Transcript 6.20-03-2015 AEC

Power amplifiers
EL= IxRL
• To deliver a large power in to a load, Large voltage and current
amplitudes are required.
• This requires reducing the base collector concentration, to avoid
punch through. This results in lower β.
Characteristics of Power Transistors.
• Lower current gain.
• Heat sinks to be used to dissipate the heat.
• When there is no input signal, large current flows through the
transistor, resulting maximum power dissipation.
Use of Transformers.
Advantages.
Better Impedance matching.
Less DC power dissipation under quiescent conditions.
Disadvantages.
Frequency Limitations.
In class-A amplifier, D.C power is wasted even in the absence of
signal. So efficiency is small.
Class-A Power Amplifier.
a) Direct Coupled
b) Transformer Coupled.
a) Direct Coupled.
Graphical representation of class A
amplifier
Applying Kirchhoff s voltage law to the circuit shown in the Fig. 4.11, we
get
Vcc – ICRL = VCE
1
D.C. Operation
The slope of the load line = -1/RL
Y intercept = Vcc /RL
The collector supply voltage VCc and resistance RB decides
the d.c. base-bias current IBQ. The expression is obtained
applying KVL to the B-E loop and with
------------------- 2.
ICQ = β IBQ
---------------------- 3.
VCEQ = VCC – ICQ RL ---------------------------- 4.
D.C. POWER INPUT.
PDC = VCC ICQ
A.C. POWER OUTPUT.
VPP = Vmax – Vmin
IPP = Imax – Imin
RMS of Pac =
Vm x Im
RMS of Pac =
Pac=
(Vmax – Vmin)(Imax – Imin)
8
Efficiency
The efficiency of an amplifier represents the amount of a.c. power
delivered or transferred to the load, from the d.c. source i.e.
accepting the d.c. power input. The generalized expression for an
efficiency of an amplifier is,
(𝑉𝑚𝑎𝑥 − 𝑉𝑚𝑖𝑛 )(𝐼𝑚𝑎𝑥 − 𝐼𝑚𝑖𝑛 )
%𝜂 =
𝑋 100
8𝑉𝐶𝐶 𝐼𝐶𝑄
Maximum voltage and current swings
%𝜂 =
(𝑉𝐶𝐶 − 0)(2𝐼𝐶𝑄 − 0)
2𝑉𝐶𝐶 𝐼𝐶𝑄
𝑋 100 =
= 25%
8𝑉𝐶𝐶 𝐼𝐶𝑄
8𝑉𝐶𝐶 𝐼𝐶𝑄
Class A Output Stage - Recap
• Class A output stage is a simple linear current amplifier.
• It is also very inefficient, typical maximum efficiency between
10 and 20 %.
• Only suitable for low power applications.
• High power requires much better efficiency.
Why is class A so inefficient ?
• Single transistor can only conduct in one direction.
• D.C. bias current is needed to cope with negative going signals.
• 75 % (or more) of the supplied power is dissipated by d.c.
• Solution : Eliminate the bias current.
Transformer-coupled transistor circuit.
TRANSFORMER COUPLED POWER AMPLIFIER.
Reflected Impedance from Secondary to Primary of theTransformer
2
X
Pacmax =
RL
Family of collector curves for power transistor.
Transformer-coupled class B push-pull amplifier.
Ic1 and Ic2 flow in opposite directions in the primary of the
transformer. So their magnetizing effects cancel and hence
no magnetic saturation problems.
Combined collector curves for push-pull operation.
Transistor characteristics and operating points:
(a) input characteristic of a silicon transistor;
(b) bias operating points on a load line.
(b)
η=
FIGURE 14-16
A class AB push-pull amplifier.
Class B Output Stage
• Q1 and Q2 form two unbiased
emitter followers
– Q1 only conducts when the input
is positive
– Q2 only conducts when the input
is negative
• Conduction angle is, therefore,
180°
• When the input is zero, neither
conducts
• i.e. the quiescent power
dissipation is zero
Class B Current Waveforms
Iout
time
IC1
time
IC2
time
Class B Efficiency
Average power drawn from
the positive supply:
P  ve   VS I C1
IC1
A/RL
A sin(q)
0
I C1 
1
2
2
 I C1 d 
0
p
2p Phase, q

1 A
A
sin



d


2 0 RL
RL
 P  ve 
VS A

RL
By symmetry, power drawn from +ve and –ve
supplies will be the same. Total power, therefore:
2VS A
PS  P  ve   P ve   2 P  ve  
RL
Load power:
2
vout
A2 sin 2 t  A2
PL 


RL
RL
2 RL
Efficiency:
PL
A2 RL
A
 

PS 2 RL 2VS A 4VS
NB. A  VS     / 4  78.5%
Power Dissipation
To select appropriate output transistors, the
maximum power dissipation must be
calculated.
2VS A A2
PD  PS  PL 

