Equation that explains the behaviour of a circuit in voltage clamp
Bill
I've already satisfied myself that the function to explain the
behaviour of a neuronal membrane (in current clamp) is
dV/dt = I/C where I is the sum of all transmembrane currents
But what is the equation for a membrane under voltage clamp? That is
how does one explain the behaviour of the holding current I, in
response to membrane capacitance and changes in membrane conductances?
Thanks for any help.
r norman
<connel...@gmail.com> wrote:+++
I don't have my copy of Hodgkin/Huxley (1952) with me at the moment,
but that is exactly where you must begin. Here, to my recollection,
is how it goes:
Iclamp = Imembrane because any longitudinal current is eliminated by
space clampl
Imembrane = Icap + Iionic because those are the only two ways current
can cross the membrane.
Icap = C dV/dt = 0 because during voltage clamp, the voltage is
ordinarily constant. Of course, there is an enormous transient
capacitative current at the time of the voltage step.
Iionic = INa + IK + Ileak because some ion has to carry the current
and Na and K are ordinarily the only significant players. Any
chloride current is hidden in the "leak".
So that reduces to Iclamp = INa + IK + Ileak.
Now throw in INa = gNa(Vm - ENa), IK = gK(Vm - EK) and Ileak =
gLeak(Vm = Eleak) where the ENa, EK and Eleak are the Nernst
equilibrium potentials dependent only on concentration, hence constant
during the clamp. You now have the clamp current as a function of
membrane potential and membrane conductances.
Once again, during the clamp the membrane potential is ordinarily
constant. Therefore, any changes in clamp current during the clamp
must necessarily be the result of changes in membrane conductance. The
trick now is to separate the total current into the separate ionic
components so you can study the separate ionic conductances; the
behavior of the ion channels.
Bill.Connelly
> Icap = C dV/dt = 0 because during voltage clamp, the voltage is
> ordinarily constant. Of course, there is an enormous transient
> capacitative current at the time of the voltage step.
Aah, that is exactly where I am having the issue. What happens during
the voltage step. Now I appreciate at the point of the step, dV/dt =
infinity (or close enough), so Icap is infinity, and the capcitor
fills instantly. However, because of the series resistance of the
electrode, peak delta I = delta V * Rs. But this is where I fall over.
Because what happens during Non-step voltages (ramps etc).
I also get that in a parallel RC circuit
Itot = V/Rm + C * dV/dt
But that leaves me in the same position, with dV/dt = infinity. So it
needs some kind of modifier, but I can't figure it out.
r norman
You just accept that there is an enormous spike of capacitative
current that the equipment can handle and recover from quickly. If
course it is not infinite, but if you integrate the very large current
over the very short duration of the interval, you get the total charge
necessary to change the voltage by the step, Cm times the step
voltage. The capacitative current is so large that it is "off screen"
and doesn't show in the clamp records except as a very short gap.
My impression of the history is that K.C. Cole, who actually invented
the voltage clamp, worried that the "backwards" inward current flow
immediately after the step was some sort of instrumental artifact
resulting from that enormous capacitative current spike and spent a
tremendous amount of time trying to figure out if it was real or not.
Hodgkin and Huxley, on the other hand, used Cole's idea of the
circuitry but didn't worry about it and just accepted the results as
real. They then went on to show that that inward component was the
Sodium inrush and the rest (including Nobel Prizes) is history.
Certainly Cole would have done the same thing, but just a little while
later.
If you use non-step clamp voltages, you just calculate the dV/dt term
and subtract the capacitative component from the total current to get
the pure ionic component. It is not hard to measure the capacitance.
For a ramp it is very simple because dV/dt is constant during the
ramp. Again, any variation in current must be caused by variation in
ion conductance.
Bill.Connelly
> If you use non-step clamp voltages, you just calculate the dV/dt term
> and subtract the capacitative component from the total current to get
> the pure ionic component.
Well this is exactly where I am stuck, how do I calculate the current
that "crosses" the capcitor, I know its somehow proportional to the
series resistor, the membrane capacitance and dV/dt, but exactly how,
I'm not sure.
r norman
Icap = C dV/dt
The resistance (conductance) is not involved.
C is a constant, easily measured. V(t) is known, hence dV/dt is
known.
Bill.Connelly
> Icap = C dV/dt
>
> The resistance (conductance) is not involved.
> C is a constant, easily measured. V(t) is known, hence dV/dt is
> known.
