How to reduce complicatd RC network

I am trying to determine the time constant for the following circuit:



V is a DC voltage source. It is 0V at t<0. At t=0 it goes to voltage V. Cs is a sampling cap. I am interested in the time constant at Vout. I'm a bit unsure whether I can treat this is a DC problem since I'm interested in the charge time which is inherently a non-constant signal. But assuming I can treat this as DC, then I short the DC source and get this:



This is where I get stuck. The C3 branch and the R2-Cs branch have the same voltage across them so they are in parallel. But I don't think I can just combine C3 and Cs since R2 is on the second branch. And even if I could, I sill have an RC on the left and an RC on the right connected by C2. How do I reduce those elements?

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I am trying to determine the time constant for the following circuit:



V is a DC voltage source. It is 0V at t<0. At t=0 it goes to voltage V. Cs is a sampling cap. I am interested in the time constant at Vout. I'm a bit unsure whether I can treat this is a DC problem since I'm interested in the charge time which is inherently a non-constant signal. But assuming I can treat this as DC, then I short the DC source and get this:



This is where I get stuck. The C3 branch and the R2-Cs branch have the same voltage across them so they are in parallel. But I don't think I can just combine C3 and Cs since R2 is on the second branch. And even if I could, I sill have an RC on the left and an RC on the right connected by C2. How do I reduce those elements?

Unclear what you want to achieve. The circuit can't be described by a single time constant and can't be effectively reduced.

It's also not clear what you mean with "charge time" of the circuit.

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This is the complete transfer function as calculated by Sapwin


After cancelling C2s, it's still a second order low pass.

You have to solve this problem by using "state variables" technique that will give you " transient response" and "steady state response".
When you do this analysis, you will find a solution like Co*exp(-a*t/b*t) for each node/branch. Then you can find every time related answer about the circuit.

This is an equivalent circuit model of a capacitive sensor. A voltage pulse is initiated at Vin and the output voltage is measured at Vout. My goal is to determine how long of a pulse needs to be generated at Vin to allow Vout to reach some percentage of it's full potential voltage. Ideally what I would like is an equation that calculates the required pulse length at Vin as a function of the component values and the desired percentage of full charge at Vout (call that p):

Vin pulse length = f(R1, R2, C1, C2, C3, Cs, p)

I was hoping I could reduce it to a simple RC circuit so I could easily calculate the desired Vin pulse length, but from the feedback so far that's not possible. My background is in embedded software, not circuit design, so circuit analysis is not my area of expertise. I'm not sure if I can use the transfer function to somehow calculate the time I'm interested in. I suppose I could translate the transfer function back to the time domain and solve it as a differential equation?

Any suggestions would be appreciated.

The step response of a low pass with two real poles is a superposition of two exponential functions, it can be of course calculated for given component values.

I presume you know that the output DC level of this circuit is undefined without setting the initial condition, e.g. by a reset switch. A real circuit with finite load resistance will involve an additional high pass.

I plugged in some typical values for the caps and resistors into the transfer function and calculated the coefficients for the Laplace terms. I came up with this:



I then plugged that into an inverse Laplace engine at www.symbolab.com and got this:



In case that's not readable, the equation is 4406849 * e^{(-18933377 * t)}* sin(6322814 * t). If I'm using this correctly, then V_out(t) = V_in * 4406849 * e^{(-18933377 * t)}* sin(6322814 * t). But if I set t to 1 second which should be long enough for the circuit to reach steady state, then V_out(1) = 0 regardless of V_in. That can't be right.

So I'm back to the same question. How can I use the transfer function to generate an equation of the form:

V_in pulse length = f(R1, R2, C1, C2, C3, Cs, p)

where p is the desired percentage of the steady state voltage at Vout?

Did you wrongly calculate the impulse response instead of the step response? The step response of a low pass must have a nonzero asymptotic value, either if the poles are real or complex.

According to a my brief hand calculation, your transfer function has complex poles which isn't possible for a passive RC circuit. So there's probably something wrong in the transfer function derivation.

