Equivalent Resistance for Repetitive Buss / Load Network

I perform VD calcs all the time. I use an excel spreadsheet. Works like a charm.

Is not so tricky i think.Let's say the "entrance" impedance of some "n+1" sector of such transmission line is Zin,n+1=A+(1/((1/B)+(1/Zin,n)),so at the lim, assumming that

lim Zin,n+1=lim Zin,n Then at the limit that equation is Z=A+(1/((1/B)+(1/Z)) Which is a second degree polinomial equation.I think if this is not exactly what you were looking for you will adapt the reasoning.-

Make R1 = 2Rb + Rl

R2 will be then R1//R2 meaning R2 = R1/2

R3 will be then R2//R1 meaning R3 = R1/3

R4 will be then R3//r1 meaning R4 = R1/4

RN will be then RN-1//r1 meaning RN = R1/N

Best Regards

Equivalent Resistance for Repetitive Buss / Load Network

Make R1 = 2Rb + Rl

R2 will be then R1//R2 meaning R2 = R1/2

R3 will be then R2//R1 meaning R3 = R1/3

R4 will be then R3//r1 meaning R4 = R1/4

RN will be then RN-1//r1 meaning RN = R1/N

Best Regards

There is a theoretical method for this - but you may find it to be more trouble than it is worth.

I'll outline the method for both uniform and lumped lines.

But first I need suitable definitions:
For the uniform line:
. Series resistance per unit length = R
. Parallel conductance per unit length = G
For the lumped line, we need to work with a basically symmetrical element structure. In the hope of minimising communication issues, I'll rename the resistances so that the limiting case is the same as the uniform line - each stage is a TEE with:
. Input resistor = R/2
. Conductor to ground = G, and
. Output resistor R/2.
Note that you can easily correct for a different-value first-stage resistor once you have the solution for this specific special case

I'll deal with the basics of the continuous case first:
The resistance Z of an infinite (one-sided) length of line can be found by equating the resistance of an infinite length to the length with a miniscule incremental length added, which gives:
Z = √(R/G)
The next step is to look at the matched attenuation (per unit length):
Vout=exp-(2.x.√(R.G)), where x is the length of the line.

For the lumpped case, the basics are:
Z = √(R²/4 + R/G), and
attenuation per section is given by
Vout = R/2.(R/2+Z)/G.Z/(R+Z+1/G)/Z+R/2)
. (There could be mistakes, and I expect this can be much simplified - but it's the principle I'm trying to show)

So we now have an impedance and an attenuation. There are many ways forward from here. One simple way forward is to substitute a minimal lumped attenuator (say a TEE-type arrangement) with the same impedance and attenuation. There are many calculators for this on the web, but you can find derive the equations directly or find them here for example.

Good luck

Fyz

Simplified equations for single TEE attenuators (using same notation as #5)
Calculating the stage attenuation:
Continuous:
Z = √(R/G) (unchanged)
Vx_out = exp(-x.√(R.G)) (corrected)
Calculating the equivalent discrete attenuator:
Requc = 2.Z.(1-Vx_out)/(1+Vx_out)
Gequc = Requc/(Z²-(Requc/2)²)
Input resistance = 1/Gequc = (Z²-(Requc/2)²)/Requc

Discrete:
Z = √(R/G+R²/4) (unchanged)
V1out(matched) = [(1-(R/2)/Z)/(1+(R/2)/Z)] (single stage) (corrected)
VNout(matched) = [(1-(R/2)/Z)/(1+(R/2)/Z)]^N ' (N stages)
Calculating the equivalent attenuator:
Requd = 2.Z.(1-VNout)/(1+VNout)
Gequd = Requd/(Z²-(Requd/2)²)
Input resistance of actual cable with N stages:
= (Z²-(Requd/2)²)/Requd + (resistance of actual input series element)

I'm reasonably confident that these are ready for assembly into a single equation; nevertheless, I would strongly suggest you make some sample checks against your spread-sheet results before assuming they are correct.

Fyz

There's a term missing above. I'll come back with a correction and using your values when I have time.

