3. POWER AMPLIFIERS/ OUTPUT STAGES
-
Power amplifiers or large-signal amplifiers provide sufficient power to an output load to
drive a speaker or other power device, typically a few watts to tens of watts.
-
The main features of a large-signal amplifier are the circuit’s power efficiency, the
maximum amount of power that the circuit is capable of handling, and the impedance
matching to the output device.
-
In this, amplifiers are classified by class. Classes represent the amount the output signal
varies over one cycle of operation for a full cycle of input signal.
-
The amplifier classes are: Class A, Class B, Class AB, Class C and Class D.
-
These are employed as output stages of op amps and audio power amplifiers.
-
Class A amplifiers have very low distortion, as the collector current flows through-out the
cycle. However, they are very inefficient and are rarely used for high power designs. The
Class A amplifier is more suitable for outdoor musical systems, since the transistor
reproduces the entire audio waveform without ever cutting off. As a result, the sound is
very clear and more linear, that is, it contains much lower levels of distortion.
-
Class B amplifiers are used in low cost designs or designs where sound quality is not that
important, e.g. in siren drivers. They are significantly more efficient than class A
amplifiers, however they suffer from bad distortion.
-
Class AB is probably the most common amplifier class currently used in home stereo and
similar amplifiers, e.g. CD players, TVs and radios and public address systems. Class AB
amps combine the good points of class A and B amps. They have the improved efficiency
of class B amplifiers and distortion performance that is a lot closer to that of a class A
amplifiers.
-
Class C amplifiers are usually employed in RF power amplification, i.e. in mobile
phones, radio and TV transmitters. These suffer tremendous distortions, but RF circuits
M. Lupupa, 2022/2023
43
where Class C amplifiers are used employ filtering so that the final signal is completely
acceptable.
-
Class D amplifiers are often found in car audio subwoofer amplifiers.
-
All the common-emitter, common-base and common-collector amplifiers fall into the
class A category.
Amplifier Efficiency
-
The efficiency of an amplifier is defined as the ratio of power output to power input. This
improves as you progress from Class A to Class D.
Class A
-
The emitter follower is the most popular class A output stage, because of its low output
resistance.
-
In this the transistor operates in the active region at all times, i.e. the collector current
flow for 360 of the AC cycle.
Operating cycle
Class A
360
Class B
Class AB
180
180 to 360
Power efficiency
25% - 50%
78.5%
M. Lupupa, 2022/2023
Between (25% 50%) and 78.5%
Class C
Less than
180
Class D
Pulse
operation or
digital
Approaches Typically
100%
over 90%
44
Series –Fed Class A Amplifier
-
The simple-fixed bias circuit below is used to discuss the main features of a class A series
fed amplifier, even though it is seldom used for power amplification due to its power
collector efficiency.
-
The only differences between this circuit and the small-signal version considered is that
the signals handled by the large-signal circuit are in the range of volts, and the transistor
used is a power transistor that is capable of operating in the range of a few to tens of
watts.
-
The beta of a power transistor is generally less than 100, the overall amplifier circuit uses
power transistors that are capable of handling large power or current while not providing
much voltage gain.
DC Bias Operation
-
The dc bias set by VCC and RB fixes the dc bias current at
IB 
VCC  0.7V
,
RB
(3.1)
with I C   I B
-
The collector-emitter voltage is then
VCE  VCC  I C RC
-
(3.2)
The transistor characteristics of the above circuit showing the load line and Q-point are as
shown:
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AC Operation
-
When an input ac signal is applied to the series-fed class A amplifier, the output will vary
from its dc bias operation voltage and current.
-
A small input signal will cause the base current to vary above and below the dc bias point,
which will then cause the collector current (output) to vary above and below the dc bias
point set.
-
As the input signal is made larger, the output will vary further around the dc bias point
until either the current or the voltage reaches a limiting condition.
-
For the current this limiting condition is either zero current at the low end or
VCC
at the
RC
high end of its swing.
-
For the collector-emitter voltage, the limit is either 0V or the supply voltage, VCC .
-
In order to achieve the maximum symmetrical swing of current and voltage, the Q point
should be located at the centre of the dc load line. In this case the operating point is
V
V
I CQ  CC and VCE  CC .
2
2 RC
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Power Considerations
-
The power into an amplifier is provided by the supply.
-
With no input signal, the dc current drawn is the collector bias current, I CQ .
-
The power drawn from the supply is then given by:
Pi ( dc )  VCC I CQ
(3.3)
-
Even with an AC signal applied, the average current drawn from the supply remains the
same.
-
A class A amplifier stage passes the same load current even when no input signal is
applied so large heatsinks are needed for the output transistors.
-
The output voltage and current varying around the bias point provide the AC power to the
load. This AC power is delivered to the load, RC .
-
The AC power delivered to the load, RC may be expressed as:
Using RMS values:
Note: Vrms 
Vp
2
Po (ac )  VCE ( rms ) I C ( rms )
M. Lupupa, 2022/2023
47
Po ( ac )  I C2 ( rms ) RC
Po ( ac ) 
2
VCE
( rms )
RC
(3.4)
Using peak values:
Po ( ac ) 
VCE ( p ) I C ( p )
2
Po ( ac ) 
Po (ac) 
I C2 ( p ) RC
2
2
VCE
( p)
2 RC
(3.5)
Efficiency
-
This is a measure of the ability of an active device to convert the dc power of supply into
the AC (signal) power delivered to the load and is given by:
% 
-
Po ( ac )
 100%
Pi ( dc )
(3.6)
A class A series-fed amplifier has a maximum efficiency of 25%.
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Example
Calculate the input power, output power and efficiency of the amplifier circuit below for an
input voltage that results in a base current of 10mA peak.
-
Below is the simplest type of class A power amplifier circuit. It uses a single-ended
transistor for its output stage with the resistive load connected directly to the collector
terminal.
-
Another simple way to increase the current handling capacity of the circuit while at the
same time obtain a greater power gain is to replace the single output transistor with a
Darlington transistor.
-
These types of devices are basically two transistors within a single package, one small
"pilot" transistor and another larger "switching" transistor.
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-
The big advantage of these devices are that the input impedance is suitably large while
the output impedance is relatively low, thereby reducing the power loss and therefore the
heat within the switching device.
-
The overall current gain, beta,  value of a Darlington device is the product of the two
individual gains of the transistors.
Transformer-Coupled Class A Amplifier
-
To improve the efficiency of a class A amplifier, a transformer is used to couple the
output signal to the load as shown in the figure below:
-
A transformer can increase or decrease voltage or current levels according to the turns
ratio. It also provides for impedance matching.
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Advantages of Transformer Coupled Class A Power Amplifier
-
No loss of signal power in the base or collector resistors.
Excellent impedance matching is achieved.
Gain is high.
DC isolation is provided.
Disadvantages of Transformer Coupled Class A Power Amplifier
-
Low frequency signals are less amplified comparatively.
Hum noise is introduced by transformers.
Transformers are bulky and costly.
Poor frequency response.
Applications of Transformer Coupled Class A Power Amplifier
-
This circuit is where impedance matching is the main criterion.
These are used as driver amplifiers and sometimes as output amplifiers.
-
The voltage transformation is given by:
V2 N 2

