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[Neuroscience] Re: Series Resistance compensation, what the hell?

r norman via neur-sci%40net.bio.net (by r_s_norman from _comcast.net)
Wed Apr 4 07:33:17 EST 2007

On 4 Apr 2007 02:05:39 -0700, "Bill" <connelly.bill from gmail.com> wrote:

>On Apr 4, 1:46 pm, r norman <r_s_norman from _comcast.net> wrote:
>> The voltage lost across Rs from an external voltage source is
>> determined by the voltage divider effect.  However the portion of the
>> input voltage caused by externally forced electrode current is simply
>> given by ohm's, E = IRs.  That portion of the voltage is what gets
>> compensated.
>Yes, I get this now: that as the input resistance drops, the holding
>current increases, which increases the voltage lost over Rs. Which is
>just another way of explaining the voltage divider concept.
>> I assume the "transient" nature of the command potential increase is
>> caused simply by the transient nature of the current pulse being
>> injected.  At 90% compensation, 90% of the voltage change is caused by
>> the series resistance and only 10% by the system under test.
>> Therefore, the command voltage has to be increased 10-fold so that the
>> 10% the system under test sees is the correct value.  At 100%
>> compensation, essentially all the voltage change is spurious, caused
>> by the series resistance and essentially zero is caused by the actual
>> system you are looking at.  That is, 100% compensation is caused by
>> the series resistance being infinitely greater than the cell
>> resistance.  The percent compensation is essentially the voltage
>> divider you mentioned previously.
>Wait. So if I say, apply a voltage command of 10mV, and I have 50%
>compensation, I will actually apply a 20mV voltage step? And at 90%,
>100mV. Surely that can not be correct.
>Looking at the Axon manual again: "Further assume that the access
>resistance (Ra; the sum of the pipette resistance and the residual
>resistance of the ruptured patch) is 5 MOhm. After compensation [80%],
>the effective value of Ra (Ra,eff) would be just 1 MOhm". So at 100%
>compensation, Ra would be 0, so you would get infinite current flow;
>i.e. dead cell. However, how does the amplifier achieve this feat?
>Sorry for being thick, I just can't figure this one out.

You are thinking about it wrong.  Suppose the resistance of the system
you are studying, is 1 MOhm and the pipette resistance (the series
resistance to be compensated out) is 4 MOhm.  The total Ra is then 5
MOhm but 80% of that needs compensation.  If you apply current through
the system, you will get a voltage of 5 mV across  the whole thing,
across Ra of which 4 mV appears across the electrode and only 1 mV
appears across your cell.  In other words, to produce a voltage
command of 1 mV on your cell, the equipment has to generate a 5 mV
command across the cell-electrode system.

If the compensation is 100%, then 100% of the voltage appears across
the electrode and 0% across your cell.  How much total voltage would
you then need to get a 10mV command on your cell?  Let's see, now ---
all you have to do is solve 10 mV is  0 % of X.  What is X?

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