Concept

Output impedance — where it appears

The voltage a source loses per ampere drawn from it at a stated frequency, and the quantity that decides how much a load disturbs it. It is measured here by driving the node with a current source and reading the voltage, which is the definition rather than a substitution into an expression.

Named by 31 essays across 6 fields — each of them below, with the objects they name alongside it.

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.

networks · Divider
The equivalent of six elements, and the heat it does not account for. computed by solving, not by drawing at 73 loads. The network reduces to 8.56032 V behind 599.768 Ω, obtained twice and agreeing to twelve figures. Into every load on the axis the two put the same voltage to 2.2e-16 of it and deliver the same power to 7.8e-16 — there is no load that can tell them apart. Inside, at a load of 5.62 kΩ, the real network is dissipating 49.168 mW and the equivalent claims 1.1349 mW, a factor of 43.3. With the port open the equivalent says nothing at all is being burned and the network is burning 48.03 mW. The slider moves the resistor bridging the two sources, which joins two ideal sources and so cannot change anything the port can reach: the equivalent stays 8.56032 V behind 599.768 Ω to the last bit at every setting, while the heat inside moves by two and a half times.

Exact outside and wrong within

Six elements reduce to one source and one resistor that no load can distinguish from them: the same voltage into every load across six decades, to the last bit of a double. The reduction is wrong about the heat by a factor of forty-three, and with nothing connected it says the network is dissipating nothing while it burns 48 milliwatts.

networks · Equivalent circuit
A five-volt regulator's output impedance, with 1.00 Ω of series resistance. 0.430 mΩ at direct current, 1.95 Ω at 10.0 kHz — a factor of 4.52e+3 — and it has already doubled by 4.81 Hz. The upper curve is the same circuit with its loop opened, and the ratio between them is the loop gain. A regulator is a voltage source below a frequency and the datasheet's milliohms are the value at the bottom of it.

A source below a frequency

A five-volt regulator's output impedance is 0.43 milliohms, which is the number a datasheet quotes. It has doubled by 4.8 hertz, is ten times worse by 27, and reaches 1.95 ohms at ten kilohertz — four and a half thousand times its own specification, and higher there than the same circuit with its feedback loop cut. Two routes to that curve, sharing only the netlist, agree to a part in ten to the thirteenth.

applied · Regulator
A copy out by 1.3% for the reason everybody names, and 11% for the one nobody does. computed by solving, not by drawing at 60 output voltages, with the Early conductance iterated to self-consistency against the current that sets it. Two base currents are stolen from the reference, so the copy is β/(β+2) of it — 1.32% low at β = 150 — and that is exact at exactly one output voltage, 0.7043 V, which is 9.39 mV under the reference's own base-emitter voltage of 0.7137 V — a displacement that goes as 1/(β+2), so that the product of the two is 1.427 V at every β the slider offers. Everywhere else the Early effect is larger: the current rises at 1.21% per volt, so moving the output from one volt to ten changes it by 11.2%. One per cent holds over 0.810 V, which is the Early voltage over a hundred and contains neither the current nor any resistor. The slider is β: it moves the first error by fifty times and the second by nothing at all.

The copy, and its two errors

Every account of a current mirror leads with the base currents: two are stolen from the reference, so the copy is beta over beta plus two, which is 1.32 per cent at beta of 150 and is what a third transistor is spent on. The Early effect is a footnote and is nine times larger over any useful swing — 11.2 per cent between one volt and ten. Moving beta from 20 to 1000 changes the first by a factor of forty-five and the second by nothing at all.

semiconductors · Current mirror
A follower's output impedance from 1 kΩ of source, bare and with 100 pF on it. computed by solving, not by drawing, on a small-signal follower at 2.0 mA with β = 150 and fT = 560 MHz. At 100 Hz the emitter presents 19.08 Ω against a textbook 1/gₘ + Rₛ/(β+1) of 19.55 Ω — the expression is an upper bound here and at every source resistance on the slider, 2.4% high at this one. What it cannot describe is the frequency axis: the β that divided the source resistance down is itself falling, so the impedance rises, and the reactance at 3 MHz is 4.3 Ω — an inductance of 0.229 µH against Rₛ/ωT = 0.284 µH. With 100 pF hung on the output that impedance peaks at 67.0 Ω at 29.3 MHz, 3.51 times its own low-frequency value: an inductive source and a capacitive load are a resonant circuit, and this one is inside a part whose output impedance is quoted as a single number.