But that can't be right, because that in response to a step, where dv/
dt = infinity, Icap = ifinity. But in reality it's Rs*V. It's the
pressence of Rs that is screwing everything up. if it isn't for Rs,
the current can be explained as Itot = V/Rm + C dV/dt
r norman
We have been over this. The current is not infinite because the step
does not have an infinitely steep rise. It takes a certain time for
the voltage to change from one level to another, a very short time,
and during that time there is a very large capacitative current
flowing.
If you have significant series resistance, then you do not have a good
voltage clamp. Ideally, you should measure the voltage with different
electrodes than the ones you use to pass current. If there is
significant series resistance then you must estimate its value and
sutract out I times Rs from the voltage where I is the total current.
Whether or not there is Rs, the membrane current is still Im = V/Rm +
C dV/dt where V is the true transmembrane voltage, not a measured
voltage contaminated by the I * Rs term.
Imre Vida
it is the other way around:
the dV/dt is defined by the current flowing through the capacitor
and the current is defined by the series resistance:
At the moment when the voltage step (Vp) is applied (or switched off),
all the current flows through the membrane capacitor, the membrane
impedance is ~0 and the full voltage drop occurs across the series
resistance. The current is calculated as Vp/Rs (actually, one measures
the initial peak I and calculate Rs as Vp/I)
As the membrane C is charging up, the voltage drop across Rs is
decreasing and the total current flowing through the pipette
into the cell (and across the membrane) will be reduced.
At the same time, as the membrane potential changes, larger and
lager part of the current will flow through the resistive elements
and smaller and smaller current will be charging the capacitor.
In steady state, all current flows through the membrane resistance
(Rin, "input resistance" of the cell). The current will be Vp/(Rs+Rin)
and the dV on the membrane is Vp*Rm/(Rs+Rm). We often assume that
dV=Vp but this is valid only if Rs is negligibly small compared to
Rin.
I hope this helps
imre
Bill.Connelly
> On Sat, 4 Apr 2009 14:57:34 -0700 (PDT), "Bill.Connelly"
> Whether or not there is Rs, the membrane current is still Im = V/Rm +
> C dV/dt where V is the true transmembrane voltage, not a measured
> voltage contaminated by the I * Rs term.
Ohhhhh... okay then.... So
Im = Vm/Rm + C dVm/dt
And Vm... well
dVm/dt=(Vcmd-Vm)/(Rs*Cm)
No, that still isn't right, because Vm tends absolutely towards Vcmd.
Somehow I need to take into account Im*Rs
I'm sorry I'm being stupid here. I'm trying to grasp what you're
saying, I'm just not very good at the math and the physics.
Bill.Connelly
>and the dV on the membrane is Vp*Rm/(Rs+Rm). We often assume that
> dV=Vp but this is valid only if Rs is negligibly small compared to
> Rin.
Wait, is that dV, or is that just the steady state value of Vm, i.e.
Rs and Rm are acting as a voltage divider.
Assuming a neglible Rs,
dVm/dt=(Vcmd-Vm)/(Rs*Cm).
Maybe, to include Rs, it should be
dVm/dt=(Vcmd-Vm-Vs)/(Rs*Cm).
Where Vs is the voltage over Rs, which is Itot*Vs
?
Bill.Connelly
> Maybe, to include Rs, it should be
> dVm/dt=(Vcmd-Vm-Vs)/(Rs*Cm).
> Where Vs is the voltage over Rs, which is Itot*Vs
> ?
>
No, I'm an idiot, ignore that.
r norman
We may be talking about different things here.
I am assuming you have a series resistance, Rs, in series with a
parallel combination of membrane resistance, Rm, and membrane
capacitance, C. The current through the series resistance is
necessarily exactly the same current that flows through the parallel
circuit of the membrane and is both membrane current and clamp
current. The membrane current is, as already described, Vm/Rm +
C dVm/dt. the voltage across the entire thing is IRs + Vm.
You should be able to measure Rs. You also measure I. Therefore you
can calculate I Rs. You know the clamp voltage. Therefore you can
calculate Vm. And you can calculate dVm/dt. You should be able to
measure C. Therefore you can calculate C dVm/dt. Therefore you
should be able to calculate Iionic = Vm/Rm. You already calculated Vm
so you should be able to calculate Rm = 1/gm. That is, from the
voltage clamp data you should be able to see the membrane conductances
change. Also knowing Iionic, you should be able to separate that into
its components, INa and IK using the same tricks Hodgkin and Huxley
used.
r norman
Why don't you read up on "single electrode voltage clamp", as in
http://www.scholarpedia.org/article/Single_electrode_voltage_clamp
I have been describing the techniqyue for what is there called
"continuous single-electrode clamp" or cSEVC. It requires the Rs be
small or that the current be small. It also requires that Rs be
constant and linear, something that is not at all true if it results
from a microelectrode.
