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.
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.

Emitter follower

  1. 1 The buffer that is not a buffer
  2. 2 The input that pushes back
  3. 3 The resistance that is below zero
  4. 4 What the cure at the base costs
  5. 5 The follower that lets go on the way down
5 essays · semiconductors An order-5 ladder driven from 2× the 1 kΩ it was designed between. computed by solving, not by drawing. A doubly-terminated Butterworth ladder is a two-port designed between two stated resistances, and the resistances are part of the design rather than the environment it happens to be used in. At match it loses 6.021 dB — exactly half the voltage — and its passband has no peak anywhere in it. Driving the same five reactances from 2× that resistance moves the shape by 2.345 dB and the insertion loss to 9.542 dB. The band inside which the shape is right to 0.5 dB runs 0.8857× to 1.1371× — a window of 25 per cent on a quantity usually written down as a round number. And the two sides are not alike: at twenty times the design resistance the departure has settled at 5.33 dB, while at a twentieth of it the passband is 15.51 dB out with 7.21 dB of peaking on it, so driving a ladder from too low an impedance is worse than driving it from too high a one.

Filter termination

  1. 1 The two resistors a ladder was designed between
  2. 2 The termination an even order cannot have
  3. 3 What a filter does not pass, it sends back
  4. 4 The heat a delay predicts
  5. 5 The load a filter is in its stopband
5 essays · filters A single-pole low-pass with its corner at 995 Hz. Solved at 209 frequencies. The straight-line sketch, drawn faintly, is 3.01 dB wrong at the corner and within a tenth of a decibel only below 152 Hz. The phase is already −5.7° a decade before the corner and −84° a decade after it.

Frequency response

  1. 1 One solve, read four ways
  2. 2 Four readings of one distance
  3. 3 The circle a peak sits on
  4. 4 A tolerance is an arrow
  5. 5 Two parts that drift together
5 essays · frequency A clean board is 14.0 mV of error, and a guard makes it 1.0 nV. computed by solving, not by drawing. Two boards differing by one wire, solved with the leakage present and again with it removed, so the number plotted is the leakage's own contribution and nothing else. Unguarded, a teraohm across the laminate from a 15 V rail through a 1 GΩ source is 13.99 millivolts — and a humid morning takes that resistance down two decades, which is the left-hand end of this axis. Guarded, the ring is held at the input's own potential by the amplifier, so what is across the leakage is the amplifier's own error and the result is 1.00 nanovolts. The guarded line is flat in the rail and proportional to the signal: the offset has become a gain error of 1.00 parts per billion, and the reading is low by it rather than high: the ring sits a little below the input, so the last of the leakage pulls the input down.

Guarding

  1. 1 The current that does not reach the input
  2. 2 The sign of what the guard gives back
  3. 3 The resistor that moves the lag
  4. 4 The source that rings against its guard
  5. 5 The second pole the ringing guard sees
5 essays · instruments A one-henry inductor with nothing magnetic in it, good for 3.6 decades. computed by solving, not by drawing. The impedance at the input of an Antoniou impedance converter, read as an inductance: a current source drives the node and the voltage is solved for, and the imaginary part divided by ω is what is plotted. Five components — four resistors of 10 kΩ and a 10 nF capacitor — behave as 1000 mH, which as a wound coil would be several henries of wire. It is that inductance to within one per cent from 1.00 Hz to 3.65 kHz, 3.56 decades, and the upper edge belongs to the amplifiers rather than to the arrangement: with ideal ones in the same netlist the inductance is exact everywhere drawn. Nothing in it stores energy in a magnetic field — the current lags because an amplifier is holding a capacitor's voltage somewhere else in the loop.

Gyrator

  1. 1 The inductor that is an amplifier
  2. 2 One inductor, and ten components
  3. 3 Eight amplifiers, and what they add
  4. 4 The boundary that improves when the part gets worse
  5. 5 The resistor that pays the amplifiers back
5 essays · filters 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
  4. 5 A sweep that never stops
  5. 6 The ring a drive leaves behind
5 essays · frequency The load that takes the most power, and the load that wastes the least. computed by solving, not by drawing at 71 load values. The power into the load peaks at a ratio of 1.000000, which is the magnitude of the source impedance to six figures, and the efficiency there is 0.500000 — the source dissipates as much as the load receives. Ninety per cent efficiency needs a ratio of 9.7 and delivers 34% of the available power.

