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

Series — page 2

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 pair at Q = 0.7071: the sketch is -3.01 dB out at the corner. computed by solving, not by drawing. The solved magnitude against the two straight lines that stand in for it. At a conjugate pair the error at the corner is 20 log Q = -3.010 dB, which is unbounded in both directions, and the worst error anywhere is 3.010 dB at 1.00× the corner. No damping brings it inside 0.770 dB: that is the minimax, at Q = 0.9152, where the corner error and an interior maximum are equal.

Asymptotic approximation

  1. 1 The straight lines, and where they are not the curve
  2. 2 The margin the straight lines report
  3. 3 The gain margin the straight lines get exactly wrong
  4. 4 The corner error a filter hides in its sections
  5. 5 The zero that lifts the lines
  6. +1 more
6 essays · frequency 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.

Cascode

  1. 1 The device that never sees the swing
  2. 2 The source that holds to the supply
  3. 3 The drop the lower device did not need
  4. 4 The noise the cascode device does add
  5. 5 The resistor the upper base is held through
  6. +1 more
6 essays · semiconductors The best shunt drops 7.75 mV, whatever the current is. computed by solving, not by drawing. Two errors on one axis, both from solved networks: the shunt's own drop, which lowers the current that was to be measured, and the amplifier's 5.0 µV of offset divided by the voltage the shunt develops. The first rises with the burden voltage and the second falls, so the worst case has an interior minimum at 7.7460 mV — the geometric mean of the offset and the 12 V supply — where the error is 0.1291%, being twice the root of the offset over the supply. Neither the shunt's resistance nor the current appears in either number: at 1 A the answer is 7.75 mΩ, and at a hundred times the current it is the same burden voltage across a hundredth of the resistance. What does depend on the current is the 7.7 mW the shunt then dissipates, and 40 K of self-heating at 50 ppm/K is 0.2000% on its own. The third curve is the shunt's own Johnson noise in a kilohertz of measurement bandwidth, as a fraction of the current: it is 4.5e-6% at the best burden and is the only line here that moves with the current at all, falling as one over its square root — so above about an ampere it leaves the bottom of these axes entirely and is drawn nowhere rather than flattened onto the floor.

Current sensing

  1. 1 The ammeter that is a resistor
  2. 2 The ammeter that is not in the circuit
  3. 3 The optimum that hands back a bandwidth
  4. 4 The rail that is an input error
  5. 5 The resistance that depends on the reading
  6. +1 more
6 essays · instruments A 4 kHz Butterworth, mapped to 48 kHz two ways. computed by solving, not by drawing. The upper panel is the digital response on the unit circle; the lower one is where each analogue frequency lands. The bilinear transform compresses the axis as it approaches half the sample rate — -3.11% at a tenth of the rate and -15.2% at a quarter — so a design mapped straight through is -3.432 dB at its own corner instead of -3.010. Pre-warping puts that one frequency back exactly and no other: above it the pre-warped curve is the further of the two from the analogue prototype. It is a choice of where to be right, not a correction.

Digital realisation

  1. 1 The corner that moved
  2. 2 The same filter, rounded twice
  3. 3 Zero in, and not zero out
  4. 4 Which section goes first
  5. 5 The word length that is not a threshold
  6. +1 more
6 essays · digital 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
  4. 4 The level the phase cannot know
  5. 5 The two mirrors that change places
  6. +1 more
6 essays · frequency One bit, oversampled — and where the quantisation noise went. computed by solving, not by drawing. A first-order modulator is marched forward one sample at a time with a one-bit quantiser inside the loop, and the noise inside the band is read out of the transform of the error. It falls by 8.99 dB for every doubling of the oversampling ratio — measured 9.33, 10.12, 6.96, 9.54 — against 3.01 dB for plain oversampling, which is drawn beside it from the same starting point. The loop does not make less noise; it moves the noise out of the band, and at a ratio of 128 one bit is worth 9.18.

Noise-shaping

  1. 1 One bit, and where the noise went
  2. 2 The loop that is worse at full scale
  3. 3 Two loops, and the mismatch between them
  4. 4 A floor, or five tones
  5. 5 The levels the loop cannot shape
  6. +1 more
6 essays · digital A 1 V step onto 1.00 m of 50 Ω line into an open circuit. computed by solving, not by drawing as a sum of 81 arrivals. The source drives 0.8333 V into the line immediately — set by 10 Ω against the line's 50 Ω, and not by the load, which it cannot yet know about. One delay of 4.83 ns later the far end reaches 1.6666 V. The staircase settles at 0.999990 V, which is what the resistive divider gives.

