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

Series — page 3

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.
What the summing junction of an inverting amplifier actually is, at 1.00 MHz of gain–bandwidth. computed by solving, not by drawing by driving a current into the node and reading the voltage. It is 100 mΩ at direct current, rises 1.000 decades per decade of frequency, and settles at 909.5 Ω — which is the 1 kΩ and 10 kΩ in parallel, with the amplifier contributing nothing. It passes one per cent of the input resistor at 995 Hz, a factor of 1,005 below the gain–bandwidth. The second route — the open-loop impedance over one plus the return ratio from the cut loop — agrees to 0.045%.

Virtual earth

  1. 1 The node that is at ground for a while
  2. 2 The node that does not care how many
  3. 3 The node that is an inductance
  4. 4 The shelf a capacitor makes
4 essays · feedback 50 Ω + j100 Ω of line, and the load angle past which the far end rises. computed by solving, not by drawing at 71 load resistances and four load angles. A source of 50 Ω + j100 Ω feeding loads of the same resistance and different power factor: at unity power factor the voltage across the load climbs towards the source's and stops there, reaching 0.9675 of it at the largest load drawn. A lagging load leaves less. A leading one leaves more, and past a computable angle it leaves more than the source has: the condition is 2Rₗ(Rₛ + Xₛ·tanφ) + |Zₛ|² < 0, which for the largest load here is 28.35° of lead — bisected on the solve at 28.35° — tending to atan(Rₛ/Xₛ) = 26.57° as the load grows. So the edge is a property of the line and the load angle together, and "voltage regulation" quoted as a percentage carries neither.

Voltage regulation

  1. 1 The far end that rises
  2. 2 The load that has two voltages or none
  3. 3 The headroom that is the line's own charge
  4. 4 The regulator that pushes past the nose
4 essays · power A 0.5 mm conductor's resistance against frequency, exact and asymptotic. computed by solving, not by drawing. The exact ratio is computed from the Kelvin functions by their series; the dashed curve is the asymptote everybody quotes, which treats the current as flowing in one skin depth of the rim and is drawn only where that annulus is inside the wire. At 17.4 kHz, where the skin depth equals the radius and the rule of thumb says the effect "starts", the asymptote says 1.0000 — no effect at all — and the exact answer is already 1.0208. The rule of thumb names a frequency the effect has passed, which is the same shape as the tenth-of-a-wavelength criterion marking a point at which the lumped model is already 30% wrong. Two decades above, the two agree to 0.00%, which is what makes it an asymptote rather than a formula.

Winding

  1. 1 The resistance that grows with frequency
  2. 2 The copper that makes it worse
  3. 3 The optimum that does not move
  4. 4 The optimum a spectrum moves
4 essays · magnetics 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
3 essays · networks 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
3 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
3 essays · power The floor an amplifier adds, against the source it is given. computed by solving, not by drawing. A part with 4.00 nV/√Hz of voltage noise and 0.60 pA/√Hz of current noise is quietest into 6.67 kΩ, where its noise figure is 1.138 dB. That resistance is the ratio of the two generators and the floor there depends only on their product. Matching the same part for maximum power into its own 1 MΩ input instead — a resistance 150 times larger — costs 12.57 dB.

Device noise

  1. 1 The floor a circuit has
  2. 2 The corner where averaging stops working
  3. 3 The bowl, and the bottom of it
3 essays · noise 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
3 essays · semiconductors 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.

Divider

  1. 2 The divider, and the thing it does not know about
  2. 3 The stage that is wrong is the far one
  3. 4 The branch the other resistance decides
3 essays · networks What each family costs, at order 5. Measured on the solved networks. The Chebyshev is 36 dB further down at three times the corner than the Bessel, and pays for it in delay: its group delay varies 49.0% across the passband against the Bessel's 0.06%.

Filter tradeoff

  1. 2 What a steep skirt costs
  2. 3 A ladder is not a cascade
  3. 4 The selectivity that is not free
3 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
3 essays · instruments 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
3 essays · feedback A gapped core: where the inductance goes, and where the energy is. computed by solving, not by drawing. Reluctance in series — the gap's lg/µ₀Ae and the core's le/µ₀µᵣAe — with the inductance N²/ℛ and the share of the stored energy in each proportional to its share of the reluctance. At two hundred microns on a µᵣ = 2000 core the inductance has fallen from 41.89 mH to 5.366, and 87.2% of the energy is in the gap — which is air. The share is lg/(lg + le/µᵣ), so it contains neither the turns nor the area and what decides it is µᵣ·lg against the path length; the slider shows the same gap holding 40% at µᵣ = 200 and 98% at 15,000. That is the reason a gap is a design parameter: it is the part of the magnetic circuit whose properties do not drift, do not saturate and do not depend on temperature.