V1 N 1
-
The current transformation is given by:
I 2 N1

I1 N 2
-
(3.7)
(3.8)
The impedance transformation is given by:
RL'  N1 


RL  N 2 
where
2
(3.9)
RL' is the impedance on the primary coil
RL is the impedance on the secondary coil
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-
In order to have maximum AC power output, the peak value of collector current due to
input AC signal alone should be equal to the zero-signal collector current.
-
To achieve this, the operating point Q is located at the centre of the ac load line. This is
achieved by adjusting the biasing circuit ( R1 , R2 and RE ).
-
When AC signal is applied, collector current fluctuates from maximum to minimum
(zero), and operating point Q moves up and down the load line.
-
At the peak of the positive half cycle of the input signal, the total collector current
I C max  2 I CQ and collector-emitter – voltage VCE min  0 .
-
While at the peak of the negative half cycle of the input signal, the collector current
I C min  0 and collector-emitter voltage VCE max  2VCC .
-
Thus collector-emitter voltage varies in opposite phase to the collector current. The
variation of collector voltage appears across primary of the transformer. Now AC voltage
is induced in the transformer secondary which in turn develops AC power and supplies to
the load.
-
Peak to peak collector emitter voltage is:
VCE ( p  p)  2VCC
-
(3.10)
Peak to peak collector current is:
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52
I C ( p  p )  2 I CQ
-
(3.11)
The maximum output power is given as:
2
'
VCE ( p  p )  I C ( p  p ) 2VCC  2 I CQ VCC I CQ I CQ RL
Po ( ac ) 



