8
Ω
)
×
(5
×
10
−
10
)
=
0.05 second
=
50 ms
When the current is turned off,
ΔΦ
0
=
30 mV and
ΔΦ
SS
= −
70 mV
From (3.21),
50
50
⎛
⎞
−
⎜
⎟
(
)
⎝
⎠
70
30
( 70)
e
33.22 mV
ΔΦ = −
+
−
−
= −
ΔΦ
ss
ΔΦ
ss
V
rest
V
rest
100
100
0
0
150
150
Time (ms)
Time (ms)
Using the equivalent circuit model, the Goldman equation can be linearized
using conductances and individual ion potentials in the resting membrane. The
conductance is related to membrane permeability but is not the same.
ΔΦ
rest
is
constant and there is no net current crossing in or out of the cell, although each
individual conductance may be carrying a net current. So, in steady state, the total
current must be zero
(
. Hence,
I
++=
I
I
0)
+
+
−
K
a
Cl
(
)
(
)
(
)
()
()
()
g
ΔΦ
− ΔΦ
t
+
g
ΔΦ
− ΔΦ
t
+
g
ΔΦ
− ΔΦ
t
=
0
K
m
K
Na
m
Na
CL
m
Cl
