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

Series

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.
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.

Capacitive load

  1. 1 The load that gets inside the loop
  2. 2 The resistor that buys the margin back
  3. 3 The path that buys the error back
  4. 4 What the second path costs at the floor
  5. 5 What the load sees looking back
  6. +5 more
10 essays · feedback Where a switch is a switch: a band, and the 6.43 MHz at which it closes. computed by solving, not by drawing. A switch of 0.5 Ω closed, 100 MΩ open and 5 pF across it is within 1.0% of being ideal only for loads between 49.5 Ω and 1.01 MΩ — 4.31 decades, and both edges are the same part. The upper edge is a frequency as well as a resistance, because the off-capacitance shunts the open switch: it falls a decade per decade above 318 Hz and meets the lower edge at 6.43 MHz, where the band closes and no load at all will do. Checked by scanning every load at 1.3 times that frequency and finding the best possible error to be 1.17%.

Ideal switch

  1. 1 A band rather than an edge
  2. 2 The width no load can change
  3. 3 Where an open switch leaks to
  4. 4 The capacitance a third switch moves
  5. 5 The floor below any load
  6. +3 more
8 essays · limits A major hysteresis loop at 9.0 A/m of coercivity, and the anhysteretic curve it closes onto. computed by solving, not by drawing. The B–H loop of a core driven sinusoidally to ±400 A/m, marched through a superposition of twenty-four play operators and drawn over the single-valued curve the two rungs below this one measured. The loop encloses 12.481 joules per cubic metre per cycle, which is the core loss and which no single-valued model can produce, because a curve has no area. The coercivity is 8.96 amperes per metre and the remanence 22.3 millitesla; both are read off the marched descending branch rather than handed in. The slider takes the threshold spread down to zero, where the two branches become one, the area falls to 9.8e-15 J/m³, and the object is exactly the core the field already had.

Magnetic loss

  1. 1 The area a curve cannot have
  2. 2 The exponent nobody put in
  3. 3 The duty cycle that costs nothing
  4. 4 Two inductances at one current
  5. 5 The current inside the iron
  6. +3 more
8 essays · magnetics What a 0.7 V constant costs, in the quantity it is used to predict. computed by solving, not by drawing by Newton's method on the exponential at 94 supplies through four resistors. The model is exact at 5.748 mA — the current at which the true drop is 0.7 V — and every curve crosses zero there, at four different supplies. Below it the model is low and above it high, and how much depends on the headroom rather than on the diode. Through the 87 Ω curve the drop is 49 mV out at 0.725 V and 147 mV out at 150.7 V — a factor of 3.0 — while the error in the current falls from -66% to 0.10%, a factor of 674, because the headroom underneath it has grown by 2017. On that curve the model is inside one per cent only above 1.12 V.

Diode model

  1. 1 The one current a constant is right at
  2. 2 The constant that is a window
  3. 3 Two currents with one name
  4. 4 The resistance a slow curve cannot see
  5. 5 The logarithm is in the collector
  6. +1 more
6 essays · limits The noise of a 1.6 kΩ resistor through a 10.0 kHz filter. computed by solving, not by drawing. A seeded white sequence of 5.06 nV/√Hz marched through the network gives 619.3 nV across six seeds, spread 1.79%. Integrating the same density against the solved |H(f)|² gives 620.6 nV — -0.21% apart, well inside the spread. The noise bandwidth is 15.03 kHz against a −3 dB point of 10.00 kHz.

Johnson noise

  1. 1 The floor a resistor sets
  2. 2 The loss in front, counted twice
  3. 3 The resistor the noise comes from
  4. 4 Only the real part is warm
  5. 5 The floor that is only a floor while nothing flows
  6. +1 more
6 essays · noise A single pole, and the brick wall that passes the same noise. computed by solving, not by drawing and integrated over 267 frequencies. The equivalent noise bandwidth is 1.5706 times the −3 dB point, and π/2 is 1.5708. A noise voltage computed with the corner frequency instead is 20.2% low. The area under the curve and the area of the rectangle are the same number.

Noise bandwidth

  1. 2 The bandwidth noise sees
  2. 3 The ratio that does not walk to one
  3. 4 The filter an average is
  4. 5 The bandwidth a bin is not
  5. 6 The window the square-root law has
  6. +1 more
6 essays · noise 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.

Regulator

  1. 1 A source below a frequency
  2. 2 Two requirements pulling one capacitor
  3. 3 What gets through from the rail
  4. 4 The ripple that arrives as a comb
  5. 5 The capacitor across the upper resistor
  6. +1 more
6 essays · applied 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 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
5 essays · frequency 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 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
5 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
5 essays · digital A threshold crossed once, in 20k samples of noise. With no hysteresis the comparator changes its mind 22.7 times on average and as many as 29, on a signal that crosses the threshold once. The vertical bars are the range over twelve seeds. The expected number of extra transitions falls below a tenth at 3.57 standard deviations — so the hysteresis a threshold needs is set by the noise under it and not by the signal over it.

Hysteresis

  1. 1 Two thresholds because there is a floor
  2. 2 The period a delay lengthens
  3. 3 The delay that is two delays
  4. 4 The decision taken where the ramp is slowest
  5. 5 The walk the core sees
5 essays · applied Where four of these models stop being true. In order: the ideal operational amplifier at 1.42 kHz, a 10 V output at full amplitude at 7.96 kHz, Kirchhoff's laws on 10.0 cm at 3.97 MHz, the ideal 100 nF capacitor at 4.69 MHz. The fifth boundary is an amplitude rather than a frequency and cannot share this axis: a small-signal model is 1% wrong above 7.3 mV, at every frequency there is.

Model edges

  1. 1 Every model has an edge
  2. 2 The edge that is a region
  3. 3 The edges that move with the room
  4. 4 The edges that are lengths
  5. 5 A boundary is a model and a tolerance
5 essays · limits The closed-loop poles at a gain of 3.05. The locus of the two poles as the amplifier's gain runs from 2.7 to 3.3. It crosses the imaginary axis at a gain of 3.000000 — bisected on the netlist, not quoted — and at 3.05 the real part is 2.500e+2 radians a second, which is an envelope multiplying by 1.17015 every cycle. The crosses are the closed form ω₀(k−3)/2 and they sit on the measured circles.

Oscillator

  1. 1 The gain that is exactly one
  2. 2 The amplitude nothing linear predicts
  3. 3 What the limiter charges for
  4. 4 The frequency that is not the formula
  5. 5 Two exponentials, and where they meet
5 essays · applied 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
5 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 ladder 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
5 essays · networks 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
5 essays · magnetics 1000 µF across a 100 Ω load, rectified from 17 V peak. The output sits at 15.69 V with 1.331 V of ripple, against the 1.569 V the expression I/2fC gives — 15.1% high, because the capacitor is being recharged for part of the cycle rather than discharging throughout it. The lower panel is why: the diode conducts for 28.8° of each half cycle and carries 2.10 A at the peak, which is 13.4 times the 157 mA the load draws.

Unregulated supply

  1. 1 The direct voltage that is a sawtooth
  2. 2 The current that sizes the transformer
  3. 3 The resistance that lowers the ripple
  4. 4 The resistance that belongs in the winding
  5. 5 Two capacitors that are one
5 essays · applied 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
4 essays · digital

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