8
8
2
2
-
The power drawn from the supply is then given by:
2
Pi (dc)  VCC I CQ  ICQ
RL'
-
(3.12)
(3.13)
For a class A transformer-coupled amplifier the efficiency can be expressed as:
% 
Po ( ac )
 100%
Pi ( dc )
-
The maximum theoretical efficiency of a class A transformer-coupled amplifier goes up
to 50%.
-
The power dissipated (loss) by the transistor as heat is expressed as:
PD  Pi ( dc)  Po ( ac )
-
(3.14)
When no signal drives an amplifier, the power dissipation of the transistor equals to the
product of dc voltage and current:
PD  VCEQ I CEQ
(3.15a)
or
PD  VCC I C
(3.15b)
Example
A class A transformer coupled power amplifier has zero signal collector current of 50mA . If
the collector supply voltage is 5V , find:
i)
The maximum AC power output
ii)
The power rating of the transistor
iii)
The maximum collector efficiency
Advantages of Class A Amplifier
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-
It has high reliability because of the output exact replica of an input signal.
It has improved high-frequency response because the active device is ON full time, i.e. no
time is required to turn on the device.
There is no crossover distortion because the active device conducts for the entire cycle of
the input signal.
Disadvantages of Class A Amplifier
-
Due to the large power supply and heat sink, class A amplifier is costly and bulky.
It has poor efficiency.
Due to the transformer coupling, frequency response is not as good.
Class B
-
This consists of a complementary pair of transistors (an npn and a pnp) connected in such
a way that both cannot conduct simultaneously. In this, each transistor conducts current
for only one half of the signal cycle.
-
To obtain output for the full cycle it is necessary to use two transistors and have each
conduct on opposite half cycles, the combined operation providing a full cycle of output
signal.
-
Since one part of the circuit pushes the signal high during one half cycle and the other
part pulls the signal low during the other half cycle, the circuit is referred to as a push-pull
circuit.
-
The class B operation of these transistors provides greater efficiency than was possible
using a single transistor in class A operation.
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54
Input (DC) Power
-
The amount of input or dc power can be calculated using:
Pi ( dc )  VCC I dc
(3.16)
where I dc is the average or dc current drawn from the power supplies, and is expressed as
I dc 
2I L ( p)


2VL ( p)
 RL
(3.17)
where I ( p ) is the peak value of the output current waveform.
-
The input power can thus be expressed as:
 2V ( p) 
 2 I ( p) 
Pi (dc)  VCC 

  VCC 
  
  RL 
(3.18)
Output (AC) Power
-
The power delivered to the load can be expressed as:
Po (ac) 
VL2 (rms ) VL2 ( p )

RL
2 RL
(3.19)
Efficiency
-
The efficiency of the class B amplifier can be calculated using the basic equation:
% 
-
Po ( ac )
 100%
Pi ( dc )
Using the above equations for input and output power, the efficiency can be expressed as:
% 
 VL2 ( p ) 


 2 RL 
Po ( ac )
 VL ( p )
 100% 
 100% 
 100%
Pi (dc )
4 VCC

 2 I ( p) 
VCC    



(3.20)
where:
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I ( p) 
-
VL ( p )
RL
(3.21)
At the maximum value, i.e. when VL ( p )  VCC , the maximum efficiency can be written
as:
% 

4
 100%  78.5%
(3.22)
Power Dissipated by Output Transistors
-
The power dissipated (loss as heat) by the output power transistors is the difference
between the input power delivered by the supplies and the output power delivered to the
load.
P2 D  Pi ( dc )  Po ( ac )
(3.23)
where P2 D is the power dissipated by the two output transistors.
-
The power dissipated by one transistor is:
PD 
P2 D
2
(3.24)
Maximum Power Considerations
-
For class B operation, the maximum output power is delivered to the load when
VL ( p )  VCC :
maximum Po (ac ) 
-
VL ( p ) VCC

RL
RL
(3.26)
The maximum value of the average or dc current drawn from the power supplies can be
written as:
maximum I dc 
-
(3.25)
The corresponding peak AC current:
I ( p) 
-
2
VL2 ( p ) VCC

