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Voltage Divider

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EngIntoHW

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Hi,

I'm trying the prove the equation for Vvictim, but I don't manage to work out this voltage divider.
I firstly tried to calculate Vx voltage, but yet to manage to get to the correct answer.
I'd appreciate your guiding on this.
Thank you.

View attachment 60463
 
Hi,

At first this looks like a 3rd order system, but it's more likely a 2nd order system because at least two capacitors are connected together at one lead each.

As such, that means the time response to a step input could be either a damped exponential or a damped sinusoidal. It's all caps though so it's probably a damped exponential.

But to clear one thing up... when you say you want to find delta Vx voltage or delta Vvictim voltage do you mean the max peak that will ever be seen with a step input?
 
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I'd appreciate your guiding on this.

I am not sure that I am the best guy to advise, but let me just waffle for a minute.

We are trying to calculate the voltage impressed on the blue line by the voltage on the red line.
If the voltage on the red line was a sine wave, and, if Rvictim was "very high", then the simple capacitive divider formed by Cadj and Cgnd-v could be used as in the top expression in the attachment.
However, if Rvictim is sufficiently small to be of a similar order to the reactance of Cgnd-v, the the simple capacitive divider model will not work.

But, the redline voltage is not a sine wave but a step, so we are trying to allow for this by looking at the rise-time of the step on the red line and the time constant of the blue line, and applying a correction factor 1/(1+k) to the simple voltage divider.

This is where my knowledge breaks down, I have never seen (or at least can never remember it) this correction for risetime applied like this.
At a first look it seems to make sense.

An agressor waveform with a rapid rise time (Taggressor = small) will have a less effect on a victim with a slow rise time (Tvictim = large) hence K is small, than it will on a victim with a fast rise time (Tvictim = small) hence k is large.

At this point it is lunch time and my poor little brain is feeling confused.:confused:
Does this diatribe help you at all?

JimB
 
Hi,

At first this looks like a 3rd order system, but it's more likely a 2nd order system because at least two capacitors are connected together at one lead each.

As such, that means the time response to a step input could be either a damped exponential or a damped sinusoidal. It's all caps though so it's probably a damped exponential.

But to clear one thing up... when you say you want to find delta Vx voltage or delta Vvictim voltage do you mean the max peak that will ever be seen with a step input?

Hi,

Yes, that is what I mean, and this is what the lecturer meant when he reached that equation of ΔVvictim
 
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I am not sure that I am the best guy to advise, but let me just waffle for a minute.

We are trying to calculate the voltage impressed on the blue line by the voltage on the red line.
If the voltage on the red line was a sine wave, and, if Rvictim was "very high", then the simple capacitive divider formed by Cadj and Cgnd-v could be used as in the top expression in the attachment.
However, if Rvictim is sufficiently small to be of a similar order to the reactance of Cgnd-v, the the simple capacitive divider model will not work.

But, the redline voltage is not a sine wave but a step, so we are trying to allow for this by looking at the rise-time of the step on the red line and the time constant of the blue line, and applying a correction factor 1/(1+k) to the simple voltage divider.

This is where my knowledge breaks down, I have never seen (or at least can never remember it) this correction for risetime applied like this.
At a first look it seems to make sense.

An agressor waveform with a rapid rise time (Taggressor = small) will have a less effect on a victim with a slow rise time (Tvictim = large) hence K is small, than it will on a victim with a fast rise time (Tvictim = small) hence k is large.

At this point it is lunch time and my poor little brain is feeling confused.:confused:
Does this diatribe help you at all?

JimB

Hi,
Thank you.
I understand your logic.
I'm still struggling with mathematically solving that problem.
I'll put more effort into it.
 
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Hi,

Yes, that is what I mean, and this is what the lecturer meant when he reached that equation of ΔVvictim


Hi again,

Ok, i thought you meant that but wanted to make sure.

Another question however...are you absolutely sure those two equations are correct in every detail? I ask because the preliminary results i got do not match up with those equations in that the peak Vvictim comes out to a different value than that as predicted by those two equations.
The way i went about to prove it was a most direct way which by its nature is an exact method:
I stated with an equation for the network that relates Vvictim to delta Vaggressor with delta Vaggressor being a simple unit step.
I then transform to the time domain so i get an equation for Vvictim(t).
I then solve for the time t where Vvictim(t) is a maximum (peak).
I then use that value of t in the equation for Vvictim(t).
The result is a voltage that does not agree with the two equations given.

Is it possible he meant something else here too, or some added detail? Like could the two equations be meant to be used as approximations rather than exact solutions (or something else like that)?
 
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Hi Mr Al,

Thank you very much for explaining it to me.
Yes, it could be that he made a certain approximation here, to make it more neat.
Unfortunately, he didn't give more details, but he hinted that it might show up in the exam.
 
Hi again Eng,


I took a closer look at this and here is what i found. It looks like you might have to ask the instructor if his assumptions follow the same as those mentioned at the end of this post. It looks like he is assuming that the value of Cadj is much greater than the values of the two caps going to ground. That would be ok for certain applications so you might want to ask about that.

