Circuit Response
every model has an edge, and the figure carries it
Series

Divider — the series

5 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. A 10 kΩ + 10 kΩ divider, solved with its load. The unloaded answer is 6.00 V. It is 1% low at a load of 495 kΩ and 4.00 V at a load equal to the divider's own resistance. The ratio does not predict any of this; the magnitude does.

    The divider, and the thing it does not know about

    A two-resistor divider's output is set by the ratio of its resistances — with nothing connected. Connect anything at all and what decides the answer is the quantity the ratio was built to discard: the magnitude. Two dividers of identical ratio give six volts and one volt into the same load.

    part 2 · networks
  2. Three dividers in a row, and where the error actually is. computed by solving, not by drawing. Three two-resistor dividers cascaded with nothing between them. The product of their ratios is 0.1250 and the solved output is 0.076923, 38.5% low at a staircase of ×1. Decomposed stage by stage with the rest of the chain in place — and the product of those three is the answer exactly — they are 0.3846, 0.4000, 0.5000: the LAST stage is exactly its own ratio because nothing is connected to it, and the error is at the front, which is the opposite of where a probe goes. Staggering each stage's impedance up by a factor takes the departure down in proportion, fitted exponent -0.993, so a decade a stage is within 4.88% and two decades within 0.50% — which is why a chain that has to be right is built as a staircase and not out of one value repeated.

    The stage that is wrong is the far one

    Three identical ten-kilohm dividers in a row give 0.076923 rather than the product of their ratios, 0.125 — 38.5 per cent low. Decomposed stage by stage with the rest of the chain in place, and the product of those three is the answer exactly, they are 0.3846, 0.4000 and 0.5000: the LAST stage is exactly its own ratio because nothing is connected to it, and the error is entirely at the front, which is the opposite end from where a probe goes. Staggering each stage's impedance up by a factor takes the departure down in proportion — fitted exponent −0.993.

    part 3 · networks
  3. The current divider, and the resistance that is not in the branch. computed by solving, not by drawing. A current source into two parallel branches, the metered one 10.0 kΩ and the other 1.00 kΩ. The metered branch takes 0.090909 of the current, which is the OTHER branch's resistance over the sum; writing the subscripts the way a voltage divider writes them gives 0.90909, a different number at every ratio but one. The dual of a voltmeter's finite resistance is an ammeter's non-zero one, and the threshold has the same shape with the roles exchanged: one per cent of error at 111.1 Ω, which is that resistance over ninety-nine to 1.7e-12%. And the headline of the loaded divider holds in the dual too — two current dividers of identical ratio read 0.04762 and 0.09090 into one hundred-ohm meter — while a perfect ammeter reads them identically.

    The branch the other resistance decides

    A voltage divider's output is set by the resistance the output is taken across; a current divider's is set by the resistance the current does not go through. The dual of a voltmeter's finite resistance is an ammeter's non-zero one, and the threshold has the same shape with one word changed: one per cent at a meter resistance of R/99 where R is what the meter looks back into — and removing an ideal current source means OPENING it, so that R is the two branches in series, 11 kΩ here, not the 909 Ω of their parallel combination — a factor of twelve in the same construction on the same network.

    part 4 · networks
  4. 4 branches read with 10 Ω ammeters: one at a time every reading is low, all at once they add up and the small branches are high. computed by solving, not by drawing. A 1 A source into 4 parallel branches of 100 Ω, 1 kΩ, 10 kΩ, 100 kΩ, each read with an ammeter of 10 Ω. With one ammeter moved from branch to branch, each reading is low — by −0.989%, −0.902%, −0.099%, −0.010% — exactly r over the resistance that meter looks back into plus r, and the four readings add to 990.275 mA rather than 1000.000. With an ammeter in every branch at once the readings add to 1000.000 mA and are wrong by −0.892%, +7.939%, +8.910%, +9.008%: the meter in the largest branch pushes current out of it, and the small branches take it.

    The readings that add up and are wrong

    Four branches of 100 Ω, 1 kΩ, 10 kΩ and 100 kΩ share a 1 A source. Read them one at a time with a single 10 Ω ammeter moved from branch to branch, and every reading is within one per cent of the truth, but the four add to 990.275 mA. Put a 10 Ω ammeter in every branch at once and the four readings add to 1000.000 mA exactly, satisfying Kirchhoff's current law to the last digit. Yet three of them are 7.9 to 9.0 per cent high. The check a reader would run on the readings is passed by the wrong set and failed by the right one.

    part 5 · networks
  5. The R–2R ladder is the chain designed for the rest of itself: every node halves exactly, in place and only in place. computed by solving, not by drawing. A current source into an 8-bit R–2R ladder, each bit's 2R returned to a virtual earth. With the rest of the ladder in place every node splits the current arriving at it exactly in half, so bit k carries 2 to the −k of the source and the step at the major carry is exactly one least significant bit. A stage alone, into a short, would send a third on rather than a half; it is right because of what is connected to it. The same ladder with every resistor equal — the three-divider chain's habit — gives 0.61803, 0.23607 and then falls away by 2.6182 a stage, which is the square of the golden ratio, 2.6180, to the fourth figure.

    The R–2R ladder, built for the rest of itself

    Three equal voltage dividers in a row deliver 0.076923 of their input instead of 0.125. Their error sits in the front stage, at 0.3846, 0.4000 and 0.5000 per stage. The dual chain, three current dividers fed by a current source, gives the same three numbers to the last digit, and its repair runs the other way: each stage's resistance must fall, not rise. The R–2R ladder is the chain whose stages are wrong alone and right in place, halving exactly at every node. It still keeps its error at the front: a one per cent error in its first resistor makes an eight-bit ladder step backwards at the major carry, and the same error in its last shunt moves that step by 0.0034 LSB.

    part 6 · networks

All series