Formula sheet

Every formula this tool uses, in the order the tool applies them: what makes a loop oscillate at all, then where each topology gets its phase and what that costs in gain, then what the op-amp's own lag does to the loop and how the component values are retuned for it, then the amplitude control. Each one is derived from the one before it, and the ones that matter were checked against LTspice running the exported circuits.

What makes a loop oscillate

Barkhausen's condition

An oscillator is an amplifier of gain A whose output is fed back to its own input through a network that returns a fraction beta. There is no input signal: whatever noise is already present goes once round the loop and comes back multiplied by A beta. It grows only if what comes back is at least as large as what left, and in step with it.

A\,\beta = 1 \quad\Longleftrightarrow\quad |A\,\beta| = 1 \ \text{ and } \ \angle(A\beta) = 0^\circ \ (\text{mod } 360^\circ)

How to use: the phase half picks the frequency, because only at one frequency does the network turn the signal by the right amount. The magnitude half then says what gain the amplifier must supply at that frequency. Both come out of the same network, so the two cannot be chosen independently: pick the network, and the required gain is decided.

\beta(f_0) = \dfrac{1}{A} \ \Rightarrow\ A = \dfrac{1}{|\beta(f_0)|}

Excess gain, and why it is needed

A gain of exactly 1/beta is a knife edge: a hair under and any oscillation decays, a hair over and it grows without bound. No real circuit holds a knife edge, so the amplifier is set deliberately high by a small excess, and a nonlinearity pulls the gain back down once the amplitude has arrived. The growth that excess buys depends on the network's sharpness: a Wien bridge, whose network has a Q of only a third, grows fast on a small excess.

A_{start} = (1 + \epsilon)\,A_{required}, \qquad \text{Wien bridge: } (sRC)^2 + (3 - A)\,sRC + 1 = 0 \ \Rightarrow\ \text{growth per cycle} = e^{\pi (A - 3)} - 1

How to use: the excess is what guarantees start-up over part tolerance and temperature; the nonlinearity that removes it again is what sets the distortion, and is the real design problem in an oscillator. Section 6 is where that happens. The tool reports the excess the rounded parts really give, not the one asked for.

Wien bridge

The Wien network

A resistor and capacitor in series feeding a resistor and capacitor in parallel to ground. The series arm blocks low frequencies and the parallel arm shorts high ones, so what gets through peaks in between; the network is a voltage divider between the two impedances.

\beta(s) = \dfrac{Z_p}{Z_s + Z_p}, \qquad Z_s = R + \dfrac{1}{sC}, \qquad Z_p = \dfrac{R}{1 + sRC}

Substituting both and collecting terms, with s = j omega:

\beta(j\omega) = \dfrac{1}{3 + j\left(\omega RC - \dfrac{1}{\omega RC}\right)}

How to use: the imaginary part is the whole design. It vanishes when omega RC = 1, and only there is the output exactly in phase with the input.

Frequency and gain

At omega RC = 1 the denominator is a real 3, so the network passes exactly a third of the signal with no phase shift at all.

\omega_0 = \dfrac{1}{RC}, \qquad f_0 = \dfrac{1}{2\pi RC}, \qquad \beta(f_0) = \dfrac{1}{3} \ \Rightarrow\ A = 3

A gain of 3 is the lowest of any topology here, and it is a non-inverting gain because the network adds no phase, so the amplifier is the ordinary non-inverting pair:

A = 1 + \dfrac{R_f}{R_g} = 3 \ \Rightarrow\ R_f = 2\,R_g

How to use: R_g is a free choice here (10 k in this tool), unlike the ladder topologies where it is forced. Section 7 shows why a gain of 3 rather than 29 is the single most useful number on this page.

RC phase-shift ladder

What the ladder has to supply

An inverting amplifier already turns the signal by 180 degrees, so the network only has to find another 180 for the loop to close in phase. One high-pass RC section approaches 90 degrees but never reaches it, so at least three sections are needed; n sections share the 180 degrees equally.