2 RL
2 RL
2 I ( p)

2VCC
 RL

The maximum value of the input power can thus be given as:
M. Lupupa, 2022/2023
(3.27)
56
maximum Pi (dc)  VCC
-
 2V
  maximum I dc   VCC  CC
  RL
(3.28)
For class B operation, the maximum power dissipated by the output transistors does not
occur at the maximum power input or output condition. The maximum power dissipated
by the two transistors occurs when the output voltage across the load is:
VL ( p ) 
-
2
 2VCC

  RL
2VCC

(3.29)
The maximum power dissipated by the two transistors is given as:
maximum P2 D 
2
2VCC
 2 RL
(3.30)
-
Among the disadvantages of a class B amplifier is that the nonlinear cut-off region is
included in the operation range.
-
Because of the biasing arrangement, class-B amplifiers are subject to a type of distortion.
-
Therefore, there is a time interval between the positive and negative alternations when
neither transistor is conduction.
-
The resulting distortion in the output waveform is quite common and is called crossover
distortion.
-
To prevent crossover distortion, both transistors will normally be biased at a level that is
slightly above cut-off.
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-
Biasing both transistors slightly above cut-off will allow the amplifier to provide a linear
output that contains no distortion.
Advantages of Class B Amplifier
-
As compared to class A amplifier, this is more efficient.
Used for very powerful outputs.
High efficiency as compared to class A
The push-pull system within this amplifier removes even order harmonics within AC
output signal.
There are no DC components within the output
Disadvantages of Class B Amplifier
-
Crossover distortion can be created in these types of amplifiers
Distortion is more as compared to class A
It is very difficult to obtain two complementary transistors with the same characteristics.
Example
For the unity gain amplifier circuit below, calculate the input power, output power and power
handled by each output transistor and the circuit efficiency for an input of 12Vrms .
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Class AB
-
In order to combine both the excellent linearity of class A and the high efficiency of class
B amplifiers, the class AB amplifier has been developed.
-
As the name suggests, a class AB amplifier works half-way between class A and class B
amplifiers. The figure below shows the operating zone of class AB amplifiers:
-
Class AB amplifiers can be biased by choosing an operating point along the blue line,
excluding the class A and class B biasing points.
-
The choice of its location depends on the desired levels of efficiency and linearity. If the
class AB operating point is closer to the class A operating point than the class B operating
point the circuit will behave more such as a class A amplifier by presenting a higher
linearity but a lower efficiency and vice-versa.
-
To eliminate crossover distortion which is experienced in the class B amplifier, both
transistors in the push-pull arrangement must be biased slightly above cut-off when there
is no signal.
-
There are different biasing methods for class AB amplifiers to create voltage intervals
based on the simultaneous conduction of complementary transistors used in this
configuration. These include:
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Voltage Biasing
-
A very easy and intuitive method is to bias directly and independently the bases of the
NPN and PNP transistors. This supply can be done by batteries or via a DC generator. But
it is not implemented for real circuits due to the packaging, and high cost.
Voltage Divider Network
-
With this configuration, a voltage drop across the two bases (across the total resistance
R2  R3 ) of 1.4V is observed. This value corresponds to 2  VBE where VBE represents the
threshold voltage of the transistors. This voltage drop will bring both transistors above the
cut-off region, they will therefore both conduct simultaneously a portion of the input
signal.
-
The main reason why this configuration cannot be used for class AB amplifiers is that the
set of values of the resistors will only work for a particular push-pull configuration.
Moreover, the set of values needs to be chosen very carefully, especially if the
complementary transistors are different.
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Resistor Biasing/ Potentiometer Biasing
-
This method is very similar to the voltage divider network. The only difference is that an
adjustable resistance (or potentiometer) is placed between the two networks. The
advantage of this solution is to closely match the biasing of the two complementary
transistors with a controllable resistance, even if the transistors have different electrical
properties.
-
However, this solution as well as the voltage divider network involve resistor that
generate heat dissipation by Joule’s effect. Since the class AB amplifiers are often biased
slightly above the cut-off point to offer a maximum efficiency, any temperature change
caused by this phenomenon can result in an inappropriate behaviour of the push-pull
configuration. This is the reason why diode biasing if often preferred
-
One common approach to remove crossover distortion is to insert a base biasing network
as shown below:
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61
-
This form of bias is called diode bias.
-
This method biases the transistors so that their turn-on/off points actually overlap. That is,
both transistors are in a state of conduction for a brief moment during the crossover
period.
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-
The addition of the resistor and diode effectively raises the base voltage a diode drop
above/below the emitter junction of the npn/pnp transistors.
-
Thus, each transistor is biased so that their quiescent currents are just on the verge of
turning on when vin  0 . This form of amplification is technically known as Class AB
rather than Class B, because each transistor is on for more than 50% of the time during a
complete waveform cycle.
-
A disadvantage is increased power consumption of the amplifier circuit.
-
The diodes, D1 and D2 , produce the required bias for the base-emitter junction of each
transistor.
-
For this bias method to work properly, the I F versus VF curve of each diode must match
the VBE versus I E curves of each transistor.
-
With diode bias, the collector current equals the current through the compensating diodes,
and is given by:
I CQ 
VCC  2VBE
2R
(3.31)
where R1  R2  R
-
To obtain a greater amount of output power, a split supply can be used as shown below:
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-
The collector bias current in both transistors can be calculated as:
I CQ 
VCC  VBE
R
(3.32)
Input (DC) Power
-
The amount of input or dc power can be calculated using:
Pi ( dc)  VCC I dc
(3.16)
where I dc is the average or dc current drawn from the power supplies, and is expressed as
I dc 
2I L ( p)


