The following circuit has some interesting properties:


The LT5400 is just there to get it to simulate. (Need an LT part). It looks like a Christmas tree.
The circuit simulation is attached.

- This creates a 'super accurate' resistor!
- In this case I wanted 1 ohm. But as long as all the resistors are scaled the same, you can make the actual resistor any value you please (In other words it is normalized).
- The test port gives 1V per ohm output.

- It could also be used to make a high power 'resistor' like with 1 ohm parts but the load is distributed accordingly.

Has anybody seen something like this before? I just came up with it while messing around.

 

In-fact you can make the tree as high as desired, and the higher it is the more insanely accurate it is:


There appears to be no limit to this process.

 

In-fact you can make the tree as high as desired, and the higher it is the more insanely accurate it is:

View attachment 306068
There appears to be no limit to this process.

That's a 385mΩ resistor?
If you put ten 10Ω 1% resistors in parallel, then you get 1Ω ±0.3%, assuming that their values are random in a normal distribution.

 

That's a 385mΩ resistor?
If you put ten 10Ω 1% resistors in parallel, then you get 1Ω ±0.3%, assuming that their values are random in a normal distribution.

You are correct in that 10 resistors of 10 ohms in parallel is 1 ohm.
The networks resistance is 1 ohm. And is one ohm no matter how large the mesh is.
But its accuracy is insanely better than simply 10 ohms in parallel the simulation shows "absolute accuracy". i.e. +/- 0% So I ran some experiments on the bench using standard old 100 ohm resistors for the network and found that the accuracy goes up with the larger the net system. I have a system with around 24 of these resistors in the mesh network and the accuracy does go up at this level I got even distribution of heat in parts and was able to created an insanely accurate heater with 2-3 Watts of steady output.
It also created like a heated bed heater because the values are spread out on the board. Here is a picture of an early attempt with a network of 14 resistors:

 

The following circuit has some interesting properties:

View attachment 306066
The LT5400 is just there to get it to simulate. (Need an LT part). It looks like a Christmas tree.
The circuit simulation is attached.

- This creates a 'super accurate' resistor!
- In this case I wanted 1 ohm. But as long as all the resistors are scaled the same, you can make the actual resistor any value you please (In other words it is normalized).
- The test port gives 1V per ohm output.

- It could also be used to make a high power 'resistor' like with 1 ohm parts but the load is distributed accordingly.

Has anybody seen something like this before? I just came up with it while messing around.

Of course your test port gives 1 V/Ω. You have a 1 A current source delivering 1 A of current into R17, which is a perfect 1 Ω resistor. What else are you going to get?

Your Christmas tree of resistors has ZERO impact on what you see at the test port. The ideal current source isolates the circuitry on the two sides.

Don't believe me?

Replace your Christmas tree with a single resistor and make it 8285.235 Ω and see what you get at the test port.

Just a glance at your Christmas tree shows that it's resistance MUST be significantly less than 1 Ω. You have a 1 Ω resistor in parallel with a 2 Ω resistor. Even ignoring everything else, this results in a resistor of 2/3 Ω. The additional branches just drive the resistance lower. With your four branches, the total resistance is 0.48 Ω.

 

In-fact you can make the tree as high as desired, and the higher it is the more insanely accurate it is:

View attachment 306068
There appears to be no limit to this process.

How is this insanely accurate?

First, you aren't doing ANYTHING that even begins to assess the accuracy of anything. Your resistors in your simulation are being treated as if they are perfectly and exactly 1 Ω.

Assuming they are perfectly accurate to begin with, the nominal resistance of this seven-layer cake is

Rtot = 1/(1/1Ω + 1/2Ω + 1/3Ω + 1/4Ω + 1/5Ω + 1/6Ω + 1/7Ω) = 0.38567493112947658402203856749311... Ω

If you want to take a bunch of resistors with a certain tolerance and combine lots of them to get a resistor of the same value but with a tighter tolerance, put them in a square grid consisting of N parallel strings, each with N series resistors in it.