The buffer that is not a buffer

An emitter follower is reached for when something has to be driven without being loaded: unity gain in, high impedance seen, low impedance presented. The last of those is a number with a range, and the range is narrow. At a kilohm of source the emitter presents 19.08 ohms at low frequency and 67 ohms at 29 megahertz, because the current gain that made it small is falling — and the peak is worst in the middle of the slider, so it cannot be avoided by making the source stiffer or softer.

semiconductors · Emitter follower
An inverting unity gain driving 2.2 nF, and the pole that is inside the loop. computed by solving, not by drawing. Two ten-kilohm resistors around a 10 MHz amplifier make a gain of 1.00, and a loop that closes against 2.01 — one plus the ratio, not the ratio. Hanging 2.2 nF on the output leaves the closed-loop gain at a kilohertz unchanged — 0.99998002 against 0.99997988 — and takes the phase margin from 90.0° to 30.1°. The mechanism is at the other end of the amplifier from the summing-junction case and the arithmetic is the same: the load works against the amplifier's own fifty ohms of output resistance, which puts a second pole in the forward path — inside the loop, where the feedback has to live with it — while the gain the loop closes against does not move at all. Forty-five degrees is reached at 905 pF, bisected on the netlist. The capacitance is not part of the signal path and does not appear in any expression for the gain.

The load that gets inside the loop

Hanging a capacitor on an amplifier's output changes nothing a reader can find in any expression for its gain, and takes the phase margin of a unity-gain inverter from ninety degrees to thirty. The mechanism is fifty ohms of output resistance that no data sheet page puts next to the stability page: the load works against it, the pole that results is in the forward path, and forty-five degrees arrives at 905 picofarads — which is a metre of coaxial cable.

feedback · Capacitive load
9.9 Ω restores 45°, 23 Ω restores 60°, and the load pays for it in ohms. computed by solving, not by drawing. Phase margin against the resistor placed between a unity-gain inverter's output and 2.2 nF of load capacitance, with the feedback taken from the amplifier's own side of it. With no resistor the margin is 30.11°; 9.90 Ω restores 45° and 23.20 Ω restores 60°. The lower curve is the same resistor with the feedback taken from the load instead, where it makes every value worse — the pole is then inside the loop rather than outside it, and at 220 Ω the margin is 13.4°. The rising curve is what it costs: the loop no longer regulates the load's node, so 1 kΩ of resistive load pulls the output down by 0.990% at 10 Ω, uncorrected, at direct current.

The resistor that buys the margin back

Two point two nanofarads takes a unity-gain inverter's phase margin from ninety degrees to thirty. Ten ohms between the amplifier and the load restores forty-five, twenty-three restores sixty, and it works for a reason that reads as a cheat: the feedback is taken from the wrong side of the resistor, so its pole is outside the loop. Take the feedback from the load instead — which is what anyone controlling the load would do — and the same resistor makes every value worse. What it costs is that the loop no longer regulates the load's node at all: ten ohms is one per cent of error into a kilohm, at direct current, uncorrected.

feedback · Capacitive load
One Sallen-Key design at 10 kΩ, and the band of impedance levels it survives. computed by solving, not by drawing. A 10.0 kHz unity-gain Sallen-Key section realised at seven impedance levels three decades apart, with every resistance multiplied and every capacitance divided by the same factor. The design is exactly invariant: the solved magnitudes agree to 1.0e-15 of each other across all seven and every frequency in the passband, which is the last bits of a double rather than a good approximation. The realisation is not, because two of its quantities do not scale — 50 Ω of amplifier output resistance and 2 pF of stray capacitance to everywhere — and they bind at opposite ends. Inside 0.1 dB the band runs 31.6 Ω to 31.6 kΩ, with the least departure of 0.0133 dB at 1000 Ω; at this setting it is 0.036 dB at 20.0 kHz. The consideration that does not appear on either axis is noise: the resistors' density goes as √R, so across the six decades drawn the noise moves by 1000 times and points at the low end of the band.

The same filter a thousand times larger

Multiply every resistance by a thousand and divide every capacitance by a thousand and the response does not change — not approximately, but to a part in ten to the fifteenth, which is the last bits of a double. So a designer has a free parameter that the design says nothing about, and what decides it is the two quantities that refuse to scale: fifty ohms of amplifier output resistance at one end and two picofarads of stray at the other. Between them the realisation survives over three decades of impedance level and nowhere else.

filters · Impedance scaling
An order-6 cascade at 10 kΩ: a band 10× wide. computed by solving, not by drawing. A 10.0 kHz unity-gain Butterworth of order 6, 3 Sallen-Key sections in cascade, realised at seven impedance levels three decades apart, with every resistance multiplied and every capacitance divided by the same factor. The design is exactly invariant: the solved magnitudes agree to 4.0e-15 of each other across all seven and every frequency in the passband, which is the last bits of a double rather than a good approximation. The realisation is not, because two of its quantities do not scale — 50 Ω of amplifier output resistance and 2 pF of stray capacitance to everywhere — and they bind at opposite ends. Inside 0.1 dB the band runs 316 Ω to 3.16 kΩ, with the least departure of 0.0763 dB at 1000 Ω; at this setting it is 0.182 dB at 12.9 kHz. The consideration that does not appear on either axis is noise: the resistors' density goes as √R, so across the six decades drawn the noise moves by 1000 times and points at the low end of the band.

