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Feedback amplifiers and oscillators

Feedback amplifiers and oscillators are one loop read two ways. With negative feedback ( $A β > 0$ in the mid-band) the closed-loop gain $A / (1 + A β)$ gives up raw gain in exchange for a stable gain, wider bandwidth, lower distortion, and input and output resistances set by the topology. With positive feedback, a loop gain of exactly 1 at zero phase sustains a sinusoidal oscillation; a relaxation oscillator instead switches between two states at a rate set by an RC time constant. These notes show how to identify a topology from a circuit, work out closed-loop gain, resistances, bandwidth and sensitivity, and find an oscillator's frequency and the minimum gain it needs.

4 previous-year questions · 4 solved · 6 practice questions

Negative feedback: the basic loop

A feedback amplifier samples its output $x_{o}$ and passes it through a feedback network with transfer ratio $β$ , so $x_{f} = β x_{o}$ . The result is subtracted from the source: the basic amplifier sees $x_{i} = x_{s} - x_{f}$ and delivers $x_{o} = A x_{i}$ . Here $x$ is a voltage or a current, depending on the topology, and $A$ and $β$ carry whatever units make the product $A β$ dimensionless.

The loop gain $T = A β$ is the gain once round the loop. To find it, set $x_{s} = 0$ , break the loop at the amplifier input and inject a test signal $x_{t}$ : the signal that comes back is $- A β x_{t}$ . Feedback is negative when $A β > 0$ in the mid-band, so that $∣1 + A β ∣ > 1$ and the gain falls. If $∣1 + A β ∣ < 1$ at some frequency, the feedback is positive there. If $1 + A β = 0$ , the circuit oscillates.

The price is the gain itself. Every improvement described below comes by the factor $1 + A β$ , and the gain drops by that same factor.

CLOSED-LOOP GAIN (NEGATIVE FEEDBACK)

A_{f} = \frac{x _{o}}{x _{s}} = \frac{A}{1 + A β}

Convention: $x_{i} = x_{s} - x_{f}$ , $x_{f} = β x_{o}$ . The feedback is negative when $A β > 0$ in the mid-band (in general, when $∣1 + A β ∣ > 1$ ).

LOOP GAIN AND AMOUNT OF FEEDBACK

T = A β, D = 1 + A β, gain reduction = 20 lo g_{10} ∣1 + A β ∣ dB

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

LARGE LOOP GAIN

A_{f} \approx \frac{1}{β} (A β ≫ 1)

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

DESENSITIVITY TO THE AMPLIFIER GAIN

\frac{d A _{f}}{A _{f}} = \frac{1}{1 + A β} \cdot \frac{d A}{A}

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

SENSITIVITY TO THE FEEDBACK NETWORK

\frac{d A _{f}}{A _{f}} = - \frac{A β}{1 + A β} \cdot \frac{d β}{β} \approx - \frac{d β}{β}

Small changes in $β$ with $A$ fixed; the approximation needs $A β ≫ 1$ . The closed-loop gain is only as stable as $β$ . That is why $β$ is built from resistors or capacitors, not from active devices.

EXAM TIP

Books use different sign conventions. With $x_{i} = x_{s} - x_{f}$ the gain is $A / (1 + A β)$ and the circuit oscillates at $A β = - 1$ . With $x_{i} = x_{s} + x_{f}$ the gain is $A / (1 - A β)$ and it oscillates at $A β = + 1$ . The physics is the same in both, so check which convention a question uses before you substitute.

EXAM TIP

Using $1/ β$ when $A β$ is not large costs accuracy. With $A β = 20$ the true gain is $20/21$ of $1/ β$ , which is 4.8% lower.

What negative feedback buys: bandwidth, distortion and noise

Take an amplifier with a single high-frequency pole, $A (s) = A_{M} / (1 + s / ω_{H})$ , and a frequency-independent $β$ . Substituting into $A / (1 + A β)$ gives $A_{f} (s) = \frac{A _{M} / ( 1 + A _{M} β )}{1 + s / [ ω _{H} ( 1 + A _{M} β )]}$ . The result still has a single pole: the mid-band gain falls by $1 + A_{M} β$ and the pole moves up by the same factor. A single low-frequency pole moves down by that factor, so the bandwidth widens at both ends.

UPPER CUTOFF WITH FEEDBACK

A_{M f} = \frac{A _{M}}{1 + A _{M} β}, ω_{H f} = ω_{H} (1 + A_{M} β)

Amplifier with a single high-frequency pole $ω_{H}$ and mid-band gain $A_{M}$ ; $β$ independent of frequency.

LOWER CUTOFF WITH FEEDBACK

ω_{L f} = \frac{ω _{L}}{1 + A _{M} β}

Single low-frequency pole, $A (s) = A_{M} / (1 + ω_{L} / s)$ ; $β$ independent of frequency.

GAIN-BANDWIDTH PRODUCT

A_{M f} ω_{H f} = A_{M} ω_{H}

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.

Negative feedback also reduces nonlinear distortion, and noise or hum, that arise inside the loop after the input, typically in the output stage. Measured at the output, such a disturbance is divided by $1 + A β$ . The comparison is fair only at the same output signal amplitude: the input must be raised by the factor $1 + A β$ to bring the output back up, and the extra gain ahead of the loop that does this is assumed to add no distortion or noise of its own. Noise that enters with the signal at the input is amplified exactly like the signal, so feedback cannot improve the signal-to-noise ratio of the input itself.

DISTORTION AND NOISE GENERATED INSIDE THE LOOP

V_{d, f} = \frac{V _{d}}{1 + A β}, V_{n, f} = \frac{V _{n}}{1 + A β}

$V_{d}$ 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.

EXAM TIP

Feedback does not clean up noise that arrives with the signal. That noise and the signal are scaled by the same $A_{f}$ , so the input signal-to-noise ratio stays the same. Only disturbances injected after the input stage are reduced.