IF the resistors are symmetrically distributed about their mean AND that mean that matches the nominal value, THEN the ensemble will have that same mean but a much smaller tolerance. Note that those are two significant assumptions.

 

You are correct in that 10 resistors of 10 ohms in parallel is 1 ohm.
The networks resistance is 1 ohm. And is one ohm no matter how large the mesh is.
But its accuracy is insanely better than simply 10 ohms in parallel the simulation shows "absolute accuracy". i.e. +/- 0% So I ran some experiments on the bench using standard old 100 ohm resistors for the network and found that the accuracy goes up with the larger the net system. I have a system with around 24 of these resistors in the mesh network and the accuracy does go up at this level I got even distribution of heat in parts and was able to created an insanely accurate heater with 2-3 Watts of steady output.
It also created like a heated bed heater because the values are spread out on the board. Here is a picture of an early attempt with a network of 14 resistors:

View attachment 306145

Of course the simulation shows absolute accuracy -- your simulation is using a resistor model that is perfectly accurate!

In the real world, the improvement in accuracy goes up as about the square-root of the number of resistors. So you get a significant improvement with just four resistors, but to see that same degree of improvement again you need sixteen and to see that same improvement yet again you need 256. Rapidly reaches a point of diminishing returns.

 

The accuracy improvement to the nominal resistor value is only if the resistance tolerance follows a standard deviation bell curve.
If you have batch of resistors from the same lot that have their resistance tolerance skewed from their listed nominal value, then the average of them in the network will also be skewed, no matter how many you use.

 

The accuracy improvement to the nominal resistor value is only if the resistance tolerance follows a standard deviation bell curve.
If you have batch of resistors from the same lot that have their resistance tolerance skewed from their listed nominal value, then the average of them in the network will also be skewed, no matter how many you use.

The distribution doesn't have to follow a standard deviation bell curve (i.e., a normal distribution). The mean of the distribution will determine the mean of the resistor networks and there's not much getting around that without screening the resistors individually (which largely defeats the purpose), but while a symmetric distribution is highly advantageous, it turns out this is not really even necessary. Even with a highly asymmetric distribution, the distribution of the networks will quickly become normally distributed with a mean equal to the mean of the original distribution and a standard deviation that goes down as the square root of the number of resistors in the network.

 

Of course the simulation shows absolute accuracy -- your simulation is using a resistor model that is perfectly accurate!

In the real world, the improvement in accuracy goes up as about the square-root of the number of resistors. So you get a significant improvement with just four resistors, but to see that same degree of improvement again you need sixteen and to see that same improvement yet again you need 256. Rapidly reaches a point of diminishing returns.

You are absolutely correct. That is why I ran actual experiments on the bench.

 

Of course your test port gives 1 V/Ω. You have a 1 A current source delivering 1 A of current into R17, which is a perfect 1 Ω resistor. What else are you going to get?

Your Christmas tree of resistors has ZERO impact on what you see at the test port. The ideal current source isolates the circuitry on the two sides.

Don't believe me?

Replace your Christmas tree with a single resistor and make it 8285.235 Ω and see what you get at the test port.

Just a glance at your Christmas tree shows that it's resistance MUST be significantly less than 1 Ω. You have a 1 Ω resistor in parallel with a 2 Ω resistor. Even ignoring everything else, this results in a resistor of 2/3 Ω. The additional branches just drive the resistance lower. With your four branches, the total resistance is 0.48 Ω.

You are missing the point, the thread shows that I not only simulated this but actually built it on a bread-board using standard old axial resistors with 100 ohm resistance and the results are still very accurate even in that situation.

Attached are my experimental results thus far:

 

Interesting to note that the sum to infinity of 1/n does not converge, unlike the sum to infinity of 1/(n^x) where x>1.
So adding more and more series chains in parallel will NOT converge on any particular value.

 

You start with a 1Ω resistor. You cannot put any resistance in parallel with it still have 1Ω. Can you really not see that?

Try your breadboard circuit again. Start with 1 100Ω resistor. Measure and record the resistance. Now put two more resistors in series and put them in parallel with the other one. Measure the resistance and record it. Then three…

You will see that it is not getting closer to 100Ω.