The band that closes with the order

One Sallen-Key section is inside a tenth of a decibel of its own design over three decades of impedance level, bounded below by fifty ohms of amplifier output resistance and above by two picofarads of stray. Give it three more sections and the band is one decade; give it four and there is no impedance level at all that meets a tenth of a decibel. Every section brings three more nodes each carrying their own stray and one more amplifier carrying its own output resistance, so the floor rises with the order until it crosses the tolerance — a boundary in the order rather than in the impedance.

filters · Impedance scaling
The load sees 10.0 Ω, 1.2e-3 Ω or 1.2e-3 Ω at direct current, and the peak is lowest for the arrangement with both paths. computed by solving, not by drawing. The impedance at the load node of all three arrangements, measured by grounding the input and driving a unit current into the load. Feedback from the amplifier leaves the load looking at the isolation resistor — 10.0 Ω, with no loop gain in it at all. Feedback from the load gives 1.2e-3 Ω, and the two-path arrangement has the same, which is what its direct-current path is for. All three resonate with the load capacitance near 3.2 MHz, and the two-path arrangement's peak is the lowest — 27.6 Ω against 37.4 and 59.2. What it gives up is between: above the 159 kHz handover it has let go of the load node.

What the load sees looking back

Four rungs of this argument have measured what the amplifier does to the signal — the margin, the settling, the error, the noise. None has asked the question from the other end. A load that draws its own current sees an impedance looking back, and with the feedback taken from the amplifier that impedance is the isolation resistor, with no loop gain in it whatever: ten ohms, and a load step leaves an error that never goes away. The two-path arrangement recovers to a thousandth of it and charges for that in a quantity none of the four rungs below measured.

feedback · Capacitive load
The step at which the output impedance stops being a number. computed by solving, not by drawing. The excursion divided by the step, against the step. The flat line is the linear model, and it is flat to 0.0 parts per million across four decades — which is what an impedance is. The rising curve is the same netlist with the differential pair's tanh in the transconductor, and it leaves at 10.6 mA: the input error there is 3.63 thermal voltages, so the boundary is an amplitude in the pair's own units rather than a current with the amplifier's name on it. At 300 mA the ratio is 54.5 Ω against the linear 23.6 — 131 per cent, and it is no longer a property of the circuit at all. The slew rate that decides it is 3.25 V/µs, which is twice the thermal voltage times the gain-bandwidth in radians, and contains no design choice.

The step too large to have an impedance

The rung below drove the load node with a current step and reported an impedance: a voltage divided by a current, which is a number only if the ratio does not depend on the current. Give the amplifier the differential pair's own tanh in place of a linear transconductor and it is a number up to 10.6 milliamps and not above — where the input error is 3.63 thermal voltages, and where the slew rate that decides it is twice the thermal voltage times the gain-bandwidth in radians, containing no design choice at all.

feedback · Capacitive load
Which limit binds is a property of the load, and they change places near 22 nF. computed by solving, not by drawing. Each limit measured on its own, as the departure of its march from the linear one, at a load step of half the output stage's rating. The input pair's departure falls with load capacitance — a bigger reservoir holds the node while the loop responds, which is the sixth rung's own result — and the output stage's does not fall nearly as fast, because what it has to supply is the charge the capacitor wants. Below about 22 nanofarads the thermal voltage decides the answer and above it the output stage does, and nothing about the amplifier changed.

The current above which there is no impedance

The sixth rung found the impedance leaving at 10.6 mA, where the input pair's own tanh takes over and the slew rate is twice the thermal voltage times the gain-bandwidth in radians, with no design choice in it. A real output stage has a second limit that is nothing but design choice, and the two do not bind at the same load: at 0.47 nF the input pair's departure is 19.4 per cent against the output stage's 4.2, at 22 nF it is 0.9 against 2.3, and above the output stage's rating the excursion does not come back at all — 2,254 Ω for a quantity that was 37.

feedback · Capacitive load
A cascode multiplies rₒ by β, not by gₘrₒ — and the two are 21× apart. computed by solving, not by drawing. The output resistance of a cascode stage, measured by driving the output node with a current source and reading the voltage, against the current gain of the upper device. The plain stage's is 80 kΩ — rₒ and nothing else. The cascode's is 11.5 MΩ at β = 150, which is βrₒ to within a tenth and is 21 times below the gₘrₒ² every reference gives. The reason is in the netlist rather than in the algebra: the upper device's base draws current, so its rπ sits from the lower device's collector to signal ground and shunts the node the feedback works through. What the arrangement buys therefore scales with β and stops when β does, and the curve is the two expressions drawn against the measurement.

