The divider, and the thing it does not know about

Two resistors in series across a supply, with the output taken from the join. It is the first circuit anybody meets, its behaviour is a single fraction, and the fraction is memorable enough that most people never write it down again. It is also, in exactly the form it is usually taught, a model with a range — and the range is not stated anywhere in the fraction.

What the fraction says, and what it leaves out #

With nothing connected to the output, the current through the two resistors is the same, so the potential at the join divides in proportion to the resistances. Two equal resistances give half the supply. That is exact, it needs no approximation, and it is true for any pair of values with the same ratio: a hundred ohms and a hundred ohms, ten kilohms and ten kilohms, a megohm and a megohm all give half.

The moment anything is connected across the output, that stops being true, and the reason is immediate: the two resistors no longer carry the same current. Some of the current arriving through the upper resistor now leaves through whatever has been attached, and only the remainder goes on through the lower one. Less current through the lower resistor means less potential across it, which means a lower output. The divider has not changed; the network has.

The interesting part is what decides how much lower, because it is not the ratio. Attaching a load across the output puts it in parallel with the lower resistor, and the answer depends on how the load’s resistance compares with that lower resistance — an absolute comparison between two quantities in ohms, not a ratio. The two equal resistances that gave half the supply are equal whether they are a hundred ohms or a megohm, and a hundred-kilohm load leaves the first almost untouched and destroys the second.

The figure above makes that concrete, and the numbers are worth having in the prose. With a twelve-volt supply and a hundred-kilohm load across the output:

• a divider of two 1 kΩ resistors gives 5.970 V — 0.5% below its unloaded six volts
• a divider of two 10 kΩ resistors gives 5.714 V — 4.8% low
• a divider of two 1 MΩ resistors gives 1.000 V

The third one is not a degraded version of the first. It is a sixth of the answer, from a circuit with the same ratio, into the same load. The specification that the divider is “a two-to-one divider” has told a reader nothing at all about what will come out of it.

The number the model is missing #

Every model on this site is drawn together with the point at which it stops being true, and the divider’s is a resistance rather than a frequency. It can be computed exactly by solving the loaded network and asking where the answer falls one per cent below the unloaded value.

For a divider of two equal resistances R, the answer is a load of about 49.5 R. Not a suspicious number and not a rule of thumb: it falls out of requiring the parallel combination of R with the load to be 0.9802 R, and it says something a reader can carry away. A divider is accurate to one per cent only when what it drives is fifty times its own resistance. Below that, the ratio has stopped being a prediction and become an upper bound.

That is the number in the caption strip of the figure above, and it moves with the slider exactly as it should: a divider of two 100 Ω resistors is good to one per cent into 4.95 kΩ, and a divider of two megohms is good to one per cent only into 49.5 MΩ, which is not a load anybody has.

There is a second, blunter number for the case where a reader wants a feel rather than a specification. A load equal to the divider’s own resistance takes a third of the answer away. Two equal resistances loaded by a third of the same value give four volts from a twelve-volt supply rather than six, whatever the value is. That figure is worth remembering because it is exactly the situation that arises when one divider is asked to drive another.

Why the error is one-sided #

There is a structural fact hiding in the curve that is worth drawing out, because it is the reason this failure mode is dangerous in a way that a symmetric error would not be.

Adding a load can only ever reduce the output. Whatever is connected across the lower resistor draws current that would otherwise have flowed through it, and there is no arrangement of passive components that can make the join sit higher than the unloaded fraction says. The error therefore has a sign, always the same sign, and it accumulates rather than averaging out.

That matters in a chain. Three dividers in a row, each perfectly adequate on its own, do not have independent errors that partially cancel; every one of them is low, and the product of three one-sided errors is worse than any of them. The ladder later in this essay is exactly that situation and its output is forty per cent below what the three fractions predict.

It also means that a circuit built with the fraction and then measured is not merely inaccurate but inaccurate in a direction that hides itself. An output that reads low is usually attributed to a resistor being out of tolerance, or to a supply that has sagged, and the loading explanation is several places down the list — even though it is typically an order of magnitude larger than either.

