02_power circuits.pdf

7/23/2019 02_Power circuits.pdf

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Amplifiers for analog electronics

In typical analog circuits (as in operational amplifiers) the amplifier must pay attentionto the following aspects:

Signal amplification (voltage or current)

Analogamplifier

input

signaldrivers

Linearity of the output signal deliveredto the load

Frequency rangeof the output signal

Input and output resistance of circuit

the power gain between input andoutput signals is not the main goal for

University of Napoli Federico IIDIETI

Ettore NapoliPower Devices and Circuits 2014-15

1

load VL

e power s age excep or power

amplifiers)

e.g. a radio system that needs to amplify the signal from the antenna but the load is anothercircuits that processes the signal.

Too much power transferred to the load could also be detrimental..


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Review of basic of power amplifiers for analog electronics

In typical output stages (as in operational amplifiers and audio systems) the poweramplifier that drives the load must pay attention to the following aspects:

Power conversion efficiency = PL/ PSdefined as the ratio between the

Power

circuit

input

signaldrivers

Supply PS

Pi

average power (in a cycle) given to theload and the one taken by the powersupply (always less than 1). Often givenas a percentage (


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Classes of operation of power amplifiers (and power circuits)The operation of the power amplifiers is defined in different classes according to the

way the active devices in the circuit are operating during the period of input signal(sinusoidal input exemplify the typical working condition).

For analog amplifiers we define 2 main classes of operation:

conducting during the entire time period of

input signal waveform (360)

Class B, where the device is conducting for

about one/half the time period of the inputsignal waveform (180). Two devices arerequired to obtain a good output linearity.

For the switchin circuits we can define a

curre

n

time

curren

t

time



3

Class D operation where the device is madeto commutate between full conduction (on)and interdiction (off) states (we will discuss itlater on)

current

time


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Class A operation

In class A operation the device (here a BJT is assumed) is biased at the middlepoint Q(Io, Vo) of the load line, and the operating point is driven by the inputsignal along the load line to a max current less or equal than Imax and min currentlarger ore equal than 0.The output power is max when the operating point reaches Imax (ideally, when the

.

Ic

Io

Imax

IpQ

Class-A BJT output stage inemitter follower configuration



4

VceVCC

Vo

Vp


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Class A power efficiency

Max power efficiency: assuming a linear operation up to the limit values one has:

Power absorbed from the positive supply voltage:

Power absorbed from the negative supply voltage:

)(1

0IVdttiT

VP CCT

CCCS == +0IVP CCS =

Total power from the supply:

Power provided to the load:

Max a.c. peak values (for maximum efficiency) :

2

,OCC

LMAX

OPCCP

IVP

IIVV

=

==

component)a.ctheofspeak value,(2

PPPP

L IVIV

P

=

2 0IVP CCS =



5

The max power efficiency is then: MAX= LMAX

PS

= 4 = 25%

The power absorbed by the supply is always constant and equal to PS.The efficiency is linearly dependent on output power PL.


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Power balance in class A amplifiers

The power balance of the circuit is:LpolDevS PPPP ++= where:

PDev is the power dissipation on the power device

AX

=P

LMAX= 25%The max power efficiency is:

OCCOCCSDev IVIVPP ==

po PL is the dc power dissipated in the load

The max power dissipation in the circuit is obtained for zero power on the load.Ppol is constant and is: OCCpol IVP =

Max 50% of PS



6

S

Lets consider the meaning of these results: To obtain a (controlled ) power output of 50 W one need a supply power of at least

200 W the device must dissipate 100 W in the steady state to transfer a max power of 50 W

to the load!


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Power balance in class A amplifiers

Exercise:Which is the power dissipation on hte device when the mximum power istransferred to the load ?

Sim le calculation show:

component)a.ctheofspeak value,(2

PPLPP

Dev IVPIV

P =

=

Exercise:Typical impedance of a 50W loudspeaker i 5 Ohms.Which is the peak voltage and current the BJT needs to handle?