\text{one high-pass section: } \dfrac{j\omega RC}{1 + j\omega RC}, \qquad \angle = 90^\circ - \arctan(\omega RC)

n \ \text{sections} \Rightarrow \ \angle_{each} = \dfrac{180^\circ}{n} \ \Rightarrow\ \omega_0 RC = \tan\!\left(90^\circ - \dfrac{180^\circ}{n}\right)

How to use: high-pass sections, not low-pass. That choice is what lets the last shunt element be a resistor, which is the point of the block after next.

Buffered: the sections multiply

A unity-gain follower between the sections stops each from loading the one before, so the sections really are independent and the ladder transfer is the product of n identical ones. Each contributes a magnitude of cos(180/n) at the phase-correct frequency.

|\beta| = \left[\cos\!\left(\dfrac{180^\circ}{n}\right)\right]^{n} \ \Rightarrow\ A = \dfrac{1}{|\beta|}

n = 3:\ \omega_0 RC = \tan 30^\circ = \dfrac{1}{\sqrt{3}},\ A = 8 \qquad n = 4 \ (\text{Bubba}):\ \omega_0 RC = 1,\ A = 4

How to use: three buffered sections cost three op-amps (two followers and the gain stage) for a gain of 8; four sections cost four op-amps for a gain of 4, better frequency stability and quadrature taps, which is why the Bubba is the better of the two buffered versions.

Unbuffered: the ladder has to be solved as one network

Without buffers each section loads the one before, and the product of three transfers is simply wrong. The ladder is a tridiagonal network: with x = omega RC, every node has a series capacitor of admittance jx to each neighbour and a shunt resistor of admittance 1 to ground. This tool solves that system numerically at every frequency rather than quoting a per-section result.

\text{node } k: \quad -jx\,v_{k-1} + \big(1 + jx\,m_k\big)v_k - jx\,v_{k+1} = 0, \qquad m_k = \begin{cases} 2 & k < n \\ 1 & k = n \end{cases}

v_0 = 1, \qquad \beta(x) = v_n, \qquad \Im\{\beta(x_0)\} = 0 \ \Rightarrow\ x_0 = \omega_0 RC, \quad A = \dfrac{1}{|\beta(x_0)|}

Solved for three unbuffered sections, that comes out exactly at:

n = 3 \ \text{unbuffered}: \quad x_0 = \dfrac{1}{\sqrt{6}} \approx 0.408, \qquad \beta(x_0) = -\dfrac{1}{29} \ \Rightarrow\ A = 29

How to use: Texas Instruments SLOA060 analyses this circuit as though the sections did not load each other and predicts a gain of 8, then measures 27. The exact solve above gives 29, which is the number to build to. Where a source's simplification and its own measurement disagree, the measurement is the one to trust.

R_g is the last ladder resistor

The detail that is easiest to get wrong. The last shunt resistor of the ladder is the inverting amplifier's own input resistor R_g, and its far end sits at the virtual ground, which is why the ladder above needs no load term and why the textbook constants come out exact.

R_g = R, \qquad A = \dfrac{R_f}{R_g} = \dfrac{R_f}{R} \ \Rightarrow\ R_f = A\,R

How to use: hanging a separate R_g on the end of a complete ladder instead adds a shunt conductance to the last node, and the required gain for three unbuffered sections climbs from 29 to nearly 40. Setting the load admittance to zero in the node equation above is exactly the statement that R_g is the ladder's own last resistor.

Quadrature loop

Two integrators and an inverter

An integrator delays a sine by exactly a quarter cycle at every frequency, not just at one. Two in series make half a cycle, which is a sign flip; an inverter flips it back and the loop is closed in phase at any frequency at all. What picks the frequency here is the magnitude condition alone.

\text{integrator: } \dfrac{V_{out}}{V_{in}} = -\dfrac{1}{sRC} \ \Rightarrow\ \text{two of them: } \dfrac{1}{(sRC)^2} = -\dfrac{1}{(\omega RC)^2} \ \text{ at } s = j\omega

\dfrac{1}{(\omega_0 RC)^2} = 1 \ \Rightarrow\ \omega_0 = \dfrac{1}{RC}, \qquad f_0 = \dfrac{1}{2\pi RC}, \qquad A = 1

How to use: a gain of 1 is the least any topology here asks of the op-amp, so this reaches the highest frequency of the group, and the two integrator outputs are a true sine and cosine pair by construction rather than by trimming. It is the Tow-Thomas two-integrator loop with the damping resistor removed, so the poles sit on the imaginary axis instead of just inside it. The price is three op-amps.

