Associate CET · Competency 11.0

Amplifiers

Competency 11.0 is the amplifier: how a small signal controls a larger one, what each part of a stage does, how much gain and power a transistor stage delivers, and why a meter can lie when you measure one. This page follows ETA's list in order: amplifier types, the parts of a stage, the classes, BJT and FET gain and power, frequency response, preamplifiers and line amplifiers, op-amps, distortion, meter and scope loading, and broadband versus narrowband amplifiers. Section 12 is job practice. Try the numbers in the Transistor & Amplifier calculator and the Op-Amp Gain tool in the Calculator Suite.

On the exam from this page

class A conducts for the whole cycle and is the least efficient; class B conducts for half and needs a push-pull pair; class AB adds a small idle current to remove crossover distortion; class C conducts for less than half and needs a tuned circuit; class D switches
a common-emitter stage inverts the signal; with the emitter resistor bypassed its voltage gain is about R_C divided by r′e, and with it unbypassed about R_C divided by R_E
a FET's gate draws almost no current, so a FET stage has a very high input impedance and a large power gain even when its voltage gain is modest
bandwidth is measured between the two −3 dB (half-power) points
a meter or probe is a resistance in parallel with the point it measures, so on a high-impedance point it reads low

Competency: 11.0
Sections: 12 + self-check
Then: Drill comp 11

What amplifiers do, and where you meet them

Competency 11.1 asks you to list common amplifier devices. An amplifier lets a small input signal control a larger output signal. The extra power comes from the DC supply, not from the input: the input only steers it. The Navy's NEETS course sorts amplifiers two ways.

By what they enlarge. A voltage amplifier makes the output voltage larger than the input voltage. A power amplifier makes the output power larger than the input power. It is usually the final stage, driving the output device, such as a loudspeaker or a transmitting antenna. Small-signal stages come first and power stages last.

Amplifier	Frequency range (NEETS)	Where you meet it
Audio amplifier	15 Hz to 20 kHz	Preamplifiers for microphones and other weak sources, line amplifiers, and power amplifiers driving speakers in PA systems, receivers and two-way radios.
Video amplifier	10 Hz to 6 MHz	Picture signals in cameras, monitors and video distribution. NEETS calls them wide-band amplifiers, and notes they give less gain than narrower-band amplifiers.
RF amplifier	10 kHz to 100,000 MHz	Receiver front ends and IF stages, and transmitter drivers and final stages.

Gain is output divided by input. ETA's sheet gives the decibel forms. Use 20 log for a voltage ratio and 10 log for a power ratio:

Gain: dB = 20 × log₁₀(V_out / V_in) · dB = 10 × log₁₀(P_2 / P_1)

Worked example

Voltage, current and power gain: a stage takes 10 mV RMS into an input impedance of 1 kΩ and delivers 1.0 V RMS into a 10 kΩ load. The voltage gain is 1.0 / 0.01 = 100, or 20 log(100) = 40 dB. The input power is 0.01² / 1000 = 0.1 µW and the output power 1.0² / 10,000 = 100 µW, so the power gain is 100 / 0.1 = 1000, or 10 log(1000) = 30 dB. The current gain is (1.0 / 10,000) / (0.01 / 1000) = 10, and the power gain is the voltage gain times the current gain: 100 × 10 = 1000.

Trap

Exam trap: the voltage-gain and power-gain decibel figures agree only when the input and output impedances are equal. In the example the voltage gain is 40 dB but the power gain is 30 dB, because the output works into ten times the input's impedance.

The parts of an amplifier stage

Competency 11.2 asks for the purpose of each component in an amplifier circuit. The common-emitter stage with voltage-divider bias is the standard example. Every part either sets the DC operating point (the bias) or steers the AC signal.

Part	Its purpose
R1 and R2, the bias divider	Set the base voltage, and so the DC operating point (the Q point). For class A, Kuphaldt sets the bias halfway between cutoff and saturation, so the output can swing both ways without clipping.
R_C, the collector load resistor	The changing collector current flows through it, and the voltage across it is the output signal.
R_E, the emitter resistor	Sets the emitter current from the base voltage, and gives temperature stability (NEETS).
C_E, the emitter bypass capacitor	An open circuit to DC, so R_E still sets the bias, but close to a short for the signal, so R_E no longer cuts the gain. Leave part of R_E unbypassed (a swamping resistor) and the gain falls but the distortion falls too.
C_in and C_out, the coupling capacitors	Pass the AC signal in and out while blocking DC, so the source, this stage's bias and the next stage's bias don't upset each other. They also set the lowest frequency the stage passes (section 6).
The transistor	The active device: a small base current controls a collector current about β times larger.

