Classes of Operation Class A operation – the transistor operates in the active

region at all times the collector current flows for 360 of the ac cycle the designer usually tries to locate the Q point

somewhere near the middle of the load line. the signal can swing over the maximum possible range

without saturating or cutting-off the transistor, which would distort the signal.

Class B Operation the collector current flows for only half the cycle

(180) a designer locates the Q point at cut-off only the positive half of ac base voltage can

produce collector current reduces heat in power transistors

Class C Operation the collector current flows for less than

180 of the cycle only part of the positive half-cycle of ac

base voltage produces collector current brief pulses of collector current

Types of Coupling Capacitive Coupling

the coupling capacitor transmits the amplified ac voltage to the next stage

ac coupling blocks the dc voltage

Transformer Coupling the ac voltage is coupled through a

transformer to the next stage ac coupling blocks the dc voltage

Direct Coupling there is a direct connection between the collector

of the first transistor and the base of the second transistor

both the dc and ac voltages are coupled there is no lower frequency limit dc amplifier

Ranges of Frequency Audio Amplifier – an amplifier that operates

in the range of 20Hz to 20kHz Radio-Frequency (RF) Amplifier – one that

amplifies frequencies above 20 kHz, usually much higher

Narrowband Amplifiers – works over a small frequency range

Tunes RF Amplifiers – their ac load is a high-Q resonant tank tuned to a radio station or television channel

Wideband Amplifier – operates over a large frequency range

Untuned – their ac load is resistive

Tuned RF Amplifiers

Signal Levels Small-signal Operation – the peak-to-peak

swing in collector current is less than 10% of quiescent collector current

Large-signal Operation – a peak-to-peak signal uses all or most of the load line

Stereo System : the small signal from a radio tuner, tape player, or compact disc player is used as the input to a preamp, an amplifier that produces a larger output suitable for driving tone and volume controls. The signal is then used as the input to a power amplifier, which produces output power ranging from a few hundred milliwatts up to hundreds of watts

DC Load Line One way to move the Q point is by varying

the value of R2

large values of R2 : IC(sat) = VCC / (RC + RE)

very small values of R2 : VCE(cutoff) = VCC

AC Load Line RE has no effect on the ac operation the ac collector resistance is less than the dc collector

resistance when an ac signal comes in, the instantaneous operating

point moves along the ac load line the peak-to-peak sinusoidal current and voltage are

determined by the ac load line

MPP VCCwhere: MPP – maximum peak-to-peak output voltage

Equation of the ac load line:

IC = ICQ + VCEQ/rc + VCE/rc When the transistor goes into saturation:

ic(sat) = ICQ + VCEQ/rc When the transistor goes into cut-off:

vce(cutoff) = VCEQ + ICQ rc where: ic(sat) = ac saturation current

ICQ = dc collector current VCEQ = dc collector-emitter voltage

rc = ac resistance seen by the collectorvce(cutoff) = ac cut-off voltage

Clipping of Large Signals when the Q point is at the center of the dc

load line, the ac signal cannot use all of the ac load line without clipping

if the ac signal increases, a cut-off clipping will result

If the Q point is moved higher, a large signal will drive the transistor into saturation

a saturation clipping will occur

A well-designed large-signal amplifier has a Q point at the middle of the ac load line

results in a maximum peak-to-peak unclipped output

ac output compliance

Maximum Output Q point below the center of the ac load line:

maximum peak (MP) = ICQ rc

Q point above the center of the ac load line:maximum peak (MP) = VCEQ

Maximum Peak-to Peak Output MPP = 2MP

When the Q point is at the center of the ac load line:

ICQ re = VCEQ

The circuit’s emitter resistance can be adjusted to find the optimum Q point:

RE = (RC + rc) / (VCC/VE) -1

Power Gain – equals the ac output power divided by the ac input power

Ap = pout / pin

Output Power in rms volts: pout = vrms

2 / RL

in peak-to peak volts: pout = vout2 / 8RL

maximum output power occurs when the amplifier is producing the maximum peak-to-peak output voltage

vout = MPP2 / 8RL

Transistor Power Dissipation Quiescent Power Dissipation : PDQ = VCEQ

ICQ

When signal is present: the power dissipation of a transistor decreases

worst case: quiescent power dissipation power rating must be greater than PDQ

Current Drain dc source has to supply a dc current Idc to

the amplifier dc current is called current drain Idc has two components:

the biasing current through the voltage divider collector current through the transistor

Efficiency dc power supplied to an amplifier by the

source:Pdc = VCCIdc

efficiency – used to compare the design of power amplifiers

= pout / Pdc x 100% ac output power divided by the dc input

power

Efficiency a way to compare two different designs

because it indicates how well an amplifier converts the dc input power to ac output power

between 0 and 100 percent the higher the efficiency, the better the

amplifier is at converting dc power to ac power

important in battery-operated equipment

Efficiency in Class A Amplifier the maximum efficiency of a class A

amplifier with a dc collector resistance and separate load resistance is 25%

in some applications, the low frequency of class A is acceptable

Example: If the peak-to-peak output voltage is 18V and the input impedance of the base is 100, what is the power gain? What is the transistor power dissipation and efficiency?