Why gain does nothing here, and damping does everything

Give the inverter a gain g instead of 1 and the characteristic equation still has purely imaginary roots: the frequency rises as the square root of g, and the amplitude neither grows nor decays. So "excess gain to start" is meaningless for this loop, and a limiter that only compresses gain (diodes across the inverter's feedback resistor) cannot hold its amplitude either: LTspice shows it climbing without end. The only thing that moves these poles off the axis is damping across an integrator capacitor.

(sRC)^2 + g = 0 \ \Rightarrow\ s = \pm j\,\dfrac{\sqrt{g}}{RC} \quad \text{(purely imaginary for any } g)

A resistor R_n from the inverter output into the second integrator's input puts a conductance across that capacitor whose sign is negative, because the inverter output is minus the integrator output. That is a designed start-up growth:

(sRC)^2 - \dfrac{R}{R_n}\,(sRC) + 1 = 0 \ \Rightarrow\ \sigma = \dfrac{1}{2 R_n C}, \qquad \text{growth per cycle} = e^{\pi R / R_n} - 1 \ \Rightarrow\ R_n = \dfrac{\pi R}{\ln(1 + \text{growth})}

How to use: 10 percent per cycle wants R_n about 33 R. The clamp that stops the growth is a positive conductance in the same place that appears only above a threshold; it is in section 6, and the op-amps' own lag, which is a small negative damping of its own, is folded into the same balance.

From frequency to components

One constant per topology, then the retune

Every topology above ends with the same shape of answer: a dimensionless constant k = omega_0 RC that the network fixes, and a required gain. That gives the textbook R C product. The tool then retunes it for the op-amp's lag (section 7) before choosing any part, so the product it actually uses is a few percent smaller at high frequency.

RC_{textbook} = \dfrac{k}{2\pi f_0}, \qquad k = \begin{cases} 1 & \text{Wien, quadrature, Bubba} \\ 1/\sqrt{6} & \text{3 sections, unbuffered} \\ 1/\sqrt{3} & \text{3 sections, buffered} \end{cases}, \qquad RC = (1 + r)\,RC_{textbook}

How to use: r is the retune, negative, found by solving the loop with the op-amp model in it until its zero-phase frequency lands on the target. It is a few tenths of a percent at 1 kHz with a 3 MHz op-amp, and about minus 8 percent for a Wien bridge at 55 kHz.

Picking C, then solving for R

Capacitors come in far fewer values than resistors, so C is chosen first from whatever series is stocked and R is solved from it, then rounded to the nearest available resistor. Every candidate capacitor is tried and the pair scored, with the frequency error that the rounding leaves weighted far above how comfortable R is to buy.

R_{target} = \dfrac{RC}{C}, \qquad 1\,\text{k} < R_{target} < 1\,\text{M}, \qquad 100\,\text{pF} \le C \le 1\,\mu\text{F}

\text{score} = 100\ln^2\!\left(\dfrac{R}{R_{target}}\right) + 0.05\ln^2\!\left(\dfrac{R}{10\,\text{k}}\right) \ \rightarrow\ \text{minimize}

How to use: the two logarithms are ratios, so the score does not care about absolute size, and the factor of 2000 between them means a comfortable R is only ever a tie-breaker. The loop is then solved once more with the rounded parts, and the frequency the tool prints is that one, not the textbook value of the parts.

Amplitude control

Why something has to hold the amplitude

An amplifier set above the gain the loop needs makes the amplitude grow until the op-amp reaches its supply rails. It then clips, and clipping is a gain below 1 for the peaks, so the circuit does settle: it settles into a rounded square wave. SLOA060 measured 2.8 percent distortion on a Wien bridge doing exactly that. Something gentler has to take the gain from A_start back to A_required before the rails do.

