Concept

Input impedance — where it appears

The impedance a stage or an instrument presents at its own terminals, which is what the circuit in front of it is loaded by. It is a property of the connection rather than of either part alone, so a reading taken through it measures the pair and not the thing intended.

Named by 6 essays across 3 fields — each of them below, with the objects they name alongside it.

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.

The current that does not reach the input

A teraohm across a board from a fifteen-volt rail is fourteen millivolts of error through a gigohm source, and a humid morning takes that resistance down two decades. A ring held at the input's own potential leaves a nanovolt — proportional to the signal rather than to the rail, so an offset has become a gain error of one part in a billion — and the same wire multiplies the input resistance by the loop gain, which makes it 10¹⁸ Ω at direct current and 10¹² Ω at a megahertz.

instruments · Guarding

Four wires against a 10 MΩ voltmeter. computed by solving, not by drawing at 81 resistances, twice each, with a voltmeter of 10 MΩ and 50 mΩ in every lead. The four-wire error is not zero: it is the voltmeter's own divider, −(R + 2R_lead)/(R + 2R_lead + R_m), which grows with the resistance being measured rather than shrinking. The two-wire error is that same quantity plus the leads, so it passes through zero at 1000 Ω — where the reading is right to 1.8e-12 while the four-wire reading is 0.0100% low — and above 707 Ω the two-wire arrangement is the more accurate of the two at every resistance.

The voltmeter four wires do not remove

A four-terminal measurement is described everywhere as removing the leads from the answer. It moves them. What is left is the voltmeter's own input resistance, and it grows with the resistance being measured rather than shrinking: with a ten-megohm voltmeter and fifty milliohms of lead, the four-wire reading is 0.0100 per cent low at a kilohm, where the two-wire reading is exactly right — 1.8 × 10⁻¹² — because its lead error and its loading error cancel. Above 707 Ω the two-wire arrangement is the more accurate of the two at every resistance.

instruments · Four-terminal

The guard leaves a negative resistance, and it reaches −1.59 kΩ. computed by solving, not by drawing. The magnitude of the conductance a source sees looking into the input, guarded and not, with 100 pF of cable and a 1.00 MHz amplifier. The unguarded input's conductance is positive everywhere — a capacitance to ground and a leakage to a rail are both losses. The guarded one is negative above 0.0404 Hz, and its magnitude rises as the square of frequency: −15.9 MΩ at 10 kHz, −161 kΩ at 100 kHz, −3.18 kΩ at a megahertz. Above the amplifier's gain-bandwidth product it flattens at ωₜ·C, which is −1.59 kΩ. That is the same input the guard raises to 10¹⁸ Ω at direct current, and nothing about the leakage the guard was installed for appears in it: the negative resistance is a product of the amplifier's bandwidth and the cable it is driving.

The sign of what the guard gives back

A guard ring is sold on two numbers and they are both about magnitudes: a teraohm of leakage multiplied to 10¹⁸ ohms, and a hundred picofarads of cable bootstrapped out of the way. The guard is also driving that capacitance with a copy of the input that lags it, and a capacitance driven by a lagging copy of its own voltage takes current out of phase with the voltage across it. What the guarded input presents is a negative conductance rising as the square of frequency — −15.9 megohms at ten kilohertz, −3.18 kilohms at a megahertz, flattening at the gain-bandwidth product times the capacitance — and a faster amplifier makes it worse.

instruments · Guarding

In the stopband the series-first ladder filter is an inductance — 5.24 R₀ at three times the corner — and its dual a capacitance, 0.191 R₀: the same filter, opposite loads. computed by solving, not by drawing. The input impedance of the order-5 0.5 dB Chebyshev ladder filter realised with a series inductor first and with a shunt capacitor first, both terminated in the design resistance R₀, against frequency from a tenth of the corner to ten times it. In the passband both sit near R₀ and are nearly resistive. Above the corner they part: at 1.2 times the corner 1.33 R₀ at 89.4° against 0.749 R₀ at −89.4°; 2 times the corner 3.19 R₀ at 90.0° against 0.314 R₀ at −90.0°; 3 times the corner 5.24 R₀ at 90.0° against 0.191 R₀ at −90.0°; 5 times the corner 8.90 R₀ at 90.0° against 0.112 R₀ at −90.0°. The product of the two is R₀² at every frequency to 2e-15, which is what duality means for an input impedance. The slider is the order.

The load a filter is in its stopband

A fifth-order Chebyshev ladder filter can begin with a series inductor or, as its dual, with a shunt capacitor, and between its design resistances the two transmit identically. In the stopband they are opposite loads: at three times the corner the first is 5.24 R₀ of inductance and the second 0.191 R₀ of capacitance, their product R₀² at every frequency. That stops mattering only while the source is the design resistance. From a tenth of R₀ the shunt-first ladder filter rejects 55.9 dB at three times the corner instead of 64.0, draws 14 times the passband current at five times the corner against the series-first ladder filter's 0.24, and with a lead inductance of a tenth of R₀ over the corner's angular frequency grows a hole in its stopband — −33.8 dB where a matched source gives −54.0. A soft source does the mirror image to the series-first ladder filter. The first element has to be the one the source cannot short or open.

filters · Filter termination

Five readings of one passive ladder network, each taken forwards and backwards. computed by solving, not by drawing; every reading built as its own netlist, never converted from another. Transimpedance (current in, open-circuit voltage out), transadmittance (voltage in, short-circuit current out) and the reverse voltage gain against minus the forward current gain all give a ratio of one, to 1.0e-13. The voltage gain forwards against backwards does not: 1.000 at 100 Hz, 2.129 at 1 MHz. Nor does the current gain.

The readings that come back the same

Reciprocity is usually stated as 'swap the source and the meter and the reading is unchanged', and that is true of some readings and false of others. On one passive ladder network the transimpedance, the transadmittance, and the reverse voltage gain against the forward current gain each come back the same to 10⁻¹³, while the voltage gain forwards and backwards differ by 13.1 times at 82.5 kHz — exactly z₂₂/z₁₁. A 600-to-50-ohm pad has voltage gains of 0.0426 and 0.511 and one power gain both ways. A controlled source breaks all three reciprocal pairs by the same factor.

networks · Reciprocity

A terminated ladder network's output noise three ways: one solve per resistor, one solve from the output, and 4kT·Re{Z}. computed by solving, not by drawing, at 290 K. The line is 4kT·Re{Zout}. The dots sum each resistor's 4kT/R through its own transfer to the output, one solve per resistor. The rings get the same transfers from a single solve with an ampere injected at the output — reciprocity — and agree to 4.4e-16; the sum agrees with Nyquist's expression to 4.4e-16. 4.60 nV/√Hz at 100 Hz, 3.82 at 1 MHz.

The resistance colder than the room

In a passive network the noise at a port is 4kT times the real part of its impedance, and reciprocity is why: the transfer from each resistor to the port equals the transfer from the port to that resistor, so one solve from the output gives every resistor's share and the shares add to Nyquist's expression, here to 4.4 × 10⁻¹⁶. With gain in the network both fail. A 50-ohm termination made by feedback around an amplifier of gain A presents 50.000 Ω and has the noise of a 50-ohm resistor at 290/(1 + A) kelvin — 2.90 K at a gain of 99 — but only where the amplifier still has its gain.

networks · Reciprocity

Named alongside it

The objects these essays reach for when they reach for this one.

Verification Source impedance Bootstrapping Controlled source Guarding Loop gain Model range Reciprocity Transfer impedance Design tradeoff Doubly terminated ladder filter Duality

All concepts