too many possibilities

How many divider tolologies can be made from 8 resistors?

For each topology, I can put any one of 19 resistor values at any location.

Even the 3-resistor cases (I think there are five) are a nuisance to calculate.

Sure, but I need a program first. I can't do that in my sleep.

There are more

Dividers are trivial if I'm willing to buy new E192 parts. Less trivial to make dividers from parts in stock, but my Rugrat program does that. Downright tedious to do all my circuit functions on a

Oh well, my board design is done. But this math puzzle will be back.

Normally you don't need to, assuming you want to get by with just two components. All you need to achieve is a ratio. Well, disregarding the error term from the regulator adjust pin current.

If three parts then adding 100ohms or whatever by memory is easy.

I should have brought you my vellum when I was down there. Next time I sure should.

I still use an HP-11C. When I sat for the US ham radio exams I asked around if friends had a spare calculator with trig functions for the unlikely event that the HP-11C quits on me. None did. None! Since cell phones are prohibited for obvious reasons I brought my slide rule. And yeah, again some people stared at that thing.

Early adopters always bear the brunt. Our family was first in town with PCs. Dad always had the latest and greatest from IBM , starting with this one:

The model with 64k RAM cost a lot more than a car back in the 70's.

How many dividers can be made from two identical quad r-packs, namely

Then how many using two different-value quad r-packs?

1. There is only one topology composed of one resistor.
2. Any topology composed of N+1 resistors can be created by connecting a single resistor to any N-resistor topology either in series with it as a whole or to any resistor it contains or in parallel to any two different nodes of the N-resistor circuit.

With this recursive rule you will be able to enumerate all possible arrangements.

But it does not make any practical sense. You either have N Even the 3-resistor cases (I think there are five) are a nuisance to

Oh, come on. For 3 resistors you have only 6 topologies to consider. R1+R2+R3, R1+(R2|R3), (R1|R2)+R3, (R1+R2)|R3, R1|(R2

Indeed, with this kind of attitude only a magic wand would help.

Fewer values make the problem simpler, not harder.

Best regards, Piotr

I have already given you a bottom-up enumeration algorithm, so code it and count.

Or is it a commercial inquiry?

Best regards, Piotr

Maybe the Mission Saloon will be open by then. You'll like Jessica.

My PDP-8I cost $12K, with a teletype and 4K of 12-bit core. That would have bought a house.

OK, it's simple. What's the answer?

And 19 candidate values. I got my 20 volt supply to 20.25 with available parts, good enough.

The limiting case is "impossible", which is admittedly simple.

No idea, I guess it is in the 50e3-1e6 ballpark. Giving an exact answer would be only barely simpler than solving your general problem. And solving your general problem would be a commercial activity.

One of the six enumerated is not even a proper divider, but detecting this kind of outliers could be more costly than incorrectly assuming it is a valid arrangement and performing the calculations blindly. Barely

The limiting case is the ratio 0.5, which you may like or not. How close you need to be is an input parameter.

Best regards, Piotr

Oh. I thought it was simple and you'd know the answer.

At this point, it's just interesting to me, exploring an enormous solution space in a finite time. (I prefer overnight, but this isn't that sort of problem.)

But if there was a good program that found resistors and dividers, from a list of available values, I'd pay for it.

It is not trivial and I cannot give you the answer off the tip of my head. The three first values are 1, 2, 6 -- looks very much like the factorial function. If it is indeed the case, the answer would be 40320. Validating that would require more effort than I am willing to spend at the moment. Anyway, what I am trying to say is that the state space will be smaller than you suggest and perfectly amenable to automated exploration for the value of 8.

Best regards, Piotr

I think the answer to that is 16!/8!/9! but some of them are a complete waste of parts - at least if the resistors all have the same value.

eg for N=5

--R---R-- | R |

--R---R--

As N increases you get other fairly pointless ones that look like regular N-gons with one resistor spanning a pair of internal nodes and fishing nets. None of these make much sense for a potential divider.

The best buy solution is to work out all the two terminal series parallel combinations out to some fixed depth like 4 with your list of chosen parts. It probably isn't worth the effort of going any deeper.

BTW Resistors in the approximate ratio between 1:phi (5:8) and 1:3 look like about the best buy for spanning a decent range values with the smallest number of parts in modest combinations hence E12, E24 etc.

x 19^8

Simulated annealing is probably the way to go for this sort of tedious combinatorial problem keeping the best 10 result(s) found so far in a safe place until you get bored. It quickly finds a good local optimum at relatively low computational cost but can also drift away again.

It is tedious to do by hand but essentially instant for a brute force attack given a list of available R and a choice of network size N.

The problem that you ought to be wanting an answer to is given a pick and place machine with M resistor feeds and a BOM requiring N>M distinct resistor values what is the optimum way to obtain all those values with specified tolerance from P

With 5 identical resistors each Ru in value, the best one can get is

With 8 identical resistors, the best one can get is 3% error, many different ways.

With 9 identical resistors with value Ru, you can get 20V exactly. At the bottom of the divider is 1Ru, top of the divider is 5.6666Ru, which can be made from 6 x Ru resistors in series, with 2Ru in parallel with one of these unit resistors. That is probably fairly obvious though.

That one is pretty constrained by one half of the divider being inside the module. Without knowing your BOM, I can't really help you with the other half.

LTM8078 is cool, a dual switcher with magnetics in a tiny BGA. It has an internal 0.8 volt reference and a 250K resistor.

Vout = 0.8 * (1 + 250K/Rext)

It will also switch positive-to-negative, if one is careful.

I'm out of coffee. Gotta go.