Passive Membranes - Quantitative Neurophysiology - page 12

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Figure 2.5: Charging and discharging of a passive membrane.

2.3 STRENGTH-DURATIONRELATIONSHIP
To perform our analysis of the passive membrane we assumed that V m <V t m . It would therefore be
helpful to know in what situations our assumption is valid. Examining Eq. (2.15), V m (t) during the
stimulus is dependent on V

R m C m . Consider that I stim may be small such that
V ∞ <V t m . In this case, the stimulus may remain on forever and V m will charge up to a steady state value
below V th . In other words, the membrane will remain passive. If I stim is steadily increased, eventually V

∞ =

I stim R m and τ m =

∞
will be equal to V t m . In this case, the steady state will level out exactly at the threshold but it may take
a very long time for V m

will clearly be greater than
V t m . As the right side of Fig. 2.5 indicates, however, even if V ∞ >V t m the stimulus may be on for such
a short time that the membrane does not have time to charge to the threshold. It follows that there is an
interplay between the strength of I stim and the duration of I stim , which can be represented graphically in
a strength-duration plot. Given a constant value for V t m , Fig. 2.6 shows the combinations of strength and
duration that exactly charge the membrane to the threshold. Mathematically, we can set V m = V t m

to reach threshold. If I stim is increased further, V ∞

and

substitute into Eq. (2.15)

V t m e − D/τ m )

= −

I stim R m ( 1 (2.17)

where D is the duration of the stimulus. One interesting limit is to find the minimum current that can
bring the membrane to V t m .If D →∞

V t m = I stim R m (2.18)

V t m
R m

I stim = I rhe = (2.19)

which explains the value of asymptote in Fig. 2.6. This value is called rheobase , I rhe , and is a measure of
current. Another important measure can be derived by assuming that we apply two times the rheobase
current to find the corresponding time to charge the membrane to V t m .

V t m e − D/τ m )

= 2 I rhe R m ( 1 − (2.20)