The device that never sees the swing

A second transistor standing between the first and the load does two things that every reference gives one expression each for, and one of the two expressions has no ceiling in it. The output resistance is not gₘrₒ² — that is 248 megohms here, and the measurement is 11.5 — it is βrₒ, because the upper device's base draws current and shunts the very node the feedback works through. The bandwidth really is fourteen times better, and what it costs is two volts of a five-volt supply.

semiconductors · Cascode
One channel's load step reaches another through the supply, and the compensation decides by 4497×. computed by solving, not by drawing. Two identical amplifiers on one rail — sharing no signal node — with an ampere of load step pulled from the first and the second's output read. With the wiring left out the coupling is exactly zero, which is what the seven rungs below this one computed. With 30 nanohenries and fifty milliohms of rail and 10 microfarads of decoupling it is not: 3.84 microvolts per ampere at 271 kHz if the compensation capacitor returns to ground, and 17.29 millivolts per ampere at 2.33 MHz if it returns to the rail. That is a factor of 4497 decided by a modelling choice, which is why both are drawn. The channel that caused the step is unaffected: its own loop corrects the disturbance along with everything else, and the crosstalk is entirely a problem for the channel that did not.

The rail the load moves

Seven rungs of this ladder end by saying the same thing: the supply is an ideal voltage source, so a load step is drawn from a node that cannot be disturbed. Giving the rail an impedance turns out to change the disturbing channel's own output impedance by three parts in ten million — its loop corrects the supply along with everything else — and to open a path to a second amplifier that shares nothing with it but a wire. How large that path is is a modelling choice: 3.84 microvolts per ampere with the compensation capacitor returned to ground, 17.3 millivolts with it returned to the rail, a factor of four and a half thousand.

feedback · Capacitive load
The capacitor across the upper resistor: 90.9° of margin at 836.5 pF, and less ripple past it. The regulator's phase margin against a capacitor across the upper divider resistor, with the output ripple the 1000 µF reservoir leaves beside it. With no capacitor the margin is 46.47° at a crossover of 9.73 kHz, the output impedance at 10 kHz is 1.95 Ω, the worst rail rejection is 5.79 dB and the ripple 60.62 mV. The margin is greatest, 90.929°, at 836.5 pF — a zero at 6.34 kHz and a pole at 25.4 kHz around a crossover moved to 16.9 kHz — where the output impedance at 10 kHz is 828 mΩ, the worst rail rejection -1.50 dB and the ripple 52.33 mV. At 10 nF the margin has fallen back to 66.34° and the ripple is 35.40 mV.

The capacitor across the upper resistor

The rejection essay said a regulator reproduces its reference times its divider's four, and that a capacitor across the lower divider resistor brings that down to one at high frequency. Measured, the loop peaks the reference's gain to 5.68 near its crossover before any capacitor is added; a nanofarad across the lower resistor raises the peak to 11.2; and the capacitor that brings it down belongs across the upper resistor, where a nanofarad keeps the gain from ever exceeding four. The same capacitor is a lead pair in the loop: 836.5 picofarads takes the phase margin from 46.5 to 90.9 degrees, and ten nanofarads, past that optimum, still holds 66 while cutting the output ripple from 60.6 millivolts to 35.4.

applied · Regulator
A cascoded mirror is 90× the output resistance, and 43% of it goes back into the reference. computed by solving, not by drawing. The output resistance of a two-transistor mirror and of the same mirror with a cascode on each branch, measured by moving the output a little either side of its operating point and reading the current, against the current gain of every device. The plain mirror sits at rₒ = 89 kΩ and does not move. The cascoded one reaches 7.39 MΩ at β = 150 and rises with β until β stops being the smaller of the two quantities, where it saturates on gₘrₒ² = 268 MΩ. The third curve replaces the diode-connected upper device with a held voltage at the same potential and recovers 1.76 times the resistance, which is the upper device's base current being charged a second time — to the reference branch, where it moves the mirror's own bias.

The source that holds to the supply

Putting a second transistor on each branch of a current mirror is always described as buying output resistance and costing headroom, and both halves of that are measured here rather than repeated. The resistance goes from 82 kΩ to 7.39 MΩ, the floor rises by 0.71 volts — and the range over which the current is actually what it was set to goes from 1.70 volts to 9.09, because a plain mirror's current never stops climbing. Forty-three per cent of the resistance that should be there is missing, and it is in the reference branch.

semiconductors · Cascode
A follower fed through 100 nH has an output resistance of -21.9 Ω. computed by solving, not by drawing. The real part of the impedance looking into the emitter, driven by a current source and read, at every frequency. At direct current it is 5.50 ohms, which is the same rₛ/(β+1) + 1/gₘ the follower's first essay computed. Between 110 MHz and 301 MHz it is negative: rπ and Cπ delay the current the transistor sources into the emitter, and past a quarter of a cycle of delay pushing the emitter up makes the device push it up as well. The dashed curve is the same follower with no inductance between the source and the base, and it never goes below zero — the sign belongs to the wire and the transistor together, and to neither alone.

