power amplifier and differential amplifier
TRANSCRIPT
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?