Reflections

  1. 1 The staircase in time
  2. 2 The dip whose area is fixed
  3. 3 The via two lines can weigh
  4. 4 The via that is a piece of line
  5. 5 The pad the stub comes out of
  6. +1 more
6 essays · lines The return under 10.0 cm of track, 200 µm above the plane. computed by solving, not by drawing, as an estimate from the geometry: a low-frequency path assumed to be three track-widths wide, 83.3 mΩ, and the parallel-plate inductance µ₀h/w, 125.7 nH, which is the limit for a track much wider than its height. The estimated resistance and reactance are equal at 106 kHz. Solved across the plane instead of assumed, the return does not change path at one frequency: it gathers beneath the track across a band about three decades wide, and this corner falls inside that band. Above the band the loop is the track's length times its height, 20.0 mm², and a milliamp round it at 100 MHz radiates -1.1 dBµV/m at three metres.

Return path

  1. 1 Where the current comes back
  2. 2 The corner that is three decades wide
  3. 3 Where the plane runs out
  4. 4 Two returns in one plane
  5. 5 The slot that does not separate
  6. +1 more
6 essays · lines The reverse peak reaches the forward current at 12.6 A/µs. computed by solving, not by drawing. The peak current a diode conducts backwards while its stored charge is removed, against the rate the external circuit drives its current down. The charge equation is integrated in closed form and marched trapezoidally, and the two agree to a part in a thousand. Drawn over it are the two expressions that describe the limits: √(2·IF·a·τ), which is what a reference gives and which is 16.7% high at this slope, and aτ, which contains no forward current at all and is what the peak approaches when the ramp is slow. At 100 A/µs the junction goes on conducting for 48.3 ns and reaches 3.83 A backwards — 3.83 times the current it was carrying forwards — and the two are equal at 12.6 A/µs.

Reverse recovery

  1. 1 The diode that conducts backwards
  2. 2 The heat a recovery leaves behind
  3. 3 Two loops, and one heatsink
  4. 4 The pulse the heatsink does not feel
  5. 5 Two ladders the terminals cannot tell apart
  6. +1 more
6 essays · transients A difference quotient is best at a step of 1e-5, and is 1e+6 times worse at 10⁻¹¹. computed by solving, not by drawing. The worst disagreement between the adjoint network's derivatives and a central difference quotient of the same quantities, against the fractional step the quotient is taken with, on a 7-element LC ladder filter at 1000 Hz. The curve has a minimum because two errors pull opposite ways: the curvature the quotient neglects falls as the square of the step, and the digits its subtraction destroys rise as one over the step. The best it reaches is 1.4e-9, against the 2.3e-11 that the two-thirds power of the machine epsilon predicts. The exact route costs 2 solves against 15, and has neither error term.

Sensitivity

  1. 1 Every derivative, and the one that is zero
  2. 2 The derivative of a root
  3. 3 The tolerance that can only take away
  4. 4 The three tolerances that do nothing
  5. 5 The direction a response is most sensitive to
  6. +1 more
6 essays · networks The heat a core makes against the heat its path removes, and the two temperatures where they are equal. computed by solving, not by drawing. The rising straight line is what the thermal path can carry away at a temperature — (T − 25)/45 watts, a line because a thermal resistance is a resistance. The curve is what the wound part actually dissipates at that temperature, marched from a hysteresis loop at a material whose saturation flux and permeability both move with temperature. They cross twice. The lower crossing at 88.8 degrees is the operating point and its loop gain is -0.192 — negative, so the core is a stabilising feedback and not a destabilising one. The upper crossing at 191.1 degrees is an ignition temperature: above it the part cannot get rid of what it makes. The slider moves the thermal resistance.

Thermal feedback

  1. 1 The loss that depends on what it causes
  2. 2 The protection that is gone by the second time
  3. 3 The sensor inside its own answer
  4. 4 The boundary that is a starting point
  5. 5 The degrees a thermocouple cannot see
  6. +1 more
6 essays · magnetics A 1:1 transformer at k = 0.99, and the band it is a turns ratio over. computed by solving, not by drawing. Two 10 mH windings coupled at 0.99, driven from 50 Ω into 50 Ω, with 0.5 Ω of winding resistance and 100 pF across the secondary. The response is flat at 0.4901 — which is 98.02% of the 0.5000 an ideal transformer of this ratio would give, and that shortfall is the coupling itself: the flat part is k times the turns ratio, times what the two winding resistances leave of the loop, to four figures at every k on the slider — between 400 Hz and 81.3 kHz, which is 2.31 decades. Both edges are bisected on the solved network. Below the first, the magnetising inductance is a short across the source; above the second, the leakage inductance is in series with the load. The slider moves the coupling, and it moves the upper edge only.

Transformer

  1. 1 The band a turns ratio holds over
  2. 2 What coupling buys, and where it does not
  3. 3 Which picture sets the upper edge
  4. 4 Where the band goes entirely
  5. 5 The inductance that is a shape
  6. +1 more
6 essays · magnetics A photodiode's own capacitance sets the bandwidth, as its -0.50 power. computed by solving, not by drawing. The bandwidth of a 1.0 MΩ transimpedance stage against the capacitance of the diode driving it, with the feedback capacitor at each point bisected to give exactly forty-five degrees of phase margin on the solved loop. The classical expression √(GBW/2π·rf·cd) is drawn over it: the right shape, and conservative by about a fifth at every capacitance. The bandwidth falls as the -0.497 power of the capacitance — a square-root law, so a diode of four times the area costs half the bandwidth rather than three quarters of it. At 30 pF the compensation is 0.598 pF against the expression's 0.725, and the bandwidth 295 kHz against 220.