 

You start with a 1Ω resistor. You cannot put any resistance in parallel with it still have 1Ω. Can you really not see that?

Try your breadboard circuit again. Start with 1 100Ω resistor. Measure and record the resistance. Now put two more resistors in series and put them in parallel with the other one. Measure the resistance and record it. Then three…

You will see that it is not getting closer to 100Ω.

For 1Ω resistors it tends towards
\(
\frac{1}{\ln(n)+\frac{1}{\sqrt(3)}}
\)
where n is the number of strings

 

You are missing the point, the thread shows that I not only simulated this but actually built it on a bread-board using standard old axial resistors with 100 ohm resistance and the results are still very accurate even in that situation.

Attached are my experimental results thus far:

Your attached schematic shows that you are still making the same mistake. You change the values of the resistors in your tree, but then you measure the value of a DIFFERENT resistor that is PERFECT and claim that this tells you something about the behavior of your tree. It tells you NOTHING about it, because the current source completely isolates the tree from the the measurement you are making.

Change all of the resistors in the tree to be perfect 1000 Ω resistors and you will still find that the resistance determined by your measurement is 1 Ω. Now, change all of them to perfect 1 Ω resistors and change the value of R17 to 10 Ω. Guess what? Now your measurements will show that the resistance is 10 Ω.

Your physical experiment would seem to directly contradict your claim that this creates a "super accurate" resistor. You claim that this tree topology you created out of a bunch of resistors of value R leads to a super accurate resistance of R. Yet you put five layers of 1 kΩ resistors and you measure something that is "around 500 Ω". How is that experimental evidence that your circuit results in a super accurate 1 kΩ resistor? If your resistors were perfectly 1 kΩ, the total resistance would be about 437.96 Ω.

And what's the nonsense about trying to explain away the observed behavior on some lame claim about leakage current in the resistors because they are old?

 

Your attached schematic shows that you are still making the same mistake. You change the values of the resistors in your tree, but then you measure the value of a DIFFERENT resistor that is PERFECT and claim that this tells you something about the behavior of your tree. It tells you NOTHING about it, because the current source completely isolates the tree from the the measurement you are making.

Change all of the resistors in the tree to be perfect 1000 Ω resistors and you will still find that the resistance determined by your measurement is 1 Ω. Now, change all of them to perfect 1 Ω resistors and change the value of R17 to 10 Ω. Guess what? Now your measurements will show that the resistance is 10 Ω.

Your physical experiment would seem to directly contradict your claim that this creates a "super accurate" resistor. You claim that this tree topology you created out of a bunch of resistors of value R leads to a super accurate resistance of R. Yet you put five layers of 1 kΩ resistors and you measure something that is "around 500 Ω". How is that experimental evidence that your circuit results in a super accurate 1 kΩ resistor? If your resistors were perfectly 1 kΩ, the total resistance would be about 437.96 Ω.

And what's the nonsense about trying to explain away the observed behavior on some lame claim about leakage current in the resistors because they are old?

Yes, I know all that. The experiments still have yielded some interesting results. The resistance goes up closer to the target the more bottom rows you add to it. Also the power distribution gets evenly distributed throughout the resistor network. I was able to pull well over 3W of power into the network. You seem to have something against experimentation, but I am not sure what it is? Is not science about experimentation?

 

Interesting to note that the sum to infinity of 1/n does not converge, unlike the sum to infinity of 1/(n^x) where x>1.
So adding more and more series chains in parallel will NOT converge on any particular value.

Actually it does converge. But you have to perform the experiment to know that. LTSpice is not going to show you that. I have and convergence does occur.

 

The accuracy improvement to the nominal resistor value is only if the resistance tolerance follows a standard deviation bell curve.
If you have batch of resistors from the same lot that have their resistance tolerance skewed from their listed nominal value, then the average of them in the network will also be skewed, no matter how many you use.

Absolutely correct! You seem to be the only one on the forum agreeing with me that such circuits can be useful.