A FET stage has the same layout with different names: a gate resistor R_G (which sets the input impedance), a drain load R_D, and a source resistor R_S with its own bypass capacitor. Try bias values in the Transistor & Amplifier calculator.

Classes of amplifier

Competency 11.3 asks for the usages and classes of amplifiers. The class says how much of each input cycle the active device conducts for, its conduction angle. That one choice trades fidelity against efficiency. Efficiency is the load power divided by the DC power taken from the supply.

Class	Conducts for	Maximum efficiency (theory)	Used for
A	The whole cycle, 360°	25% with a resistive (RC-coupled) load; 50% with an inductor or transformer load	Small-signal and low-power stages where fidelity matters more than efficiency. It dissipates the most with no signal.
B	Half the cycle, 180°, one device for each half (push-pull)	78.5% (π/4)	Push-pull output stages. Each device is cut off for half the time, but near zero volts neither conducts, which causes crossover distortion.
AB	More than 180°, less than 360°	Between class A and class B	Audio output stages: a small idle current keeps both devices just on, which removes the crossover notch.
C	Less than 180°	Higher than class B, approaching 100% as the angle narrows, though the power it can deliver falls	RF stages, especially high-power transmitters, where a tuned (LC tank) circuit restores the sine wave. Too distorted for audio.
D	Switches fully on or fully off (pulse-width modulation); an LC low-pass filter is often fitted before the speaker to pass the audio and reject the switching frequency	Ideally 100%; ADI quotes about 90% at the onset of clipping (full output)	Audio power amplifiers, especially battery-powered and compact products, where the efficiency means longer battery life and less heat.

The class B figure is π/4 ≈ 0.785, about three times the resistive class A limit. A class D device wastes little because it is either off (no current) or fully on (almost no voltage across it).

Trap

Exam trap: the efficiency figures are theoretical maximums at full output. At lower output every class does worse: ADI's example amplifier, delivering 1 W from a 10 W design, which is closer to normal listening, runs about 3% efficient in class A, 28% in class B and 78% in class D.

On the exam

Exam point: class A conducts for the whole cycle and is the least efficient; class B conducts for half and needs a push-pull pair; class AB adds a small idle current to remove crossover distortion; class C conducts for less than half and needs a tuned circuit; class D switches. The less of the cycle a device conducts, the more efficient and the less faithful the stage.

BJT amplifiers: gain and power

Competency 11.3.1 asks you to analyze BJT amplifier circuits for gain and power. A bipolar transistor's current gain, β, is the collector current divided by the base current. Its AC emitter resistance r′e depends on the emitter current, and it sets the gain of a bypassed stage. ETA's 2025 sheet prints none of these three formulas, so learn them as given here (the forms Kuphaldt and Fiore use, and the same as the Transistor & Amplifier calculator; some texts use 25 mV instead of 26 mV, about 4% different, so on a multiple-choice question pick the option nearest your result):

β = I_C / I_B · r′e = 26 mV / I_E · A_v = −R_C / (r′e + R_E)

In the gain formula R_E is only the unbypassed part of the emitter resistance. Fully bypassed, R_E drops out and the gain is −R_C/r′e, which is high but changes with the emitter current. With all of R_E unbypassed, R_E is usually much larger than r′e and the gain is close to −R_C/R_E, which is low but steady. A load in parallel with R_C lowers the gain: use R_C in parallel with R_L in place of R_C. The minus sign means the output is inverted, 180° out of phase with the input.