Magnetic path

  1. 1 The energy is in the gap
  2. 2 The gap that is bigger than it is
  3. 3 The gap that is three gaps
3 essays · magnetics 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
3 essays · lines One magnitude curve, two phase curves, and the one of them the magnitude decides. computed by solving, not by drawing. A passive lead network — a resistor with a capacitor across it, over a second resistor, its zero at 1.00 kHz — and the same network followed by a first-order all-pass. The two magnitudes agree to the last bits of a double at every one of the 1601 frequencies sampled, and the phases differ by as much as 180°. Bode's gain–phase integral, fed the magnitudes alone with their phases discarded, returns 39.29° at the corner against a solved 39.29°, and tracks the minimum-phase curve to 0.02° across the band — while being wrong about the second network by the all-pass's own phase, which is what excess phase means. The edge here is the span of the sweep rather than a frequency of the circuit: ±3 decades of magnitude carries 99.92% of the integral's weight, and what is left out is the tail of a logarithm.

Minimum-phase

  1. 1 The phase the magnitude already knows
  2. 2 The phase a decibel buys
  3. 3 The energy that arrives first
3 essays · frequency A sum that is exact, and the bandwidth estimate that is not. computed by solving, not by drawing, at 28 spreads of the three capacitor values in a resistor chain. The sum of the open-circuit time constants — each capacitor's own value times the resistance seen at its terminals with the other two removed — is 600.00 µs here, and it equals the ratio of the first two coefficients of the denominator to 2.0e-9 and the sum of the negated reciprocal poles to 2.0e-9. That much is a theorem. What is an estimate is the bandwidth: one over 2πΣτ gives 265.3 Hz against a measured 309.2 Hz, low by 14.2%. It is low at every spread on the axis — the estimate is never optimistic — and comes within ten per cent only once one of the three time constants is 7.48 times the others.

Open circuit time constants

  1. 1 A sum that is exact, and the estimate that is not
  2. 2 Shorted instead of opened, and the error changes sign
  3. 3 Where the estimate stops being a bound
3 essays · transients Three element voltages closing on one source, at 1.59 kHz. Solved at 1.59 kHz, which is 1.00× the frequency at which the two reactances cancel. The three phasors add head to tail to the 1 V source exactly; their magnitudes sum to 5.26 V, which is not the same statement.

Phasors

  1. 2 Three voltages that close on one, and the steady state they assume
  2. 3 The arrow that goes past where it settles
  3. 4 How long a sweep waits at each step
3 essays · frequency Two probes on a 2.0 kΩ source. computed by solving, not by drawing twice per frequency: the node alone, and the node with the probe's elements across it. The one-to-one probe's 115.0 pF makes the reading one per cent wrong at 6.79 kHz. The ten-to-one probe puts 12.8 pF in series with the cable, so its tip sees 11.5 pF and the same error arrives at 69.2 kHz — 10 times further up, bought with a factor of ten in signal — the two edges stand in the ratio of the tip capacitances, 10.00. At direct current neither probe is capacitive at all and the ten-to-one still reads 0.02% low, because 10 MΩ across 2.0 kΩ is a divider.

Probe loading

  1. 2 The probe is part of the circuit
  2. 3 The instrument's own rise time
  3. 4 The probe that takes a tenth
3 essays · instruments An amplifier a hundred times the corner moves the unity-gain section's Q by 2.0%. computed by solving, not by drawing. A unity-gain Sallen–Key section designed for Q = 2 at 1.00 kHz, built with a one-pole amplifier, and its poles recovered by rooting the determinant. The section as drawn has two poles; as built it has three, and the pair is not where it was put. At a gain-bandwidth of a hundred times the corner — which is the rule of thumb — the quality factor is 1.97 per cent high and the pole frequency is 1.97 per cent low. The two are the same number with opposite signs over the window where the amplifier is well clear of the section and its own pole is still resolvable, and the number is the designed quality factor, 2, divided by the ratio: 2.00 per cent at 100 times the corner.

Q enhancement

  1. 1 The Q the amplifier decides
  2. 2 The ripple that is a temperature
  3. 3 Where the Q comes from
3 essays · filters A resonant circuit of Q = 8, and its measured bandwidth. The half-power points are 1.50 kHz and 1.69 kHz, a bandwidth of 198.9 Hz. The components predict f₀/Q = 198.9 Hz. They differ by 0.000%.

Resonance

  1. 2 Resonance, and the bandwidth it sets exactly
  2. 3 The Q the components allow
  3. 4 Two parasitics, and the resonance neither of them has
3 essays · frequency

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