EXAM TIP

With a frequency-independent $β$ , an amplifier with one or two poles (all in the left half-plane) and no zeros stays stable, because the phase of $A β$ reaches $- 18 0^{\circ}$ only as $ω \to \infty$ , where $∣ A β ∣ \to 0$ . Two poles can still cause gain peaking in the closed-loop response. With three or more poles the phase passes $- 18 0^{\circ}$ at a finite frequency. If $∣ A β ∣ \geq 1$ there, the negative-feedback amplifier oscillates.

GAIN MARGIN AND PHASE MARGIN

GM = - 20 lo g_{10} ∣ A β (j ω_{180}) ∣ dB, PM = 18 0^{\circ} + ∠ A β (j ω_{1})

Negative-feedback convention, with $∠ A β = 0^{\circ}$ in the mid-band. $ω_{180}$ is the frequency at which $∠ A β = - 18 0^{\circ}$ , and $ω_{1}$ the frequency at which $∣ A β ∣ = 1$ . For an amplifier whose open-loop poles all lie in the left half-plane and whose $∣ A β ∣$ falls through 1 only once, the feedback amplifier is stable when both margins are positive.

The four feedback topologies

At the output, the feedback network either samples the output voltage (it is connected in shunt across the load) or the output current (it is connected in series with the load). At the input, it either subtracts a voltage (in series with the source) or a current (in shunt at the input node). The name gives the input mixing first and the output sampling second.

Series connections add impedances, so series mixing raises the input resistance and series (current) sampling raises the output resistance. Shunt connections lower them. The factor is always $1 + A β$ , with $A$ and $β$ of the type that matches the topology. Each topology pushes the amplifier towards its own ideal: a voltage amplifier wants high $R_{in}$ and low $R_{o u t}$ , a transconductance amplifier wants both high, a transresistance amplifier wants both low, and a current amplifier wants low $R_{in}$ and high $R_{o u t}$ . In a real circuit the $β$ network also loads the basic amplifier. The standard method folds that loading into $A$ (the A circuit) and then applies the ideal formulas.

Topology (mixing–sampling)	Older name	Samples	Mixes	Gain type of $A$ and $A_{f}$	Units of $β$	$R_{in}$	$R_{o u t}$
Series-shunt	Voltage-series	Output voltage	Voltage, in series	Voltage gain (V/V)	V/V	× $(1 + A β)$	÷ $(1 + A β)$
Series-series	Current-series	Output current	Voltage, in series	Transconductance (A/V)	V/A = Ω	× $(1 + A β)$	× $(1 + A β)$
Shunt-shunt	Voltage-shunt	Output voltage	Current, at a node	Transresistance (V/A)	A/V = S	÷ $(1 + A β)$	÷ $(1 + A β)$
Shunt-series	Current-shunt	Output current	Current, at a node	Current gain (A/A)	A/A	÷ $(1 + A β)$	× $(1 + A β)$

INPUT RESISTANCE, SERIES MIXING

R_{i f} = R_{i} (1 + A β)

The feedback voltage is in series with the source. $R_{i}$ is the input resistance of the basic amplifier.

INPUT RESISTANCE, SHUNT MIXING

R_{i f} = \frac{R _{i}}{1 + A β}

The feedback current is added at the input node.

OUTPUT RESISTANCE, VOLTAGE (SHUNT) SAMPLING

R_{o f} = \frac{R _{o}}{1 + A β}

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

OUTPUT RESISTANCE, CURRENT (SERIES) SAMPLING

R_{o f} = R_{o} (1 + A β)

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

EXAM TIP

The factor $1 + A β$ applies only to a resistance measured at a point inside the loop. A bias resistor that shunts the input of a series-mixed stage, or the collector resistor $R_{C}$ in parallel with the output of a current-sampled stage, is outside the loop and is combined in parallel afterwards. For example, in a CE stage with unbypassed $R_{E}$ the resistance looking into the collector rises, but the output resistance of the stage is still about $R_{C}$ .

WORKED EXAMPLE · WORKED EXAMPLE 1: SERIES-SHUNT AMPLIFIER

A voltage amplifier has open-circuit voltage gain $A = 2000$ , input resistance $R_{i} = 4 k Ω$ , output resistance $R_{o} = 1 k Ω$ and a single high-frequency pole at 25 kHz. It is placed in a series-shunt negative-feedback loop with an ideal, non-loading $β$ network, no load is connected, and the closed-loop gain is 40. Find (a) $β$ , (b) $R_{i f}$ and $R_{o f}$ , (c) the closed-loop upper 3-dB frequency, and (d) the change in closed-loop gain if $A$ falls by 10%.

Solution

(a) $A_{f} = A / (1 + A β) = 40$ , so $1 + A β = 2000/40 = 50$ , $A β = 49$ and $β = 49/2000 = 0.0245$ .
(b) Series mixing: $R_{i f} = R_{i} (1 + A β) = 4 k Ω \times 50 = 200 k Ω$ . Voltage sampling ( $A$ is the open-circuit gain, as given): $R_{o f} = R_{o} / (1 + A β) = 1000/50 = 20 Ω$ .
(c) Single pole: $f_{H f} = f_{H} (1 + A β) = 25 kHz \times 50 = 1.25 MHz$ . Check: $2000 \times 25 kHz = 40 \times 1.25 MHz = 50 MHz$ .
(d) Differential estimate: $d A_{f} / A_{f} \approx (- 10%) /50 = - 0.2%$ .
Exact: with $A = 1800$ , $1 + A β = 1 + 1800 \times 0.0245 = 45.1$ and $A_{f} = 1800/45.1 = 39.911$ , a change of $- 0.222%$ .

Answer: $β = 0.0245$ ; $R_{i f} = 200 k Ω$ , $R_{o f} = 20 Ω$ ; $f_{H f} = 1.25 MHz$ . A 10% drop in $A$ moves $A_{f}$ from 40 to 39.91, about $- 0.22%$ (the differential estimate gives $- 0.2%$ ).

