AC analysis and the Bode plot

What AC analysis computes

AC analysis first finds the operating point. It then replaces every nonlinear device by its small-signal model: the tangent at that point, as in E1.2. Finally it solves the resulting linear circuit for a sine wave at each frequency of the sweep.

The answer at each node and frequency is a complex number: a magnitude and a phase. Because the circuit is now linear, nothing clips and nothing distorts, whatever the amplitude. Doubling the source doubles every output, so the source amplitude carries no information.

Write the source as V1 in 0 dc 0 ac 1. The dc 0 part is its value for the operating point, and ac 1 makes it the small-signal input with magnitude 1. Then read vdb(out), the magnitude in decibels (20·log₁₀ of the magnitude), and vp(out), the phase. With set units=degrees the phase prints in degrees; otherwise it prints in radians.

A sweep is written ac dec 1000 1k 100meg: 1000 frequencies per decade from 1 kHz to 100 MHz. Bode plots use a log frequency axis, so points per decade is the natural unit.

One RC section: a pole

The RC low-pass passes low frequencies and blocks high ones. Its transfer function, output over input, is 1/(1 + j·2πf·RC). The corner frequency f_c = 1/(2πRC) = 159.2 kHz is where the magnitude is 1/√2, which is −3.01 dB, and the phase is −45°.

Above the corner the magnitude falls 20 dB per decade, a factor of ten in voltage for every factor of ten in frequency. That single corner with a 20 dB per decade slope is one pole. A Bode plot stacks magnitude in dB and phase against log frequency, so a pole's magnitude reads as two straight lines that bend at f_c.

Two sections load each other

Two equal sections in a row do not give two poles at 159.2 kHz. The second section loads the first, and the transfer function becomes 1/(1 + 3sτ + s²τ^{2}), with s = j·2πf and τ = RC. Solved for −3 dB, that puts the corner at 0.3742·f_c = 59.56 kHz, and the lab measures f3two = 5.956201e+04.

Put an ideal buffer between the sections and the loading disappears. Each section is then a pole at f_c, and two equal poles reach −3 dB at f_c·√(√2 − 1) = 102.4 kHz: the lab measures f3buf = 1.024312e+05. Loading is the first reason a hand analysis and the simulator disagree: the simulator never forgets a load.

AC sees the bias

A diode's small-signal resistance is r_d = n·U_T/I, from E1.2. So a diode can be the resistor of a low-pass, and its DC current sets the corner. The figure biases the same diode at 100 µA and at 10 µA, each with 1 nF.

The formula 1/(2π·r_d·C) gives 615.3 kHz and 61.53 kHz. The 100 µA run lands on it; the 10 µA run is 0.05 % high. That gap comes from the operating point: the solver stopped once the answer moved by less than its 0.1 % tolerance, so the diode current is a little off. With .options reltol=1e-6 the 10 µA corner moves to 61.53 kHz.

Try it

Run it, read the six meas lines, then open Plots for the Bode plots of every node. WHEN vdb(o1)=-3.0103 finds the −3 dB frequency, and FIND ... WHEN reads the phase there.

Now break the buffer: change E1 b3 0 m3 0 1 to a wire by replacing it with R6 m3 b3 0.001. Predict f3buf first. Then change ac 1 to ac 1m and predict which meas lines still print: every gain drops by 60 dB, so a line that looks for −3.0103 dB finds no crossing.

Check yourself

* the RC low-pass driven with ac 5
V1 in 0 dc 0 ac 5
R1 in out 1k
C1 out 0 1n
.control
ac dec 10 10 1meg
meas ac vlow FIND vm(out) AT=10
.endc
.end