2VL ( p )
 RL
(3.17)
where I ( p ) is the peak value of the output current waveform.
-
The input power can thus be expressed as:
 2V ( p ) 
 2I ( p) 
Pi (dc)  VCC 

  VCC 
  
  RL 
(3.18)
Output (AC) Power
-
The power delivered to the load can be expressed as:
VL2 (rms ) VL2 ( p )
Po (ac) 

RL
2 RL
(3.19)
Efficiency
-
The efficiency of the class AB amplifier can be calculated using the basic equation:
% 
-
Po (ac )
 100%
Pi ( dc )
Using the above equations for input and output power, the efficiency can be expressed as:
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 VL2 ( p ) 


2 RL 
P ( ac )
 VL ( p )
%  o
 100%  
 100% 
 100%
Pi ( dc )
4 VCC

 2 I ( p ) 
VCC    



(3.20)
Power Dissipated by Output Transistors
-
The power dissipated (loss as heat) by the output power transistors is the difference
between the input power delivered by the supplies and the output power delivered to the
load.
P2 D  Pi (dc )  Po ( ac )
(3.23)
where P2 D is the power dissipated by the two output transistors.
-
The power dissipated by one transistor is:
PD 
P2 D
2
(3.24)
Advantages of Class AB Amplifier
-
The main advantage of the class AB amplifier is the elimination of cross-over distortion.
Provides high efficiency and amplification of the signal than class A amplifier.
It is a linear amplifier because the amplitude and the phase of the output signal are the
same as the input.
Provides high efficiency and high-frequency response.
Less harmonic distortion.
Disadvantages of Class AB Amplifier
-
Practically the efficiency of the class AB amplifier is less when compared to the
efficiency of the class B amplifier
Construction of circuit is complex
Cost is high
Applications of the Class AB Amplifier





