[Neuroscience] Re: Equation that explains the behaviour of a
circuit in voltage clamp
r norman
via neur-sci%40net.bio.net
(by r_s_norman from comcast.net)
Sat Apr 4 17:51:49 EST 2009
On Sat, 4 Apr 2009 14:57:34 -0700 (PDT), "Bill.Connelly"
<connelly.bill from gmail.com> wrote:
>On Apr 5, 3:26 am, r norman <r_s_nor... from comcast.net> wrote:
>
>> Icap = C dV/dt
>>
>> The resistance (conductance) is not involved.
>> C is a constant, easily measured. V(t) is known, hence dV/dt is
>> known.
>
>But that can't be right, because that in response to a step, where dv/
>dt = infinity, Icap = ifinity. But in reality it's Rs*V. It's the
>pressence of Rs that is screwing everything up. if it isn't for Rs,
>the current can be explained as Itot = V/Rm + C dV/dt
We have been over this. The current is not infinite because the step
does not have an infinitely steep rise. It takes a certain time for
the voltage to change from one level to another, a very short time,
and during that time there is a very large capacitative current
flowing.
If you have significant series resistance, then you do not have a good
voltage clamp. Ideally, you should measure the voltage with different
electrodes than the ones you use to pass current. If there is
significant series resistance then you must estimate its value and
sutract out I times Rs from the voltage where I is the total current.
Whether or not there is Rs, the membrane current is still Im = V/Rm +
C dV/dt where V is the true transmembrane voltage, not a measured
voltage contaminated by the I * Rs term.
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