How Does Adding a Series Resistor Affect Current Calculation in an RC Circuit?

[bill.connelly]

Hi,

I'm just a humble biologist (writting up a PhD Thesis in Neuroscience) and I'm trying to get a feel for how my equipment works. The circuit for monitoring the current that flows across a cells membrane (in response to an imposed voltage step) can essentially be seen as a resistor, connected to a resistor and capacitor in parallel (Rs, Rm and Cm)
(the full circuit, ignore the op-amps and such if you can)

I understand that for a resistor and capacitor in parallel, the current that flows in response to a voltage can be explained by
$$i(t)=\frac{v(t)}{R}+C\frac{dv(t)}{dt}$$
However this means that in response to a step change in voltage, I = infinity (which is of course what you expect). With a resistor in series to that parallel circuit, I also appreciate that at after a step of voltage V,
$$\Delta I=V.R_{s}$$
and then I recays exponentially to
$$\Delta I=V.(R_{s}+R_{m})$$
However, I want to be able to calculate I and any time, in response to an arbitrary voltage. Can anyone explain, or even just give me the equation to calculate that.

 

[The Electrician]

You just convolve the impulse response of your circuit with the arbitrary input.

See:

http://www.swarthmore.edu/NatSci/echeeve1/Ref/Convolution/Convolution.html

Also:

http://en.wikipedia.org/wiki/Convolution

 

[bill.connelly]

Thanks The Electrician,

I'm sorry, but someone is going to have to dumb it down for me.

 

[The Electrician]

Do I understand correctly that you will be applying some sort of stimulus function (you mentioned a step function) to the CMD input of your circuit, and you want to be able to calculate the response current?

If your input is truly arbitrary, meaning it can't be described as some simple mathematical function such a sine wave, or a step function, then how would you describe it for the purpose of making mathematical computations? Some sort of a time series, an array of values for small increments of time?

 

[bill.connelly]

What I'm ACTUALLY trying to do is show that for and given Vcmd, VI is = If*Rf. I was able to do that if I ignore Rs and Cm (ignoring Rs is okay because Rs is a finy fraction of Rm). So essentially I was working in the steady state case, where the capacitance didn't mater.

What I'm looking for is an equation that relates the voltage output of op amp A1, to the current that flows, If. Previously I was just using Ohms law, so If = V1/(Rf+Rm). But I assume there is an equation that will take into account that capacitor and Rs.

 

[The Electrician]

If the steady state condition is that a steady DC voltage is applied to Vcmd, then capacitors don't matter. In that case it is possible to derive an expression for what you want that doesn't include capacitors. Of course, the value of the various components will determine the settling time; how long you must wait until the currents and voltages are stable.

The way you have drawn Rs makes it look like there is a distributed capacitance associated with. If you have a time varying voltage applied to Vcmd, that distributed capacitance would make for a more complicated expression for the impulse response. But if all you care about is the steady state response to a DC stimulus, then the various capacitances don't matter.

So, is the circuit response to a steady DC stimulus what you want?

 

[bill.connelly]

If the steady state condition is that a steady DC voltage is applied to Vcmd

Indeed, and I have written up equations to explain the behaviour of the circuit in the steady state case.

What I want is an equation, given an applied voltage V, not a step or steady response, but just any arbitrary V, what is equation that explains the current that is drawn by Rm, Cm and Rs.

Exactly in the same manner that if I had an arbitrary current I, applied to a similar circuit, I could explain the voltage as dV/dt = I/C

or how if it weren't for the resistor Rs, I could explain the current by
$$i(t)=\frac{v(t)}{R}+C\frac{dv(t)}{dt}$$

But how do I modify that equation to account for Rs.

 

[Phrak]

I'm completely confused. The circuitry that monitors[i/] the voltage consists of the op amps. Somewhere in this monitoring circuit are two points that connect across the cell membrane. Where are they?

 

[bill.connelly]

Hi Phrak,

The Voltage Vi is measured relative to earth; does that answer your question? But all that part of the circuit is not neccesary...

Lets just consider this circuit
https://www.physicsforums.com/attachment.php?attachmentid=18291&stc=1&d=1238836893
A variable voltage Vcmd is supplied, and causes Itot to flow. What is the equation that relates Vcmd, Rs, Rm and Cm to Itot.
I know that As DVcmd/Dt approaches infinity, Itot = Vcmd/Rs, and when DVcmd/Dt = 0, Itot tends towards Vcmd/(Rs+Rm) and decays towards that value with the time constant of Cm*Rs.
BUT what I want is the single equation that explains how Itot varies over time, in response to voltage? It is presumbly some way proportional to dVcmd/dt, but I can't figure out the exact details.

 

[The Electrician]

There isn't any expression for i(t) as simple as:
$$i(t)=\frac{v(t)}{R}+C\frac{dv(t)}{dt}$$

You can't escape using convolution if the applied voltage is arbitrary.

See the attachment.

 

[bill.connelly]

*sigh* that's a shame. Thanks anyway.

 

[The Electrician]

It's really not a big deal. If you have your arbitrary stimulus in the form of a data file, a sampled waveform in other words, you can use the built-in convolution function that most modern mathematical software such as Matlab include.

