[Neuroscience] Re: Equation that explains the behaviour of a circuit in voltage clamp

Bill.Connelly via neur-sci%40net.bio.net (by connelly.bill from gmail.com)
Sat Apr 4 20:40:50 EST 2009

```Hi Imre,

>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
> Rin.

Wait, is that dV, or is that just the steady state value of Vm, i.e.
Rs and Rm are acting as a voltage divider.
Assuming a neglible Rs,
dVm/dt=(Vcmd-Vm)/(Rs*Cm).

Maybe, to include Rs, it should be
dVm/dt=(Vcmd-Vm-Vs)/(Rs*Cm).
Where Vs is the voltage over Rs, which is Itot*Vs
?

On Apr 5, 1:30 pm, "Bill.Connelly" <connelly.b... from gmail.com> wrote:
> On Apr 5, 10:51 am, r norman <r_s_nor... from comcast.net> wrote:
>
> > On Sat, 4 Apr 2009 14:57:34 -0700 (PDT), "Bill.Connelly"
> > 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.
>
> Ohhhhh... okay then.... So
>
> Im = Vm/Rm + C dVm/dt
> And Vm... well
> dVm/dt=(Vcmd-Vm)/(Rs*Cm)
>
> No, that still isn't right, because Vm tends absolutely towards Vcmd.
>
> Somehow I need to take into account Im*Rs
>
> I'm sorry I'm being stupid here. I'm trying to grasp what you're
> saying, I'm just not very good at the math and the physics.

```