Simple calculation show: Vce_max32V, Ic_max5A



7

Exercise:Calculate the efficiency for a costant output voltage


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Class B operation

In class B operation the device (here it is assumed a BJT) is biased at zero current

point Q(0, VCC) of the load line. As a result the power dissipation in the quiescentstate is zero.Two devices (and two power supplies) are needed to obtain an output signal analog tothe input one. The NPN device operates as an emitter follower for positive signal

, .

Class-B output stage with 2 BJTin push-pull configuration

Ic

VceVCC

Io

Vp

Ip

Q



8

Vo


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Class B power efficiency

Max power efficiency:Assuming linear operation up to the limit values, and sinusoidal input

IMAXFor the power Psabsorbed from the supply,

Q1 Q2

tTT/2

Q1

given by the two supplies:

PS=

2VCC

TI

MAXsint dt

0

T/2

=2V

CCI

MAX

),:arespeak valuemaxthecase,that(in2

MAXPCCPPP

L IIVVIV

P ===

V



9

The max power efficiency is then: MAX =P

LMAX

PS

=V

CC MAX

2

2VCC

IMAX

=

4 78.5%

,LMAX

=

2


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Power balance in class B amplifiers

From the previous results on power efficiency and power dissipation it comes out that:

The max power conversion from power supply to load is 78% the total power dissipation (on both device) is 40% of the max output power: then

Lets consider the meaning of these results:

To obtain with class-B a power output of 100 W one need a supply power of at least130 W

To transfer a max power of 100 W to the load, each device must be able to dissipate20 W (at the IP* rated)



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onc us on: ass s etter t an c ass n power convers on, we pay t s w t some

degradation in linearity), but this is still not sufficient if we need power conversion aboveseveral kW.For a 10 kW output power we need a power dissipation on each device of more than2kW and this is not feasible with usual power packages, as we will see later on.


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What is the Package

It is the plastic, metallic or ceramic box the surounds the piece of silicon (chip) wherethe power device is manufactured.

The package provides:

protection from environmentl humidity, chemicals, dust and various pollution



12


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Class D operation

To increase the power available at the output of a power circuit, one must decrease thepower dissipation of the active devices (the circuit)

This is to increase the efficiency and to afford less expensive and bulky power devices

will come back on that point later on).

The best way of reducing the power dissipation on the device is to let it operate into twolimit operating points:

a) OFF state, where the power dissipation is zero because the device current is null.b) ON state, at the minimum voltage drop allowed by the operation of the device (often

indicated as saturation voltage)



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This is the Class-D operation: the device operates as a switch, that is either open (OFFstate) or closed (ON state). In this way, the device, driven by input pulses capable tobring it either in ON or OFF state, can operate at a power much less than theavailable output power, thus increasing both the power output and the powerefficiency.


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Class D operation

Ideally the power dissipation of the device working in the ON-OFF state is

OFF state: P = Vcc * Ileakage -> NegligibleON state: P = Imax * VON -> VON is a fraction of Vcc. The power dissipation is small

- - .

transition the higher the power dissipation (more time spent far from ON and OFF states)

In class D operation the chosen power device can easily handle a significant power asthe limitation is NOT on the maximum static power (there is a limitation on the dynamicpower however, but this is less stringent) but on the maximum current and voltage.

Example: designing a circuit with 50V-2A supply voltage and load current (100W)In Class A operation need to chose a power device with the ratings of 50V, 2A, and 50Wof max. ower dissi ation.



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In Class D operation the max. power dissipation can also be as low as 10W since VON isabout 2V.


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Class D operation

In other words, as schematically shown below with reference to a BJT device, theoperating load line (red line) can overcome the max power dissipation locus (greenhyperbolic line), because in the ON state (point B) the dissipated power is much lessthan the maximum power dissipation PDMAX, and in the OFF state (point A) is almost zero



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However, one must pay attention on the time required by the device to switch between

ON and OFF states: we can define an average steady-state power dissipation PDS andan averagedynamic power dissipation PDd :

Class D operation

PDS: the average power dissipationin the ON state assumin

TTON

T1

P =T

ONV

negligible the one in the OFFstate)

PDd: the average power dissipationduring the switching transitions

T1 and T2 between ON andOFF states T2



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PDd=

1

Ti(t)v(t)dt

T1+T2

The total power dissipation PD is the sum of the two components PDS and PDd indicated above.