 

strings​	Parallel resistance​	​
1​	1.00000​	​
2​	0.66667​	​
3​	0.54545​	​
4​	0.48000​	​
5​	0.43796​	not converging​
6​	0.40816​	​
7​	0.38567​	​
8​	0.36794​	​
9​	0.35349​	​
10​	0.34142​	still not converging​
11​	0.33114​	​
12​	0.32225​	​
13​	0.31445​	​
14​	0.30754​	​
15​	0.30137​	still not converging​
16​	0.29579​	​
17​	0.29074​	​
18​	0.28611​	​
19​	0.28187​	​
20​	0.27795​	still not converging​
21​	0.27432​	​
22​	0.27094​	​
23​	0.26779​	​
24​	0.26483​	​
25​	0.26206​	still not converging​
26​	0.25944​	​
27​	0.25697​	​
28​	0.25464​	​
29​	0.25242​	​
30​	0.25031​	still not converging​
31​	0.24831​	​
32​	0.24640​	​
33​	0.24457​	​
34​	0.24282​	​
35​	0.24115​	still not converging​
36​	0.23955​	​
37​	0.23801​	​
38​	0.23652​	​
39​	0.23510​	​
40​	0.23372​	still not converging​
41​	0.23240​	​
42​	0.23112​	​
43​	0.22989​	​
44​	0.22869​	​
45​	0.22753​	still not converging​
46​	0.22641​	​
47​	0.22533​	​
48​	0.22428​	​
49​	0.22325​	​
50​	0.22226​	still not converging​
51​	0.22130​	​
52​	0.22036​	​
53​	0.21945​	​
54​	0.21856​	​
55​	0.21769​	still not converging​
56​	0.21685​	​
57​	0.21603​	​
58​	0.21523​	​
59​	0.21444​	​
60​	0.21368​	still not converging​
61​	0.21294​	​
62​	0.21221​	​
63​	0.21149​	​
64​	0.21080​	​
65​	0.21012​	still not converging​
66​	0.20945​	​
67​	0.20880​	​
68​	0.20816​	​
69​	0.20753​	​
70​	0.20692​	still not converging​
71​	0.20632​	​
72​	0.20573​	​
73​	0.20515​	​
74​	0.20458​	​
75​	0.20403​	still not converging​
76​	0.20348​	​
77​	0.20294​	​
78​	0.20242​	​
79​	0.20190​	​
80​	0.20139​	still not converging​
81​	0.20089​	​
82​	0.20040​	​
83​	0.19992​	​
84​	0.19944​	​
85​	0.19898​	still not converging​
86​	0.19852​	​
87​	0.19806​	​
88​	0.19762​	​
89​	0.19718​	​
90​	0.19675​	still not converging​
91​	0.19633​	​
92​	0.19591​	​
93​	0.19550​	​
94​	0.19509​	​
95​	0.19469​	still not converging​
96​	0.19430​	​
97​	0.19391​	​
98​	0.19353​	​
99​	0.19315​	​
100​	0.19278​	still not converging​
101​	0.19241​	​
102​	0.19205​	​
103​	0.19169​	​
104​	0.19134​	​
105​	0.19099​	still not converging​
106​	0.19064​	​
107​	0.19031​	​
108​	0.18997​	​
109​	0.18964​	​
110​	0.18931​	still not converging​
111​	0.18899​	​
112​	0.18867​	​
113​	0.18836​	​
114​	0.18805​	​
115​	0.18774​	still not converging​
116​	0.18744​	​
117​	0.18714​	​
118​	0.18684​	​
119​	0.18655​	​
120​	0.18626​	still not converging​
121​	0.18597​	​
122​	0.18569​	​
123​	0.18541​	​
124​	0.18513​	​
125​	0.18486​	still not converging​
126​	0.18459​	​
127​	0.18432​	​
128​	0.18406​	​
129​	0.18379​	​
130​	0.18353​	still not converging​
131​	0.18328​	​
132​	0.18302​	​
133​	0.18277​	​
134​	0.18252​	​
135​	0.18228​	still not converging​
136​	0.18203​	​
137​	0.18179​	​
138​	0.18155​	​
139​	0.18131​	​
140​	0.18108​	still not converging​
141​	0.18085​	​
142​	0.18062​	​
143​	0.18039​	​
144​	0.18016​	​
145​	0.17994​	still not converging​
146​	0.17972​	​
147​	0.17950​	​
148​	0.17928​	​
149​	0.17907​	​
150​	0.17885​	still not converging​
151​	0.17864​	​
152​	0.17843​	​
153​	0.17822​	​
154​	0.17802​	​
155​	0.17781​	still not converging​
156​	0.17761​	​
157​	0.17741​	​
158​	0.17721​	​
159​	0.17701​	​
160​	0.17682​	still not converging​
161​	0.17662​	​
162​	0.17643​	​
163​	0.17624​	​
164​	0.17605​	​
165​	0.17586​	still not converging​
166​	0.17568​	​
167​	0.17549​	​
168​	0.17531​	​
169​	0.17513​	​
170​	0.17495​	still not converging​
171​	0.17477​	​
172​	0.17459​	​
173​	0.17442​	​
174​	0.17424​	​
175​	0.17407​	still not converging​
176​	0.17390​	​
177​	0.17373​	​
178​	0.17356​	​
179​	0.17339​	​
180​	0.17322​	still not converging​
181​	0.17306​	​
182​	0.17289​	​
183​	0.17273​	​
184​	0.17257​	​
185​	0.17241​	still not converging​
186​	0.17225​	​
187​	0.17209​	​
188​	0.17193​	​
189​	0.17177​	​
190​	0.17162​	still not converging​
191​	0.17146​	​
192​	0.17131​	​
193​	0.17116​	​
194​	0.17101​	​
195​	0.17086​	still not converging​
196​	0.17071​	​
197​	0.17056​	​
198​	0.17042​	​
199​	0.17027​	​