Class A Power Amplifier class A power amplifier driving a loudspeaker uses voltage-divider bias the ac input signal is transformer-coupled to the

base produces voltage and power gain to drive the

loudspeaker through the output transformer

Class A Power Amplifier the load resistance is also the ac collector

resistance the efficiency of this class A amplifier is

higher Impedance-reflecting ability of a

transformer:( Np/Ns)2

the maximum efficiency increases to 50%

Emitter-Follower Power Amplifier locate the Q point at the center of the ac

load line to get maximum peak-to-peak output

large values of R2 – saturate the transistor

IC(sat) = VCC / RE

small values of R2 – drive the transistor into cut-off

VCE(cutoff) = VCC

ac load line end points:ic(sat) = ICQ + VCE/re

VCE(cutoff) = VCE + ICQre

MPP VCC

Q point is below the center of the ac load line:

maximum peak (MP) = ICQre

Q point is above the center of the load line:

maximum peak (MP) = VCEQ

Push-Pull Circuit clips off half a cycle use two transistors in a push-pull

arrangement push-pull – one transistor conducts for

half cycle while the other is off and vice versa

Advantages and Disadvantages there is no current drain when the signal is

zero the maximum efficiency of a class-B push-

pull amplifier is 78.5% the uses of transformer

Class B Push-Pull Emitter Follower an npn emitter follower and a pnp

emitter follower connected in push-pull arrangement

DC Equivalent Circuit select biasing resistors to set the Q-point

at cut-off this biases the emitter diode of each

transistor between 0.6 and 0.7 ICQ = 0

VCEQ = VCC / 2

DC Load Line the dc saturation current is infinite dc load line is vertical difficult thing: setting up a stable Q

point at cut-off

AC Load Line operating point moves up along the ac

load line voltage swing of the conducting transistor

can go all the way from cut-off to saturation

MPP = VCC

AC Analysis almost identical to a Class-A emitter

follower AV 1

zin(base) = RL

Crossover Distortion distorted output signal crossover distortion – the clipping occurs between the

time one transistor cuts-off and the other one comes on there is a need to apply a slight forward bias to each

emitter diode ICQ from 1 to 5% of IC(sat)

Class AB a conduction angle between 180 and 360

Power Formulas

Transistor Power Dissipation ideally: power dissipation is zero when

there is no input signal input signal: PD(max) = MPP2 / 40 RL

Example: The adjustable resistor sets both emitter diodes on the verge of conduction. What is the maximum transistor power dissipation? The maximum output power? If the adjustable resistance is 15, what is the efficiency?

Biasing Class B/AB Amplifiers Voltage-Divider Bias thermal runaway – then the temperature

increases, the collector current increases, the junction temperature increases even more, reducing the correct VBE

Diode Bias compensating diodes – produces the bias

voltage for the emitter diodes Ibias = (VCC – 2VBE) / 2R

ICQ has the same value as Ibias

Class C Operation the collector current flows for less than

half a cycle parallel resonant circuit: can filter the

pulses of collector current and produce a pure sine wave of output voltage

tuned RF amplifiers maximum efficiency can be 100%

Tuned RF Amplifier resonant frequency: fr = 1 / 2 LC always intended to amplify a narrow band

of frequencies ideal for amplifying radio and television

signals

Load Lines

DC Clamping of Input Signal input signal: drives the emitter diode amplified current pulses: drives the resonant tank circuit the input capacitor is part of a negative dc clamper the signal appearing across the emitter diode is negatively

clamped

Filtering Harmonics harmonics - multiples of the input

frequency; equivalent to a group of sine waves with frequencies of f, 2f, 3f,..., nf

resonant tank circuit – has a high impedance only at the fundamental frequency f

Class C Formulas tuned class C amplifier – a narrowband

amplifier Bandwidth : BW = f2 –f1

BW = fr / Q a large sinusoidal voltage at resonance

with a rapid drop-off above and below resonance

Current Dip at Resonance tune a resonant tank: look for a decrease in the dc current