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Class D amplifier

In class D amplifier, the information content of the signal cannot modify the amplitude ofthe pulses, because these latter are of constant amplitude, but it can be transferred to theoutput by a modulation of the width of the pulses.

n o er wor s, we nee a u se o u a on o r ve e ev ce an otransfer this information to the (amplified) output, i.e. to the load.

The simplest PWM modulation technique is done by using a signal comparator tocompare the analog signal with a triangular waveform.

The output will be made of a pulse train having an amplitude equal to the supply voltageof the comparator, and ON (OFF) duration defined by the time interval where thetriangular waveform is lower (higher) than the one of the modulation signal.



17


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PWM Modulation of Class D amplifiers

An example of PWM modulation, made by a sinusoidal signal fS using a signal comparatorand a triangular waveform fM, is reported in the following plot.



18


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Example of analog Class D amplifier

Driver



19

FilterPower devices


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Class D power block

Powersupply

Class DPWMsignal frequency fS

circuit

signaldemodulation

(filtering)

load

modulationcarrier frequency fM



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To reconstruct the output signal after the class D operation we need to demodulatethe signal by a low pass filter, that will cut off the carrier frequency fM, while leavingunaltered the signal frequency fS .The filter must be realized with only L, C components to minimize the power losses.


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Assuming that the L,C components of the filters are lossless (in real cases at least theinductance will have some series resistance that will dissipate some power), thepower efficiency of the circuit is basically linked to the power losses of the

semiconductor devices used to realize the switching elements.

We can then define the power efficiency =P

L

PS

=P

SP

D

PS

=1P

D

PS

.

For each device one can define, as seen before, the steady state power dissipationPDS as:

and the dynamic (or switching) power dissipation PDd as:

PDS

=T

ON

T

ION

VMIN



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We will use these expressions of PDS and PDd to evaluate the power efficiency ofsome basic switching power circuits.

PDd=

T

i(t)v(t) tT1+T2


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The filter cut-off frequency fF must be:fF


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Circuit simulation of a class D power Amplifier

A SWCAD analysis of a push-pull power amplifier operated in class D with a PWMmodulation with a voltage comparator and an LC filter at the output, and twocomplementary Power MOS is done using the following schematics:



23


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Discussion on the LC filter

The pole of the LC circuit is at 7.9kHz:

The signal frequency is 5kHz while the modulating signal is at 50kHz

LCf =

2

1



24


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Discussion on the LC filter

For analog amplifiers this kind of dimensioning of the LC circuit is adequate.

LCf

=

2

1

,

circuits, it is more convenient to see the LC tank as an energy storage circut taking careof providing constant voltage and currrent to the load.



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Power Circuits


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Power Circuits

DC Powersupply

DC/DC

converterload

unregulatedDC

regulatedDCThe basic power circuits are:

DC/DC converters, that control the d.c. poweron the load, by variable control signals

DC Power

supply

DC/AC

converter

load

unregulatedDC

regulatedAC

control

DC/AC converters (Inverters), that generatea regulated a.c power from a d.c. powersupply, and control the a.c. power delivered



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AC lineAC/AC

converterload

unregulated

AC

regulated

AC

control

AC/AC converters, that generate acontrolled a.c power (both in frequency andamplitude) from the line a.c. power supply

Power Circuits


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Power Circuits

Apart from the above circuits, we can consider also the power circuits that transform a AC

power (usually the one of the grid main supply) into a DC one (unregulated) : these circuits arenamed Rectifiers,

unregulated unregulated

AC Powersupply

DC/AC

converterload

control

AC/DC converters (Rectifiers), that generatean (unregulated) d.c power from a a.c.power supply

These circuits are usually made of transformers and diodes. They are needed if theelectronic s stem has to be su lied b the mains; in that case the rectifier is used to ive



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the unregulated DC input power that is assumed for the power circuits presented above.

02_power circuits.pdf

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