Circuit Response
every model has an edge, and the figure carries it
Networks, and how a solve is checked

The source that is not a source

An ideal voltage source holds its voltage at any current, which makes it the flattest line in the subject and the most commonly assumed model in it. Its edge is a current, set by one resistance nobody draws — and past that current the model is not approximately right, it is describing a different object.
14 min read 6 figures The parasitic is the componentEvery model has an edgeRefused, not extrapolated

An ideal voltage source is defined by what it refuses to do. Whatever current is drawn from it, its terminal voltage is unchanged: one ampere, a thousand amperes, the answer is the same. Drawn on a graph of voltage against current it is a horizontal line, and it is the flattest line anywhere in this subject.

It is also an idealisation with a boundary, and the boundary is a current.

A 9 V source with 500 mΩ inside itThe ideal source is the flat line. The solved terminal voltage leaves it at a rate set entirely by the internal resistance: 1% low at 180 mA, half gone at 9.0 A.024681010m100m110current drawn from the source (amperes)terminal voltage, solvedthe ideal source: 9 V at any current1% low at 180 mAthe model9 Vrloadsolved, then checked — the load swept over four decadesthe ideal source is 1% wrong above 180 mA
Fig. 1 A nine-volt source with half an ohm inside it, solved with a load that draws each current on the axis. The flat line is the ideal source. The curve is what the terminals do. The two are indistinguishable up to about a hundred milliamperes and have nothing to do with each other by ten. The slider is the internal resistance, and every number in the figure moves with it.

Where the idealisation comes from

Nothing that stores or converts energy can supply current without something inside it opposing the flow. A battery has electrolyte with a resistance; a bench supply has a regulator with a finite output impedance and wires from it to the terminals; a signal generator has a deliberate fifty ohms in series with its output. In every case there is a resistance between the mechanism that sets the voltage and the terminals where it is measured, and current through that resistance produces a potential difference that the outside world never sees separately.

The consequence is a straight line rather than a flat one: the terminal voltage falls linearly with the current drawn, at a rate equal to the internal resistance. Extrapolate that line to zero volts and the current there is the short-circuit current, which for the nine-volt, half-ohm source in the figure is eighteen amperes.

That single resistance carries all the information. Two sources of the same voltage and different internal resistance are different objects in every way that matters, exactly as two dividers of the same ratio and different magnitude are, and for the same reason: the quantity that decides the behaviour is the one the simple description discards.

The edge, as a number

The ideal-source model is one per cent wrong when the current drawn multiplied by the internal resistance reaches one per cent of the source’s voltage. For the source in the figure, that is 180 milliamperes — a current an indicator lamp would draw without anyone thinking about it.

Three points on the same curve are worth having:

  • at 180 mA the terminal voltage is 8.91 V, and the ideal answer is 1% high
  • at 1.8 A it is 8.10 V, and the ideal answer is 10% high
  • at 9 A — half the short-circuit current — it is 4.50 V, and the ideal answer is double the truth

The last one is the sharpest way to put it. At half the short-circuit current, an ideal source is not slightly optimistic; it is wrong by a factor of two, and any calculation built on it has stopped describing the circuit in front of it.

The same arithmetic as the divider, wearing a different hat

A reader who has met the loaded divider will recognise all of this, and the recognition is exact rather than an analogy. A source with internal resistance driving a load is a divider, with the internal resistance as its upper arm and the load as its lower one. The terminal voltage is the supply times the load over the sum, which is the same fraction, and the one-per-cent boundary is the same computation with the arms relabelled.

This is worth making explicit because it means the two rules collapse into one. A source is accurate to one per cent when the load it drives is about a hundred times its internal resistance; a divider is accurate to one per cent when the load it drives is about fifty times its own two resistances in parallel. The factors differ by two only because the divider’s fraction is a half rather than one, and the underlying statement is identical: an interface is faithful only when the impedance on one side of it is far larger than the impedance on the other.

That single sentence covers loading, measurement error, filter section interaction and probe compensation, and it is the reason all of them behave the same way when they go wrong.

A 10 kΩ + 10 kΩ divider, solved with its loadThe 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.02461001k10k100k1M10Mload resistance across the output (ohms)output voltage, solved with the load in place6.0 V with nothing connected1% low at 495 kΩthe circuit12 VR₁R₂R_Lsolved, then checked — the load swept over six decadesthe ratio is 1% wrong below 495 kΩ
Fig. 2 The same statement as the previous essay drew it. A divider is a source of half the supply behind a resistance equal to the two arms in parallel, and this curve is that source’s terminal characteristic drawn against load resistance instead of against current. The two figures are the same figure in different coordinates.

Why maximum power is at the worst place

There is a classical result here that is usually presented as a curiosity and is better understood as a warning.

The power delivered to a load is the product of the terminal voltage and the current. At very high load resistance the current is negligible and so is the power; at very low load resistance the voltage has collapsed and so, again, is the power. Somewhere in between there is a maximum, and it sits exactly where the load resistance equals the internal resistance.