Used in Audio Frequency (AF) amplifiers in TVs
Used in Public addressing systems
Used in TV receivers
Used in radio receivers
Used in CD players
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Used satellite and other wireless communication systems
Example
For the circuit below, determine:
a) I CQ
b) For the 5Vrms input, determine the power delivered to the load resistor.
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Example
For the circuit below, determine:
a) I CQ
b) Assuming the input voltage is 10V p , determine the power delivered to the load resistor.
Class C
-
Despite their differences, we have seen in the previous discussions that the class A, class
B and class AB amplifiers are linear or partially linear.
-
We will now discuss on a non-linear amplifier namely, the class C amplifier.
-
Class C amplifiers are not defined by a single operating point but rather an operating
zone. This fact is illustrated in the figure below:
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-
Since this operating zone extends beyond the class B operating point, which represents a
78.5 % efficiency and a 180° conduction angle, class C amplifiers are therefore
characterised by a very high efficiency between 78.5% and 100%.
-
Class C amplifiers are capable of providing large amounts of power.
-
They are often used for transmitter power stages, such as radio or communications, where
a tuned circuit is included to eliminate the higher harmonics in the output signal.
-
The resonant frequency of the tuned circuit is determined by the formula:
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f 
1
2 LC
(3.33)
-
The tuned circuit in the output will provide a full cycle of output signal for the
fundamental or resonant frequency of the tuned circuit of the output.
-
This type of operation is therefore limited to use at one fixed frequency, as occurs in a
communications circuit, for example.
-
Operation of a class C circuit is not intended primarily for large-signal or power
amplifiers.
-
Inappropriate biasing produces spikes in the output signal, resulting in crossover
distortion.
Example
What is the resonant frequency of a tank circuit with L  10mH and C  0.001 F .
Class D
Class D amplifiers generally consists of three different modules, namely a modulator, a
switching stage and a low pass filter. The signal path along with the succession of these
different modules is presented in figure below:
-
While classic amplifiers accept as an input a sinusoidal signal, Class D amplifiers
previously transform it through a modulator into a rectangular signal.
-
The switching stage is where the amplification takes places thanks to transistors. The
transistors work in a complementary configuration in order to amplify correctly the
rectangle signal.
-
Finally, a low-pass filter is used in order to restore the sinusoidal shape of the signal.
Moreover, this final stage eliminates undesirable harmonics that may have been generated
during the amplification process.
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Amplifier Distortion
-
Any pure sinusoidal signal has a single frequency at which the voltage varies positive and
negative by equal amounts. Any signal varying over less than the full 360 cycle is
considered to have distortion.
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Amplifier Distortion can take on many forms such as amplitude, frequency and phase
distortion due to clipping.
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When distortion occurs, the output will not be an exact duplicate (except for magnitude)
of the input signal.
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One technique for describing distorted waveforms is by using Fourier analysis. This
method describes any periodic waveform in terms of its fundamental frequency
component and frequency components at integer multiples.
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These components are called harmonic components or harmonics.
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Harmonic distortion can be explained as any distortion or corruption in the output
waveform due to the generation of harmonics.
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For example, a signal that is originally 1kHz could result after distortion in a frequency
component at 1kHz and harmonic components at 2kHz , 3kHz , 4kHz and so on.
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The original frequency of 1kHz is called the fundamental frequency, and the rest
harmonics.
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The 2kHz component is called the second harmonic, 3kHz , the third harmonic and so on.
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Harmonic distortion in power amplifiers are mainly caused by the non linearities of the
active elements (transistors).
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Fourier analysis does not allow for fractional harmonic frequency, but only integer
multiples of the fundamental.
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When an output signal has a number of individual harmonic distortion components, the
signal can be seen to have a total harmonic distortion based on the individual elements as
combined by the relationship of the following equation:

THD 
V22  V32  V42  ...

(3.34)
V1
where THD is the total harmonic distortion, V1 is RMS value of the fundamental
component and Vn is RMS value of the nth harmonic.
-
Total harmonic distortion or THD is a very important parameter in the audio amplifier
domain.
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Total harmonic distortion is a measurement of the harmonic distortion present in a signal
and is defined as the ratio of the sum of the powers of all harmonic components to the
power of the fundamental frequency.
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If the fundamental frequency has an amplitude A1 and the nth frequency component has
an amplitude An , then a harmonic distortion, Dn can be defined as:
% nth harmonic distortion = % Dn 
-
An
A1
 100%
(3.35)
When an output signal has a number of individual harmonic distortion components, the
signal can be seen to have a total harmonic distortion based on the individual elements as
combined by the relationship of the following equation:
%THD 


