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
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.