Biomedical Engineering Reference
In-Depth Information
As the current pulse magnitude is increased, depolarization of the membrane increases,
causing activation of the
Na
þ
and K
þ
voltage-time-dependent channels. For sufficiently
large depolarizations, the inward
Na
þ
current exceeds the sum of the outward
K
þ
and leak-
age currents
ð
I
Na
>
I
K
þ
I
l
Þ
. The value of
V
m
at this current is called
. Once the
threshold
Na
þ
and
K
þ
voltage-time channels are engaged and run
membrane reaches threshold, the
to completion, as shown in Figure 12.27.
If a slow rising stimulus current is used to depolarize the cell membrane, then thresh-
old will be higher. During the slow approach to threshold, inactivation of
G
Na
channels
occurs and activation of
G
K
channels develops before threshold is reached. The value of
V
m
,where
I
Na
>
I
K
þ
I
l
is satisfied, is much larger than if the approach to threshold occurs
quickly.
12.6.2 Equations Describing G
Na
and G
K
The empirical equation used by Hodgkin and Huxley to model
G
Na
and
G
K
is of
the form
D
t
Þ¼
A
þ
Be
Ct
G
(
ð
12
:
41
Þ
Values for the parameters
were estimated from the voltage clamp data
that were collected on the squid giant axon. Not evident in Eq. (12.41) is the voltage
dependence of the conductance channels. The voltage dependence is captured in
the parameters as described in this section. In each of the conductance models,
A
,
B
,
C
,and
D
is
selected as 4 to give a best fit to the data. Figure 12.27 was actually calculated using
SIMULINK, a simulation package that is partofMATLAB,andtheparameterestimates
found by Hodgkin and Huxley. Details concerning the simulation are covered later in
this section.
D
Potassium
The potassium conductance waveform shown in Figure 12.26 is described by a rise to a
peak while the stimulus is applied. This aspect is easily included in a model of
G
K
by using
the general Hodgkin-Huxley expression as follows.
G
K
¼
G
K
n
4
ð
12
:
42
Þ
where
G
K
K
þ
conductance and
is thought of as a rate constant and given as
the solution to the following differential equation:
is maximum
n
dn
dt
¼
a
n
ð Þ
1
b
n
n
ð
12
:
43
Þ