The resistance that is below zero

An emitter follower's output resistance is 5.5 Ω at direct current and −21.9 Ω at 257 MHz, and the sign is not the transistor's: with an ideal source at the base there is no negative band at all, and a hundred nanohenries of wire between the source and the base produces one from 110 to 301 MHz. A capacitance resonating inside that band is a resonator with loss of the wrong sign, so 4.7 to 100 pF on the emitter oscillates while 1 pF and 470 pF do not — a band of load capacitance with quiet ground on both sides of it.

semiconductors · Emitter follower
One gain takes the equivalent resistance from 5 kΩ through infinity to negative. computed by solving, not by drawing. A 5 V source drives a node through 10 kΩ; the node also reaches 10 kΩ whose far end is held at A times the node's own voltage. The Thévenin resistance looking into that node is r1 in parallel with r2/(1 − A), which the solve returns to a part in a billion without being told: 5 kΩ at no gain, 10 kΩ at unity where r2 takes no current at all, and unbounded at A = 2.00 where the two conductances cancel. Above that it is negative. The open-circuit voltage follows it, because the short-circuit current is 500.0 µA at every gain — a short across the controlling node leaves the dependent source nothing to be controlled by — so the open-circuit voltage is simply the short-circuit current times whatever the resistance is, and reaches 150.0 volts from a five-volt source inside the range drawn.

The resistor that is not made of the resistors

Exact outside and wrong within reduced six elements to one source and one resistor and found the resistor two ways that agreed to the last bit. Put a dependent source in the network and one of those routes stops working, because setting the sources dead kills the independent ones and leaves the dependent one where it is. On a bootstrap of two ten-kilohm resistors the Thévenin resistance runs from five kilohms through infinity to minus ten, the open-circuit voltage of a five-volt source reaches 225, and above one gain the equivalent's resistor is negative — which the netlist refuses to stamp, correctly, because a negative resistance is a controlled source and not a resistor.

networks · Equivalent circuit
Two large-signal limits, each alone and then both, at 20 mA and half of it. computed by solving, not by drawing. Four marches of one netlist at each load: neither limit, the input pair's tanh alone, the output stage's 20 mA alone, and both, driven by a 10 mA step. The three curves are each limit's departure from the linear march and the departure with both present; the faint line is the two singles added. Both lies on the sum and a little above it — 1.112 times it at 0.47 nF and 1.022 at 47 nF — so the limits are present together rather than taking turns. Where the two singles cross, near 10 nanofarads, the pair costs 1.88 times what the worse of them costs alone.

The load that neither limit owns

Nine rungs of this argument asked which of an amplifier's two large-signal limits binds, and drew the load capacitance where the answer changes hands. Both are present at every load: the excursion with both in the netlist is the two departures added and between 2 and 12 per cent more, never the larger of them. So the crossing is not a handover but a maximum — at 12 nanofarads the pair costs 2.084 times what the worse of them costs alone, against 1.35 at 2.2 nanofarads and 1.07 at 47 — and the same peak sits on the resistance axis at 20 ohms and the gain-bandwidth axis at 50 megahertz.

feedback · Capacitive load
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.

networks · Divider
The resistance that just stabilises it is 17 times smaller than the one that damps it. computed by solving, not by drawing. Two consequences of one base resistor, against how much of it there is, for a follower fed through 100 nH of wire with 47 pF on its emitter. The falling curve is the Q of the worst pole pair, rooted from the determinant so that no frequency grid is involved; the rising one is the output impedance the stage presents at low frequency. Bisecting on the SIGN of the pole's real part returns 69.7 Ω — at which the pair is stable with a Q of 1.7e+15, which is to say no damping and a peak whose height belongs to the arithmetic. A Q of one needs 1.15 kΩ, 16.6 times more, and that resistor takes the output impedance from 5.66 to 13.10 Ω — exactly R/(β+1) added, which is the bare follower's own expression with the base resistance in the place of the source resistance.