Identifying the topology from a circuit

Find the feedback network: the element or elements that carry a signal from the output side back to the input side. This may be a resistor $R_{F}$ , an unbypassed emitter or source resistor shared by the input and output loops, or a divider feeding an op-amp input.
Sampling: short the output ( $v_{o} = 0$ ). If the feedback signal disappears, the output voltage is sampled (shunt, or voltage, sampling). Open the output loop ( $i_{o} = 0$ ). If the feedback signal disappears, the output current is sampled (series, or current, sampling). In practice, feedback taken from the output node itself is voltage sampling, and feedback taken from a resistor that carries the output current is current sampling.
Mixing: if the feedback returns to the same node that receives the input signal, currents add at that node and the mixing is shunt. Examples: the base, when the signal drives the base; the inverting input, when the signal enters through $R_{1}$ at the inverting input. If it returns to the other terminal of the input port, the feedback voltage subtracts around the input loop and the mixing is series. Examples: the emitter or source, when the signal drives the base or gate; the inverting input, when the signal drives the non-inverting input.
Check the sign by tracing a small increase at the input around the loop: the feedback must oppose it. For a BJT, feedback to the base must come from a point in antiphase with the base, and feedback to the emitter must come from a point in phase with the base.
Name the topology mixing first and sampling second, then apply the matching $A$ , $β$ and resistance rules.

Circuit	Topology	Feedback factor $β$	Closed-loop result (large loop gain)
Non-inverting op-amp amplifier ( $R_{1}$ to ground, $R_{2}$ feedback)	Series-shunt	$R_{1} / (R_{1} + R_{2})$	$v_{o} / v_{s} = 1 + R_{2} / R_{1}$
Voltage follower, emitter follower, source follower	Series-shunt	$1$	$v_{o} / v_{s} \approx 1$
Two CE stages, $R_{F}$ from $Q_{2}$ collector to $Q_{1}$ emitter ( $R_{E 1}$ to ground)	Series-shunt	$R_{E 1} / (R_{E 1} + R_{F})$	$v_{o} / v_{s} \approx 1 + R_{F} / R_{E 1}$
CE stage with unbypassed $R_{E}$	Series-series	$R_{E}$ (Ω)	$i_{o} / v_{s} \approx 1/ R_{E}$ ( $i_{o}$ = collector current), $v_{o} / v_{s} \approx - R_{C} / R_{E}$
Op-amp voltage-to-current converter (signal at the non-inverting input, load as the feedback element, $R$ from inverting input to ground)	Series-series	$R$ (Ω)	$i_{L} = v_{s} / R$
Inverting op-amp amplifier ( $R_{1}$ input, $R_{2}$ feedback)	Shunt-shunt	$- 1/ R_{2}$ (S)	$v_{o} / i_{s} \approx - R_{2}$ with $i_{s} = v_{s} / R_{1}$ , so $v_{o} / v_{s} = - R_{2} / R_{1}$
CE stage with $R_{F}$ from collector to base, source resistance $R_{s}$	Shunt-shunt	$- 1/ R_{F}$ (S)	$v_{o} / i_{s} \approx - R_{F}$ with $i_{s} = v_{s} / R_{s}$ , so $v_{o} / v_{s} \approx - R_{F} / R_{s}$
Two CE stages, $R_{F}$ from $Q_{2}$ emitter (unbypassed $R_{E 2}$ to ground) to $Q_{1}$ base	Shunt-series	magnitude $R_{E 2} / (R_{E 2} + R_{F})$	$∣ i_{o} / i_{s} ∣ \approx 1 + R_{F} / R_{E 2}$ ( $i_{o}$ = $Q_{2}$ collector current)

CE STAGE WITH UNBYPASSED EMITTER RESISTOR

A_{v} = \frac{v _{o}}{v _{s}} = - \frac{h _{f e} R _{C}}{r _{π} + ( 1 + h _{f e} ) R _{E}} \approx - \frac{R _{C}}{R _{E}}

Series-series feedback. Hybrid-π model with $g_{m} = I_{C} / V_{T}$ ( $V_{T} = 25 mV$ at room temperature if a number is needed) and $r_{π} = h_{f e} / g_{m}$ ; the current gain is written $h_{f e}$ to keep $β$ 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 $(1 + h_{f e}) R_{E} ≫ r_{π}$ and $h_{f e} ≫ 1$ . Input resistance looking into the base: $r_{π} + (1 + h_{f e}) R_{E}$ .

EXAM TIP

Two naming systems are in use. Series-shunt (mixing, then sampling) is the same topology as voltage-series (sampled quantity, then mixing). Likewise series-series = current-series, shunt-shunt = voltage-shunt, and shunt-series = current-shunt. If a question uses the words voltage or current, it is using the older system, and the first word is the output sampling.

Oscillators: the Barkhausen criterion and amplitude control

An oscillator is a feedback loop that supplies its own input. In the positive-feedback convention $x_{i} = x_{s} + x_{f}$ , the closed-loop gain is $A / (1 - A β)$ . At a frequency where $A β = 1$ this gain is infinite, which means an output can exist with no input at all.

CLOSED-LOOP GAIN WITH POSITIVE FEEDBACK

A_{f} = \frac{A}{1 - A β}

Convention: $x_{i} = x_{s} + x_{f}$ . The gain becomes infinite, meaning an output with no input, when $A β = 1$ .

BARKHAUSEN CRITERION

A (j ω_{0}) β (j ω_{0}) = 1 \Leftrightarrow ∣ A β ∣ = 1 and ∠ A β = 36 0^{\circ} n (n = 0, 1, 2, \dots)

The steady-state condition for a linear, sinusoidal oscillation at $ω_{0}$ (positive-feedback convention). The phase condition fixes $ω_{0}$ , and the magnitude condition fixes the minimum gain. For reliable start-up the small-signal $∣ A β ∣$ is made slightly greater than 1.

