# [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)
Sun Apr 5 01:23:58 EST 2009

```On Sat, 4 Apr 2009 18:30:36 -0700 (PDT), "Bill.Connelly"
<connelly.bill 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.

Why don't you read up on "single electrode voltage clamp", as in
http://www.scholarpedia.org/article/Single_electrode_voltage_clamp

I have been describing the techniqyue for what is there called
"continuous single-electrode clamp" or cSEVC.  It requires the Rs be
small or that the current be small.  It also requires that Rs be
constant and linear, something that is not at all true if it results
from a microelectrode.

If none of that is true, then you really need to consider the
discontinuous technique, dSEVC, which is described in some detail in
that citation.  Or else you need to use a two electrode clamp, one
electrode for passing current, the other for measuring voltage.  That
eliminates the series resistance problem.

```