What the cure at the base costs

Bisecting for the smallest base resistor that stops an emitter follower oscillating gives 8 to 79 ohms. That bisection stops at the sign change, so at the value it returns the pole pair sits on the imaginary axis with a real part of 10⁻⁷ per second and a quality factor of 1.7 × 10¹⁵ — stable, and undamped. A quality factor of one needs 1.15 kΩ at 47 pF, sixteen times more, and that resistor takes the output impedance from 5.66 to 13.10 ohms. The other cure the model has always accepted and nothing has ever used is a resistor at the emitter: it reaches the same damping with 5.68 ohms and costs half the signal.

semiconductors · Emitter follower
Held one drop and 300 mV up, the cascode's floor falls from 0.832 V to 0.403 V. computed by solving, not by drawing. The output current of two cascode mirrors against their output voltage near the bottom of the supply, each divided by its own value at 5 V, on a model whose transistors saturate. The stacked mirror holds its upper device's base two drops up and stays within one per cent down to 0.832 V; the wide-swing mirror holds it one drop and 300 mV up, which leaves the lower device 0.302 V of collector-emitter voltage, and stays within one per cent down to 0.403 V. Its output resistance at 5 V is 13.2 MΩ against 7.39 MΩ, and its copy ratio 0.9758 against 0.9747.

The drop the lower device did not need

A cascode mirror's floor is a whole base-emitter drop plus the upper device's own minimum, because the stack that biases the upper base holds it two drops above ground. The lower device needs only a few hundred millivolts. Hold the upper base one drop and 300 mV up instead, and on a model whose transistors saturate the one-per-cent floor falls from 0.832 V to 0.403 V. The output resistance does not fall with it: it rises from 7.39 to 13.2 MΩ, because the upper device's base current no longer comes out of the reference. The cost has moved into the margin itself. Below about 150 mV the lower device saturates, and at 50 mV the resistance is 0.51 MΩ and the copy ratio 0.7665.

semiconductors · Cascode
A 1 kΩ bias resistance at the upper base is attenuated to 7.97 × 10⁻⁶ at low frequency and passes the lower device's noise at 15.2 MHz. computed by solving, not by drawing. The output noise current of the cascode stage at 1 mA and β = 150, driven from 1 kΩ, source by source, with the upper device's base fed through 1 kΩ instead of held at signal ground. At low frequency the resistance's noise reaches the output as 7.97 × 10⁻⁶ of the lower collector's, about 2R/(gₘrₒ²): its voltage is followed to the node between the devices and becomes a current only through the lower device's rₒ. The upper base's own shot noise, flowing in the same resistance, is 0.683%. The resistance's noise passes the lower device's at 15.2 MHz, and at 100 MHz the stage's noise figure is 6.07 dB worse than with the base held.

The resistor the upper base is held through

A cascode's upper base is drawn at signal ground, and on a board it is held by a bias network with a resistance of its own. At low frequency that resistance's noise is harmless for the reason usually given — the upper device follows its base and turns the voltage into a current only through the lower device's output resistance — so 1 kΩ reaches the output at 8 × 10⁻⁶ of the lower device's noise. But the attenuation belongs to rₒ, and rₒ is shunted by the node's capacitance from 23.5 kHz, three hundred times below the frequency at which the upper collector's own noise stops cancelling. Unbypassed, 1 kΩ costs 6.07 dB at 100 MHz, more than 100 Ω or 100 kΩ does, and 100 pF across it makes any resistance free.

semiconductors · Cascode
A mirror, cascoded or not, delivers two transistors' shot noise at low frequency — and cascoding it does not add a third. computed by solving, not by drawing. The output noise current of a 1 mA NPN mirror into a short, in units of one device's collector shot noise 2qI, against frequency, for a plain two-transistor mirror, the stacked cascode mirror and the wide-swing cascode with its upper base held by a separate diode at 100 µA. At 1 kHz they deliver 1.986, 1.960 and 1.965: the reference transistor's shot noise, copied, and the output transistor's own. At 100 MHz the mirror's own pole and the capacitance at the output transistor's collector have begun to take noise away, and the three read 1.681, 1.196 and 1.512.

The base current the mirror counts twice

A current mirror delivers two transistors' shot noise — the reference device's, copied, and the output device's own — and cascoding it does not add a third: 1.960 and 1.965 of one device's noise for the stacked and wide-swing cascodes against 1.986 for the plain mirror. What the cascode adds is its upper output transistor's base noise, and there the two arrangements differ by a factor of four. In the wide-swing mirror the base current goes to a separate bias and counts once, one part in β. In the stacked mirror it goes into the reference branch, which mirrors it back into the node it left, and it counts 3.82 times at β = 150 — the same base current that took 43 per cent of that mirror's output resistance. The reference resistor's noise, meanwhile, adds almost nothing even at 1.8 V, because it takes its place in the diode's node rather than adding to it.

semiconductors · Cascode
An unregulated supply's output against its load: 3.38 Ω of slope with perfect parts, 3.64 with 0.3 Ω in the winding, 3.32 with it in the capacitor. computed by solving, not by drawing, marched with the diodes in the netlist: a 17 V peak centre-tapped rectifier into 1000 µF, loaded from 200 Ω down to 25 Ω. The mean output with perfect parts falls from 15.99 V at 80 mA to 14.33 V at 573 mA, an average slope of 3.380 Ω against the textbook's 1/(4fC) = 5 Ω, drawn dashed without its diode drop. With 0.3 Ω in each half-winding the slope is 3.642 Ω; with the same 0.3 Ω in the capacitor it is 3.317 Ω. The slider is the resistance.