Worked example

A common-emitter stage: V_CC = 12 V, R1 = 10 kΩ, R2 = 2.2 kΩ, R_C = 3.3 kΩ, R_E = 1 kΩ, β = 100. The base voltage is 12 × 2.2 / 12.2 ≈ 2.16 V, so V_E ≈ 2.16 − 0.7 = 1.46 V and I_E ≈ 1.46 V / 1 kΩ = 1.46 mA. The collector sits at 12 − (1.46 mA × 3.3 kΩ) ≈ 7.18 V, so V_CE ≈ 7.18 − 1.46 = 5.72 V, near the middle of the 12 V supply. The base current is about 1.46 mA / 100 ≈ 14.6 µA. Then r′e = 26 mV / 1.46 mA ≈ 17.8 Ω. With R_E bypassed the gain is −3300 / 17.8 ≈ −185; unbypassed it is −3300 / (1000 + 17.8) ≈ −3.24. The transistor dissipates P = V_CE × I_C ≈ 5.72 × 1.46 mA ≈ 8.35 mW at rest.

On the exam

Exam point: a common-emitter stage inverts the signal; with the emitter resistor bypassed its voltage gain is about R_C divided by r′e, and with it unbypassed about R_C divided by R_E. It has both voltage and current gain, so it gives the most power gain of the three configurations: the follower has current gain only, and the common base voltage gain only.

Configuration	Voltage gain	Current gain	Phase	Impedances
Common emitter	High	β	Inverted	Intermediate input, output about R_C
Common collector (emitter follower)	Just under 1	β + 1	In phase	High input, low output: a buffer that lets a weak source drive a heavy load
Common base	High	Less than 1	In phase	Low input (tens of ohms), high output

Power. Power gain is the output power divided by the input power, which equals the voltage gain times the current gain (section 1). The transistor itself dissipates V_CE × I_C as heat. A class A stage dissipates the most at idle: with full signal, Fiore shows the transistor dissipating only half its idle power, because the rest now goes to the load.

FET amplifiers: gain and power

Competency 11.3.2 asks you to analyze FET-based amplifier circuits for gain and power. A FET is voltage-controlled: its gate-source voltage sets the drain current. A JFET's gate is a reverse-biased junction and a MOSFET's is insulated, so the gate draws practically no current, and the input impedance is set by the gate resistor R_G, often 1 MΩ or more. The FET's figure of merit is its transconductance g_m (datasheets also call it g_fs): the change in drain current per volt of gate-source voltage, in siemens. It is not constant; it changes with the bias. ETA's sheet does not print FET formulas either. For the common-source stage (Fiore's forms):

A_v = −g_m × r_L (source bypassed) · A_v = −g_m × r_L / (1 + g_m × r_S) (swamped)

Here r_L is R_D in parallel with the load, and r_S is the unbypassed source resistance. Like the common emitter, the common source inverts. The common-drain stage (source follower) is the FET's emitter follower: a gain just under 1, in phase, with a high input and a low output impedance.

Worked example

A common-source stage: g_m = 4 mS, R_D = 3.3 kΩ, a 10 kΩ load, R_G = 1 MΩ. Then r_L = 3.3 kΩ ∥ 10 kΩ ≈ 2.48 kΩ, and with the source bypassed A_v = −0.004 × 2480 ≈ −9.92. Leave 220 Ω of the source resistor unbypassed and the gain falls to −9.92 / (1 + 0.004 × 220) = −9.92 / 1.88 ≈ −5.28. Now the power, with the source bypassed: 10 mV RMS across 1 MΩ is 0.01² / 1,000,000 = 0.1 nW in; the output is 9.92 × 10 mV = 99.2 mV RMS, and 0.0992² / 10,000 ≈ 0.984 µW out. The power gain is about 9,840, or 10 log(9,840) ≈ 39.9 dB, from a voltage gain of under 10.

On the exam

Exam point: a FET's gate draws almost no current, so a FET stage has a very high input impedance and a large power gain even when its voltage gain is modest. FET stages generally give less voltage gain than BJT stages.

Frequency response

Competency 11.4 asks you to explain an amplifier's frequency response and why it matters. Every amplifier has a midband where its gain is flat. Below it the gain falls because the reactance of the coupling capacitors and the emitter bypass capacitor rises; above it the gain falls because of small, unavoidable capacitances in the transistor and the wiring. The collector-base capacitance matters most, because the stage's own gain multiplies it (the Miller effect). The edges are the lower and upper cutoff frequencies, where the output power has fallen to half its midband value: −3 dB, or 0.707 of the midband voltage. The bandwidth is the difference between them.