Where the deciding quantity actually lives #

The reason a ratio cannot answer this question is that the divider has a property the ratio does not record, and it is worth naming because it turns up everywhere afterwards.

Looking back into the output of an unloaded divider, what is seen is the two resistances in parallel. That combination — not the ratio, not either resistance alone — is the quantity that decides everything about loading. A divider of two 10 kΩ resistors looks like a 6 V source behind 5 kΩ. A divider of two 1 MΩ resistors looks like a 6 V source behind 500 kΩ. The two are the same divider by every rule taught with the fraction and completely different objects by the only measure that matters when something is connected.

This is Thévenin’s observation, and it is not being derived here so much as pointed at: any network of sources and resistances, seen from a pair of terminals, behaves exactly like one source behind one resistance. What makes it more than a convenience is that it identifies the missing quantity. The divider’s fraction gives the source; the parallel combination gives the resistance; and the fraction alone is half a description.

That ladder is the argument in its most compact form. Anyone building it from three applications of the divider fraction would predict five volts, two and a half, and one and a quarter. The solved answers are 3.846, 1.538 and 0.769 — a factor of a third each time. The fraction is wrong at the first step by twenty-three per cent, and the error compounds.

Two ways to get the right answer, and why the second one is better #

The loaded divider has a closed form. Replace the lower resistance by its parallel combination with the load and apply the fraction again. It works, it is exact, and it is what most treatments offer.

It also does not scale. The ladder above cannot be done that way without working backwards from the far end, and a network with a bridge in it — the one in the previous essay — cannot be done that way at all, because no two of its resistors are purely in series or purely in parallel. The moment a circuit stops being a chain, the reduction technique runs out.

The nodal solve does not care. It writes one equation per node and solves them all at once, and a bridge, a ladder and a lone divider are the same problem at different sizes. Every number in this essay came out of it, which is also why they can be trusted: the solve checks itself twice before returning, once by rebuilding the branch currents from the element laws and summing them at each node, and once by requiring the resistors’ dissipation to equal the sources’ delivery.

The figures here take the closed form and the solve as two routes to the same answer, and require them to agree. That is not an abundance of caution. It is the only way to find out that a figure is right, because a plot of the wrong divider looks exactly like a plot of the right one.

What this generalises to #

Loading is not a resistive-divider phenomenon. It is what happens whenever one part of a circuit is described without reference to what is attached to it, and the pattern recurs through the whole subject in forms that look unrelated:

• A signal source has an internal resistance, and the voltage at its terminals falls as current is taken. That is the next essay, and it is the same arithmetic with the divider’s upper resistor playing the part of the source’s insides.
• An amplifier’s gain is specified with a stated load, and changes with a different one.
• A filter designed as a cascade of sections behaves as designed only if the sections do not load one another — which is why the filters elsewhere in this collection have followers between their sections, and why that decision is stated rather than hidden.
• A probe placed on a circuit to measure it changes the circuit it is measuring, by exactly this mechanism, which is why an oscilloscope probe’s input resistance and capacitance appear on its label.

The honest statement of the model #

The divider fraction is not wrong. It is a statement about a particular network — one with nothing connected to its output — and it is exactly right about that network. What goes wrong is the silent extension of it to a network that has something connected, which is every network anybody actually builds.

Written out completely, the model reads: the output is the supply times the ratio of the lower resistance to the sum, provided that what is connected across the output has a resistance at least fifty times the parallel combination of the two. The proviso is not a footnote and it is not conservative. It is the boundary of the claim, it is computable from the divider’s own values, and it is the first instance on this site of the rule the whole collection runs on.

The choice that has no good answer #

Since a divider is more faithful the lower its resistances are, an obvious question is why anybody would ever build one out of megohms. The answer is that the boundary drawn above is one of two, and they pull in opposite directions.

