Biomedical Engineering Reference
In-Depth Information
P
K
d
þ
qV
½
K
þ
o
J
K
e
q
KT
¼
d
KT
ð
12
:
26
Þ
P
K
d
þ
qV
½
K
þ
i
J
K
KT
d
Solving for
J
K
in Eq. (12.26) gives
0
@
1
A
qV
KT
½
K
þ
o
½
K
þ
i
e
J
K
¼
qVP
K
KT
ð
12
:
27
Þ
qV
KT
e
1
Chlorine Ions
The same derivation carried out for
K
þ
can be repeated for
Cl
, which yields
0
@
1
A
qV
KT
½
Cl
i
½
Cl
o
e
J
Cl
¼
qVP
Cl
KT
ð
12
:
28
Þ
qV
KT
1
e
Cl
.
where
P
Cl
is the permeability for
Summarizing for Potassium and Chlorine Ions
Using space charge neutrality,
J
K
¼
J
Cl
, and Eqs. (12.27) and (12.28) gives
qV
KT
qV
KT
½
Cl
i
P
K
½
K
þ
o
½
K
þ
i
e
¼
P
Cl
½
Cl
o
e
ð
12
:
29
Þ
Solving for the exponential terms yields
KT
¼
P
K
½
K
þ
o
þ
P
Cl
½
Cl
i
P
K
½
K
þ
i
þ
P
Cl
½
Cl
o
qV
e
ð
12
:
30
Þ
Solving for
gives
V
P
K
½
K
þ
o
þ
P
Cl
½
Cl
i
P
K
½
K
þ
i
þ
P
Cl
½
Cl
o
V
¼
v
o
v
i
¼
KT
q
ln
ð
12
:
31
Þ
or in terms of
V
m
P
K
½
K
þ
o
þ
P
Cl
½
Cl
i
P
K
½
K
þ
i
þ
P
Cl
½
Cl
o
V
m
¼
KT
q
ln
ð
12
:
32
Þ
This equation is called the
Goldman equation
. Since sodium is also important in membrane
K
þ
,
Cl
, and
Na
þ
can be derived as
potential, the Goldman equation for
P
K
½
K
þ
o
þ
P
Na
½
Na
þ
o
þ
P
Cl
½
Cl
i
P
K
½
K
þ
i
þ
P
Na
½
Na
þ
i
þ
P
Cl
½
Cl
o
V
m
¼
KT
q
ln
ð
12
:
33
Þ
Na
þ
. To derive Eq. (12.33), first find
where
is the permeability for
J
Na
and then use space
P
Na
charge neutrality
J
K
þ
J
Na
¼
J
Cl
. Equation (12.33)
then follows.
In general, when the