Feedback & oscillators formulas

Convention: $x_i=x_s-x_f$, $x_f=\beta x_o$. The feedback is negative when $A\beta >0$ in the mid-band (in general, when $|1+A\beta |>1$).

$T$ is the gain once round the loop. Every benefit of negative feedback, and the loss of gain, is by the factor $D$.

Exact form: $A_f=\frac{1}{\beta }\cdot \frac{A\beta }{1+A\beta }$. So $1/\beta$ overstates the true gain by the fraction $1/(A\beta )$; equivalently, the true gain lies below $1/\beta$ by the fraction $1/(1+A\beta )$.

A differential result, valid for small changes in $A$ with $\beta$ fixed; $1+A\beta$ is also called the desensitivity factor. For a large change, work out $A_f$ exactly before and after.

Amplifier with a single high-frequency pole ${\omega }_H$ and mid-band gain $A_M$; $\beta$ independent of frequency.

Single low-frequency pole, $A(s)=A_M/(1+{\omega }_L/s)$; $\beta$ independent of frequency.

Constant for a single-pole amplifier: gain is traded one-for-one for bandwidth. It is not exact when the amplifier has more than one significant pole.

$V_d$ and $V_n$ are the distortion and noise components at the output without feedback, generated inside the loop after the input (for example in the output stage). $V_{d,f}$ and $V_{n,f}$ are the same components with feedback, compared at the same output signal amplitude.

The feedback voltage is in series with the source. $R_i$ is the input resistance of the basic amplifier.

The feedback current is added at the input node.

Here $A$ is the open-circuit (load removed) gain of the basic amplifier, because its output is modelled as a Thevenin source.

Here $A$ is the short-circuit (load shorted) gain of the basic amplifier, because its output is modelled as a Norton source.

Series-series feedback. Hybrid-π model with $g_m=I_C/V_T$ ($V_T=25\text{mV}$ at room temperature if a number is needed) and $r_{\pi }=h_{fe}/g_m$; the current gain is written $h_{fe}$ to keep $\beta$ for the feedback factor; $r_o$ neglected; no source resistance; output at the collector, with $R_C$ the total collector load. The approximation holds when $(1+h_{fe})R_E\gg r_{\pi }$ and $h_{fe}\gg 1$. Input resistance looking into the base: $r_{\pi }+(1+h_{fe})R_E$.

The steady-state condition for a linear, sinusoidal oscillation at ${\omega }_0$ (positive-feedback convention). The phase condition fixes ${\omega }_0$, and the magnitude condition fixes the minimum gain. For reliable start-up the small-signal $|A\beta |$ is made slightly greater than 1.

Three identical high-pass sections, ideal op-amp, third $R$ being the inverting amplifier's input resistor. The inverting gain magnitude $R_f/R$ must be at least 29.

Three identical low-pass sections under the same conditions (zero-impedance drive, unloaded output). The minimum inverting gain is still 29.

Each section is isolated by an ideal buffer, so it does not load the one before it. Each then supplies exactly $60^{\circ }$ at $f_0$ and attenuates by $\cos 60^{\circ }=1/2$, so the three give $(1/2{)}^3=1/8$ and the inverting amplifier needs a gain magnitude of at least 8.

$R_1$ and $C_1$ form the series arm; $R_2$ and $C_2$ (in parallel) form the shunt arm, across which the output is taken. The phase of $\beta$ is zero at ${\omega }_0$. Minimum non-inverting gain: $1/\beta ({\omega }_0)$.

$R_1=R_2=R$ and $C_1=C_2=C$, with an ideal non-inverting op-amp amplifier.

Pure (lossless) reactances. $X_o$ and $X_i$ must be the same kind of reactance, and $X_f$ the opposite kind. Two capacitors and an inductor make a Colpitts oscillator; two inductors and a capacitor make a Hartley oscillator.

$C_1$ runs from the amplifier output to ground, $C_2$ from the amplifier input to ground, and $L$ between them; $\beta =-C_1/C_2$. The required gain is (input-side capacitance)/(output-side capacitance).

$L_1$ is on the output side, $L_2$ on the input side, and $C$ is across both; $M$ is the mutual inductance of a tapped coil (aiding). For separate, uncoupled coils put $M=0$, which gives $|A_v|\ge L_1/L_2$.

Resonance of the motional arm alone; here the crystal's impedance is at its minimum (resistive, equal to $R$). $Q$ is the quality factor of the motional arm, with ${\omega }_s=2\pi f_s$.

The motional arm resonating with $C_s$ in series with $C_p$; here the crystal's impedance is at its maximum. The approximation uses $C_s\ll C_p$.

Ideal op-amp whose output saturates symmetrically at $\pm V_{sat}$ and switches instantly. Each half-period is $RC\ln [(1+\beta )/(1-\beta )]$: the capacitor starts at $\mp \beta V_{sat}$ and charges towards $\pm V_{sat}$. With $R_1=R_2$, $\beta =1/2$ and $T=2RC\ln 3\approx 2.197RC$. The duty cycle is 50%; with unequal saturation levels the two half-periods differ.