The first signal comes from noise and the switch-on transient, which contain every frequency. If $∣ A β ∣ > 1$ at the frequency where the loop phase is $0^{\circ}$ , that component grows on every trip round the loop while all others die away. It keeps growing until some nonlinearity reduces the effective gain to $∣ A β ∣ = 1$ . Controlling that step is called amplitude stabilisation.

Design the small-signal loop gain slightly above 1 (for example, a Wien-bridge amplifier gain slightly above 3) so that oscillation starts from noise.
With no deliberate limiter, the amplifier's own saturation (clipping) limits the amplitude, and the output is then distorted.
Diode limiting: two diodes connected in antiparallel (one for each polarity) across part of the feedback resistor conduct on large swings and lower the gain as the amplitude rises.
Thermal limiting (Wien bridge, gain $1 + R_{f} / R_{g}$ ): use an NTC thermistor as $R_{f}$ , or a small lamp filament (whose resistance rises with temperature) as $R_{g}$ . A larger amplitude heats the element and reduces the gain.
Automatic gain control: a JFET works as a voltage-controlled resistor in the gain-setting network, with its gate driven by the rectified and filtered output amplitude.
In steady state the effective (large-signal) loop gain at the oscillation frequency is exactly 1, and the frequency is still set by the phase condition.

EXAM TIP

The frequency comes from the phase condition and the minimum gain from the magnitude condition. Extra gain above the minimum does not set the frequency. It makes the waveform grow until it clips, unless an amplitude limiter is present.

EXAM TIP

Count the amplifier's own phase. An inverting amplifier supplies $18 0^{\circ}$ , so the network must supply another $18 0^{\circ}$ (the RC phase-shift oscillator, and Colpitts and Hartley oscillators built around an inverting stage). A non-inverting amplifier needs a network with $0^{\circ}$ at $f_{0}$ (Wien bridge).

EXAM TIP

A loop gain of exactly 1 is not a design value. Below 1 the oscillation dies away, so the start-up gain must be above 1, and a nonlinearity then brings the effective gain back to 1.

RC oscillators: phase-shift and Wien bridge

RC phase-shift oscillator. A single RC section shifts the phase by less than $9 0^{\circ}$ at any finite, nonzero frequency, and two sections reach $18 0^{\circ}$ only in the limit where their output falls to zero, so at least three sections are needed. With an inverting amplifier supplying $18 0^{\circ}$ , the ladder supplies the other $18 0^{\circ}$ at exactly one frequency. In the op-amp version, the op-amp output (a near-zero source impedance) drives the ladder, and the ladder's third resistor is the inverting amplifier's input resistor, which returns to virtual ground. The ladder therefore sees exactly the load it was analysed with. Op-amps here are ideal: infinite gain, infinite input resistance and zero output resistance.

RC LADDER TRANSFER FUNCTION (THREE IDENTICAL HIGH-PASS SECTIONS)

β (j ω) = \frac{1}{( 1 - 5 α ^{2} ) - j ( 6 α - α ^{3} )}, α = \frac{1}{ω R C}

Series $C$ and shunt $R$ in each section, driven from a zero-impedance source, with the output unloaded (last $R$ to ground or to a virtual ground). The imaginary part vanishes at $α^{2} = 6$ .

OP-AMP RC PHASE-SHIFT OSCILLATOR

f_{0} = \frac{1}{2 π 6 R C}, β (f_{0}) = - \frac{1}{29}, R_{f} \geq 29 R

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.

LOW-PASS LADDER (SERIES R, SHUNT C)

f_{0} = \frac{6}{2 π R C}, β (f_{0}) = - \frac{1}{29}

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

THREE BUFFERED (NON-INTERACTING) RC SECTIONS

f_{0} = \frac{1}{2 π 3 R C} (high-pass), f_{0} = \frac{3}{2 π R C} (low-pass), β (f_{0}) = - \frac{1}{8}

Each section is isolated by an ideal buffer, so it does not load the one before it. Each then supplies exactly $6 0^{\circ}$ at $f_{0}$ and attenuates by $cos 6 0^{\circ} = 1/2$ , so the three give $(1/2)^{3} = 1/8$ and the inverting amplifier needs a gain magnitude of at least 8.

BJT RC PHASE-SHIFT OSCILLATOR

f_{0} = \frac{1}{2 π R C 6 + 4 k}, h_{f e} \geq 23 + \frac{29}{k} + 4 k, k = \frac{R _{C}}{R}

Three high-pass sections feeding the base. Approximate h-parameter model with $h_{r e}$ and $h_{oe}$ neglected and bias resistors neglected; the third resistor is chosen so that it plus $h_{i e}$ equals $R$ ; $R_{C}$ is the collector resistor. The minimum of $23 + 29/ k + 4 k$ is about 44.5, at $k \approx 2.7$ .

Wien-bridge oscillator. The Wien network is a series $R C$ arm followed by a parallel $R C$ arm, with the output taken across the parallel arm. At low frequencies the series capacitor blocks the signal, and at high frequencies the parallel capacitor shorts it. In between, at $ω_{0}$ , the transfer peaks with zero phase. The network is therefore used with a non-inverting amplifier, typically an op-amp with gain $1 + R_{f} / R_{g}$ . The Wien network and the gain-setting divider form the two arms of a bridge.

WIEN NETWORK, GENERAL COMPONENTS

ω_{0} = \frac{1}{R _{1} R _{2} C _{1} C _{2}}, β (ω_{0}) = \frac{1}{1 + \frac{R _{1}}{R _{2}} + \frac{C _{2}}{C _{1}}}

$R_{1}$ 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 $β$ is zero at $ω_{0}$ . Minimum non-inverting gain: $1/ β (ω_{0})$ .

WIEN BRIDGE, EQUAL COMPONENTS

f_{0} = \frac{1}{2 π R C}, β (f_{0}) = \frac{1}{3}, A = 1 + \frac{R _{f}}{R _{g}} \geq 3 (R_{f} \geq 2 R_{g})

$R_{1} = R_{2} = R$ and $C_{1} = C_{2} = C$ , with an ideal non-inverting op-amp amplifier.