Power transfer

  1. 1 The load that takes the most
  2. 2 The load that may be complex
  3. 3 The reactance that is not free
  4. 4 The load for a source that will not hold still
  5. 5 What each element buys against the limit
5 essays · power Where a converter stops measuring the signal and starts measuring the resistor. computed by solving, not by drawing. The quantisation floor is q/√12 and halves with every bit; the Johnson floor of a 1 kΩ source in 100 kHz is 1.266 µV and does not move. They cross at 18.80 bits. Below that the converter is the limit; above it the resistor is, and a further bit buys a more precise measurement of thermal noise. A resolution quoted without a source impedance and a bandwidth is not a resolution — which is the same sentence the instruments field makes about a probe.

Quantisation

  1. 1 The floor a converter sets
  2. 2 Six decibels a bit, and the half step blamed on it
  3. 3 When a floor stops being a floor
  4. 4 The dither that is a decision
  5. 5 One step of dither per moment
5 essays · digital 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
  4. 5 Five readings of one Q
  5. 6 The reading a coil leaves nearly exact
5 essays · frequency Two meters, one current, and neither of them measuring the heat. computed by solving, not by drawing, at 35 conduction angles. The first curve is an average-responding meter: it rectifies, averages and multiplies by 1.1107, which is exactly right for a sinusoid — -7.8e-5% here — and exactly 11.07% high on a square wave, because the error is the ratio of two form factors and contains neither the amplitude nor the frequency. On a rectifier drawing its 100 W in sixty degrees of conduction it is -35.90% low. The second curve is a true-RMS meter that reaches 9 harmonics, which has no shape assumption in it and a bandwidth instead: it returns the root-sum-square of the lines it can see, and is one per cent low below every angle here of conduction. The crest factor at sixty degrees is 1.733, which is inside every instrument's rating — neither meter is failing because the peak is large. One is failing because the shape is not a sinusoid and the other because the spectrum is wider than it is.

Rms and average

  1. 1 What a meter multiplies by
  2. 2 The average a square root pulls low
  3. 3 The noise a true-RMS meter reads low
  4. 4 The cycle a converter has to know
  5. 5 The window that forgives the period
5 essays · power A 9.0 kHz input sampled at 10 kHz arrives as 1.0 kHz. computed by solving, not by drawing. The dots are the samples. The input at 9.00 kHz is above half the 10 kHz rate, and every dot also lies on the 1.00 kHz curve drawn beside it — the two sample sequences differ by 9.3e-15, which is the arithmetic and not a small effect. Nothing is attenuated and nothing is distorted: the samples are the samples of a different signal, at full amplitude, and there is no measurement of them that could say which one was there.

Sample rate

  1. 1 The frequency a sample rate invents
  2. 2 An exact answer to a different question
  3. 3 The rate each reading needs
  4. 4 The peak that is a corner
  5. 5 The samples a rise time needs
5 essays · digital The voltage a winding may carry, which is a volt-second limit read at a frequency. computed by solving, not by drawing. The dots are bisections on a marched flux — the voltage integrated sample by sample until the peak excursion reaches 0.35 T — and the line is N·Ae·Bsat·2πf. They agree to 0.001% over three decades, and the fitted slope is 1.000000: exactly proportional, because flux is the integral of voltage and nothing else. The quantity that belongs to the core is the 3.500 mWb-turn, which has no frequency in it. A transformer "rated for 50 Hz" is a transformer whose volt-second product was divided by 2π × 50 once.

Saturation

  1. 1 A boundary in volt-seconds
  2. 2 The flux that walks
  3. 3 The inductance the current decides
  4. 4 The walk that stops
  5. 5 The flux the reset leaves behind
5 essays · magnetics 25% compensation: 4.26% regulation, and a resonance at 25.0 Hz. computed by solving, not by drawing. The current a 230 V feeder draws, swept from two hertz to a hundred and fifty, with 25% of the line's reactance cancelled by a series capacitor. The regulation falls from 5.17% to 4.26%, which is what the capacitor was fitted for. What comes with it is a series resonance at 9.59 Hz with the load connected and 24.99 Hz with the far end shorted — the latter being exactly f₀√k = 25.00 Hz, a frequency the compensation fraction chooses on its own. The line's inductance does not appear in it and neither does the voltage. It is always below the fundamental, which is where the machines are.

Series compensation

  1. 1 The reactance cancelled, and the resonance it buys
  2. 2 The drop a series capacitor can reach
  3. 3 Sixty degrees, and the limit compensation stops raising
  4. 4 The reactance the capacitor cannot reach
  5. 5 The peak a sagging receiver keeps
5 essays · power A 100 nF capacitor with 100 mΩ in series, written the other way round. computed by solving, not by drawing. At 100 kHz the series pair and the parallel pair are the same impedance to 8.7e-19 of itself — the arithmetic's floor, not a tolerance — with Rp = 2.533 kΩ against Rs = 0.100 Ω and Cp = 99.996 nF against Cs = 100 nF. Away from it they part company at a rate set by Q = 159.2: the substitution costs one per cent below 48.2 kHz and above 207 kHz, a band of 4.3 to one.

