## When I(inward) = I(outward) dV/dt = 0 right?

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Bill Connelly
Posts: 86
Joined: Thu May 22, 2008 11:54 pm
Location: Australian National University

### When I(inward) = I(outward) dV/dt = 0 right?

Hi.

I made a simple soma with Cm and HH. I added an AlphaSynapse. Then I summed all the currents and compared this with the soma.v. I was assuming that when the total current was outward, soma.v would be depolarizing. When the total current was inward, soma.v would be hyperpolarizing and when the total current was zero soma.v would be stable (i.e. the peak of the EPSP).

However, this was not the case.

The currents I summed were
soma.i_cap(0.5)
soma.ik(0.5)
soma.il_hh(0.5)
soma.ina(0.5)
AlphaSynapse[0].i

In hindsight, the reason for this effect is that all those 'currents' are actually in mA/cm^2 apart from AlphaSynapse right? So how do I convert those currents into whole cell currents? (A way that didn't just use 4 . pi . r^2 would be good)

Thanks.
ted
Posts: 5810
Joined: Wed May 18, 2005 4:50 pm
Location: Yale University School of Medicine
Contact:
Seems easier to convert the single current that is in "absolute" units into a value in "density"
units. See the Programmer's Reference via its link at
http://www.neuron.yale.edu/neuron/docs
for documentation of the area function, which returns the area of the currently accessed
segment in um^2.
Bill Connelly
Posts: 86
Joined: Thu May 22, 2008 11:54 pm
Location: Australian National University
So a soma with a length of 10, and a diam of 10 has an area of 314?
That means that neuron is modelling it as a cylinder without the ends. For a section where Length = radius doesn't that mean neuron is 50% out when calculating currents given in current/cm^2?
ted