BW = f_high − f_low · f_c = 1 / (2π × R × C) (one coupling RC)

The cutoff formula is not printed on ETA's sheet, but it comes straight from the sheet's X_C = 1 / (2π × f × C): it is the frequency where the capacitor's reactance equals the resistance it feeds. Half power is 10 log(0.5) ≈ −3.01 dB, and 0.707 of the voltage is 20 log(0.707) ≈ −3.01 dB, the same point.

Worked example

The low-frequency corner: a 1 µF coupling capacitor feeds a stage with a 10 kΩ input impedance. Using the sheet's 2π = 6.28, f_c = 1 / (6.28 × 10,000 × 0.000001) ≈ 15.9 Hz, and at that frequency X_C = 1 / (6.28 × 15.9 × 0.000001) ≈ 10 kΩ, equal to the resistance. A smaller capacitor raises the corner and cuts the bass. If the upper corner is 50 kHz and the lower 20 Hz, the bandwidth is 50,000 − 20 = 49,980 Hz, about 50 kHz.

On the exam

Exam point: bandwidth is measured between the two −3 dB (half-power) points. Coupling and emitter bypass capacitors set the low-frequency limit, and the transistor's and the wiring's capacitance set the high one.

Why it matters. The amplifier must pass every frequency in its signal at the same gain, or the output is no longer a faithful copy: NEETS gives an audio amplifier about 15 Hz to 20 kHz and a video amplifier up to about 6 MHz. An amplifier that is too wide also passes noise it doesn't need, and noise rises with bandwidth. Fiore notes that noise gets worse as temperature, resistance and frequency range increase.

Preamplifiers and line amplifiers

Competency 11.5 asks you to explain the terms preamplifier and line amplifier and where they are used. Audio equipment interconnects at standard levels. Professional mixers and processors use +4 dBu line level, consumer players use −10 dBV, both around a volt, while a microphone's output is usually between −60 and −40 dBu: tens of decibels lower (see the Audio & Video guide for the levels).

Unit	What it does	Where it is used
Preamplifier	Raises a weak signal, such as a microphone's, up to line level before any other processing.	Microphone inputs: mixers, stand-alone mic preamps and mic-to-line amplifiers.
Line amplifier	Takes a signal that is already at line level and boosts, buffers or splits it, so it can drive long cables or several inputs without losing level. Radio Design Labs' line amplifier, for example, gives unity to +24 dB of gain and drives long balanced lines into 600 Ω loads.	Long balanced audio lines, and distribution amplifiers, which send one line-level input to several isolated outputs.

Operational amplifiers

Competency 11.6 asks how operational amplifiers are used and how they differ from other amplifiers. An op-amp is a differential amplifier with an extremely high voltage gain: it amplifies the difference between its two inputs. It is DC-coupled, so it amplifies DC as well as AC, and it has a very high input impedance and a fairly low output impedance. Its open-loop gain is far too high and too variable to use directly as an amplifier (a comparator is the open-loop exception), so as an amplifier it is always used with negative feedback: two resistors set the gain, and the op-amp's own gain hardly matters as long as it is very large. Kuphaldt's examples include voltage followers (buffers), summers, differential and instrumentation amplifiers, integrators and comparators.

Competency 11.6.1 asks you to analyze op-amp circuits for gain and power. ETA's sheet gives both gain formulas:

A_V = −(R_f / R_I) (inverting) · A_V = 1 + (R_f / R_I) (non-inverting)

Worked example

Gain, then power: a non-inverting amplifier has R_f = 39 kΩ and R_I = 3.9 kΩ, so A_V = 1 + 39/3.9 = 11, or 20 log(11) ≈ 20.8 dB. A 0.25 V peak input gives 2.75 V peak out, in phase. Into a 1 kΩ load that is 2.75 mA peak and (2.75 / 1.414)² / 1000 ≈ 3.78 mW. Now try to drive 1 W into an 8 Ω speaker: that needs a peak of √(2 × 8 × 1) = 4 V and 4 / 8 = 0.5 A. TI's LM358B sources about 30 mA typically, so an op-amp alone cannot do it: it drives a power stage, or you use a power amplifier IC.