The resistance a supply's regulation counts

An unregulated supply's output falls as its load rises, and the slope is its output resistance: 4.17 Ω at 80 mA and 2.80 at 575 for a thousand microfarads, below the textbook's 1/(4fC) = 5 Ω and falling, because the diodes conduct for longer as the load grows. A winding resistance adds to that slope at about its own value, 0.94 to 1.18 of it; a capacitor's adds −0.14 to −0.25. The guess that each should count by the share of the cycle it carries current in — a sixth and five sixths — is wrong in both places, and the right weighting is the mean current each carries: all of the load's through the winding, none through the capacitor.

applied · Unregulated supply
The bootstrap reduced both ways: Thévenin's resistance diverges at A = 2, Norton's conductance crosses zero there. computed by solving, not by drawing at 121 gains, R₁ = R₂ = 10 kΩ. The Norton current is Vs/R₁, 0.500 mA, at every gain to a part in a billion (dashed, at 1), because shorting the node also shorts the bootstrap's input. The Norton conductance times R₁ is a straight line, 2 at A = 0 through zero at A = 2 to −1 at A = 3 (solid). The Thévenin resistance over R₁ is its reciprocal (faint, clipped at ±3): 0.5, 1 at A = 1, infinite at A = 2, −1 at A = 3. One network, two coordinate systems, and only one of them is singular exactly where the network is interesting.

The reduction that stays finite

The bootstrap's Thévenin pair diverges at A = 2: its resistance and its open-circuit voltage both pass through infinity and the solve that finds them refuses. Its Norton pair does not. The short-circuit current is Vs/R₁, 0.5 mA, at every gain to a part in a billion, and the conductance is a straight line through zero. A bench that drives a current and reads a volt loses the network for every gain from 1.992 to 2.008; a bench that drives a volt and reads a current sees it to a nanosiemens throughout. A 0.1% error in the gain moves the conductance by a constant 0.1 µS and the resistance by an amount that diverges at the pole — and four hundred bootstraps built from 0.1% parts to sit on it have conductances within ±0.3 µS of zero and resistances that are a median 10.6 MΩ, 196 of them negative.

networks · Equivalent circuit
The supply's mean output against its load, with no leakage inductance, 0.3 mH and 1 mH in the winding. computed by solving, not by drawing, marched with the diodes in the netlist: 17 V peak into 1000 µF, 50 mΩ of winding throughout. Without inductance the output falls 1.69 V from 200 Ω to 25 Ω, an average slope of 3.43 Ω. With 0.3 mH it falls 0.62 V, 1.14 Ω, from a higher start; with 1 mH, 0.42 V, 0.77 Ω. An element that carries the load's current and drops no mean voltage makes the regulation better, not worse.

The inductance that flattens the supply

An unregulated supply's output resistance counts each resistance by the mean current it carries, and an inductance in the winding carries the whole mean current and drops no mean voltage — so by that rule it should count for nothing. Marched, it counts for less than nothing: 0.3 mH of leakage takes the output resistance at 100 Ω from 3.88 Ω to 0.67, and the mean output falls 0.62 V from 200 Ω to 25 Ω instead of 1.69. The effect has an interior optimum — 0.03 mH makes it slightly worse, 0.3 mH best, 3 mH back to 2.86 Ω — because the inductance first rings the reservoir up and then turns the rectifier toward choke input.

applied · Unregulated supply
A 1 µF capacitor at 5 V joined through 100 Ω to an empty one of the same size. computed by solving, not by drawing, from the netlist's poles. Charge is conserved and both settle at 2.500 V. Energy is not: the pair holds 50.0% of the 12.50 µJ the first one started with, and the resistor has taken 50.00% — C2/(C1 + C2), with no resistance in it.

The energy two capacitors cannot share

Join a charged capacitor to an empty one and the charge is shared exactly while the energy is not: the pair loses C₂/(C₁ + C₂) of what the first one held, through 1 Ω or 10 kΩ, with an inductor ringing for hundreds of cycles or without one. An inductor and a diode turn the sharing into an exchange — equal capacitors swap 96.9% of the energy at a loop Q of 50. And a charge pump is the sharing loss billed once a cycle: a doubler's efficiency is its output voltage over twice its input, to four parts in ten million, at every load and every clock.

transients · Switching energy
The droop of the ceramic regulator against the rise time of a 100 mA load, with the capacitor at full value, half and a fifth. computed by solving, not by drawing, marched from the operating point: 10 µF, 5 mΩ of series resistance, 630 pF across the upper divider resistor, the 100 mA arriving as a straight ramp. A step droops 103 mV at 100%, 137 mV at 50%, 203 mV at 20% of the capacitance. A ramp of a millisecond droops 2.98 mV, 2.56 mV, 2.56 mV, and the slowest ramps approach the loop's inductance, 25.2 µH, times the ramp's rate (dashed) — a quantity with no capacitance in it — times 1.18, 1.03, 1.03, the share the loop's ringing adds. The full-value and fifth-value curves cross at 23 µs; slower than that, the larger capacitor droops more.