I'm not sure what you mean by "calculate the impulse instead of the step response". I thought there was only one transfer function. I calculated the transfer function by going through the circuit current loops and voltages from Vout back to Vin. I came up with this transfer function:



I checked this against the transfer function you got from Sapwin and they match. I then created an Excel spreadsheet to calculate the three terms given values for C1, C2, C3, Cs, R1, and Rx:



The cap values are 4.91 pF, 1.95 pF, 8.34 pF, and 18pF. The resistors are 6030 Ohms and 8775 Ohms. The calculated Laplace coefficients are 3.26E-14, 1.02E-6, and 14.51 (note: these values are just slightly different from the values I used in the original Laplace inverse because I did the first round of calculations with a calculator instead of Excel. The differences don't make a significant different in the inverse Laplace.)

I just double checked everything as I typed this response and it all looks right. So I must be missing something fundamental.

Tony614 said:

I plugged in some typical values for the caps and resistors into the transfer function and calculated the coefficients for the Laplace terms. I came up with this:

View attachment 142012

I then plugged that into an inverse Laplace engine at www.symbolab.com and got this:

View attachment 142014

In case that's not readable, the equation is 4406849 * e^{(-18933377 * t)}* sin(6322814 * t). If I'm using this correctly, then V_out(t) = V_in * 4406849 * e^{(-18933377 * t)}* sin(6322814 * t).

Click to expand...

Here is what you did wrong.
The transfer function is G(s)=V_out(s)/V_in(s)=what FvM shown.

When you take the inverse Laplace transform, you need to take the inverse laplace transform of V_out(s)=G(s)*V_in(s)

For Vin(t)=K => Vin(s)=K/s, so multiply this by G(s) and then take the inverse laplace transform of the whole.

I'm not sure what you mean by "calculate the impulse instead of the step response"

Click to expand...

The inverse laplace transform is the system response to a dirac pulse, but you are asking for the step response.

Starting from the laplace transfer function you can find the time-domain differential equation, that solved gives you the behaviour of the circuit over the time.
I've calculated Vo(t) = Vi*f(R1, R2, C1, C2, C3, Cs, t). where Vi is constant for t>=0 and 0 for t<0



Now, if I've understood well you want to know after how many time the ratio Vo/Vi will reach a given value.

I can see three ways:

1. Try to represent the exp with its series development, stopping at the second order that is exp(x)=1+x+X^2/2 and then solve the second degree equation. The problem is that you will have a probably quite high error
2. Inserting your component values have a look if one of the two exponential can be reasonably neglected
3. Implement an algorithm to find the unknown time from the function. for instance Newton-Raphson (you know the first derivative): https://en.wikipedia.org/wiki/Newton's_method.

albbg,

Thanks for the detailed derivation! I can't say I followed the entire derivation (it's been decades since I worked on Laplace transforms and differential equations), but I followed enough to see how you got to the solution for Vo/Vi. But the solution seems to have a fundamental issue that doesn't make sense to me. Time only appears in the e exponent multiplied by lambda. Given the typical pF capacitors and < 10k resistors in this circuit, the values for lambda1 and lambda2 are very, very small, on the order of 10^-44. That means that until t gets into the opposite range, i.e. 10^+44, the values of both exponential terms will be 1, which means that the circuit output will be constant. In other words, the circuit won't reach the steady state value of (Vin * C2/K0) until 10^44 seconds have passed. That can't be right.

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albbg,

Ignore my last post! I was working through this in Excel and made a mistake in the lambda formula. Once I corrected that, the result came out as expected! Here's a graph of Vout vs. time for typical values used in this circuit:



So the circuit reaches steady state in about 0.15 us. That's in the range of the expected value.

Thanks for your help!!!!

So the circuit reaches steady state in about 0.15 us.

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Depends on what you consider steady state. It reaches 0.999 of final value in about 0.5 us. Everything under the ideal assumption of infinite output load resistance.

Yes, the calculation is made under the condition of infinite load resistance. However is not difficult to include it in the calculation. After the laplace transfer function is extracted (you can use sapwin) then you can follow the same derivation I did, passing through the differential equation.
When a load resistance is applied, the voltage for t--> infinite will be 0 regardless the value of the load.
However I expect the rise of the signal will not change very much if Cs*Rload >> unloaded rise time. That is, with your value roughly Rload >> 10k that means Rload 100k or greater. However this is my feeling.

I think you will condition the signal by means of an high impedance op-amp, so there shoudn't be problems (just be sure the op-amp is fast enough and take into account noise and voltage offset introduced by the op-amp itself)