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 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. +4 more
9 essays · applied A 350 Ω bridge, with ideal leads. The straight line is the expression every textbook gives, Vδ/4; the curve is the solve. They part company at half the fractional change — 0.498% at 1.0% — so the tangent is worth one per cent only up to 2.020%. Driving the bridge from a current source instead halves the departure at every point and moves that edge to 4.040%.

Bridge

  1. 1 The bridge that is linear near one point
  2. 2 The leads that are in the bridge
  3. 3 The two leads nobody counts
  4. 4 The carrier a cable allows
  5. 5 The capacitance that is already a difference
  6. +3 more
8 essays · applied 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. +3 more
8 essays · limits 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
  6. +3 more
8 essays · applied 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 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. +3 more
8 essays · noise 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 magnetisation curve with no hysteresis in it. 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 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
  6. +3 more
8 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
  6. +3 more
8 essays · applied Linearising an exponential at 27 °C, and what it costs. The linear model understates the gain by 1% at 7.30 mV and by 10% at 22.8 mV. The thermal voltage at this temperature is 25.9 mV, so "small compared with Vₜ" is not the criterion — 28% of Vₜ is already 1% wrong.

Small-signal

  1. 2 How small is small signal
  2. 3 Where the mechanisms are one mechanism
  3. 4 Three amplitudes, all of them one per cent
  4. 5 Two boundaries removed, and one moved
  5. 6 Symmetry and a resistor together
  6. +3 more
8 essays · limits 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
  4. 4 The source resistance the flicker moves
  5. 5 Two generators with one cause
  6. +2 more
7 essays · noise Five resistances, five corner frequencies, and one total on the capacitor. computed by solving, not by drawing. The noise density at a 1 pF capacitor charged through 100 Ω, 1 kΩ, 10 kΩ, 100 kΩ and 1 MΩ, at 290 K. The densities are 100 times apart and the noise bandwidths 1.0e+4 times apart, in opposite directions, so the area under every curve is the same: 63.2762 µV against 63.2762 µV, and √(kT/C) is 63.2762 µV. The resistance has cancelled out of the answer, and the reason is that ½C⟨v²⟩ is the ½kT a degree of freedom in contact with a bath holds — which no arrangement of resistors can change. The claim is about the whole frequency axis and nothing less: inside a 15.9 MHz band the same five networks give 5.05 µV to 63.07 µV, a factor of 12.5.

kT/C noise

  1. 1 The total that has no resistor in it
  2. 2 The noise a clock does not make
  3. 3 The amplifier inside the sample
  4. 4 The sample that is subtracted
  5. 5 The average that is not N times quieter
  6. +2 more
7 essays · noise 10.0 cm of track, solved as a lumped circuit and as a line. The two agree to 0.030% at 3.97 MHz, where the track is one degree long, and to 30.1% at 143 MHz, where it is a tenth of a wavelength. Above that the lumped model is not approximately right; it is describing a different object.

Lumped element

  1. 2 Kirchhoff's own frequency
  2. 3 The boundary that is a rise time
  3. 4 The load that rings with the line
  4. 5 How long a track may be
  5. 6 The heavier input that allows a longer track
  6. +2 more
7 essays · limits 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. +2 more
7 essays · noise What a mistuned arm leaves at 1300 Hz. computed by solving, not by drawing. Neither curve is the null's depth — the null is still bottomless, it has simply moved — but the depth at the frequency the notch was designed for, which is the number a filter is bought for. One per cent components leave -32.9 dB if both errors go the same way and -78.8 dB if they oppose, a factor of 197 from the same tolerance on the same two parts. The slopes are 20.0 and 40.0 decibels per decade: first order in the error on the product LC, second order in the error on the impedance level.

Null depth

  1. 1 What actually fills a null
  2. 2 The null that is stationary in nothing
  3. 3 The sign the null leaves behind
  4. 4 The peak the amplifier moves
  5. 5 The residue that has only one sign
  6. +2 more
7 essays · filters The droop a zero-order hold imposes at 48 kHz. computed by solving, not by drawing. Holding each sample for a clock period convolves the output with a rectangle, so the spectrum is multiplied by a sinc: -0.143 dB down at a tenth of the sample rate, -0.912 at a quarter and -3.922 at half — which is exactly 20 log(2/π) and contains no design decision at all. The dots are the amplitude of the fundamental read out of the transform of the staircase itself, agreeing with the closed form to 0.008%. There is also half a sample of delay, 10.417 µs here, which is the reason a held reconstruction is not a droopy copy of the signal but a droopy copy that has moved.

Reconstruction

  1. 1 The staircase on the way out
  2. 2 The nulls are where nothing is
  3. 3 Flatness, and the two currencies it is bought in
  4. 4 One knob, and the two exponents it turns
  5. 5 The pole a straight line is worth
  6. +2 more
7 essays · digital Two floors on one axis, and the 50.0 mV between them. computed by solving, not by drawing. The flat line is the Johnson current noise of 1 kΩ, √(4kT/R), which has no current in it. The rising line is shot noise, √(2qI), which has no resistance in it. They cross at 50 µA — and the direct voltage across the resistance there is 49.981 mV, which is 2kT/q and contains neither quantity. The slider moves the resistance over six decades; the crossing moves with it and the voltage at the crossing does not move at all, to the last bit of a double.

Shot noise

  1. 1 The floor a current sets
  2. 2 The junction that is a resistor at zero volts
  3. 3 The resistor in the same loop
  4. 4 The two generators that are one current
  5. 5 The count under the density
  6. +2 more
7 essays · noise 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
  6. +2 more
7 essays · applied What 10 ps of aperture jitter is worth, in bits. computed by solving, not by drawing. Samples are taken at instants displaced by a seeded Gaussian of 10 ps and the error is measured against the same sinusoid sampled exactly. The line is −20 log(2π f × jitter), which the measurement matches to 0.12 dB across three decades. The penalty is exactly twenty decibels a decade of input frequency, because the error is the signal's slope times the timing error and nothing else — so a converter holds 16 bits only up to 199 kHz and 12 bits up to 3.18 MHz. An aperture figure quoted without an input frequency states no resolution at all.

Aperture jitter

  1. 1 A picosecond, read as bits
  2. 2 A floor, or a line
  3. 3 The frequency a band moves at
  4. 4 A jitter that grows with the look
  5. 5 The loop a reference allows
  6. +1 more
6 essays · digital

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