A_{start} > A_{required} > A_{limited} \quad \text{(the gain must straddle what the loop needs)}

How to use: this inequality is the one test that decides whether an amplitude control works at all. If A_limited is still above A_required with the limiter fully engaged, nothing brings the amplitude back and the output clips on the rails instead.

The diode is a curve, not a switch: its describing function

Every diode limiter here is sized from the diode's real exponential curve, because at the tens of microamps a limiter runs it a 1N4148 drops 0.48 V at 100 uA and 0.38 V at 10 uA, not 0.6 V, and the hard-switch picture put the settled amplitude 15 to 70 percent off in LTspice. For a sinusoid of amplitude V across an anti-parallel pair, the current it draws, averaged the way the loop averages it, is a conductance: the fundamental of the current divided by V.

i(v) = I_s\left(e^{v/nV_T} - e^{-v/nV_T}\right), \qquad G(V) = \dfrac{1}{\pi V}\int_0^{2\pi} i(V\sin\theta)\,\sin\theta\,d\theta

How to use: G rises steeply with V once V passes a drop or so, which is what makes the amplitude definite. The tool integrates this numerically with the same I_s and n it writes into the LTspice model, so the simulation and the page agree to a few percent.

Diodes across part of the feedback resistor (Wien bridge and the ladders)

The feedback string is R_f1 + R_f2 with the pair across R_f2. With the diodes off it presents R_s = R_f1 + R_f2, which sets the start gain; at balance it must present R_t, what the loop needs; so the diodes' conductance has to bring R_f2 down to x = R_t - R_f1. The voltage across R_f2 is a fixed fraction of the output: (A-1)/A of it for the non-inverting Wien stage, all of it for an inverting stage whose input is a virtual ground.

\dfrac{1}{R_{f2}} + G(V_2) = \dfrac{1}{x}, \qquad V_2 = \text{fraction} \times V_{out} \times \dfrac{x}{R_t}, \qquad x = R_t - R_{f1}

\text{Wien: } R_t = (A_{req} - 1)R_g, \ \text{fraction} = \dfrac{A_{req}-1}{A_{req}}; \qquad \text{ladder: } R_t = A_{req}\,R, \ \text{fraction} = 1

How to use: to size the parts, solve for x at the wanted amplitude (one bisection: the left side rises with x through V_2, the right side falls), then R_f2 = x + (R_s - R_t) and R_f1 = R_t - x. To predict the amplitude of given parts, run it backwards: G(V_2) = 1/x - 1/R_f2 fixes V_2, hence V_out. Because the amplitude hangs on R_t - R_f1, a small difference of two large numbers, R_f1 is taken from the 1 percent series and the pair is chosen among stock neighbours for the closest amplitude.

The clamp on the quadrature loop

A divider R_d1 R_d2 from the cosine output, with the pair from its tap into the second integrator's input. Above the divider's threshold the diodes conduct into the virtual ground, and averaged over a cycle that is a positive conductance across the integrator capacitor. It has to cancel the negative one from R_n and the small extra growth the op-amps' lag contributes:

G_{clamp}(A) = \dfrac{1}{R_n} + 2\,\sigma_{lag}\,C, \qquad G_{clamp}(A) = \dfrac{1}{\pi A}\int_0^{2\pi} i_{clamp}(A\sin\theta)\,\sin\theta\,d\theta

How to use: R_d2 is a round 1 k so the conductance rises steeply above the threshold, and R_d1 is solved so the balance lands on the wanted amplitude. Because the threshold is a resistor ratio and a diode drop, the amplitude does not hang on the exact value of R_n or of the diode current, which a plain series resistor would make it do. The clamp current is integrated before it reaches either output, so its harmonics arrive divided by their order: under a percent of distortion in LTspice.

Incandescent lamp

A small filament lamp in the lower feedback leg does the same job with heat: its resistance rises as the current warms the filament, so a larger output means a larger R_g and a smaller gain. Nothing clips and nothing has a kink in it anywhere, which is why this is the cleanest of the three, under 0.1 percent measured.

