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  • 1. POWER SUPPLIESEquipment containing electronic circuits require voltages of between 5 and 30 volts dc. Wetherefore need to convert the 240 volt ac mains supply to a suitable value before it can be used topower the equipment, this is the job of the power supply.TRANSFORMERThe transformer converts the hazardous ac mains supply voltage to an isolated low voltagesuitable for rectification into a dc voltage used by the electronic circuits.Transformers used in power supplies are of two types of construction,1. Laminated iron core and bobin for 50Hz linear power supplies.2. Solid ferrite core and bobin for switch mode power supplies.Comprising at least two windings mounted in the bobin, primary and secondary winding. Theprimary winding connects to the mains input voltage and the secondary winding provides theoutput voltage and are isolated from each other providing a safety barrier.RECTIFIER CIRCUITSThere are three main classes of rectifier,1. Half-Wave.2. Full-Wave.3. Full-Wave Bridge.electronics-powersupplies-130517160825-phpapp02.docPage 1transformer rectifier smoothingcircuitstabilizingcircuitdcoutputacinputPower Supply Block Diagram
  • 2. Half Wave RectificationDuring the first half cycle diode D conducts since its anode is positive with respect to it cathode.During the second half cycle its anode is negative with respect to the cathode and the diodeblocks the flow of current. Therefore only the positive half cycles will be available to the load atthe frequency of the mains supply.The voltage Vo will have an amplitude equal to the peak value of the secondary voltage Vi minusone diode volt drop (0.7V), however only half the power will be available due to the missing halfcycle of the negative part of the input voltage.Assuming the secondary voltage Vi to be 12V rms then Vo will be,electronics-powersupplies-130517160825-phpapp02.docPage 2IDViVo0 time+--ViVoacmainsR+Half-WaveVVVV Dio 268.167.0)12414.1(2 =−×=−×=
  • 3. Full Wave RectificationThe full wave rectifier circuit uses two diodes and a transformer with a centre-tapped secondary.With the centre tap being the reference point at 0V, then when A is at its positive peak B will be atits negative peak diode D1 will conduct. When A is at its negative peak B will be at its positivepeak and diode D2 will conduct.In full wave rectification both halves of every cycle of input voltage are available to the load at afrequency of twice the supply frequency. This circuit consists of two half-wave rectifiers supplyingthe load on alternate half cycles of input voltage Vi.The voltage Vo will have an amplitude equal to the peak value of the secondary voltage Vi minusone diode volt drop (0.7V), however double the power of the half-wave version will be availabledue to the additional half cycle from the negative part of the input voltage.Assuming the secondary voltage Vi to be 12V rms then Vo will be,electronics-powersupplies-130517160825-phpapp02.docPage 3VVVV Dio 268.167.0)12414.1(2 =−×=−×=B-timeVi Vo0+-+ViacmainsD2D1oRVoFull-WaveAII
  • 4. Full Wave Bridge RectificationThe full wave bridge rectifier circuit uses four diodes and, for a single rail output, an untappedtransformer secondary. When A is at its positive peak B will be at its negative peak diodes D1 andD2 will conduct. When A is at its negative peak B will be at its positive peak and diodes D3 and D4will conduct.In full wave rectification both halves of every cycle of input voltage are available to the load at afrequency of twice the supply frequency.The voltage Vo will have an amplitude equal to the peak value of the secondary voltage Vi minustwo diode volt drops(1.4V), due to the current path through out going and return. As with the half-wave the output will fluctuate at twice the mains frequency.Assuming the secondary voltage Vi to be 12V rms then Vo will be,electronics-powersupplies-130517160825-phpapp02.docPage 4ViVoacmainsD2 D3D1D4RFull-Wave BridgetimeVi Vo0+-ABVVVV Dio 57.154.1)12414.1(2 =−×=−×=
  • 5. Dual Rail OutputAn adaptation of the full-wave bridge circuit is that of the dual rail power supply commonly usedto power analogue electronic systems and computer circuits since they usually require bothpositive and negative supply rails.With the addition of a centre-tapped secondary two supply rails can be achieved using the centretap as the 0V reference.Smoothing CircuitsThe varying dc output voltage from the rectifier although adequate for charging a battery is notsuitable for electronic equipment and must be smoothed to a steady dc level.The simplest way to obtain a smoothed output is to connect a high value capacitor across it,called a reservoir capacitor. Its value can range from 100μF to 10,000μF depending on thecurrent and smoothing required.electronics-powersupplies-130517160825-phpapp02.docPage 5-ViRD2+-0V+VoacmainsD3D1D4BA-VoR+ViDual Rail Power SupplyPositive Output Negative OutputtimeVi Vo0+-time0+ Vi-Vo
  • 6. During the part of the positive half-cycle C is charging to near the peak value of the ac secondaryvoltage. During the rest of the cycle C is supplying the load by partly discharging into it and Vofalls until the next half-cycle tops up the charge on C.The small variation in the smoothed dc output is called the ripple voltage and in the half-wave hasthe same frequency as the mains supply.Since the half-wave uses only the positive half-cycles the discharge period is twice as long as thefull-wave causing the ripple voltage to be greater.Capacitor and Diode RatingsCapacitor. The smoothing action of the reservoir capacitor is due to its large capacitance makingthe time constant CR large, (R is the resistance of the load), compared with the time for one cycleof ac mains. The larger C is the better the smoothing but the peak charging current pulses aregreater.electronics-powersupplies-130517160825-phpapp02.docPage 6Connection of Reservoir CapacitorID-ViVoacmainsR+CC charging C discharging0 timeVO-Half-Wave Output0 timeVoC charging C dischargingFull Wave output
