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

Series — page 4

A field says what an essay is about. A series follows one idea essay by essay — from the question that introduces it to the one that assumes all the others.
A divider of 4 equal 1.0% resistors, solved 3000 times. computed by solving, not by drawing. Every resistor drawn from its tolerance band and the divider solved, 3000 times. The worst case is ±1.000% — the part tolerance itself, and it does not improve when the divider is built from more parts — while the measured spread is 0.2944% and the worst of 3000 draws reached 82% of the bound. The root-sum-square, offered as though it were a standard deviation, is 1.70 of one here: for uniformly distributed parts it is √3 σ, a coverage of about 92%.

Component tolerance

  1. 1 The tolerance that is not on any part
  2. 2 The shape a tolerance does not print
  3. 3 What a ratio does not see
  4. 4 The small leg of a large ratio
4 essays · networks Fourteen decades of imbalance, fourteen digits gone, and a matrix in perfect health. computed by solving, not by drawing. A Wheatstone bridge walked towards balance, with the relative error of the solved output against a closed form that cannot lose digits. The condition number of the nodal matrix is 505.0 at every imbalance and the smallest pivot is 2.0e-3 of the matrix norm — orders above the 1e-12 at which this solver refuses to answer at all. Neither number moves, and the answer still loses one digit per decade of imbalance, reaching 33% at δ = 1e-15. The bound drawn over it is the round-off divided by the imbalance, which the measurement stays under at every point. The third curve is the same closed form written as ½ − 1/(2+δ) — algebraically identical, and it loses its digits at the same rate, which is where the loss lives: in the subtraction of two nearly equal numbers, not in the matrix.

Conditioning

  1. 1 The matrix that is ill, and the answer that is not
  2. 2 The digits the arithmetic did not have
  3. 3 Where the matrix is worst, and where the answer is not
  4. 4 The step that returns the digits
4 essays · networks Two tracks, and a far end that cancels exactly when the field is all in one material. computed by solving, not by drawing, on 12 coupled sections of a 100 mm pair terminated in 50 Ω at all four ends. A mutual capacitance injects a current proportional to dV/dt and splits it towards both ends of the quiet track; a mutual inductance injects a voltage proportional to dI/dt and drives the two ends in opposite directions. So the near end goes as Cm/Ct + Lm/Lt and the far end as their difference, with the same constant in front of both — measured here as 1.048e-2 either way, over a slider that moves the ratio by five times. The consequence is that the far end is not a smaller effect but a cancellation: at a ratio of one it is 7.52e-19 of the drive, which is zero to the last bits of a double, while the near end is 1.048e-3. That is why a stripline has no far-end crosstalk and a microstrip has some — what shows up there measures the field that is in air, not the spacing. The model is lumped and stops where it says: a section is one degree long at 50.0 MHz.

Crosstalk

  1. 1 The far end that cancels
  2. 2 How wide a null is
  3. 3 The near end that stops growing
  4. 4 The far end that is a derivative
4 essays · lines 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.

Current mirror

  1. 1 The copy, and its two errors
  2. 2 The error that is a distribution
  3. 3 The refusal, and what it was protecting
  4. 4 The mismatch that cancels itself
4 essays · semiconductors How long a second-order step takes to arrive inside ±2%. computed by solving, not by drawing from the residue expansion at 260 damping ratios. The fastest is ζ = 0.780 at 3.60/ω₀; critical damping takes 5.83/ω₀, which is 62% longer. Between ζ = 0.775 and 0.780 the time falls by 33% in one step of the sweep, because which excursion is the last one outside the band changes there — the overshoot at the fastest damping is 1.99%, which is the band itself, and one step to the left it is larger. The faint curves are the other bands, each with its own step in a different place.

Damping

  1. 1 The cliff before the fastest settling
  2. 2 Three cliffs, and where they are
  3. 3 The best damping is not the one to build
  4. 4 The damping a zero asks for
4 essays · transients A bulk capacitor and a ceramic, and the peak between them at 6.52 MHz. computed by solving, not by drawing. Each capacitor is three elements — its capacitance, its series resistance and its series inductance — and a one-amp source drives the node, so the node voltage is the impedance. Alone, each dips to its own series resistance at its own self-resonance and rises on either side. Together they do not: between the two resonances the bulk part is an inductor and the ceramic is still a capacitor, and an inductance across a capacitance is a parallel resonance. The pair reaches 1.187 Ω at 6.52 MHz, where the bulk alone would give 0.2023 Ω and the ceramic alone 0.2055 — 5.87 times worse than either. The dashed curves are the two parts on their own; the solid one is what the load actually sees.