WIEN NETWORK RESPONSE, EQUAL COMPONENTS

β (j ω) = \frac{1}{3 + j ( ω R C - \frac{1}{ω R C} )}

A band-pass response: a maximum of $1/3$ at $ω = 1/ (R C)$ , with phase leading below that frequency and lagging above it.

EXAM TIP

Check which way round the ladder is. Series $C$ with shunt $R$ gives $1/ (2 π 6 R C)$ ; series $R$ with shunt $C$ gives $6 / (2 π R C)$ . Both need a gain of 29, but only for three identical, unloaded sections driven from a zero-impedance source. If the sections are buffered from one another, the factor becomes $3$ and the gain needed is 8.

EXAM TIP

The gain of 29 belongs to the phase-shift oscillator, not the Wien bridge. The Wien network needs $A \geq 3$ with equal components, i.e. $R_{f} \geq 2 R_{g}$ . With unequal components, use $A \geq 1 + R_{1} / R_{2} + C_{2} / C_{1}$ .

LC and crystal oscillators

Colpitts and Hartley oscillators share one structure: an inverting amplifier of open-circuit voltage gain $- A_{v}$ ( $A_{v} > 0$ ), a reactance $X_{o}$ from its output to ground, a reactance $X_{i}$ from its input to ground, and a reactance $X_{f}$ from output to input. The ideal analysis assumes the amplifier's input resistance is large enough not to load the network, and that its output resistance is nonzero (the output resistance is what makes the loop frequency-selective). Neither result below depends on the value of that output resistance.

THREE-REACTANCE (LC) OSCILLATOR

X_{o} + X_{i} + X_{f} = 0, β = - \frac{X _{i}}{X _{o}}, ∣ A_{v} ∣_{m i n} = \frac{X _{o}}{X _{i}}

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.

COLPITTS OSCILLATOR

f_{0} = \frac{1}{2 π L C _{e q}}, C_{e q} = \frac{C _{1} C _{2}}{C _{1} + C _{2}}, ∣ A_{v} ∣ \geq \frac{C _{2}}{C _{1}}

$C_{1}$ runs from the amplifier output to ground, $C_{2}$ from the amplifier input to ground, and $L$ between them; $β = - C_{1} / C_{2}$ . The required gain is (input-side capacitance)/(output-side capacitance).

HARTLEY OSCILLATOR

f_{0} = \frac{1}{2 π ( L _{1} + L _{2} + 2 M ) C}, ∣ A_{v} ∣ \geq \frac{L _{1} + M}{L _{2} + M}

$L_{1}$ 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 $∣ A_{v} ∣ \geq L_{1} / L_{2}$ .

CLAPP OSCILLATOR

f_{0} = \frac{1}{2 π L C _{e q}}, \frac{1}{C _{e q}} = \frac{1}{C _{1}} + \frac{1}{C _{2}} + \frac{1}{C _{3}}, f_{0} \approx \frac{1}{2 π L C _{3}} (C_{3} ≪ C_{1}, C_{2})

A Colpitts circuit with a third capacitor $C_{3}$ in series with $L$ , under the same ideal model. When $C_{3}$ is much smaller than $C_{1}$ and $C_{2}$ it sets the frequency, so the device capacitances that appear across $C_{1}$ and $C_{2}$ have little effect on it. The gain condition is unchanged: $∣ A_{v} ∣ \geq C_{2} / C_{1}$ .

Crystal oscillators. A quartz crystal is a piezoelectric resonator. Its electrical equivalent is a series arm (motional inductance $L$ , capacitance $C_{s}$ and a small resistance $R$ ) in parallel with the holder, or shunt, capacitance $C_{p}$ , where $C_{p} ≫ C_{s}$ . The series arm resonates at $f_{s}$ , where the crystal looks like the small resistance $R$ . The loop formed by the series arm and $C_{p}$ resonates at $f_{p}$ , where the impedance is very high. Because $C_{s} ≪ C_{p}$ , $f_{p}$ lies only slightly above $f_{s}$ . Between the two the crystal is inductive; below $f_{s}$ and above $f_{p}$ it is capacitive. Its $Q$ is far higher than that of a practical LC tank, which pins the oscillation frequency very tightly to the crystal's own resonance. The frequency formulas below neglect $R$ .

In parallel mode the crystal replaces the inductor of a Colpitts circuit (the Pierce oscillator) and works between $f_{s}$ and $f_{p}$ , where it is inductive. The exact frequency is set by the capacitance the crystal sees. In series mode the crystal sits in the feedback path, and the loop closes only near $f_{s}$ , where the crystal's impedance is lowest and resistive.

CRYSTAL: SERIES RESONANCE

f_{s} = \frac{1}{2 π L C _{s}}, Q = \frac{ω _{s} L}{R} = \frac{1}{ω _{s} C _{s} R}

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 $ω_{s} = 2 π f_{s}$ .

CRYSTAL: PARALLEL RESONANCE

f_{p} = \frac{1}{2 π L \frac{C _{s} C _{p}}{C _{s} + C _{p}}} = f_{s} 1 + \frac{C _{s}}{C _{p}} \approx f_{s} (1 + \frac{C _{s}}{2 C _{p}})

The motional arm resonating with $C_{s}$ in series with $C_{p}$ ; here the crystal's impedance is at its maximum. The approximation uses $C_{s} ≪ C_{p}$ .

CRYSTAL REACTANCE (R NEGLECTED)

X (ω) = - \frac{ω ^{2} - ω _{s}^{2}}{ω C _{p} ( ω ^{2} - ω _{p}^{2} )}

$Z = j X$ . $X = 0$ at $ω_{s}$ , $∣ X ∣ \to \infty$ at $ω_{p}$ , $X > 0$ (inductive) for $ω_{s} < ω < ω_{p}$ , and $X < 0$ (capacitive) elsewhere.

