On Sat, Apr 04, 2009 at 08:26:04AM -0700, r norman wrote:
> On Sat, 4 Apr 2009 02:34:37 -0700 (PDT), "Bill.Connelly"
> <connelly.bill from gmail.com> wrote:
>> >On Apr 4, 3:32 am, r norman <r_s_nor... from comcast.net> wrote:
> >> If you use non-step clamp voltages, you just calculate the dV/dt term
> >> and subtract the capacitative component from the total current to get
> >> the pure ionic component.
> >Well this is exactly where I am stuck, how do I calculate the current
> >that "crosses" the capcitor, I know its somehow proportional to the
> >series resistor, the membrane capacitance and dV/dt, but exactly how,
> >I'm not sure.
>> Icap = C dV/dt
>> The resistance (conductance) is not involved.
> C is a constant, easily measured. V(t) is known, hence dV/dt is
it is the other way around:
the dV/dt is defined by the current flowing through the capacitor
and the current is defined by the series resistance:
At the moment when the voltage step (Vp) is applied (or switched off),
all the current flows through the membrane capacitor, the membrane
impedance is ~0 and the full voltage drop occurs across the series
resistance. The current is calculated as Vp/Rs (actually, one measures
the initial peak I and calculate Rs as Vp/I)
As the membrane C is charging up, the voltage drop across Rs is
decreasing and the total current flowing through the pipette
into the cell (and across the membrane) will be reduced.
At the same time, as the membrane potential changes, larger and
lager part of the current will flow through the resistive elements
and smaller and smaller current will be charging the capacitor.
In steady state, all current flows through the membrane resistance
(Rin, "input resistance" of the cell). The current will be Vp/(Rs+Rin)
and the dV on the membrane is Vp*Rm/(Rs+Rm). We often assume that
dV=Vp but this is valid only if Rs is negligibly small compared to
I hope this helps