  • 7. The capacitor should have a voltage rating at least equal to the peak value of the transformersecondary voltage at maximum line input.Diode. Consideration must be given to the peak inverse voltage which is the maximum voltagethe diode can block when reverse biased and should be at least twice the peak to peak value ofthe transformer secondary voltage at maximum line input. Also the peak current rating must begreater than that required by the reservoir capacitor during charging.Stabilizing CircuitsThe circuits discussed have one major draw back in that they do not provide any form of controlon the output voltage which can change with mains and load variation,To overcome this and provide a stable output voltage of a known value we need to add astabilizing circuit, or regulator.electronics-powersupplies-130517160825-phpapp02.docPage 7dcoutputloadbadgoodLoad Regulation
  • 8. Zener Diode RegulatorA basic form of regulator can be achieved using a zener diode and a resistor to limit the zenercurrent. When a zener diode is reverse biased the voltage across it is almost constant. Thisproperty can be used to provide a stable voltage for a wide range of input voltage.The total current I is divided between the zener current IZ and the load current IL.Since R and VZ are constant then I is constant, if IL reduces due to increase in RL then IZ willincrease by the same amount. At IL = 0, then IZ = I, the zener diode must has the power rating tosafely carry this current.Zener Diode Power rating,andMaximum Current rating,electronics-powersupplies-130517160825-phpapp02.docPage 8+-RILVLZ RLIZIViZener Diode RegulatorIVVR Zi −=LZ III += andZZZVPIIVPmaxmaxmax==
  • 9. Transistor RegulatorAdding a transistor to the zener diode circuit will remove the problems associated with the simplezener regulator since the zener current is constant, neglecting small variations in IB. Also the loadcurrent capability is increased by the HFE of the transistor.Type No Pol. Vce (V) Ic (max) PTOT hFE (min) CaseBD135 npn 45 1.5A 12.5W 40@1.5A TO-126BD136 pnp 45 1.5A 12.5W 40@1.5A TO-126TIP31A npn 60 3A 40W 10@1A TO-220TIP32A pnp 60 3A 40W 25@1A TO-2202N3055 npn 60 15A 117W 20@4A TO-3TIP2955 pnp 100 15A 90W 20@4A TO-218electronics-powersupplies-130517160825-phpapp02.docPage 9VZ-VBEIZener Stabilized Transistor RegulatorIBIZ+IBVOVIRLILVZExample of Power Transistor Data
  • 10. Integrated Circuit Voltage RegulatorIntegrated circuit voltage regulators are available as three terminal devices with internalprotection from overload and overheating. They are available in both positive and negative fixedvoltages from ±5 to ±15 volts with current ratings from 100mA to 3 Amps. These devices havecomplex internal circuitry and should be mounted on a heat sink to ovoid overheating and thermalshutdown.Type No Voltage Current Polarity Style78L, 05, 12, 15 5, 12, 15 100mA positive TO9279L, 05, 12, 15 5, 12, 15 100mA negative TO9278, 05, 12, 15 5, 12, 15 1A positive TO22079, 05, 12, 15 5, 12, 15 1A negative TO22078S, 05, 12, 15 5, 12, 15 2A positive TO220LM323K 5 3A positive TO3LM387A 5 500mA positive TO220LM338T 1.2 – 37 variable 5A positive TO3LM337T 1.2 – 37 variable 5A negative TO3HeatsinksA common method of voltage regulation in linear power supplies uses either a transistor or threeterminal regulator to control the output voltage.The process of regulationdissipates power, which, inturn generates heatwithin the regulation device.To remove this heat quicklyand efficiently the regulator ismounted on a heatsink.electronics-powersupplies-130517160825-phpapp02.docPage 10Example of Heat SinksIntegrated Circuit RegulatorRLacmainsViVo7805Heat SinkCommon types of Voltage Regulator
  • 11. Heat sinks are manufactured from aluminium and usually painted black so as to dissipate theheat energy more efficiently and prevent overheating. They are rated in degrees per watt knownas the thermal resistance. For example a 2°CW-1the temperature of the heatsink rises by 2°Cabove its ambient temperature for every watt of power it has to remove. To dissipate 10W, therise will be 20°C.Switch Mode Power SuppliesThe power supplies described so far are known as linear power regulators since they operate ona stable dc level and the control voltage is a continuously varying level. Although cost effectiveand simple in their design and operation they are bulky and inefficient, as low as 50% andrequiring large heatsinks on the regulator.Switch mode supplies work by chopping the dc level at a high frequency, 100 – 200kHz using fastswitching transistors and then pass this to a small high frequency ferrite cored transformer. Theswitching frequency is pulse width modulated meaning that the ratio of ‘on’ time to ‘off’ timeelectronics-powersupplies-130517160825-phpapp02.docPage 11
  • 12. changes as the load or line input changes. These power supplies although complex in theirdesign are much smaller than the linear types and up to 85% efficient.electronics-powersupplies-130517160825-phpapp02.docPage 12Example of Switch Mode Power Supply (SMPS)