Series-parallel

  1. 1 The same part written two ways
  2. 2 The match with no knob
  3. 3 The efficiency a fixed Q costs
  4. 4 The part that decides the match
  5. 5 The Q a pi network makes a designer choose
5 essays · frequency One step response, computed twice: from the poles, and by walking the network forward. A damping ratio of 0.22, so the overshoot is 49.2%. The two curves are drawn on top of each other; the panel below is the difference between them, which is the trapezoidal rule's error at 500 steps and reaches 1.70e-3 V.

Step response

  1. 1 One step, computed twice
  2. 2 The ringing that belongs to the rule
  3. 3 The phase the rule loses
  4. 4 The rule between the two
  5. 5 The Q a rule can show
5 essays · transients Superposition holds for the solution and not for its square. computed by solving, not by drawing, at 81 settings of the second source. Two ten-volt sources reach one hundred-ohm load through a hundred ohms each. Every node voltage and every branch current is the sum of the two single-source solves to 3.8e-16 of itself — superposition, exactly. The power is not: the difference is 2·Re(I₁·conj(I₂))·R to 1.0e-15, and at 0° between the sources and equal size it is 444.4 mW against the 222.2 mW that adding gives. Adding the two powers is within one per cent of the truth only when one source is below 0.00505 of the other. The slider is the angle between them: at 90° the two curves coincide to the last bit, and at 180° the true power falls to zero while the sum does not.

Superposition

  1. 1 Two solves that add, and the one that does not
  2. 2 The peak that only has a bound
  3. 3 The source that must not be zeroed
  4. 4 The cross terms that outnumber the sources
  5. 5 How many sources the cross terms count
5 essays · networks A step on a series RLC at ζ = 0.079, and the two numbers read off H(s). computed by solving, not by drawing. A 50.3 kHz series RLC driven by a one-volt step, with the capacitor voltage and the inductor voltage drawn together. Two limits of the transfer function are two points of the waveform and neither needs the waveform: H(0) = 1.000000 is where the capacitor ends up, and H(∞) across the inductor is 1.0000, which is what it does at the first instant — the expansion gives 1.000000 for it at t = 0. Here the damping ratio is 0.0791, the response is inside ±2% after 7.6 cycles, and 100.0% of the last twenty-four cycles sit there. The poles are at a real part of -7.91e-2 of ω₀, which is the condition the final-value theorem actually has — not a property of H but of where sY(s) has its poles.

Value theorems

  1. 1 Two numbers without solving for the waveform
  2. 2 The start a step takes from infinity
  3. 3 The threshold a delay is quoted at
  4. 4 The parasitic that moves the start
  5. 5 The delay an inductance does not add
5 essays · transients 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
  5. 5 The rule that reads the sign of the gain
5 essays · power The millivolts in the wire, which are nobody's signal. computed by solving, not by drawing at 145 frequencies. A 10 mm run of one-ounce copper carries the return of a 100 mA load and the reference of a 10 mV sensor. Its 5.00 mΩ and 10.0 nH put 500.0 µV in series with the sensor at low frequency — 5.00% of the reading — rising a decade per decade above 79.6 kHz until at 1.59 MHz the error is the whole signal. The full solve and the interfering current times the shared impedance agree to 4.4e-7. The slider is the length of the shared run: both the resistance and the inductance are proportional to it, so every point on the curve moves down together and the corner stays at 79.6 kHz.

Common-impedance

  1. 1 The millivolts in the wire
  2. 2 Every tooth the same height
  3. 3 One voltage added to two readings
  4. 4 A star is half a millimetre long
4 essays · instruments A 10:1 divider with 12.8 pF across its top resistor. computed by solving, not by drawing. The divider's resistors set a ratio of 0.10000 and its capacitors 0.10000; the step starts at the second and relaxes to the first over 115 µs. The balance R₁C₁/R₂C₂ is 1.0000, and the edge lands 0.00% away from where it settles. At 12.8 pF the two ratios are the same number and the response is flat.

Compensation

  1. 1 A divider with two ratios
  2. 2 The time constant one trimmer cannot reach
  3. 3 The source the hook hides behind
  4. 4 The branches a square root needs
4 essays · instruments

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