D22  D32  D42  ...  100%
(3.36)
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When distortion occurs, the output power calculated for the undistorted signal is no
longer correct.
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When distortion is present, the output power delivered to the load resistor due to the
fundamental components of the distorted signal is:
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I12 RC
P1 
2
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The total power due to all the harmonic components of the distorted signal can then be
calculated using:
P   I12  I 22  I 32  ...
-
(3.37)
RC
2
(3.38a)
The total power can also be expressed in terms of the total harmonic distortion as:
I12 RC
P  1  D  D  D  ...
 1  THD 2  P1
2
2
2
2
3
2
4
(3.38b)
Example
Given the following information, calculate the total harmonic distortion:
Power BJTs
Power Transistor Heat Sinking
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As power transistors handle large currents, they always heat up during operation.
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Since a transistor is a temperature dependent device, the heat generated must be
dissipated to the surroundings in order to keep the temperature within permissible limits.
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Generally, a power transistor is fixed on a metal sheet so that additional heat is transferred
to the metal sheet.
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The ability of any heat sink to transfer heat to the surroundings depends upon its material,
volume, area, shape, contact between case and sink and movement of air around the sink.
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The maximum power handled by a particular device and the temperature of the transistor
junctions are related since the power dissipated by the device causes an increase in
temperature at the junction of the device.
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The junction temperature TJ must not be allowed to exceed a specified maximum, TJmax
otherwise the transistor could suffer permanent damage. For silicon devices, TJmax is in
the range 150C to 200C .
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For many applications when the input signal is applied, the average power or the total
power dissipated may be approximated by:
PD  VCE I C
(3.39)
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This power dissipation, however, is allowed only up to a maximum temperature.
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Power transistors are mounted in large metal cases called heat sinks to provide a large
area from which the heat generated by the device may radiate.
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Since even a good heat sink cannot hold the transistor case temperature at ambient (air), it
is necessary to derate the amount of maximum power allowed for a particular transistor as
a function of increase case temperature.
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A device derating factor can be given as:
PD (temp1 )  PD (temp0 )  (temp1  temp0 )  ( derating factor )
where
(3.40)
temp0 : temperature at which derating should begin.
temp1 : particular temperature of interest (above the value of temp0 ).
PD (temp0 ) and PD (temp1 ) : are the maximum power dissipations at the
specified temperatures.
derating factor: is the value given by the manufacturer in units of watts per
degree of temperature.
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The transistor power dissipation, PD is the same as the rate at which heat flows out of the
transistor.
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-
In thermodynamics, the rate of heat flow is analogous to current, thermal resistance to
resistance, and temperature difference to voltage.
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The thermal resistances can be written as:
 JA   JC  CS   SA
(3.41)
where  JA : thermal resistance (junction to ambient).
 JC : transistor thermal resistance (junction to case).
CS : insulator thermal resistance (case to heat sink).
 SA : heat sink thermal resistance (heat sink to ambient).
where CA  CS   SA
-
The analogy can also be used in applying Kirchhoff’s law to obtain:
TJ max  PD JA  TA
(3.42)
where TJ max : is the maximum junction temperature.
TA : is the ambient (air) temperature.
TC : is the case temperature.
TS : is the heat sink temperature.
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Using the electrical analogy of the above equation we obtain the series circuit model for
the power flow:
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The temperature difference between the case and the junction is given as:
TJ max  TC  PD JC
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(3.43)
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The temperature difference between junction and ambient is:
TJ max  TA  PD JA
-
(3.44)
The temperature difference between junction and heat sink is:
TJ max  TS  PD JS
(3.45)
where  JS   JC  CS
Example
A BJT is specified to have a maximum power dissipation PD  2W at an ambient
temperature TA  25C and a maximum junction temperature TJ max  150C . Find the
following:
a) The thermal resistance  JA .
b) The maximum power that can be safely dissipated at an ambient temperature of 50C .
c) The junction temperature if the device is operating at TA  25C and is dissipating 1W .
Example
A BJT is specified to have TJmax  150C and to be capable of dissipating maximum power as
follows:
40W at TC  25C
2W at TA  25C
Above 25C , the maximum power dissipation is to be derated linearly with  JC  3.12C / W
and  JA  62.5C / W .
Find the following:
a) The maximum power that can be dissipated safely by this transistor when operated in free
air at TA  50C .
b) The maximum power that can be dissipated safely by this transistor when operated at an
ambient temperature of TA  50C , but with a heat sink for which CS  0.5C / W and
 SA  4C / W . Find the temperature of the case and of the heat sink.
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Power Values of Power Transistors
Owing to their large geometry and high operating currents, power transistors display typical
parameter values that can be quite different from those of small-signal transistors. The
important differences are as follows:
1. At high currents, the exponential iC  vBE relationship exhibits a factor of 2 reduction in the
exponent: that is, iC  I S evBE /2VT .
2.  is low, typically 30 to 80, but can be as low as 5. Here, it is important to note that  has
a positive temperature coefficient.
3. At high currents, r becomes very small (a few ohms) and rx becomes important.
4. fT is low (a few megahertz), C is large (hundreds of picofarads), and C is even larger.
5. I CBO is large (a few tens of microamps) and, as usual, doubles for every 10 C rise in
temperature.
6. I C max is typically in the ampere range but can be as high as 100 A.
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