That point is worth locating precisely because of where it is: it is the point at which the terminal voltage is half its unloaded value. The condition for getting the most power out of a source is therefore the condition under which the ideal-source model is wrong by a factor of two, and the efficiency there is fifty per cent — half the energy is being dissipated inside the source.

Nothing about that is contradictory, but it does explain why the result is misapplied so often. It is the right criterion for a receiver extracting a weak signal from an antenna, where the signal is all there is and the source’s dissipation is irrelevant. It is entirely the wrong criterion for anything supplying power, where the objective is efficiency and the correct arrangement is the opposite one — a load resistance far above the source’s, in the region where the ideal model still holds.

Measuring the resistance that nobody drew

The internal resistance is not marked on anything and cannot be measured directly, since it sits behind terminals that only ever present the two quantities together. It can, however, be inferred in one step, and the method is worth stating because it is the same method the site uses everywhere: take two measurements the model relates, and solve for the parameter that relates them.

Measure the terminal voltage at two different known currents. The difference in voltage divided by the difference in current is the internal resistance — the slope of the straight line, obtained without ever knowing where the line starts. For the source in the figure, readings of 8.91 V at 180 mA and 8.10 V at 1.8 A give a slope of 0.81 V per 1.62 A, or half an ohm, which is the value that went in.

Two details make the difference between that working and not working. The two currents must be far enough apart that the difference in voltage is well above the resolution of the meter, which is why a single reading near no load tells nobody anything. And the source must be genuinely linear over the range used — a battery is not, once its chemistry starts to matter, and a regulated supply is emphatically not once its current limit engages, at which point the terminal characteristic stops being a straight line and turns down almost vertically.

That last case is worth naming because it is where the linear model ends. A bench supply in current limit is a current source: it holds the current constant and lets the voltage be whatever the load requires. It is the same instrument, described by a different model, on the other side of a boundary that is drawn on its own front panel.

The measurement problem, which is the same problem

Any instrument connected to a circuit is a load on it. A voltmeter of ten megohms across the output of a divider made from two 1 MΩ resistors changes that output by a fifth, and the reading is a correct measurement of a circuit that no longer exists.

The rule follows directly from the boundary above and requires no separate theory: an instrument disturbs a measurement by less than one per cent when its input resistance is about a hundred times the output resistance of what it is measuring. A ten-megohm meter is therefore trustworthy on anything below about a hundred kilohms, and progressively less so above.

There is a version of this that catches people at high frequency rather than high impedance, and it is worth flagging here because the mechanism is identical: an oscilloscope probe has capacitance as well as resistance, and the capacitance loads the circuit more and more as frequency rises. The resistance sets a boundary in ohms; the capacitance sets one in hertz. The same sentence, in different units.

Impedance of a series RLC of Q = 4, measured by driving itOne ampere is forced into the terminals at each frequency and the resulting voltage is the impedance. The minimum is 7.91 Ω at 5.03 kHz.1101001k10k1001k10k100k1Mfrequency (hertz)impedance magnitude (ohms)reactances cancel at 5.03 kHz7.91 Ωsolved, then checked — one ampere in, 201 frequenciesnot a component value: what the pair does
Fig. 3 Where that goes. The impedance a network presents is a function of frequency, not a number, so the resistance in every rule above is really the low-frequency value of something that moves. This is a resonant pair, whose impedance falls to a minimum where its two reactances cancel and rises steeply either side.

What a solver has to refuse

The ideal source is also the element that produces the most instructive failures, and a solver that does not refuse them is not to be trusted with the cases that do have answers.

Two ideal sources in parallel, of different voltages. Each insists on the potential difference between the same pair of nodes and they disagree. There is no set of currents that satisfies both, and there is nothing to solve for: the current circulating between them is not determined by anything in the description. A real pair of batteries connected that way does have an answer, and the answer comes entirely from their internal resistances — which is to say, from the very quantity the ideal model removed.

A loop of ideal sources. The same failure with more steps in it. Going round the loop, the voltage law either restates something already known or contradicts it, and either way one equation has been lost.

An ideal source short-circuited. The current is infinite, which is a way of saying the model has no answer. Add the internal resistance and the answer is eighteen amperes.

Every one of those failures is fixed by the same thing: the resistance the ideal model left out. That is a general pattern in this subject and it is worth stating as such. The idealisation that makes a model tractable is usually the same one that makes it fail, and the failure is usually cured by putting back exactly what was removed.

Four networks the solver refusesEach has no answer, for a reason that is a fact about the circuit rather than about the arithmetic. The solver names the reason; it does not return a number.no path to grounda node whose potential nothing fixesrefused: floatingtwo sources in a loop5 V3 Vtheir currents are not determinedrefused: vloopa resistance of zero0 Ωan element with no law of its ownrefused: valuean output that is not a node?a response asked for where there is nothingrefused: nodesolved, then checked — each refusal was runa solver that never declines is untested
Fig. 4 Networks with no answer, refused by name. Two of the four are voltage-source failures: a loop of ideal sources, whose currents nothing determines, and a resistance of zero, which is not an element but a constraint. Each message here was produced by running the case.
The 0.7 volt constant, solved over eight decades of currentThe forward voltage moves 59.5 mV for every factor of ten in current, so over the range drawn here it runs from 0.298 V to 0.774 V. The three marked points are solutions for 1 V, 5 V and 12 V through a kilohm, found by Newton's method; they span 88 mV.0.400.600.8011.21e-61e-51e-41m10m100m110100current through the diode (milliamperes)forward voltage across it (volts)the constant everybody is taught1 V in: 0.629 V5 V in: 0.693 V12 V in: 0.717 Vsolved, then checked — Newton's method on the exponentialthe constant moves 59.5 mV per decade
Fig. 5 Where the linear source model runs out entirely. A diode’s forward voltage is often treated as a constant in exactly the way a source’s terminal voltage is, and it is a constant in exactly the same sense: it moves 59.5 mV for every factor of ten in current, so the three marked operating points span 88 mV between them.

