I( elect ) = I(ion) + I ( cap )

I(cap) = C dv/dt ( C capacitance)

I(ionic) = G ( Erev - V) avec G : conductance et Erm the voltage

The capacitive component can be eliminated by keeping the membrane voltage constant during the measurement. Such a procedure is called voltage clamp. the capacitive current is zero if the derivative of the voltage is zero.

Then you will have only I(elec) = I(ion)http://en.wikipedia.org/wiki/Voltage_clamp

Cheers

Yes. There are two methods, often used concurrently:

1) capacitance compensation circuitry

2) P/4 subtraction

Both serve to eliminate the capacitance charging transient, so that you can study the ionic currents alone.

The first approach is built into most modern commercial patch clamps. The capacitance charging transient is the current that flows in response to a voltage step applied from the patch pipette to the interior of the cell. The current enters the cell via the narrow aperture (so-called "series resistance") where the circular rim of the pipette connects to the cell interior, in whole-cell patch clamp configuration. Normally this current is applied by the measurement circuitry, so it is recorded along with any ionic (transmembrane) current. When the capacitance compensation circuitry is used and a voltage step command is applied, it is the compensation circuit that applies the current needed to charge up the membrane capacitor, relieving the measurement circuitry from having to supply that current. This results in the measurement circuitry reporting only the ionic current, without the capacitance charging current.

The second approach is based on the fact that the membrane capacitor roughly approximates a linear capacitor, which means that the capacitance transient that flows to charge the capacitor should have the same shape, but differ in size in proportion to the voltage step size. Simply put, to use this approach, one applies one or more small voltage steps -- too small to activate voltage-gated ion channels, and adds them up to get the predicted full sized transient -- and subtracts that full-sized transient from the current transient that results when a full-sized voltage step is applied. In the case of applying a full-sized voltage step, the current transient is the sum of both the ionic current and the linear capacitance transient, the latter of which can be subtracted out.

I will add some references later...

Thank you very much Eric Finot and Robert Chow for your response and needful information.

Just a small addition to Roberts comment: Make sure you apply the P/N pre-steps using hyperpolarizing potentials rather than depolarizing. Then you can be sure to completely avoid acitvation of the sodium channels.

Also found this on google books:

http://books.google.co.uk/books?id=2ybw9UoRVBUC&pg=PA194&lpg=PA194&dq=P\N+to+hyperpolarizing+potentials&source=bl&ots=OP_NH8vi0C&sig=lw1utdxfnO_0RfjWbUi_i_f-Ga4&hl=en&sa=X&ei=qpNsU52JNOag0QXe7ICACA&ved=0CEcQ6AEwBg#v=onepage&q=P\N%20to%20hyperpolarizing%20potentials&f=false

Good luck

Anne

Here are some useful references:

Axon Guide: http://www.psychiatry.wustl.edu/zorumski/Axon_Guide.PDF

See page 33 on capacitance compensation

For the P/4 method, see

http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2203568/pdf/533.pdf

See figure 11

Hi! Try to use TTX in addition to the answers listed above.

Run your protocol for sodium current measurement then introduce TTX to the bath solution and repeat protocol again. Substract one trace from the other and you will get a trace without capacitive current but with sodium current in it.

Best,

Miroslav