For brevity, renaming the following variables:
Ra=Raggressor
Rv=Rvictim
Cga=Cgnd-a
Cgv=Cgnd-v
Cadj=Cadj (keeps the same name)

we end up with this exact equation (normalized for a unit step input):

Vvictim_peak=(-a*W/c)*(A^(W+1)-A^(W-1))

where:
a=Cadj*Rv
b=Cga*Cgv*Ra*Rv+Cadj*Cgv*Ra*Rv+Cadj*Cga*Ra*Rv
c=Cgv*Rv+Cadj*Rv+Cga*Ra+Cadj*Ra
w=(sqrt(c^2-4*b))/(2*b)
W=-c/(2*b*w)
A=sqrt((-2*b*w-c)/(2*b*w-c))
(and Vvictim is to be multiplied by the actual step input height)

With both resistors=100 ohms in the two formulas we get:
Cga and Cgv=2uf, Cadj=2uf, peak=0.19245v, dV=0.25v
Cga and Cgv=2uf, Cadj=20uf, peak=0.409v, dV=0.455v
Cga and Cgv=20uf, Cadj=2uf, peak=0.0335v, dV=0.0455v
Cga and Cgv=2uf, Cadj=200uf, peak=0.484v, dV=0.495

where dV is from the original set of formulas given in the link:
k=Ra*(Cga+Cadj)/(Rv*(Cgv+Cadj))
dV=Cadj/(Cgv+Cadj)*1/(1+k)

and since 'peak' is the exact we see we have a difference for those values shown.

To simplify a little just to see what we basically have here, if
we make Cgv=Cga and Rv=Ra in the original formula we get:
dV=Cadj/(2*(Cga+Cadj))

but if we do the same with the exact formula we get:
Vvictim_peak=(Cadj/Cga*((2*Cadj)/Cga+1)^(-Cga/(2*Cadj)-1))

so we can see a big difference in the two formulas. One involves a simple division while the other involves a power and is much more complicated.

Most notably, if we make both resistors the same and all three caps the same we get for the original formula:
dV=1/4 which is a constant equal to 0.25,

and the exact:
Vvictim_peak=1/3^(3/2) which is also a constant and approximately 0.19245

so we immediately see a difference.

However, if we make the resistors the same again and Cga and Cgv the same and Cadj equal to 1000 times Cga in the original formula we get:
dV=500/1001 which is approximately 0.49950

and in the exact formula we get:
Vvictim_peak=1000/2001^(2001/2000) which is approximately 0.49785

so now we only see about a 0.3 percent difference.

Thus the approximation assumption is probably that Cadj is much greater than both Cga or Cgv.

You can verify the 'exact' formula above with a circuit simulator but beware, be sure to set
the pulse rise time manually to override the default setting for pulses.

We dont really need the time to peak, but just in case you want to look at this more yourself here is the time to the peak:
t=ln(-(2*b*w)/(2*b*w-c)-c/(2*b*w-c))/(2*w)

Note that i have not investigated the limits of these formulas so there may be some combination of values that dont work as well as expected. You can check that out yourself if you care to. They do however seem to work for a wide range of values.
 
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Hi again,


I am happy you find this stuff interesting. I always did too. I hope you get something out of it and it helps.
 
Hi Mr Al,
Thank you again :)

I'd like to ask you something about it please.
Can you tell me please how did you reach the exact solution of such a complex circuit?
 
Hi again Eng,


There are a number of ways to go about this. Since there is no initial energy storage,
i went with an impedance transformation method to get to the frequency domain equation. This
involves transforming the impedances of the caps into the frequency domain, then combining with
the resistors and forming a voltage divider with Vvictim the output and delta Vagressor the input.

The frequency domain equation is then generalized to a second order system with lumped constants with step input.
This is then transformed to the time domain using an Inverse Laplace Transform method. This
gives rise to three general solutions, but it was believed that only the solution with real
poles would be the choice of interest because the network only contains resistors and capacitors,
and a network like that probably can not produce a sinusoidal output with only a step input.
This however looked a little difficult to prove algebraically, so i used a Monte Carlo method
to try to prove that the poles would always be real. Testing about 100 million combinations
proved that the solution could never be sinusoidal with a reasonable range of values of resistors
and capacitors, so there were only two possible solutions and one was dropped because it would
take an extremely rare combination of values to obtain, so only one true solution for the time domain
representation of the output Vvictim remained.

Having the time domain solution to Vvictim now allowed solving that using calculus for a value of
time t that would cause a max in the response of Vvictum. Two solutions emerged, but only one was
valid.

Now having that equation for t allowed solving for the actual equation for max Vvictum in the time
domain, which was the required solution sought after. This solution was comprised of only the
constants from the circuit elements in the original circuit, but combining them in certain ways
allowed for a more presentable formula which was posted earlier. I did not investigate
further simplifications to the final equation posted so there could be reductions possible.

Also, the solution posted is to be multiplied by the actual step input height (it's normalized
for a unit step) and i made a note in the original post to reflect this.
I also gave the time to peak equation too just in case anyone was interested in looking at
the values of t that result using different values for the resistors and capacitors.


So the steps are:
1. Tranform caps to the frequency domain impedances using 1/(s*Cn) for each cap.
2. Combine impedances to form the output frequency equation, lump constants.
3. Transform the frequency equation to the time equation, prove only one solution valid if possible.
4. Solve the time equation(s) for a time value equation that results in a max response of the output.
5. Use that solution of t to form an equation for the max (peak) output, the required solution.
6. Clean up the form of the peak output equation if possible.


If you have done any analysis in the past we could look at this more closely if you care to do so.
 
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Hi Mr Al,

I thank you so much!
It's been so helpful :)
I really appreciate your help.

Sorry for the late comment, it took me a while to digest :)
 
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