CRYSTAL WITH LOAD CAPACITANCE

f_{L} = f_{s} 1 + \frac{C _{s}}{C _{p} + C _{L}}

The same expression holds for a load capacitance $C_{L}$ in series with the crystal (where it is the frequency of zero reactance of the combination) and in parallel with it (where it is the frequency of infinite reactance). $C_{L} \to \infty$ gives $f_{s}$ and $C_{L} = 0$ gives $f_{p}$ , so a parallel-mode crystal oscillator runs between them.

Oscillator	Frequency	Amplifier needed	Network phase at $f_{0}$
RC phase-shift (three identical high-pass sections)	$1/ (2 π 6 R C)$	Inverting, $∣ A ∣ \geq 29$	$18 0^{\circ}$
RC phase-shift (three identical buffered high-pass sections)	$1/ (2 π 3 R C)$	Inverting, $∣ A ∣ \geq 8$	$18 0^{\circ}$
Wien bridge (equal $R$ , equal $C$ )	$1/ (2 π R C)$	Non-inverting, $A \geq 3$	$0^{\circ}$
Colpitts ( $C_{1}$ at the output, $C_{2}$ at the input)	$1/ (2 π L C_{1} C_{2} / (C_{1} + C_{2}))$	Inverting, $∣ A ∣ \geq C_{2} / C_{1}$	$18 0^{\circ}$
Clapp ( $C_{3}$ in series with $L$ )	$1/ (2 π L C_{e q})$ , $1/ C_{e q} = 1/ C_{1} + 1/ C_{2} + 1/ C_{3}$	Inverting, $∣ A ∣ \geq C_{2} / C_{1}$	$18 0^{\circ}$
Hartley ( $L_{1}$ at the output, $L_{2}$ at the input)	$1/ (2 π (L_{1} + L_{2} + 2 M) C)$	Inverting, $∣ A ∣ \geq (L_{1} + M) / (L_{2} + M)$	$18 0^{\circ}$
Crystal, parallel mode (crystal as the inductor)	Between $f_{s}$ and $f_{p}$ , set by the load capacitance	As for the LC circuit it replaces	$18 0^{\circ}$ (crystal with the capacitors)
Crystal, series mode (crystal in the feedback path)	$f_{s}$	Depends on the circuit; the crystal passes the feedback only near $f_{s}$	$0^{\circ}$ across the crystal

WORKED EXAMPLE · WORKED EXAMPLE 2: COLPITTS OSCILLATOR

A Colpitts oscillator uses an inverting amplifier of open-circuit voltage gain $- A_{v}$ . Its input resistance is large enough not to load the tank, and its output resistance is nonzero. The tank has $C_{1} = 2 nF$ from the amplifier output to ground, $C_{2} = 8 nF$ from the amplifier input to ground, and $L = 50 μ H$ between output and input. Find the oscillation frequency and the minimum $A_{v}$ . What changes if $C_{1}$ and $C_{2}$ are interchanged?

Solution

The phase condition $X_{C 1} + X_{C 2} + X_{L} = 0$ means $L$ resonates with the series combination $C_{e q} = C_{1} C_{2} / (C_{1} + C_{2}) = (2 \times 8) /10 = 1.6 nF$ .
$f_{0} = \frac{1}{2 π 50 \times 1 0 ^{- 6} \times 1.6 \times 1 0 ^{- 9}} = \frac{1}{2 π \times 2.828 \times 1 0 ^{- 7}} \approx 562.7 kHz$ .
At $f_{0}$ the tank is in parallel resonance, so it presents an open circuit to the amplifier's output resistance and the full amplifier output appears across $C_{1}$ . The tank returns $v_{in} / v_{o} = - X_{C 2} / X_{C 1} = - C_{1} / C_{2} = - 1/4$ , which is a $18 0^{\circ}$ shift.
Barkhausen: $(- A_{v}) (- 1/4) = A_{v} /4 \geq 1$ , so $A_{v} \geq 4 = C_{2} / C_{1}$ .
Interchanged ( $8 nF$ at the output, $2 nF$ at the input): $C_{e q}$ is still $1.6 nF$ , so $f_{0}$ is unchanged. The gain condition is always (input-side capacitance)/(output-side capacitance) $= 2/8$ , so in this ideal model $A_{v} \geq 0.25$ , because the tank now steps the voltage up by 4.

Answer: $f_{0} \approx 562.7 kHz$ and $A_{v, m i n} = C_{2} / C_{1} = 4$ . Interchanging the capacitors leaves $f_{0}$ unchanged and changes the minimum gain to $2/8 = 0.25$ in the ideal model.

EXAM TIP

Colpitts uses the series combination $C_{1} C_{2} / (C_{1} + C_{2})$ , not $C_{1} + C_{2}$ . With $C_{1} = C_{2} = C$ , $C_{e q} = C /2$ and $f_{0} = 1/ (2 π L C /2)$ . Hartley is the other way round: its inductors add in series, $L_{1} + L_{2} + 2 M$ .

EXAM TIP

The two crystal resonances are close together: $f_{p} - f_{s} \approx f_{s} C_{s} / (2 C_{p})$ . The crystal is inductive only in that narrow band, which is why a parallel-mode crystal oscillator's frequency stays between $f_{s}$ and $f_{p}$ .

Relaxation oscillator: the op-amp square-wave generator

Not every oscillator is sinusoidal. In the op-amp square-wave generator (astable multivibrator), $R_{2}$ from the output to the non-inverting input and $R_{1}$ from there to ground make the op-amp a Schmitt trigger with positive feedback, while $R$ from the output to the inverting input and $C$ from the inverting input to ground form a timing network. The output sits at $+ V_{s a t}$ or $- V_{s a t}$ . The capacitor charges towards the output level until its voltage reaches the threshold $\pm β V_{s a t}$ , where $β = R_{1} / (R_{1} + R_{2})$ , and the output then switches. The frequency is set by the RC charging time, not by a phase condition, so the Barkhausen criterion (a condition for linear, sinusoidal oscillation) is not used here.