The load that rises slower than the loop

Every droop a regulator is sized against is a load step with an infinitely fast edge, and a ceramic regulator droops 103 millivolts on its full 10 µF and 203 on a fifth of it. Give the same 100 mA a rise time and the ordering does not survive: past 23 microseconds the full capacitor droops more than the fifth, and a slow ramp's droop is the loop's own 25.2 µH times the load's rate of change — a number with no capacitance in it, raised 18 per cent on the full capacitor by the impedance peak its slower loop leaves. For a 10 mV budget the load must take 298 µs to arrive on the full capacitor and 253 on the fifth.

applied · Regulator
A cascode is 15.9 times wider than the plain stage from 100 kΩ and 1.30 times from 10 Ω, with a 5 kΩ load. computed by solving, not by drawing, each bandwidth bisected on the solved magnitude. The common-emitter stage of 1 mA, Cπ 20 pF and Cμ 2 pF into 5 kΩ, alone and with a second device stacked on its collector, across four decades of source resistance. The plain stage falls from 12.1 MHz to 110 kHz as its Miller-multiplied input node slows. The cascode runs from 15.8 MHz, where its output node sets the bandwidth — Cμ2 against 5 kΩ, 15.9 MHz — to 1.75 MHz, where its input node does: rπ in parallel with the source, against Cπ and twice Cμ, 1.78 MHz. Neither single-node figure is ever below the stage's own bandwidth. The faint line is the cascode with the upper device's own Cπ left out of the middle node, and it is never below the solid one and at most 0.74% above it.

The nodes a cascode hands the bandwidth to

A cascode is said to remove the Miller effect, and into 5 kΩ it does: from a 100 kΩ source it is 15.9 times wider than the plain stage. From 10 Ω it is 1.30 times wider, because there the bandwidth was never the input's — it belongs to the output node, Cμ2 against the load, 10.0 ns from every source. From a high source it belongs to the input, which is now rπ against Cπ and twice Cμ rather than against a multiplied Cμ, 74.7 ns of Cπ beside 15.4 of Cμ. The node the cascode adds holds 0.545 ns and moves the bandwidth less than one per cent. Its three poles cross rather than split, the sum of time constants is worst — 24% low — at the source where two nodes are equally slow, and the removal is not unconditional: the middle node's gain is 1.01 at a 1 kΩ load, 12.3 at 1 MΩ, and above 3.11 MΩ the cascode is narrower than the stage it replaced.

semiconductors · Device bandwidth
One loop, cut at its output wire with 160 nF on it: the voltage injection crosses unity twice, the current injection reads 57 dB too much at low frequency, and together they are the loop gain. computed by solving, not by drawing. A three-pole amplifier closed for a gain of a hundred, with 100 Ω of output resistance and 160 nF of load. The loop gain read at the controlled source's own input, which draws no current (solid); read by a voltage injected in series with the output wire (dashed); and by a current injected into the same wire (dotted). The true loop crosses unity at 5.31 kHz with 8.9° of margin. The voltage reading first crosses at 5.04 kHz with 15.1°, and rises through it again at 10.6 kHz, because above 9.95 kHz the load is a lower impedance than the source. The current reading crosses at 5.82 kHz with 19.8°. Combined through 1/(1 + T) = 1/(1 + Tv) + 1/(1 + Ti) (dots), the two give the controlled source's reading to 3e-9 at every frequency.

Two wrong readings and one loop gain

A loop gain measured on a bench is measured at a wire, and a wire has an impedance on each side. Cut an amplifier's output with 100 Ω behind it and 160 nF ahead, inject a voltage in series, and the reading is the loop gain plus (1 + T)·Zs/Zl — 15.1° of margin where the loop has 8.9°, and a second unity crossing at 10.6 kHz that the loop does not have. Inject a current into the same wire and the error is (1 + T)·Zl/Zs: 19.8°, and 57 dB too much loop gain at low frequency. The two errors are reciprocal, so the readings combine exactly: 1/(1 + T) = 1/(1 + Tv) + 1/(1 + Ti) returns the loop gain at the controlled source's own input to a few parts in a billion, at every frequency and every load.

feedback · Loop gain

Named alongside it

The objects these essays reach for when they reach for this one.

Model rangeDesign tradeoffCapacitive loadLoop gainCascodeThevenin equivalentCurrent mirrorLoad regulationLoadingPhase marginHeadroomNegative resistance

All concepts