Nerves, the heart, and the brain are electrical. How do these things work? This course presents fundamental principles, described quantitatively.

Loading...

来自 Duke University 的课程

Bioelectricity: A Quantitative Approach

33 个评分

Nerves, the heart, and the brain are electrical. How do these things work? This course presents fundamental principles, described quantitatively.

从本节课中

Axial and Membrane Current in the Core-Conductor Model

This week we will examine axial and transmembrane currents within and around the tissue structure: including how these currents are determined by transmembrane voltages from site to site within the tissue, at each moment. The learning objectives for this week are: (1) Select the characteristics that distinguish core-conductor from other models; (2) Identify the differences between axial and trans-membrane currents; (3) Given a list of trans-membrane potentials, decide where axial andtrans-menbrane currents can be found; (4) Compute axial currents in multiple fiber segments from trans-membrane potentials and fiber parameters; (5) Compute membrane currents at multiple sites from trans-mebrane potentials.

- Dr. Roger BarrAnderson-Rupp Professor of Biomedical Engineering and Associate Professor of Pediatrics

Biomedical Engineering, Pediatrics

So hello again. This is Roger Coke Barr for the

bioelectricity course. And today we're going to, to talk about

the axial current. You're thinking, well, finally this guy

has gotten around to talking, about what this whole week is supposed to be about.

Good news, it really is true, we are going to talk about it today.

Earlier, we were talking about making a voltage slope around four segments, a, b,

four, points, a, b, c and d. So, I've picked that thought up and put it

on the top line. Let's go through carefully, and determine

what each one of these, segments is going to be like.

The first, this segment V, A, B. Is Phi A minus Phi B,

Which means it's also equal, by Ohms Law, to IE1, R1.

That is to say, here's IE1, the extracellular current, in segment one.

It's going through the resistance in between A and B.

Here we have said the resistances are the same.

The fiber is uniform. Resistance from one segment to the next is

RE. So by Ohm's law, we have this

relationship. Vab is at IE1 times RE Now let's do the

second one. Vbc.

Well, here we're calling that Vbc, But at other moments we call that the

transmembrane potential, so this would also be Vm2.

So Vm2,, though, has the opposite side. Because Vm2 would be measured with the

plus here and the minus there, Vbc. Has the plus here and the minus there.

It is Vbc, it's 5b minus 5c. Vm2 does it the other way around.

5c minus 5b. So the one is the negative of the other

but otherwise they're the same. So I've written that over here.

Okay, I'll erase and we'll do the third one.

Now we want this segment Vcd. Vcd.

It's going to be. Only for it oh, right here.

The intracellular current times the intracellular resistance.

Notice how similar it is to the relationship we wrote for Vab.

It's almost the same thing. Except, of course, we use RI instead of

RE. And we use the intracellular current II,

rather than the extracellular current. But both are expressions of Ohm's Law.

Now for the fourth one. Vda.

That will be Vta, plus will be here, minus will be there.

Tm1. Sorry.

If I was moved over one it would be VM1. Plus here, minus there.

Should actually be drawn the, over here. The N1.

The very same thing. So Vda is equal to Vm1 So now we have

equations established for each, that is to say more detail established, for each of

these four different terms. Let's substitute each of those terms.

This is equation one, two, three, four, five, and substitute 235 back into

equation one. So when we have done that, we get this

result. It's not quite in the same order, but I

think if you'll look at this line compared to this line, you'll see that they are the

very same thing. So that the IE1, RE for Vab.

Here it is down here. So this one went down there.

This one, I believe went over here, and so forth for

each of the four terms. This is the very same thing.

Now I can't tell you how many times I've worked with people who are doing this

derivation, and they got to this line. And they were stuck, stuck, stuck, stuck.

And the problem was they have two unknowns, I1.

And IE1, two unknowns, and just one equation.

Stuck, stuck, stuck. Of course you are not stuck because you

know, from an earlier segment, that the one is the negative of the other.

And when you use that fact, that the one is the negative of the other.

The equation simplifies, and we get it in a new form, which is like this.

And we arrange it. We'll get to this wonderful form that

tells us what we want to know. The intracellular axial current in segment

one, this current. I, I1, can be found by taking.

This VM. The M1.

Subtracting this VM, VM2, and dividing by the sum of the intracellular and the

extracellular resistance. If you think about that, you'd say this

makes perfect sense. We have a driver, in terms of voltage in

this loop. And that driver is this difference in

potential. Vm1 minus Vm2, So that current is going

round and round. And the driver for this current is VM1

minus VM2. The resistance to that current flow is r I

plus r e. And that determines what I, I one, is.

You might say look, look, look, look, what happened to the membrane resistance? After

all it's in the loop too. Where did that membrane resistance go?

Well. We don't have to worry about it.

Because whatever the membrane resistance is, the result is that we have the M1.

Whatever this other resist, membrane resistance is over here, the result of

that membrane resistance is that we have VM2.

Once we know VM1 and, and VM2, we don't have to worry about the membrane

resistance on those paths. Let's do a little simplification.

Let's suppose that RE is very small compared to RI.

If RE is very small compared to RI, that term can be ignored.

Notice we're not, not saying that r e is equal to zero, we're just saying its small

enough compared to r I, r I is usually a big number, r e is small enough compared

to r I, that it can be ignored in comparison.

So in that sense such a, think about the statement, r e is virtually zero, relative

to r I, relative to r I is an important condition.

But if that is so, then we have the, have the equation that we want.

I, I1 is VM1 minus VM2 over RI. There's something kind of remarkable about

this equation, just was noticing. There is a directional shift.

So if we look at the direction, the M1 is in this direction.

The M2 is in that direction, but II1. Is in a different direction.

This is the way that II1 goes. So the result that we have, this equation

is not a simple application of Ohm's law. Even though the units seem the same.

Because the current II1 is not in the same location as the voltages VM1 or VM2.

And in fact it's in a different direction. Now of course we know it's okay because

we've done the derivation, and when you go around the loop everything is accounted

for. But it still is a little bit strange.

Usually, when you do these voltage over resistance things, the currents are in the

same direction, not a different direction. There's one final point worth noting.

Let me go to another slide to have a little bit more space.

Suppose we say, I, I1 is equal to VM1 plus VM2.

Divided by RI. Let's just think for a moment about the

units. Ri is in ohms.

It is the actual resistance for this particular fiber because we got it from

RI, which took in count, into account, the fiber diameter, as well as the bulk

resistivity. Vm1 - Vm2, this is in volts or more often

in millivolts. Ri is in ohms, but more often written in

kiloohms. That means that the resulting current.

One divided by the other is in amperes or, more frequently, in nanoamperes, nanovolts

divided by kiloamps. One has to ask in any particular problem,

are these units okay, or do I want to have some different units?

Well, I won't try to answer that, because I don't know which particular problem

you're interested in, but it is worth making a little note that our result is

associated with some particular units, and if we don't like those units, further

conversion might be required. Well thank you for watching this segment.

I'll talk to you again in a little bit, about membrane current, which we can

determine once we know the axial currents. Thank you for watching.