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Power supplies 2357.2.7 EfficiencyThe efficiency of a power supply module is its output power divided by its input power.The difference between the two quantities is accounted for by power losses in thevarious components in the power supply.Efficiency η = P out /P in = P out /(P out + P loss )The efficiency normally worsens as the load is reduced, because the various losses andquiescent operating currents assume a greater proportion of the input power. Therefore,if you are concerned about efficiency, do not use a power supply that is heavily overratedfor its purpose. Linear supply efficiency also varies considerably with its inputvoltage, being worst at high voltages, because the excess must be lost across theregulator. Switch-mode supplies do not have this problem.Normally efficiency is not of prime concern for mains power supplies, since it is notessential to make optimum use of the available power, although at higher powers theheat generated by an inefficient unit can be troublesome. It is far more important that apower converter for a portable instrument should be efficient because this directlyaffects useable battery life.Linear power supplies are rarely more than 50% efficient unless they can bematched to a narrow input voltage range, whereas switch-mode supplies can easilyexceed 70% and with careful design can reach 90%. This makes switch-mode suppliesmore popular, despite their greater complexity, at the higher power levels and forbattery-powered units.Sources of power lossThe components in a power supply which make the major contribution to losses are• the transformer: core losses, determined by the operating level andcore material, and copper losses, determined by I 2 Rwhere R is the winding resistance• the rectifiers: diode forward voltage drop, V F , multiplied byoperating current; more significant at low outputvoltages• linear regulator: the voltage dropped across the series pass elementmultiplied by the operating current; greatest at highinput voltages• switching regulator: power dissipated in the switching element due tosaturation voltage, plus switching losses in this andin snubber and suppressor components, proportionalto switching frequency.If you sum the approximate contribution of each of these factors you can generallymake a reasonable forecast of the efficiency of a given power supply design. The actualfigure can be confirmed by measurement and if it is wildly astray then you should belooking for the cause.7.2.8 Deriving the input voltage from the outputIn a linear supply with a series pass regulator element, the design must proceed fromthe minimum acceptable output voltage at maximum load current and minimum inputvoltage. These are the worst-case conditions and determine the input voltage step-down
236 The Circuit Designer’s CompanionI rmsI rmsV rmsI dc I dcV rmsV rmsI rmsFull-wave centre tap: I rms ≅ 1.2 · I dcFull-wave bridge: I rms ≅ 1.8 · I dcFigure 7.9 Rectifier configurationrequired. The minimum DC input voltage is given by the minimum output voltage plusall the tolerances and voltage drops in series:V in,dc =V out(min) + V tol,reg + V series,reg + V series, CS ...whereV out(min) is the minimum acceptable output voltageV tol,reg is the regulator voltage tolerance, assuming it is not adjustableV series,reg is the voltage drop across the regulator series pass elementV series, CS is the voltage drop across the current sense element if fittedAll the above parameters are specified at full load current. This value for V in,dc is thenthe minimum input voltage allowed for a dc input supply, or it is the voltage at theminimum of the ripple trough for a rectified and smoothed ac input supply. This isrelated to the transformer secondary voltage as follows:V tx = (V in,dc + V ripple + V D )/0.92 · (V ac(nom) /V ac(min) ) · 1/√2whereV ripple is the peak ripple voltage across the reservoir capacitorV D is the voltage drop across the rectifier diode(s)V tx is the RMS transformer secondary voltageV ac(nom) is the specified transformer input voltageV ac(min) is the minimum line input voltageAll parameters at full load currentThe figure of 0.92 is an approximate allowance for full-wave rectifier efficiency with asingle-capacitor reservoir. It can be more accurately derived using curves published bySchade † . Complications set in because the current drawn through the secondary is notsinusoidal, but occurs at the crest of the waveform (see section 7.2.5). The extra peakcurrent reduces the peak secondary voltage from its quoted value, if this value isspecified for a resistive load. You can get around this either by knowing thetransformer’s losses in advance and allowing for the extra IR drop, or by specifying thetransformer for a given circuit and letting the transformer supplier do the work for you,if you’re buying a custom component. The transformer secondary RMS current ratingis determined by the rectifier configuration (Figure 7.9).† O.H.Schade, Analysis of Rectifier Operation, Proc.IRE, vol 31 1943, pp 341–361.
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The Circuit Designer’s Companion
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NewnesAn imprint of ElsevierLinacre
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vi Contents2.2 Design rules 452.2.1
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viii ContentsChapter 5Analogue inte
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x Contents7.5.3 Secondary cells 256
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xii Contents
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2 The Circuit Designer’s Companio
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10 The Circuit Designer’s Compani
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Grounding and wiring 25Cable type R
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Grounding and wiring 27Shielding an
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Grounding and wiring 29successive h
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Grounding and wiring 31(a) Two audi
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Grounding and wiring 331. Side-by-s
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Grounding and wiring 35ABV pZ out =
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Grounding and wiring 37times will d
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Grounding and wiring 39of Z o . Thu
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Printed circuits 41material. The co
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Printed circuits 43board cost you s
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Printed circuits 45Board ABoard ABo
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Printed circuits 47which they can w
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Table 3.3 Survey of capacitor types
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334 Indexde-rating 310dielectric ab
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336 IndexFlash ADC 209Foldback curr
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338 IndexMultilayer PCB constructio
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340 Indexlimiting element voltage 7
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