Decoupling

  1. 1 The pair that is worse than either
  2. 2 The capacitor that is not where the load is
  3. 3 The floor and the ceiling move apart
  4. 4 The peak a load step does not see
4 essays · power What 1.5 mm of length mismatch does to a differential pair. computed by solving, not by drawing. Two lines of the same impedance and different lengths, driven differentially. The solid rising curve is what arrives as common mode; the dashed one beside it is sin(ωΔτ/2), which is what a lossless pair gives and is the same curve until the null. The flat curve at the top is the differential signal, and it is the point: at 3.04 GHz a tenth of the launched amplitude is common mode and the differential has lost 5011 parts per million of itself. The conversion is first order in the skew and the loss is second order, so the error is not missing from the signal — which is why a pair can pass its own eye and fail an emissions test. At 95.3 GHz the closed form has a null and the real pair does not: the longer conductor is also the lossier one, and an amplitude imbalance has no null in it.

Differential line

  1. 1 The millimetre that becomes common mode
  2. 2 The common mode that has three slopes
  3. 3 The skew in the glass
  4. 4 The angle two weaves allow
4 essays · lines Below 910 kHz a trace is a diffusion, not a line — and its velocity goes as √f. computed by solving, not by drawing. The phase velocity of an ordinary FR-4 trace against frequency, computed from γ = √((R + jωL)(G + jωC)) with a series resistance that rises as √f above its skin-effect corner and a shunt conductance proportional to frequency. Above 910 kHz the velocity is 0.4767c and does not move, which is the number every other essay in this field uses. Below it the series resistance dominates the reactance, the line is a diffusion, and the velocity falls as the square root of frequency — measured at the 0.467 power. The characteristic impedance is not a constant down there either: 1508 Ω at a kilohertz against 50.0 Ω at ten gigahertz.

Dispersion

  1. 1 The delay that is not one number
  2. 2 The permittivity a loss forbids
  3. 3 The conductor made causal too
  4. 4 The skew that moves with frequency
4 essays · lines An exponential driven 10.0 mV either side of its bias. computed by solving, not by drawing. A sinusoid in, and out comes a waveform whose peaks are taller than its troughs are deep. The second harmonic is 9.61% of the fundamental, measured by transforming 512 samples and predicted independently as I₂(0.387)/I₁(0.387) = 9.61%. The two routes agree to 5e-10 over the 5 harmonics that stand above the arithmetic's own floor, and share nothing but the amplitude.

Distortion

  1. 1 The distortion a linear model cannot have
  2. 2 The point the device is never at
  3. 3 The product that is not the third
  4. 4 The fifth a resistor adds
4 essays · semiconductors The usable range of one stage, in a 10 kHz measurement. computed by solving, not by drawing. The floor is the Johnson noise of a 1 kΩ source in the measurement's own noise bandwidth — 501.6 nV, using 15.7 kHz rather than the 10 kHz corner. The ceiling is the drive at which an exponential's distortion reaches one per cent, 1.03 mV. Between them is 66.3 dB, and nothing a designer does moves either number without changing the circuit.

Dynamic range

  1. 3 A floor and a ceiling
  2. 4 The ceiling is not at the output
  3. 5 The ceiling a gain gives back
  4. 6 The floor the gains bring
4 essays · noise An exponential with 6× of degeneration: both edges move, and not together. computed by solving, not by drawing at 61 amplitudes. Total harmonic distortion reaches one per cent at 36.6 mV and the gain falls one per cent short of its small-signal value at 85.0 mV. An emitter resistor dividing the gain by 6 moves the distortion edge by 35.4 times — the square of the factor, because the resistor both divides the drive reaching the junction and linearises what the junction does with it — while the gain edge moves by only 11.6 times. So the two edges close up: 2.32 times apart here against 7.06 bare, and a well-degenerated stage stops being limited by its linearity and starts being limited by how accurately its gain is known.

Emitter degeneration

  1. 1 What a resistor in the emitter buys
  2. 2 Where the two exponents come from
  3. 3 What the fourth order says about the third
  4. 4 The order that stops helping
4 essays · semiconductors 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.

Equivalent circuit

  1. 1 Exact outside and wrong within
  2. 2 The resistor that is not made of the resistors
  3. 3 The pole that crosses with the resistance
  4. 4 The reduction that stays finite
4 essays · networks Three filter families at order 5, all with the same half-power point. At three times the corner the Chebyshev is -64.0 dB down, the Butterworth -47.7 dB and the Bessel -28.3 dB. The inset is the passband at forty times the vertical magnification, which is the only place the Chebyshev's half-decibel of ripple is visible at all.