If none of that is true, then you really need to consider the
discontinuous technique, dSEVC, which is described in some detail in
that citation. Or else you need to use a two electrode clamp, one
electrode for passing current, the other for measuring voltage. That
eliminates the series resistance problem.
Bill.Connelly
clamp, it's just I was hoping to get an equation that relates dVcmd/dt
to Itotal, on the basis of Rm, Rs and Cm. I'm starting to assume such
an equation doesn't exist. If you think it does, I would deeply
appriciate if you could give it to me. But some people think you can't
analytically explain I for any arbitraty Vcmd.
http://www.physicsforums.com/showthread.php?p=2144713#post2144713
http://www.physicsforums.com/attachment.php?attachmentid=18292&d=1238839219
But I completely agree
I=Vm/Rm + C dVm/dt.
the voltage across the entire thing is (Vcmd).... Vcmd = IRs + Vm.
But what explains Vm?
Ohh... so
So its Vm=Vcmd - Rs * Vm/Rm+C dVm/dt
Is that right? And is that even solveable?
r norman
Yes, if you have actual data.
Your final equation is wrong, that is the problem.
Vcmd = I Rs + Vm -- you have this correct.
Now just write Vm = Vcmd - I Rs.
When doing a voltage clamp, you get a plot (or a data listing) of I
vs. t and of Vcmd vs t. So if you know Rs, you can easily calculate
Vm.
The first post you cite says : "However, I want to be able to
calculate I and any time, in response to an arbitrary voltage." That
is the problem. You dont CALCULATE I, you MEASURE it in a voltage
clamp. There is no way in general to calculate it for a real membrane
because the membrane is active and Rm (or gm) changes as a function of
V.
Bill.Connelly
> Vcmd = I Rs + Vm -- you have this correct.
>
> Now just write Vm = Vcmd - I Rs.
And as I = Vm/Rm + C dVm/dt, then
Vm = Vcmd - (Vm/Rm + C dVm/dt) * Rs.
r norman
That gives you a differential equation you can solve for Vm knowing
Vcmd.
BUT.....
The real question is just why are you doing all this algebra (and
calculus)?
Are you really interested in voltage clamp on real cells? What type
of cell, what type of activity? A real cell is NOT a passive RC
circuit. Any interesting cell has active spots, whether synaptic or
electrically excitable in which case Rm varies outside your control.
The usual purpose of voltage clamping is so you can study just how Rm
(more usually written 1/Rm = gm) varies with Vm and with time. When
Rm changes, you can NOT solve the equation you just wrote down.
So just what are you after with all these voltage clamp questions?
Are you actually attempting to do voltage clamp experiments or are you
trying to solve homework problems?
Bill.Connelly
> Are you actually attempting to do voltage clamp experiments or are you
> trying to solve homework problems?
More general curiosity than anything. What I was REALLY trying to do
was understand the behaviour of the op amp circuitry of your basic
headstage, i.e. the current to voltage converter and then the
differential amp that subtracts Vcmd. I thought I needed to know the
equation for the feedback current, but I didn't. I was trying to prove
to myself that even with a significant series resistance, the op amp
circuitry still behaved properly (even if it doesn't clamp the cell
properly).
I've gotten quite obsessive about trying to understand every detail
about e-phys recently. I was at a neuroscience course, where a lot of
the other students (some of whom were post docs and had far 'fancier'
papers than I do, doing whole-cell voltage clamp) really had no idea
about what they were actually doing during voltage clamp. They just
kinda assumed that the entire cell was clamped exactly where they set
the dial on the preamp, and the reported current was exactly equal to
the transmembrane current.
It just kinda reminded me of how anyone can be taught to use a
calculator, and 9 times out of 10, they'll get the right answer, but
if they don't understand math, that 10th time, when they times 2 by 2,
and they get 46873, they wont think it is odd.
It was really quite interesting looking back on Hodgkin and Huxley's
papers, when they 'invented' series resistance compensation... and the
resistance they were compensating for was only a couple of kiloOhms
(though the whole fact that they use opposite directions for current
and voltage makes it a bit harder to read than it should be).
But I just really like seeing those equations. I think dV/dt = I/C is
really quite nice, and while not quite as succinct Vm = Vcmd - (Vm/Rm
+ C dVm/dt) * Rs is pretty cool too.
r norman
You are right in thinking that too many biologists don't understand
their equipment and how it works and the many, many potential pitfalls
in blindly assumping that the numbers they get are real. On the other
hand, there is the story I related of KC Cole, who worried about
whether the data was rea,l and Hodking/Huxley, who interpreted the
data. Only one of these two groups walked off with the prize. You can
go overboard on the fine details.