A = 1 + \dfrac{R_f}{R_{lamp}} = A_{req} \ \Rightarrow\ R_{lamp,hot} = \dfrac{R_f}{A_{req} - 1}, \quad \text{with } \dfrac{V_{out}}{A_{req}} \text{ across it}

The design cannot set the amplitude, only state what the lamp must do; the exported circuit carries a lamp that does exactly that, so the simulation means something:

R = R_{cold}(1 + \alpha\theta), \qquad C_{th}\dfrac{d\theta}{dt} = \dfrac{V^2}{R} - \dfrac{\theta}{R_{th}}, \qquad R_{cold} = \dfrac{R_{hot}}{3}, \ \theta_{hot} = \dfrac{2}{\alpha}, \ R_{th} = \dfrac{\theta_{hot}}{P_{hot}}, \ C_{th} = \dfrac{\tau}{R_{th}}

How to use: the catch on a bench is thermal: the amplitude drifts with room temperature, the loop takes about a second to settle and can overshoot on switch-on, and at frequencies low enough that one cycle is not short next to the filament time constant the resistance follows the waveform itself and distortion climbs. Lamps of a suitable rating are also no longer easy to buy.

JFET automatic gain control

A JFET with a small drain-source voltage is a resistor whose value the gate sets, in series with a fixed resistor as the lower leg: closing the channel raises the leg and lowers the gain all the way to 1, opening it takes the gain above balance so the loop starts. A peak detector turns the output's negative peaks into a steady negative voltage, a divider scales it, and two equal resistors average it with the drain voltage into the gate, which cancels the channel's curvature exactly.

r_{DS} = \dfrac{1}{2\beta\,(V_{GS} - V_P)}, \qquad A = 1 + \dfrac{R_f}{R_{ser} + r_{DS}}, \qquad V_{GS} = \dfrac{V_c + V_{DS}}{2} \ \Rightarrow\ I_D = 2\beta\left(\dfrac{V_c}{2} - V_P\right)V_{DS}

\text{at balance: } R_{ser} + r_{DS} = \dfrac{R_f}{A_{req} - 1}, \qquad V_c = -(V_{pk} - V_f)\,\dfrac{R_b}{R_a + R_b}, \qquad \tau_{det} = (R_a + R_b)\,C_{det} \approx 50 \text{ cycles}

How to use: the channel at balance is kept to about a tenth of the leg so it sees little voltage, the divider is solved so the gate reaches the V_GS that gives that channel, and the amplitude the loop settles at is where the gain is exactly A_req. The detector's voltage is the only one available, so a JFET whose pinch-off is large next to the output cannot be controlled at all: a J111 needs about 9 V of output, a small generic part about 3 V. The tool says so instead of sizing a leg that cannot work.

The op-amp in the loop

The single-pole op-amp and the lag of a stage

A real op-amp is a single pole: its open-loop gain is A(s) = wt / s with wt = 2 pi GBW, so a stage built on it with noise gain N has a closed-loop pole at GBW / N and lags by atan(f N / GBW) at frequency f. The loss of gain is small; the lag is what matters.

f_{-3dB} = \dfrac{GBW}{N}, \qquad \phi = \arctan\dfrac{f\,N}{GBW}, \qquad N = \begin{cases} 1 + R_f/R_g & \text{non-inverting} \\ 1 + R_f/R_g & \text{inverting} \\ 1 & \text{follower} \end{cases}

How to use: the loop oscillates where its total phase is zero, so any phase the amplifier adds is phase the network must now supply, and it does so by moving off its own zero-phase point. The frequency shifts, the network attenuates differently there, and the gain needed shifts too.

Wien bridge: the closed form

For the Wien bridge the phase condition with the lag included is a quadratic in x = omega RC, and the answer shows how a small lag becomes a large frequency error, because the Wien network's Q is only a third: one degree of lag moves the frequency by about 1.2 percent.

\dfrac{x - 1/x}{3} = -\tan\phi \ \Rightarrow\ x = \dfrac{-3k + \sqrt{9k^2 + 4}}{2}, \quad k = \dfrac{f\,N}{GBW} \quad (\approx 1 - 1.5k \text{ for small } k)

How to use: 55 kHz, N = 3.15 and a 3 MHz op-amp give k = 0.058 and x = 0.918: the textbook values run 8 percent low, which LTspice confirms. The Wien bridge's saving grace is that N is 3; a phase-shift ladder's amplifier runs at N of 30 and lags ten times as much at the same frequency.