 

strings​	Parallel resistance​	​
1​	1.00000​	​
2​	0.66667​	​
3​	0.54545​	​
4​	0.48000​	​
5​	0.43796​	not converging​
6​	0.40816​	​
7​	0.38567​	​
8​	0.36794​	​
9​	0.35349​	​
10​	0.34142​	still not converging​
11​	0.33114​	​
12​	0.32225​	​
13​	0.31445​	​
14​	0.30754​	​
15​	0.30137​	still not converging​
16​	0.29579​	​
17​	0.29074​	​
18​	0.28611​	​
19​	0.28187​	​
20​	0.27795​	still not converging​
21​	0.27432​	​
22​	0.27094​	​
23​	0.26779​	​
24​	0.26483​	​
25​	0.26206​	still not converging​
26​	0.25944​	​
27​	0.25697​	​
28​	0.25464​	​
29​	0.25242​	​
30​	0.25031​	still not converging​
31​	0.24831​	​
32​	0.24640​	​
33​	0.24457​	​
34​	0.24282​	​
35​	0.24115​	still not converging​
36​	0.23955​	​
37​	0.23801​	​
38​	0.23652​	​
39​	0.23510​	​
40​	0.23372​	still not converging​
41​	0.23240​	​
42​	0.23112​	​
43​	0.22989​	​
44​	0.22869​	​
45​	0.22753​	still not converging​
46​	0.22641​	​
47​	0.22533​	​
48​	0.22428​	​
49​	0.22325​	​
50​	0.22226​	still not converging​
51​	0.22130​	​
52​	0.22036​	​
53​	0.21945​	​
54​	0.21856​	​
55​	0.21769​	still not converging​
56​	0.21685​	​
57​	0.21603​	​
58​	0.21523​	​
59​	0.21444​	​
60​	0.21368​	still not converging​
61​	0.21294​	​
62​	0.21221​	​
63​	0.21149​	​
64​	0.21080​	​
65​	0.21012​	still not converging​
66​	0.20945​	​
67​	0.20880​	​
68​	0.20816​	​
69​	0.20753​	​
70​	0.20692​	still not converging​
71​	0.20632​	​
72​	0.20573​	​
73​	0.20515​	​
74​	0.20458​	​
75​	0.20403​	still not converging​
76​	0.20348​	​
77​	0.20294​	​
78​	0.20242​	​
79​	0.20190​	​
80​	0.20139​	still not converging​
81​	0.20089​	​
82​	0.20040​	​
83​	0.19992​	​
84​	0.19944​	​
85​	0.19898​	still not converging​
86​	0.19852​	​
87​	0.19806​	​
88​	0.19762​	​
89​	0.19718​	​
90​	0.19675​	still not converging​
91​	0.19633​	​
92​	0.19591​	​
93​	0.19550​	​
94​	0.19509​	​
95​	0.19469​	still not converging​
96​	0.19430​	​
97​	0.19391​	​
98​	0.19353​	​
99​	0.19315​	​
100​	0.19278​	still not converging​
101​	0.19241​	​
102​	0.19205​	​
103​	0.19169​	​
104​	0.19134​	​
105​	0.19099​	still not converging​
106​	0.19064​	​
107​	0.19031​	​
108​	0.18997​	​
109​	0.18964​	​
110​	0.18931​	still not converging​
111​	0.18899​	​
112​	0.18867​	​