That is the practical difference from the discrete stages above: an op-amp is a complete, high-gain building block whose gain you set with two resistors, but its output current is limited to tens of milliamps, and its output cannot swing all the way to the supply rails (the LM358B's high side stays about 1.4 V below V+). Check both in the datasheet. Try the gain formulas in the Op-Amp Gain tool in the Calculator Suite or the op-amp mode of the Transistor & Amplifier calculator.

Distortion and how to reduce it

Competency 11.7 asks how distortion occurs and how to reduce it. Distortion is any change in the waveform's shape. Most distortion adds new frequencies, harmonics at whole multiples of the input frequency, and THD (total harmonic distortion) measures them as a percentage of the output.

Kind	How it happens	How to reduce it
Clipping	The output tries to swing past what the stage can deliver (its compliance) and the peaks flatten. Too much input drive clips both peaks; a Q point too close to cutoff or saturation clips one side first.	Less input drive, a stage with more headroom, and bias in the middle of the swing.
Crossover distortion	In class B, neither device conducts while the signal is within about ±0.7 V of zero, leaving a notch at each zero crossing. It hurts quiet signals most.	Class AB bias: a small idle current, often set by diodes, keeps both devices just on.
Nonlinear gain	The device's gain changes as the signal swings (a bypassed BJT stage's r′e changes with current), so one half-cycle is amplified more than the other.	Negative feedback. A swamping resistor is local feedback: Fiore's simulation cut a stage's THD from about 8% to under 0.6%, at the cost of gain.
Frequency distortion	Some frequencies get less gain than others because they fall outside the bandwidth (section 6).	Enough bandwidth: larger coupling capacitors for the lows, less stray capacitance for the highs.

Negative feedback is the general cure: it trades some gain for gain that is set by resistors, and so is steadier and more linear.

Meter and scope loading

Competency 11.8 asks how meter or scope loading gives inaccurate measurements, and 11.8.1 how to overcome it. A voltmeter or a scope probe is connected in parallel with what it measures, so its own input resistance becomes part of the circuit. On a low-impedance point that makes no difference; on a high-impedance point, such as a bias divider, a FET gate or an oscillator, the instrument pulls the voltage down and reads low. That is loading.

Worked example

Loading a 1 MΩ divider: two 1 MΩ resistors across 12 V put 6 V at the middle. A 10 MΩ DMM across the lower resistor makes it 1 MΩ ∥ 10 MΩ ≈ 909 kΩ, so it reads 12 × 909 / 1909 ≈ 5.71 V, about 4.8% low. A scope probe on 1× (1 MΩ) makes it 500 kΩ and reads 12 × 500 / 1500 = 4.0 V. A 20,000 Ω/V analog meter on its 10 V range (200 kΩ) makes it about 166.7 kΩ and reads 12 × 166.7 / 1166.7 ≈ 1.71 V. The circuit didn't change; each meter did.

At high frequencies the probe's capacitance loads the circuit too. Tektronix's P2220 probe is 1 MΩ and 110 pF on 1× but 10 MΩ and 17 pF on 10×. Using the sheet's X_C formula, at 1 MHz 110 pF is only about 1.45 kΩ and 17 pF about 9.37 kΩ. The 1× position is also limited to 6 MHz of bandwidth, against 200 MHz on 10×.

On the exam

Exam point: a meter or probe is a resistance in parallel with the point it measures, so on a high-impedance point it reads low. The higher the instrument's input impedance, the less it loads the circuit.

Ways to overcome loading (11.8.1):

Use a high-impedance instrument. A 10 MΩ DMM rather than an analog meter on a low range, and the scope probe on 10× rather than 1×.
Use an amplified (buffered) input where the point is very high impedance: an active or FET-input probe, or an electronic voltmeter. Kuphaldt also describes the null-balance (potentiometric) method, which draws no current at balance.
Measure at a low-impedance point. Read the emitter or the output of a follower stage rather than the base or a gate, or work it out from points that load easily.
Correct the reading. If you know the meter's resistance, work out the loaded divider as in the example, or compare two meters: if the reading changes with the meter, the meter is loading the circuit.

The Test Equipment guide covers meter sensitivity (ohms per volt) and the scope in more depth.