Transimpedance

  1. 1 Where the trouble is at the input
  2. 2 The factor the expression leaves out
  3. 3 The error a bigger resistor cannot help
  4. 4 The tee that charges for its own compensation
  5. 5 The compensation that costs nothing
  6. +1 more
6 essays · feedback 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
  5. 5 The sensor an integrator does not see
  6. +1 more
6 essays · feedback A 100 ms pulse through a 0.159 Hz corner, 9.52% shorter by the end of it. computed by solving, not by drawing. Marched. The dashed line is the pulse that was sent. The solid line is what a 1 MΩ input with 1 µF in front of it receives: the top decays as exp(−t/RC) for the whole 100 ms, ending 9.515% down, and the trailing edge undershoots by exactly the same amount. The closed form for the same two components gives 9.516%. The input's specification is a corner at 0.159 Hz; a top flat to one per cent needs a pulse shorter than 10.1 ms, which is a rate 625.2 times the corner — a constant with no component in it, and the reason a low-frequency specification says nothing useful about an edge.

AC coupling

  1. 1 The corner that says nothing about an edge
  2. 2 The rejection two capacitors lose at their corner
  3. 3 The second capacitor, and the tail that changes sign
  4. 4 Two coupling stages without a buffer
  5. 5 The ring a blocking capacitor buys
5 essays · instruments The sample rate each anti-alias filter demands for 80 dB. computed by solving, not by drawing. A 20 kHz passband, and each filter must be 80 dB down by the frequency that folds back into it. The required rate follows, and it is a property of the filter rather than of the converter: Bessel 3.53× Nyquist, Butterworth 2.08×, Chebyshev 1.53×. The elliptic design at a selectivity of 0.8 is refused: its equiripple stopband has a floor at -74.1 dB, which is above the requirement at every frequency, so no sample rate satisfies it. At a selectivity of 0.5 the same order needs 1.34×. The floor is the selectivity's, not the order's.

Anti-alias

  1. 1 What the filter in front costs
  2. 2 The floor every filter has
  3. 3 Two thirds of a bit for a factor of twenty-eight
  4. 4 The clock that is too fast
  5. 5 The windows a real filter leaves
5 essays · digital The same millivolts, subtracted — and what four resistors leave behind. 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. The second curve is the same measurement made differentially: a unity-gain difference amplifier across the sensor, its four resistors carrying the worst-case skew 0.1% allows. It divides the error by 500.5 — 53.99 dB against a closed-form (1 + G)/(4·tolerance) of 53.98 dB — at every frequency, because four resistors have no frequency in them. What is left is 998.9 nV, which is not zero: the amplifier subtracts the interference and the resistors decide how much of it survives.

Common-mode rejection

  1. 1 The rejection four resistors decide
  2. 2 The four resistors that decide, and the two that do not
  3. 3 The rejection the parts have
  4. 4 The corner the instrument has no part in
  5. 5 The errors that arrive before the gain
5 essays · instruments How much of the amplifier reaches the answer, at 35° of phase margin. computed by solving, not by drawing by multiplying the forward path's gain by 1 + 10⁻⁵ and solving again. At low frequency 0.0999% of a fractional change in the device reaches the closed-loop gain, which is one over the loop gain of 995.04. It changes sign at 258 Hz — above there a better amplifier gives a smaller gain — and reaches 1.61 at 8.08 kHz, which is worse than having no feedback, and settles at exactly 1 above crossover (5.73 kHz), where there is no loop gain left to spend. The second route, the real part of 1/(1 + T) from the cut loop, agrees to 9.6e-6.

Desensitivity

  1. 1 How much of the amplifier gets through
  2. 2 Half to the amplifier, half to the resistors
  3. 3 The area that feedback only moves
  4. 4 The phase a delay never stops adding
  5. 5 The band a sampled loop repays in
5 essays · feedback 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
  4. 5 The readings that add up and are wrong
  5. 6 The R–2R ladder, built for the rest of itself
5 essays · networks A 1.0% doublet: 0.078 dB in the magnitude, 36× the settling time. computed by solving, not by drawing. Above, the magnitude of a fast circuit followed by a pole and a zero that were meant to cancel and miss by 1.00%, against the same circuit with the cancellation exact: the worst disagreement anywhere up to the fast corner is 0.0777 dB. Below, the error left in the step response, in units of the tail's own amplitude of 0.909%. Settling to 0.10% takes 245.2 fast time constants against 6.9 with the cancellation exact, and the closed form τ·ln(A/B) gives 245.2 — a time that contains nothing of the fast circuit at all.

Doublet

  1. 1 The cancellation that leaves a tail
  2. 2 The slowest one decides it
  3. 3 The tail the source decides
  4. 4 How many relaxations a settling curve holds
  5. 5 The tail with no slowest relaxation
5 essays · transients

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