113​	0.18836​	​
114​	0.18805​	​
115​	0.18774​	still not converging​
116​	0.18744​	​
117​	0.18714​	​
118​	0.18684​	​
119​	0.18655​	​
120​	0.18626​	still not converging​
121​	0.18597​	​
122​	0.18569​	​
123​	0.18541​	​
124​	0.18513​	​
125​	0.18486​	still not converging​
126​	0.18459​	​
127​	0.18432​	​
128​	0.18406​	​
129​	0.18379​	​
130​	0.18353​	still not converging​
131​	0.18328​	​
132​	0.18302​	​
133​	0.18277​	​
134​	0.18252​	​
135​	0.18228​	still not converging​
136​	0.18203​	​
137​	0.18179​	​
138​	0.18155​	​
139​	0.18131​	​
140​	0.18108​	still not converging​
141​	0.18085​	​
142​	0.18062​	​
143​	0.18039​	​
144​	0.18016​	​
145​	0.17994​	still not converging​
146​	0.17972​	​
147​	0.17950​	​
148​	0.17928​	​
149​	0.17907​	​
150​	0.17885​	still not converging​
151​	0.17864​	​
152​	0.17843​	​
153​	0.17822​	​
154​	0.17802​	​
155​	0.17781​	still not converging​
156​	0.17761​	​
157​	0.17741​	​
158​	0.17721​	​
159​	0.17701​	​
160​	0.17682​	still not converging​
161​	0.17662​	​
162​	0.17643​	​
163​	0.17624​	​
164​	0.17605​	​
165​	0.17586​	still not converging​
166​	0.17568​	​
167​	0.17549​	​
168​	0.17531​	​
169​	0.17513​	​
170​	0.17495​	still not converging​
171​	0.17477​	​
172​	0.17459​	​
173​	0.17442​	​
174​	0.17424​	​
175​	0.17407​	still not converging​
176​	0.17390​	​
177​	0.17373​	​
178​	0.17356​	​
179​	0.17339​	​
180​	0.17322​	still not converging​
181​	0.17306​	​
182​	0.17289​	​
183​	0.17273​	​
184​	0.17257​	​
185​	0.17241​	still not converging​
186​	0.17225​	​
187​	0.17209​	​
188​	0.17193​	​
189​	0.17177​	​
190​	0.17162​	still not converging​
191​	0.17146​	​
192​	0.17131​	​
193​	0.17116​	​
194​	0.17101​	​
195​	0.17086​	still not converging​
196​	0.17071​	​
197​	0.17056​	​
198​	0.17042​	​
199​	0.17027​	​

Well that looks good on the surface, but you made one major fault, you only used 5 significant digits for the accuracy. You must use something more modern like a double precision floating point numbers. This is in excel and proves it is converging on 0.5 and very quickly at that it only took 21 steps in excel and about 1 minute to make the chart. I took it out to a total 3221 Entries and solid as a rock at 0.5