Lowering the resistances improves the loading behaviour and costs current. A two-to-one divider of 1 kΩ resistors across twelve volts draws six milliamperes continuously, does nothing useful with any of it, and dissipates seventy-two milliwatts as heat. The same divider in megohms draws six microamperes. For anything running from a battery, the second is the only tenable choice, and the loading problem then has to be solved elsewhere — by putting an amplifier with a very high input resistance immediately after the divider, which is precisely what such amplifiers are for.

There is a third consideration that decides the matter at high impedance, and it belongs to a later field: a divider made of megohms is loaded not only by whatever is deliberately connected to it but by stray capacitance, and stray capacitance loads it more at every increase in frequency. A megohm divider with five picofarads of stray capacitance across its lower arm has already lost a tenth of its output at sixty hertz — which is the frequency of the mains wiring in the room. The low-frequency loading rule and the capacitive one are the same statement about impedance, and at high resistance the second arrives first.

So there is no value of resistance that is simply correct. There is a range bounded below by the current that can be spared and above by what is connected and by the stray capacitance, and the range can be empty, at which point the divider is the wrong circuit and something active has to be used instead. Stating the boundary as a number is what makes that decidable rather than a matter of habit.

A note on what “one per cent” is doing #

The boundary above was drawn at one per cent, and the choice deserves defending because an arbitrary-looking threshold is exactly the sort of thing that makes a computed number feel like a quoted one.

One per cent is not a tolerance anybody has to accept; it is a reporting threshold, chosen because it is small enough to be inside what a reader would call agreement and large enough to be well outside the noise of the computation. The solve here is exact to about fifteen digits, so the boundary could equally have been drawn at a part in a thousand or a part in a million; every such boundary on this site would simply move to a different, equally computable place.

What must not happen is the boundary being drawn where it is convenient. The figures state the fraction they used, the fraction is the same across a whole field wherever that is possible, and the solve underneath is the same one that produced the curve. A reader who prefers a tenth of a per cent can move the marker with one number and the argument does not change: a divider is ten times more demanding about its load at that threshold, needing a load about 500 times its own resistance, and the point that the ratio does not predict any of this is untouched.

The circuits that exist because of this #

A divider that cannot be loaded is the reason for several arrangements elsewhere in this collection, and naming them says what the alternatives actually buy.

The bridge that is linear near one point is the oldest: two dividers subtracted, so that everything common to them cancels and what is left is the difference. It does not remove the loading — both halves are still loaded by whatever reads the output — but it makes the loading common-mode, so what survives is the difference between two loadings rather than a loading. That is a much smaller quantity and it is the whole reason the arrangement has lasted.

The node that is at ground for a while is the modern one: put an amplifier’s summing junction where the tap was, and the loading is removed by a loop rather than by arithmetic. What that costs is measured there — a tenth of an ohm at direct current, ten ohms at a kilohertz, and 909 ohms above a megahertz — so the escape from this essay’s problem is itself a model with a frequency on it.

And the probe is part of the circuit is the case where the load is an instrument rather than a circuit, and is the reason an oscilloscope probe is a ten-to-one divider rather than a wire: a hundred and fifteen picofarads across a two-kilohm source is one per cent wrong at 6.8 kHz, and dividing by ten before the cable divides the loading by ten as well.

The three are the same repair at three levels — cancel it, regulate it, or reduce it — and none of them is available to somebody reading a ratio off a schematic, which is what this essay is about.

The same fact, in the rest of the collection #

A divider is the smallest circuit in which a model omits something, and the omission recurs at every scale. What a network answers, and how the answer is checked is the machinery that puts the load back in and checks the result twice. Exact outside and wrong within is the reduction that survives loading exactly and still throws something away. The probe is part of the circuit is this page’s argument with an instrument as the load, and Two terminals measure the leads is the one arrangement in the collection that removes the error rather than bounding it. The source that is not a source is the same statement one element smaller, and A band rather than an edge is where the loading argument acquires a second side.