The dual, and why it is not a curiosity

Everything above has an exact counterpart with voltage and current exchanged. An ideal current source holds its current whatever voltage is required, its imperfection is a resistance in parallel rather than in series, and it fails when the load resistance becomes comparable with that parallel resistance rather than with a series one.

The two descriptions are not competing accounts. A source of nine volts behind half an ohm and a source of eighteen amperes across half an ohm are indistinguishable from outside: they produce the same voltage at every current, the same current at every voltage, and no measurement made at the terminals can tell them apart. That is Norton’s observation sitting beside Thévenin’s, and the equivalence is exact for any linear network.

Which description to use is therefore a question about convenience rather than about truth, and the useful guidance is that each is clumsy at the other’s easy end. A source with a very small internal resistance is naturally described as a voltage source, since the resistance is a small correction; as a current source it would be an enormous current across an almost-short-circuit, with the answer emerging as a small difference between two large numbers. The reverse holds for a transistor’s collector, which is naturally a current source with a large parallel resistance and awkward as a voltage source of hundreds of volts.

That last observation is not aesthetic. Computing a small difference between two large quantities loses precision, and a solver handed the wrong form of an equivalent source can produce a visibly worse answer to an identical question. The equivalence is exact in mathematics and not in floating point, which is a distinction this collection returns to whenever a model and a machine disagree.

Sources with a direction

One more kind of source belongs in this essay, because it is the reason the whole apparatus is worth building rather than a curiosity.

A controlled source produces a voltage or a current determined by what is happening somewhere else in the circuit. That is what a transistor does, what an operational amplifier does, and what a transformer does. In the matrix it looks almost exactly like an independent source, with one difference: the pattern it contributes is not symmetric, because its output does not affect its input.

A network solved, and checked: a network with a controlled source in itNode potentials from modified nodal analysis. The branch currents are then recomputed from each element's own law and summed at every node; the residual is 4.2e-16 of the largest current in the circuit, which is floating-point rounding and nothing else.a network with a controlled source in itnode b0.6803 Vnode c-34.0136 Vnode out-30.9215 Vcurrent law, rebuilt from the element laws4.21e-16 of the largest branch currentpower delivered against power dissipated5.27e-16 apart · 105.2 mWsolved, then checked — 6 elementsa linear network has no edge: this one is exact
Fig. 6 A network with a controlled source of gain −50 in it, solved. The output is −30.92 volts for a one-volt input, not −50. Nothing has gone wrong: the input is divided down before it reaches the controlling node and the output is divided down again on its way out, and both divisions are the loading argument of this essay applied at the two ends of an amplifier.

The number in that caption is the essay’s conclusion in a single measurement. A device specified as a gain of fifty delivers thirty-one, and neither the device nor the specification is wrong: the missing factor is entirely the interface impedances at its two ends. That gap between the specification and the circuit is computable, it is the reason a nodal solve exists, and it is what the rest of this collection spends its time measuring.

Where this goes next

02461001k10k100k1M10Mload resistance across the output (ohms)output voltage, solved with the load in place6.0 V with nothing connected1% low at 495 kΩthe circuit12 VR₁R₂R_Lsolved, then checked — the load swept over six decadesthe ratio is 1% wrong below 495 kΩ 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. a bridge, which no series-parallel reduction reachesnode a7.5566 Vnode b4.7993 Vcurrent law, rebuilt from the element laws2.71e-16 of the largest branch currentpower delivered against power dissipated4.33e-16 apart · 48.07 mWsolved, then checked — 6 elementsa linear network has no edge: this one is exact What a network answers, and how the answer is checked A circuit has exactly one answer and a matrix finds it. The part that matters is not that the answer exists but that it can be checked twice, by routes that share no arithmetic — and that a circuit with no answer is refused by name rather than returned as a large plausible number. 1101001k10k100k1M10M100M1Gfrequency (hertz)impedance looking into 10.0 cm of track (ohms)the lumped model: one L, one C1° long at 3.97 MHza tenth of a wavelength at 143 MHzthe 200 Ω at the far endsolved, then checked — the line against a two-element modelKirchhoff's laws run out at 143 MHz Kirchhoff's own frequency The current law says the current entering a node equals the current leaving it at the same instant, which assumes the signal crosses the circuit in no time. It crosses at about two-thirds the speed of light, so the law has a frequency of its own — set by nothing but the physical size of the board.