Every loop, exactly: the nodal solve

In the ladder oscillators the network is part of the amplifier's own input impedance (the last capacitor and R_g feed the virtual ground), so the amplifier's lag depends on the network and no closed form separates them. The tool writes each loop as a homogeneous nodal system in its node voltages, every conductance in units of 1/R so that p = sRC, with the op-amp as its single-pole model, and opens it where the amplifier output feeds the network, the amplifier's own feedback left intact. The limiter diodes' junction capacitance, 8 pF across R_f2, is in it too; across a 50 k R_f2 that is 8 degrees at 55 kHz.

\text{op-amp row: } \dfrac{s}{\omega_t}\,v_{out} - v_+ + v_- = 0, \qquad L(s) = \dfrac{v_{out}}{v_{in}}\Big|_{\text{loop opened at the network input}}

\text{zero-phase frequency: } \Im\{L(j\omega_z)\} = 0, \quad \text{balance gain: } |L(j\omega_z)| = 1, \quad \text{pole: } \det M(s) = 0, \ s = \sigma + j\omega, \ \text{growth per cycle} = e^{2\pi\sigma/\omega} - 1

How to use: the zero-phase frequency at the start gain is where a diode-limited loop runs (the diodes are off for most of every cycle); the balance gain is what the parts are sized to; the pole's real part says whether it starts. This solve reproduces LTspice's own AC analysis of the open loop to four figures on every topology, which is the check script's job to keep true.

Retune, and where the model stops

The R C product is scaled until the zero-phase frequency lands on the target (a fixed-point iteration, since the frequency scales almost exactly as 1/RC), the parts are rounded, and the loop is solved once more with them. The tool trusts the result while the amplifier's lag stays under about 25 degrees; past that the single-pole picture drifts from what LTspice and a real part do, and it says so.

RC \leftarrow RC \cdot \dfrac{\omega_z}{\omega_{target}} \ \text{ until } \omega_z = \omega_{target}, \qquad \text{trusted while } \phi \le 25^\circ

How to use: the "10 percent point" in the comparison is the frequency at which the textbook values would land 10 percent low with this op-amp, a fair measure of how hard the retune is working. Slew rate is the other ceiling, and the one that bites at large amplitudes rather than small ones:

\left.\dfrac{dV}{dt}\right|_{max} = 2\pi f_0 V_{pk} \ \le \ \dfrac{SR}{2}, \qquad V_{pk} \le V_{swing}

Distortion and output taps

Where the distortion figures come from

A diode limiter bends the peaks in proportion to the excess loop gain it has to absorb, so the tool reports the distortion as a rate per unit of excess, measured on LTspice runs of the exported circuits: about 0.6 percent of third harmonic per percent of excess for the Wien bridge, about 0.1 for the ladders, whose limiter sits behind a network that filters the amplifier's harmonics. The lamp and the JFET control sit outside the signal path and measure 0.1 and 0.2 percent (TI SLOA060, and LTspice agrees); the quadrature clamp's current is integrated before it reaches the outputs, about 0.25 percent at 1 kHz.

\text{THD}_{diodes} \approx \kappa \times \text{excess}, \qquad \kappa \approx 0.6 \ (\text{Wien}), \ 0.1 \ (\text{ladders})

How to use: only the Wien bridge's distortion can genuinely be driven down, because its amplitude control sits in a leg of its own rather than inside the signal path.

Taking the output after a section

In a phase-shift loop each RC section attenuates, so a tap further along the ladder is smaller than the amplifier's output and also cleaner: the limiter's kink is at the amplifier, and every section after it is a filter.

V_{tap} = \dfrac{V_{out}}{A} \qquad \left(\dfrac{1}{29},\ \dfrac{1}{8},\ \dfrac{1}{4} \ \text{for the three ladders}\right)

How to use: the export reports the distortion at the last ladder node as well as at the output, and names the Bubba's quadrature tap. When distortion matters more than amplitude, take the later tap and make up the level with a fixed gain stage after it.