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Super Accurate Resistor Fun
- Thread starter dcbingaman
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Another interesting aspect of this circuit approach is even distribution of current to a set of LEDs:
With my power supply connected to the network:
So I am powering a large number of LEDs at 15V and around .2 Amps, that is 3 Watts of power.
With my power supply connected to the network:
So I am powering a large number of LEDs at 15V and around .2 Amps, that is 3 Watts of power.
I understand what you did but most calculators do not have the accuracy that a powerful spread sheet program like Excel has. The double precision numbers have enough accuracy to verify the convergence. Also did you let the number 'go' and not round them? Rounding will also cause a problem with convergence, you cannot round the numbers, it invalidates any formula trying to achieve convergence. I might if I get some time wright a small C program to verify that Excel is accurate. And send you the program. Also did you perform actual board testing with a real circuit? I have and in confirms by calculations.
That is I simply applied the Scientific Method. After the concept I had to prove or disprove it. That required an experiment and the experiment confirms the calculations.
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The power is NOT evenly distributed.
Let's take your five layers of 1 kΩ resistors. You say that you put well over 3 W into it. So let's use 3 W for discussion.
The total resistance is about 438 Ω. But you claim it somehow converges on 500 Ω. Fine. Let's use that. How much voltage across 500 Ω does it take to get 3 W into it? Would agree that it takes about 39 V? Is that what you are applying?
How much current is flowing in that top layer consisting of a single 1 kΩ? Easy, 39 V / 1 kΩ = 39 mA. How much power does that translate to? Again, easy. 1.521 W. So more than half of the 3 W is being dumped into that single resistor.
How about each of the resistors in the fifth layer, consisting of five 1 kΩ resistors in series? Still not difficult. The current in this string is 39 V / 5 kΩ = 7.8 mA, resulting in a total power dissipating in the entire string of 0.304 W and just 61 mW in each resistor. This is just 4% of the power dissipated in the top resistor.
How can you say that the power is evenly distributed?
What measurements back up that claim? What are the voltages at each node in the network? Are they consistent with this assertion?
No, it is not accuracy. You do not lose 7 digits with one iteration of the formula:
R = 1/(1/R1 + 1/R2.)
We have exactly the same answer for the first calculation, 2/3, but you are off in the second significant digit on the next one.
1 / (1/(2/3)+ 1/3) = 1 / (3/2 + 1//3)=
1 / (9/6 +2/6) = 1 / (11/6) = 6/11 = 0.54645454
Which is what Ian and I both got.
Show us Excel formula so we can diagnose what you are doing wrong. If you would just measure the damn resistances you would see that you are wrong.
Not only that, but your calculation also contradicts what you said before, that the total resistance is 1 Ohm.
R = 1/(1/R1 + 1/R2.)
We have exactly the same answer for the first calculation, 2/3, but you are off in the second significant digit on the next one.
1 / (1/(2/3)+ 1/3) = 1 / (3/2 + 1//3)=
1 / (9/6 +2/6) = 1 / (11/6) = 6/11 = 0.54645454
Which is what Ian and I both got.
Show us Excel formula so we can diagnose what you are doing wrong. If you would just measure the damn resistances you would see that you are wrong.
Not only that, but your calculation also contradicts what you said before, that the total resistance is 1 Ohm.
You need to show the equations you used in each cell, because they are wrong.
If you show your formulas, we can help you see the errors.
What is your definition of a "node"? If it's the number of layers in your tree, how can 2 layers have a resistance 1 Ω when that is clearly what you get with just a single layer.
Please show the exact schematic -- not some picture of a breadboard that we can't tell what half of the connections are -- of your entire circuit.
The current Windows calculator uses far more sig figs than Excel does. But even single precision floating point math is far more than sufficient for the task at hand.
But since you seem to be in awe of Excel, let's see what Excel says.
The first column is the number of resistors in that layer.
The second column is the total resistance of the resistors in that layer.
The third column is the conductance of that layer (which is the reciprocal of the resistance of that layer).
The fourth column is the sum of the conductances of that layer and all of the layers above it.
The fifth column is the corresponding resistance of that layer in combination with all of the layers above it.
As you can see, the results match what everyone, except you, is getting.
And your measurement disproved your claim right from the very start. You stated that the resistance of the mesh was 1 Ω regardless of how large the mesh was. Your measurement shows that this was not even remotely the case. So how does that experiment confirm your calculations?
You are correct but it did prove one thing, the series does converge on 1/2 of the stated value of 1 ohm.
My calculation approaches 1/2 of an ohm, yes that is my point it does converge.
Here is the spreadsheet. Let me know if you see something wrong. It shows convergence to 1/2 of an ohm like I stated.
I want to thank everyone for their contribution to this thread, it has been a great learning experience for me and my hope is that everyone else has found it interesting as well. Thanks.
Getting upset about it is not gentleman like. Please try to remain civil. The word damn is not a good word and it can hurt peoples feelings. What are you thinking?
No, it does not. You have an error in your spreadsheet.
Just do the math by hand for four strings.
The resistances of the first four strings are: 1 Ω, 2 Ω, 3 Ω, 4 Ω
The equivalent resistance is:
1/Req = (1 / 1 Ω) + (1 / 2 Ω) + (1 / 3 Ω) + (1 /4 Ω)
But these over a common denominator of 12 Ω
1/Req = (12 / 12 Ω) + (6 / 12 Ω) + (4 / 12 Ω) + (3 / 12 Ω) = 25 / 12 Ω
That makes
Req = (12/25) Ω = 0.48 Ω
There is NO round off error in this calculation AT ALL. The equivalent resistance of just the first four layers is already LESS than 0.5 Ω. There is NO mechanism by which adding additional strings in parallel with the first four is somehow going to raise the total resistance.
Your speadsheet doesn't show anything of the sort.
You have hardcoded the first two rows in the resistance column to be 1 and 0.666666666666666.
What is the basis for doing that?
Then you are somehow claiming that the total resistance of a network with N+1 layers is somehow equal to one-third of the sum of 1 Ω and the resistance of the N layer network.
What is the basis for that claim?
It makes zero sense from a few perspectives. First, it has no awareness of how many resistors are in the row that is being added to the prior network. Second, where does the magic 1/3 come from?
What do you mean by "converging"? There are
2 values beginning in 0.30
5 values beginning in 0.25
17 values beginning in 0.20
55 values beginning in 0.17
That looks pretty asymptotic -ish.
ak
The series that is actually not converging is the series
G = 1/1 + 1/2 + 1/3 + 1/4 + ...... 1/N + 1/....
The sum of this series goes to infinity, but we are not interested in G, but rather R = 1/G, which therefore does converge, but to 1/Inf = 0.
Another, more practical, way of thinking about this case is that the sense in it which it doesn't converge is that it doesn't converge to a non-zero value that is dependent on the size of the resistors. Instead, whatever resistor size you choose to use, the limit of the total resistance will always be the same, namely zero.
Unfortunately not.
It was displayed to 5 decimal places.
Calculated to the best that Open Office Calc can manage, whatever that is.
That's the sequence you get with 1/(1+1/2+1/4+1/8+1/16+1/32 etc.)
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