Broadband and narrowband amplifiers

Competency 11.9 asks how broadband amplifier specifications compare with those of common narrowband units. A broadband (wideband) amplifier passes a wide range of frequencies at an even gain; the video amplifier is one, and NEETS calls it a wide-band amplifier. A narrowband amplifier is tuned: NEETS's RF amplifier uses parallel LC tuned circuits, so it is only effective at the frequency they select. That is how receiver RF and IF stages reject the signals next door.

Specification	Broadband amplifier	Narrowband (tuned) amplifier
Frequency	A frequency range, often many octaves (NEETS: video 10 Hz to 6 MHz)	A center frequency and a bandwidth
Gain	Gain plus gain flatness: how much the gain varies across the whole range	Gain at the center frequency; usually more gain than a broadband stage, which NEETS notes gives up gain for its width
Selectivity	None: it passes everything in its range, wanted or not	The tuned circuit rejects frequencies outside its passband
Matching	Input and output VSWR or return loss across the whole range	Matched at the operating frequency
Large signals	1 dB compression point (P1dB) and third-order intercept (IP3): they matter whenever several signals pass at once and mix into intermodulation products, which in a broadband amplifier is normal	The same terms, but the tuned circuit keeps most unwanted signals out
How the width is made	RC or direct coupling; video amplifiers add peaking coils, which extend the high-frequency response but lower the gain	LC tank circuits or tuned transformers

Noise figure, the noise the amplifier adds, is specified for both. The definitions follow Mini-Circuits' application note on amplifier terms: gain flatness is the variation in gain over the frequency range, and P1dB is the output level at which gain has fallen 1 dB below its small-signal value.

On the job: a dead or distorting amplifier

Job practice, not exam content. These steps apply the sections above to an audio power amplifier on the bench, following the technicians' notes collected in Goldwasser's audio repair FAQ. Your employer's procedures and the service manual come first.

Check the parts with the power off. Shorted output transistors usually read close to 0 Ω between at least two of their three pins on the diode-test or ohms range. Check the bias and emitter resistors for opens, the rectifier and filter capacitors, and any fusible resistors.
Check what fed the failed part. When the output transistors have failed, check the driver stage too; it is often damaged as well. A low-value emitter resistor can read normal with a shorted transistor across it, so lift one end.
Power up gently. Replace the line fuse with a series lamp (about 60 W), bring the supply up slowly on a variac fed from an isolation transformer, and use dummy load resistors instead of speakers. A brief flash that settles to a dim glow is normal; a lamp that stays bright means a major short remains.
Measure DC before signal. On a small-signal stage compare V_B, V_E and V_C with the expected values (in section 4's example, about 2.16, 1.46 and 7.18 V). A collector at V_CC means no collector current; a collector near the emitter voltage means the transistor is saturated or shorted. On a push-pull output, check the DC offset at the speaker terminals: Goldwasser's contributors suspect a fault above about 100 mV, and expect about 1.2 V (two V_BE drops) between the output transistors' bases.
Then trace the signal. Inject a sine wave and follow it stage by stage with a scope, using the 10× probe so you don't load the circuit (section 10). Clipped peaks point to drive or bias; a notch at the zero crossings points to a lost class AB bias.

Safety

Power amplifiers bite: their supply rails and filter capacitors hold far more energy than a small-signal stage. Unplug and discharge the filter capacitors before you work on the board (see the Power Supplies guide), and feed the unit through an isolation transformer when you probe it live with grounded test equipment.

Check yourself

Answer in your head first, then reveal.

Which class conducts for less than half the cycle, and what does it need?
class C conducts for less than half and needs a tuned circuit
What sets the voltage gain of a common-emitter stage with its emitter resistor bypassed?
with the emitter resistor bypassed its voltage gain is about R_C divided by r′e
Why can a FET stage have a large power gain with a modest voltage gain?
a FET's gate draws almost no current, so a FET stage has a very high input impedance and a large power gain even when its voltage gain is modest
Between which points is bandwidth measured?
bandwidth is measured between the two −3 dB (half-power) points
Why does a meter read low on a high-impedance point?
a meter or probe is a resistance in parallel with the point it measures, so on a high-impedance point it reads low

Practice this topic: Amplifiers questions are in the Study Hub quiz bank, and the key terms are in Flashcards.

Drill Competency 11 Transistor & Amplifier calculator Op-Amp Gain (Calculator Suite) Decibel tool Flashcards