Fyz

Let´s assume there is a value for an infinite line,lets say "Z",so then it won't change if we add a some of the same (per unit or per meter after the case), lets say Rp in paralel and Rs in serie so Z=Rs+1/((1/Rp)+(1/Z)),may be that is written in some special notaion like this:Z=Rs+(Rp//Z).But why so long?.-

Before checking my new algebra, I was wondering:

It's all very well modelling the total load, but non-uniform heating becomes significant rather earlier than changes to the total load. Presumably you need to take this into account as well? (It's a relatively small problem to include while setting up the model)

The aspect that would worry me most is the concentrated heating in the elements Rb that are closest to the supply. By the time that the series resistance has reduced the total current by 10% (relative to the zero-series-resistance case), the power in the first pair of series elements will be about 26% of the nominal power in the parallel element. The level of risk here depends on the physical layout of the heater mat; if the length of the series elements is small compared with the dimension of an individual load we could be seeing substantial temperature rises.

However, even if the layout of the heater mat is such that this concentration is not dangerous, both the uniformity and the efficacy of the system would already be significantly compromised - the total dissipation-per-unit length at the end nearest the supply would be about 126% of its nominal value, and the dissipation at the end furthest from the supply would be reduced to about 72% of nominal (and remember that this is for a reduction in total dissipation of only about 10%

Now, what do you really need to calculate?

Fyz

Closed-form expressions for "Repetitive Buss / Load Network"

The schematics below show the conversion from your original schematic to a schematic where the modified stages correspond to TEE attenuators that would match between identical impedances on each side. The regular relationship between each stage and the next makes the circuit much easier to analyse in this form.

First, we consider the TEE attenuators without the input resistor:

If we consider a single attenuator stage, it has
Impedance (Z0) = 2.Rb.Rl + Rb²
Matched Voltage attenuation (A) = (1-Rb/Z0)/(1+Rb/Z0)

N stages of attenuation corresponds exactly to a single stage attenuator with the same input impedance and Voltage attenuation (Z0 and A^N respectively)

A single attenuator with this attenuation and impedance Z0 would have:
2 x bar resistances (RS_N) = Z0.(1-A^N)/(1+A^N)
Stem resistance (RP_N) =( Z0²-RS_N²)/2/RS_N

Clearly, the input resistance of the entire network (the original question) will be:
RinN = Rb + RS_N + RP_N

Other significant calculations are:

Single-stage (unloaded) power was Vin²/(Rb+Rl)
Power in the busses of the section closest to the power source will be 2.Rb.Vin²/RinN²
For large numbers of stages, this power can easily exceed the power in a single intended heater Rl. Even when it does not, special consideration may be required because this power can be dissipated over quite a small area (depending on the specific design of "parallel type heater cable".

We also need to note that, once the input resistance is even 10% higher than
N.(2.Rb+Rl), the heating available from the section furthest from the supply will be substantially (eventually very much) lower than that from the first stage. Its value will be:
Power(N) = (2.Rb+Rl).{Vin/RinN.RP_N-1/(RP_N-1+RS_N-1+Rb+Rl)}²

Fyz

Dear Guest; Thank you for all your input - quite impressive! I'm not real familiar with the "attenuation" approach that you took - but it sure seems that you know what you are doing. I apologize, but one of my first questions is - is the "Unbalanced equivalent" circuit that you show truly equivalent to my original circuit? Somehow, that is not totally obvious to me, but I expect if I think about it some more, it will sink in.

Secondly, I understand how you got from you "Unbalanced circuit" to the Tee-attenuator circuit - but then I get lost - is there a well-known approach to solving these types of circuits?

Best regards, Andy

First - a correction. Bits got lost when pasting from my original. The equation for Z0 should read:
Impedance (Z0)
Z0² = 2.Rb.Rl + Rb²
or
Z0 = √(2.Rb.Rl + Rb²)
So I would still recommend checking some examples against your present calculations...

Q1: The unbalanced arrangement is exactly equivalent so far as power in each resistor element (1xRb or Rl) is concerned - although (obviously) not as regards the Voltages on the nodes.
Q2: Transmission lines and matched attenuators are, as you suspect, standard stuff. [When I was younger (so much younger than today...) this stuff used to be part of the first-year "A-level" course (UK equivalent to US 11th grade) - alas no more*.]
You can find resources (including design equations for attenuators on the web) - here and here for example. Worth reading, and using to check my scribblings, but they define attenuation in decibels (dB), which is not ideal for this application, and the equations need a little manipulation (which I have done) if you want to analyse an attenuator rather than design one.

Fyz

*I'm told that syllabuses don't even require numerical handling of friction - which was previously an "O-level" topic