OP-AMP SQUARE-WAVE GENERATOR

T = 2 R C ln (\frac{1 + β}{1 - β}), β = \frac{R _{1}}{R _{1} + R _{2}}, f = \frac{1}{T}

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

EXAM TIP

The capacitor voltage is a series of exponential segments between $- β V_{s a t}$ and $+ β V_{s a t}$ , not a triangle. A linear triangle wave needs an integrator in place of the RC network.

PREVIOUS-YEAR QUESTIONS

4 questions from the EC papers

Try each question first, then open the answer. We placed each question under a topic ourselves; the exam paper does not label topics.

See these papers in full: GATE ECE 2024 · GATE ECE 2023 · GATE ECE 2018

GATE 2024 EC Q.18

GATE 2024 · ECQ.18 · 1 mark · MCQ

For the closed loop amplifier circuit shown below, the magnitude of open loop low frequency small signal voltage gain is 40. All the transistors are biased in saturation. The current source $I_{S S}$ is ideal. Neglect body effect, channel length modulation and intrinsic device capacitances. The closed loop low frequency small signal voltage gain $\frac{v _{o u t}}{v _{in}}$ (rounded off to three decimal places) is ____.

GATE 2024 EC question 18: A five-transistor CMOS differential amplifier. — Figure from the GATE 2024 EC paper. It is part of the question.

(A)
0.976
(B)
1.000
(C)
1.025
(D)
0.488

Answer and solution

GATE 2024 EC Q.18: answer and solution

GATE 2024 EC Q.50

GATE 2024 · ECQ.50 · 2 marks · MCQ

In the circuit below, the opamp is ideal.

[Figure]

If the circuit is to show sustained oscillations, the respective values of $R_{1}$ and the corresponding frequency of oscillation are _____.

GATE 2024 EC question 50: An op-amp with its inverting (-) input on top and non-inverting (+) input below. — Figure from the GATE 2024 EC paper. It is part of the question.

(A)
$29 R$ and $1/ (2 π 6 R C)$
(B)
$2 R$ and $1/ (2 π R C)$
(C)
$29 R$ and $1/ (2 π R C)$
(D)
$2 R$ and $1/ (2 π 6 R C)$

Answer and solution

GATE 2024 EC Q.50: answer and solution

GATE 2023 EC Q.22

GATE 2023 · ECQ.22 · 1 mark · MCQ

A cascade of common-source amplifiers in a unity gain feedback configuration oscillates when

(A)
the closed loop gain is less than 1 and the phase shift is less than 180°.
(B)
the closed loop gain is greater than 1 and the phase shift is less than 180°.
(C)
the closed loop gain is less than 1 and the phase shift is greater than 180°.
(D)
the closed loop gain is greater than 1 and the phase shift is greater than 180°.

Answer and solution

GATE 2023 EC Q.22: answer and solution

See the question in the paper

Tap to open full size

GATE 2023 EC Q.22: answer and solution

GATE 2018 EC Q.3

GATE 2018 · ECQ.3 · 1 mark · MCQ

A good transimpedance amplifier has

(A)
low input impedance and high output impedance.
(B)
high input impedance and high output impedance.
(C)
high input impedance and low output impedance.
(D)
low input impedance and low output impedance.

Answer and solution

GATE 2018 EC Q.3: answer and solution

See the question in the paper

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GATE 2018 EC Q.3: answer and solution

PRACTICE

6 questions to test yourself

PRACTICE 12 marks · NAT

A negative-feedback amplifier has $β = 0.02$ and open-loop gain $A = 5000$ . Through ageing, $A$ falls by 20% to 4000 while $β$ is unchanged. What is the resulting percentage decrease in the closed-loop gain, rounded to two decimal places?

Answer and solution

Answer: $0.25$ (accepted range $\pm 0.01$ )

Before: $1 + A β = 1 + 5000 \times 0.02 = 101$ , so $A_{f} = 5000/101 = 49.505$ .
After: $1 + A β = 1 + 4000 \times 0.02 = 81$ , so $A_{f} = 4000/81 = 49.383$ .
Decrease $= (49.505 - 49.383) /49.505 \times 100 = 0.247% \approx 0.25%$ .
The differential formula would give $20%/101 = 0.198%$ . It underestimates here because a 20% change is not small.

PRACTICE 21 mark · NAT

An op-amp square-wave generator has $R = 10 k Ω$ and $C = 0.1 μ F$ in its timing network (from the output to the inverting input, and from the inverting input to ground), and equal resistors $R_{1} = R_{2}$ in the divider that feeds the output back to the non-inverting input. The op-amp is ideal and saturates symmetrically at $\pm V_{s a t}$ . What is the output frequency, in Hz, rounded to the nearest integer?

Answer and solution

Answer: $455$ (accepted range $\pm 2$ )

$β = R_{1} / (R_{1} + R_{2}) = 1/2$ , and $R C = 1 0^{4} \times 1 0^{- 7} = 1 ms$ .
$T = 2 R C ln \frac{1 + β}{1 - β} = 2 \times 1 ms \times ln 3 = 2 \times 1.0986 ms = 2.197 ms$ .
$f = 1/ T = 1/ (2.197 \times 1 0^{- 3}) \approx 455 Hz$ .

PRACTICE 31 mark · MCQ

A single-stage BJT common-emitter amplifier has an emitter resistor $R_{E}$ that is not bypassed. The signal is applied at the base and the output is taken at the collector, and the transistor's finite output resistance $r_{o}$ is included. Treating $R_{E}$ as the feedback network, what are the feedback topology and its effect on the input resistance (looking into the base) and the output resistance (looking into the collector, excluding $R_{C}$ )?