The impulse response of your circuit is easy to derive as I showed. Then just convolve it with your stimulus and you have the current i(t).

 

[bill.connelly]

It can't be explained by
I=Vm/Rm + C dVm/dt,
Where Vm (the voltage over the capacitor)
Vm=Vcmd - Rs * Vm/Rm+C dVm/dt

There is a bit of a debate going over here
http://groups.google.com/group/bion...00d6b5deb6f/658ba433c2e50ff9#658ba433c2e50ff9

 

[uart]

It can't be explained by
I=Vm/Rm + C dVm/dt,
Where Vm (the voltage over the capacitor)
Vm=Vcmd - Rs * Vm/Rm+C dVm/dt

There is a bit of a debate going over here
http://groups.google.com/group/bion...00d6b5deb6f/658ba433c2e50ff9#658ba433c2e50ff9

It can but that will give you a DE instead of a closed form solution. The DE you obtain is :

$$(R_1+R_2) \, i + R_1\, R_2\, C \, di/dt = v + R_2\, C\, dv/dt$$

In general you'd need to specify the input voltage as a time function, v(t), and solve the above DE.
If however you just want to consider a simple case like a "step" input then it's fairly easy.

For a step input voltage of amplitude "V" you will get :

$$i(t) = I_f + (I_0-I_f) e^{-t/T}$$

Where,

$$T = \frac{R_1 R_2 C} {R_1+R_2}$$

$$I_0 = \frac{V}{R_1}$$ {assuming zero initial capacitor voltage}

$$I_f = \frac{V}{R_1+R_2}$$

PS. R1 is the series resistor and R2 the parallel reisitor in the above.

 

[bill.connelly]

And in a more general sense, looking at the op amps above, does that mean the output of summing amp A2 (VI)...
Vi=I*Rf-Rs(Vm/Rm+C*dVm/dt)?
Or more simply
Vi=I*Rf-Rs*If

(Again, where Vm is the voltage over the capacitor Cm)

 

[uart]

The relationship between the current in the resistor-capacitor network, which I've called i(t) above, and the output of the first op-amp is :

$$v_1(t) = v + R_f \,\, i(t)$$

Therefore the output of the second amplifier (the differential amplifier) is :

$$v_2(t) = v_1(t) - v = R_f \, \, i(t)$$

Here (and above) I've used "v" for the voltage marked "v_cmd" on your diagram.

 

[The Electrician]

It can't be explained by
I=Vm/Rm + C dVm/dt,
Where Vm (the voltage over the capacitor)
Vm=Vcmd - Rs * Vm/Rm+C dVm/dt

There is a bit of a debate going over here
http://groups.google.com/group/bion...00d6b5deb6f/658ba433c2e50ff9#658ba433c2e50ff9

In an earlier post, I showed how to get the impulse response for your circuit including Rs. That essentially solved the equations you've got here.

uart says:

"It can but that will give you a DE instead of a closed form solution. The DE you obtain is :

$$(R_1+R_2) \, i + R_1\, R_2\, C \, di/dt = v + R_2\, C\, dv/dt$$

In general you'd need to specify the input voltage as a time function, v(t), and solve the above DE."

This is quite true, but "in general" you won't be able to "solve" the DE in the traditional way for an arbitrary input voltage. But, a general "solution" for arbitrary Vcmd can be found using convolution. I showed it in formal terms in the previous post, but now let me show a practical example.

See the attached images in this post for one example, and the images in the next post for another example.

 

[The Electrician]

Here's a second example.

 

[bill.connelly]

Thanks a lot for all your help, yes, I see where I had gotten confused about the output of the opamps. I just have one question left

Are the equations
$$(R_1+R_2) \, i + R_1\, R_2\, C \, di/dt = v + R_2\, C\, dv/dt$$

and
I=Vm/Rm + C dVm/dt,
Vm=Vcmd - Rs * Vm/Rm+C dVm/dt

essentially equivalent? If not, what is the difference in what they are explaining, or how they go about explaining it?

 

[The Electrician]

Thanks a lot for all your help, yes, I see where I had gotten confused about the output of the opamps. I just have one question left

Are the equations
$$(R_1+R_2) \, i + R_1\, R_2\, C \, di/dt = v + R_2\, C\, dv/dt$$

and
I=Vm/Rm + C dVm/dt,
Vm=Vcmd - Rs * Vm/Rm+C dVm/dt

essentially equivalent? If not, what is the difference in what they are explaining, or how they go about explaining it?

If you eliminate Vm from these equations:

I=Vm/Rm + C dVm/dt,
Vm=Vcmd - Rs * Vm/Rm+C dVm/dt

You should get:

$$(R_1+R_2) \, i + R_1\, R_2\, C \, di/dt = v + R_2\, C\, dv/dt$$

The easiest way to come up with this is to solve the network. Use Kirchoff's voltage law and you get two simultaneous equations which resolve to this one.

Also, Laplace transforms can be used to get the step response of the network, which is the solution of the differential equation just above. See the attached image.

But, if you want i(t) for anything other than a mathematically simple stimulus such as a step, or a sine wave, you will probably have to use the convolution theorem.

 

[bill.connelly]

Thanks again The Electrician. I'll look into convolution, and Laplace transforms.