Filter families

  1. 1 Three families, one corner
  2. 2 The zeros that buy an order
  3. 3 The line from Bessel to Butterworth
  4. 4 The attenuation a flat delay cannot buy
4 essays · filters Measuring with 50 mΩ of lead in each wire. computed by solving, not by drawing at 61 resistances, twice each. The two-wire arrangement measures the leads too, so its error is 2×50 mΩ over whatever is being measured: one per cent at 10 Ω, and 10000% at 1 mΩ. The four-wire arrangement senses on a separate pair that carries almost no current, and its error stays under 1.0e-2% across the whole range.

Four-terminal

  1. 1 Two terminals measure the leads as well
  2. 2 Two ports from two one-ports
  3. 3 The voltmeter four wires do not remove
  4. 4 The voltage four wires carry in
4 essays · instruments Group delay across the passband, at order 5. The Bessel filter's delay varies 0.1% below 0.8 of the corner; the Chebyshev's peaks near the band edge and is several times its low-frequency value. Every family here has the same half-power frequency, so this is a difference in behaviour rather than in scaling.

Group delay

  1. 3 Flat magnitude, unflat delay
  2. 4 Flat delay, bought with more delay
  3. 5 The delay that is a centre of mass
  4. 6 The envelope that leaves early
4 essays · filters 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.

Impedance scaling

  1. 1 The same filter a thousand times larger
  2. 2 The band that closes with the order
  3. 3 The band that does not close
  4. 4 The floor that outlives the arithmetic
4 essays · filters Where an amplifier's reading comes from, against the source it is reading. computed by solving, not by drawing. Three errors with three different dependences on the source, each measured by a solve with the other two set to zero. The offset voltage is flat — 50 microvolts wherever the source is. The bias current times the imbalance is linear in the source and is what balancing removes. The offset current times the source is linear too and is what balancing leaves. Unbalanced, the current overtakes the voltage at 1.77 kΩ; balanced, at 10.0 kΩ, which is the offset voltage divided by the OFFSET current and is the ratio of the two currents further along. Below about a kilohm, balancing makes the reading worse — the feedback network is already the larger resistance, and equalising means adding to the source.

Input bias current

  1. 1 The current the instrument draws
  2. 2 The cure that becomes a different circuit
  3. 3 The current a capacitor adds up
  4. 4 The cure that needs one current
4 essays · instruments The first conduction carries 26× the repetitive peak, and the factor of 7 is the user's. computed by solving, not by drawing. The largest diode current in the first conduction after switch-on, against the phase of the mains at the instant of switching, marched from an empty capacitor through the same netlist the ripple and the conduction angle are read from. The settled circuit's largest current is 1.230 A; the first one is 32.39 A at the worst instant and 4.64 A at the best, so which of them a design has to survive is decided by nothing in the design. The estimate that treats the capacitor as a short circuit gives 34.00 A, 4.7% high, the difference being the charge the capacitor takes during the pulse itself. Holding the first peak to ten times the repetitive one would need 1.90 Ω of winding resistance.

Inrush

  1. 1 The first cycle, which no steady state contains
  2. 2 The inductance that limits, and lifts
  3. 3 The instant a transformer is worst at
  4. 4 The harmonic that names an inrush
4 essays · power Loop gain of a three-pole amplifier closed for a gain of 100. Unity loop gain at 5.73 kHz, where 34.9° of phase remains before −180°. The phase reaches −180° at 89.6 kHz, where the loop gain is 46.1 dB below unity.

Loop gain

  1. 1 What is left at crossover
  2. 2 Stable, and unstable with less gain
  3. 3 The loop that never crosses
  4. 4 Two wrong readings and one loop gain
4 essays · feedback Matching 50 Ω to 200 Ω with 51.7 mm of 100.0 Ω line. computed by solving, not by drawing at 261 frequencies. The reflection at the design frequency is 4.6e-17 — nothing, to the arithmetic — against 0.600 for the bare junction, which throws 36% of the power back. It stays under 0.1 from 0.914 to 1.086 of that frequency, a band of 17.1%.

Matching

  1. 2 A quarter wave, and the path the current takes back
  2. 3 Several sections, and the band they buy
  3. 4 The number that was wrong
  4. 5 The match with no upper edge
4 essays · lines

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