(A)
Series-series; input resistance increases, output resistance increases
(B)
Series-shunt; input resistance increases, output resistance decreases
(C)
Shunt-series; input resistance decreases, output resistance increases
(D)
Shunt-shunt; input resistance decreases, output resistance decreases

Answer and solution

Answer: (A)

Output side: the collector current $i_{c} \approx i_{e}$ flows through $R_{E}$ , so $v_{f} = i_{e} R_{E}$ is proportional to the output current. Opening the output loop ( $i_{o} = 0$ ) removes $v_{f}$ , so the sampling is current (series) sampling.
Input side: $v_{f}$ appears at the emitter, so around the input loop $v_{b e} = v_{s} - v_{f}$ . The feedback voltage is in series with the source, so the mixing is series mixing.
Series-series: series mixing raises $R_{in}$ (to about $r_{π} + (1 + h_{f e}) R_{E}$ looking into the base), and current sampling raises the resistance looking into the collector from $r_{o}$ to about $r_{o} [1 + g_{m} (R_{E} ∥ r_{π})]$ with the base driven from a zero-resistance source.
Answer: (A).

PRACTICE 41 mark · NAT

An amplifier has a mid-band gain of 1000 and a single high-frequency pole at 50 kHz. Frequency-independent negative feedback is applied, which reduces the mid-band gain to 25. What is the upper 3-dB frequency of the feedback amplifier, in MHz?

Answer and solution

Answer: $2$ (accepted range $\pm 0.02$ )

$1 + A_{M} β = A_{M} / A_{M f} = 1000/25 = 40$ .
Single pole: $f_{H f} = f_{H} (1 + A_{M} β) = 50 kHz \times 40 = 2 MHz$ .
Check: the gain-bandwidth product is unchanged, $1000 \times 50 kHz = 25 \times 2 MHz = 50 MHz$ .

PRACTICE 52 marks · MSQ

Which of the following statements is/are correct? Op-amps are ideal, and each oscillator is in steady state.

(A)
An op-amp RC phase-shift oscillator built with three identical high-pass sections (series $C$ , shunt $R$ ), whose last resistor returns to the op-amp's virtual ground, oscillates at $f_{0} = 1/ (2 π 6 R C)$ and needs an inverting gain of magnitude at least 29.
(B)
A Colpitts oscillator with $C_{1} = C_{2} = C$ and inductor $L$ oscillates at $f_{0} = 1/ (2 π 2 L C)$ .
(C)
An ideal quartz crystal (motional arm $L$ , $C_{s}$ , shunt capacitance $C_{p}$ , resistance neglected) has an inductive reactance at frequencies between its series and parallel resonant frequencies.
(D)
A Wien-bridge oscillator with equal resistors $R$ and equal capacitors $C$ needs a non-inverting amplifier gain of at least 29.

Answer and solution

Answer: (A), (C)

(A) True. For three identical high-pass sections, $β = 1/ [(1 - 5 α^{2}) - j (6 α - α^{3})]$ with $α = 1/ (ω R C)$ . The imaginary part vanishes at $α^{2} = 6$ , which gives $f_{0} = 1/ (2 π 6 R C)$ and $β = 1/ (1 - 30) = - 1/29$ , so $∣ A ∣ \geq 29$ .
(B) False. $C_{e q} = C \cdot C / (2 C) = C /2$ , so $f_{0} = 1/ (2 π L C /2) = 2 / (2 π L C)$ , not $1/ (2 π 2 L C)$ .
(C) True. $X = - (ω^{2} - ω_{s}^{2}) / [ω C_{p} (ω^{2} - ω_{p}^{2})]$ is positive (inductive) for $ω_{s} < ω < ω_{p}$ .
(D) False. With equal components the Wien network gives $β = 1/3$ at $f_{0} = 1/ (2 π R C)$ , so the amplifier needs $A \geq 3$ .
Answer: (A) and (C).

PRACTICE 62 marks · MCQ

A Wien-bridge feedback network has a series arm of $R$ in series with $C$ and a shunt arm of $R$ in parallel with $2 C$ . Its output is taken across the shunt arm, and it drives an ideal non-inverting amplifier of gain $1 + R_{f} / R_{g}$ (infinite input resistance, zero output resistance). What are the oscillation frequency $ω_{0}$ and the minimum amplifier gain?

(A)
$ω_{0} = \frac{1}{2 R C}$ , minimum gain 4
(B)
$ω_{0} = \frac{1}{2 R C}$ , minimum gain 3
(C)
$ω_{0} = \frac{2}{R C}$ , minimum gain 4
(D)
$ω_{0} = \frac{1}{R C}$ , minimum gain 3.5

Answer and solution

Answer: (A)

With $R_{1} = R_{2} = R$ , $C_{1} = C$ (series arm) and $C_{2} = 2 C$ (shunt arm): $β = \frac{Z _{p}}{Z _{s} + Z _{p}} = \frac{s C _{1} R _{2}}{1 + s ( R _{1} C _{1} + R _{2} C _{2} + R _{2} C _{1} ) + s ^{2} R _{1} R _{2} C _{1} C _{2}}$ .
The phase is zero when the real part of the denominator vanishes: $ω_{0}^{2} R_{1} R_{2} C_{1} C_{2} = 1$ , so $ω_{0} = 1/ R^{2} \cdot 2 C^{2} = 1/ (2 R C)$ .
At $ω_{0}$ : $β = \frac{R _{2} C _{1}}{R _{1} C _{1} + R _{2} C _{2} + R _{2} C _{1}} = \frac{1}{1 + R _{1} / R _{2} + C _{2} / C _{1}} = \frac{1}{1 + 1 + 2} = \frac{1}{4}$ .
Barkhausen: $A β = 1$ needs $A \geq 4$ , i.e. $R_{f} \geq 3 R_{g}$ . Answer: (A).

OFFICIAL GATE 2027 SYLLABUS WORDING

Copied from the GATE 2027 syllabus; see all 2027 syllabi.

GATE EC

Schmitt trigger and oscillators

GATE EE

oscillators and feedback amplifiers

GATE IN

feedback … oscillators

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