RL 2 RL
Just need to find the maximum value of PD to
select transistors/heat sinks
E.g. VS = 15 V, RL = 100 W
1.5
PL
PS
PD
Power [W]
1
A2
PL 
2 RL
0.5
0
0
5
10
Peak Output Amplitude, A [V]
15
Maximum Power Dissipation
PD is a quadratic function of A,
2VS A A2
PD 

RL 2 RL
dPD
maximum when: dA  0
2VS A


0
RL RL
 PD (max)
 A
2VS

4VS2
2VS2
2VS2
 2  2  2
 RL  RL  RL
Substitute A in
above equation.
Efficiency / Power Dissipation
• Peak efficiency of the class B output stage is
78.5 %, much higher than class A.
• Unlike class A, power dissipation varies with
output amplitude.
• Remember, there are two output devices so the
power dissipation is shared between them.
Design Example
Design a class B amplifier which will deliver up
to 25 W into a 4 ohm load.
A2
PL  25 
 25  A  14.1 V
2 RL
Supply voltages must be larger than Amax so
choose Vs = 15V.
2VS2
PD  max   2  11.4 W  2  5.7 W
 RL
Each of the two output transistors must be able to safely
dissipate up to 5.7 Watts. Using a TIP120 & TIP 125:
TJ  TA   JA PD  TJ   JA PD  TA
 JA PD max  TA  TJ max
 5.7 JA  25  150
  JA  22 C/W
But, with qJC = 1.92 °C/W
 JA   JC   CA  1.92   CA
 CA  20 C/W
i.e. Either two heatsinks rated at less than 20°C/W are
required or a single heatsink rated at less than 10°C/W.
Suggested heatsink
Dimensions, 50mm x 50mm x 9.5mm
Accommodates two devices
Rating 6.5°C/W
Cost 60p inc VAT
Cross-Over Distortion
• A small base-emitter voltage is
needed to turn on a transistor
• Q1 actually only conducts when
vin > 0.7 V
• Q2 actually only conducts when
vin < -0.7 V
• When 0.7 > vin > -0.7, nothing
conducts and the output is zero.
• i.e. the input-output relationship
is not at all linear.
Actual Input-Output Curve
vout
vout  vin  VBE
-VBE
+VBE
vout  vin  VBE
vin
Effect of Cross-Over Distortion
Class B Summary
• A class B output stage can be far more
efficient than a class A stage (78.5 %
maximum efficiency compared with 25 %).
• It also requires twice as many output
transistors…
• …and it isn’t very linear; cross-over
distortion can be significant.
DISTORTION IN POWER AMPLIFIERS.
The possible distortions in any amplifier are
• Amplitude
• Frequency (Not significant in Power
Amplifiers)
• Phase (Cannot be detected by ear.)
Because of the non linear dynamic
characteristics of transistors, waveform
distortion occurs. This gives raise to
Harmonic Distortion.
• The second harmonic has the highest
amplitude.
Elimination of Even Harmonic Distortion. (Push-Pull)
The amplitude of second harmonic is
the largest. As the order of harmonic
increases, it’s amplitude decreases.
ic1= a1cos𝝎t + a2cos 2𝝎t + a3cos 3𝝎t + ………………
ic2 = a1cos(𝝎t + π) + a2cos2(𝝎t + π) + a3cos3(𝝎t + π) + ………….
= -a1cos 𝝎t +a2cos2𝝎t –a3cos3𝝎t + ………………………
i0 = ic1 – ic2
So even harmonics are eliminated and the % harmonic
distortion is only due to odd harmonics given by,
• If B1 is the Amplitude of fundamental Frequency
• Bn is Amplitude of nth Harmonic Component
• % Harmonic Distortion due to nth harmonic
component
Total Harmonic Distortion
Power Transistor Derating Curve
Power transistors
dissipate a lot of power
in heat. This can be
destructive to the
amplifier as well as to
surrounding
components.
Dual Supply
Single Supply
With Diode Compensation
With Driver & Diode Compensation
Quasi Complimentary With NPN Matched Pair.
A Practical Circuit.
Thermal Analogy of Power Transistors.
θ=Thermal Resistance.
Pd = Heat due to Power Dissipation.
Heat Flow with Heat Sink.
The thermal resistance as per the above diagram
• Problem:
• Determine the power handling capacity for a 60 W
power transistor rated at 25 "C if de-rating is
required above 25 "C at a case temperature of 100
"C. The de-rating factor is 0.25 W/°C.
Thus due to de-rating at 100 deg.C, the power
handling capacity decreases to 41.25 W.
𝜂 > 98%