supplied to the circuit measure the current Idc from the power supply while tuning

the circuit at resonant frequency: the ammeter reading will dip to a

minimum value the tank has a maximum impedance at this point

AC Collector Resistance QL = XL / RS

RP = QLXL

at resonance: XL cancels XC

rc = RP//RL

Q = rc / XL

Duty Cycle D = W / T the smaller the duty cycle, the narrower the

pulses compared to the period typical class C amplifier has a small duty cycle the efficiency of a class C amplifier increases as

the duty cycle decreases

Conduction Angle equivalent way to state the duty cycle D = / 360

Transistor Power Dissipation maximum output: MPP = 2 VCC VCEQ 2VCC conduction angle is much less than 180 the collector current reaches a maximum value of IC(sat)

peak current rating IC(sat)

power dissipation –depends on the conduction angle PD = MPP2/40rc

Stage Efficiency for a conduction angle of 180, the

average or dc collector current is IC(sat) / optimum stage efficiency varies with the

conduction angle

Example: If QL is 100, what is the bandwidth of the amplifier?

Differential Amplifier two CE stages in parallel with a common emitter

resistor two input voltages: v1 (non-inverting) and v2

(inverting) two collector voltages no lower cut-off frequency differential output: vout = vc2- vc1

differential input: vout = AV(v1-v2)

Single-Ended Output floating load – neither end of the load can be

grounded single-ended – one end is grounded output voltage: vout = AV(v1-v2) the voltage gain is half as much as with a

differential output

Non-inverting Input and a Differential Output vout = AV(V1)

Non-inverting Input and a Single-ended Output vout = AV(V1) but AV will be half as much

Inverting Input and Differential Output vout = - AV (v2)

Inverting Input and Single-ended Output vout = -AV (v2) but voltage gain will be half as

much

DC Analysis of a Differential Amplifier tail current: IT = VEE / RE

emitter current of each transistor: IE = IT / 2

dc voltage on either collector: VC = VCC - ICRC

DC Analysis Second Approximation IT = (VEE – VBE) / RE

Effect of Base Resistors: IT = (VEE-VBE)/ RE + RB/2dc)

Example: Calculate the currents and voltages using ideal and second approximations.

Non-inverting Input and Single-ended Output the two halves of a differential amplifier respond in a

complementary manner to the non-inverting input Q1 acts like an emitter follower that produces an ac voltage

across the emitter resistor the amplified output sine wave is in phase with the non-

inverting input

Single-Ended Output Gain each transistor has an re

the biasing resistor RE is in parallel with the re of the right transistor

RE is much greater than re

Simplified Equivalent Circuit input voltage vi across the first re is in series with

the second re

ac voltage across the tail resistor is half of the input voltage

Single-ended output: AV=RC/2re

Differential Output Gain the output voltage is twice as much since

there are two collector resistors Differential Output: AV = RC / re

Inverting-Input Configurations the inverting input v2 produces an

amplified and inverted ac voltage at the final output

Differential-Input Configurations both inputs are active at the same time Differential Input: AV = (v1 – v2)

Table of Voltage Gains the voltage gain is maximum with a differential

output the voltage gain is cut in half when a single-

ended output is used

Input Impedance In CE stage: zin = re

In a differential amplifier: zin = 2re

Example: What is the ac output voltage? If =300, what is the input impedance of the differential amplifier?

Example: What is the ac output voltage in the figure? If =300, what is the input impedance of the differential amplifier?

Common-Mode Gain same input voltage, vin(cm) is being applied to each

base common mode signal differential amplifier does not amplify common-

mode signals

Equivalent Circuit since equal voltages vin(cm) drive both

inputs simultaneously, there is almost no current through the wire between the emitters

Right side acts like swamped amplifier with common-mode input Av(cm) = RC / 2RE

the common-mode voltage gain is usually less than 1

Common-Mode Rejection Ratio (CMRR) voltage gain divided by common-mode

voltage gain CMRR = Av / AV(cm)

the higher the CMRR the better Data sheets usually specify CMRR in

decibels CMRR(dB) = 20 log CMRR

Example: What is the common-mode voltage gain? The output voltage?

Example: In the figure, Av = 150, Av(cm)=0.5, and vin = 1mV. If the base leads are picking up a common-mode signal of 1mV, what is the output voltage?

Example: A 741 is an op-amp with Av=200,000 and CMRRdB= 90dB. What is the common-mode voltage gain? If the desired and common-mode signal each has a value of 1V, what is the output voltage?