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Expedite TransformerCalculations for FlybacksBy Kirby Creel, Senior Design Engineer, Datatronics,Romoland, Calif.When charging defibrillator capacitors, a novelapproach can be used with confidence to comeup with a more exacting design, eliminating the cumbersome aspects of many otherapproaches.Using a flyback topology to generate a highvoltage is a common approach. The voltagecan charge a capacitor for a high-energypulse. Such an approach is used in defibrillator capacitors, photoflash capacitors, strobecapacitors and ignition circuits to name a few. Using a newstep-by-step procedure, it’s possible to quickly realize an initial flyback transformer design for charging a capacitor ina stated amount of time.Following this procedure eliminates “cut and try” andover-design approaches. It also allows designers to selectcritical values with confidence, and can be used to provideinsight as to the effects of holding an element constantand varying other elements. The variables included arefrequency, voltage, pulse width, peak current, load capacitance and efficiency.Before we delve into this novel approach, we need tounderstand the pros and cons of flyback topologies. AdT1D1 C1Vdcvantages include circuit simplicity, a high-voltage outputthat’s not dependent on a large transformer turns ratio, aself-limiting circuit that can be short circuited without anydamage, and an output that can be regulated over a largerange. A flyback also can provide voltage isolation, allowsfor multiple isolated outputs where one output can be usedfor a low-ratio feedback voltage, and does not require asmoothing choke.The disadvantages of flyback topologies include the needfor a large and often bulky transformer, a fast-switchingoutput that can generate problematic EMI signals, leakageinductance that must be kept low for good efficiency and acircuit that can be damaged once a load is removed withoutusing a feedback loop.Fig. 1 shows a simplified circuit and Fig. 2 shows idealized waveforms for a frequency of 50 kHz and a pulse ontime (TON) of 45%.The flyback operates by storing energy on the “charge”portion of the cycle and delivers the stored energy to theload on the discharge cycle. In the case of a flyback, thetransformer is often described as a coupled inductor. DueR10RIMARY CURRENTRAMPS CHARGER2 Q1 MS MSU1Pulsecontrol3ECONDARY CURRENT DISCHARGE EAD TIME Fig. 1. A flyback loop for pulse control keeps a flyback transformerfrom being damaged once a load is removed.26Power Electronics Technology January 2008Fig. 2. Dead time for a flyback transformer increases as thestorage capacitor approaches a full charge.www.powerelectronics.com

voltage, ( t) is the time duration from start to finish ofthe applied pulse and ( i) is the change in current overthe same interval. If i starts at zero, i is equal to the peakcurrent (IPEAK):V ( t )I PEAK A. (Eq. 2)LEq. 2 is Eq. 1 solved for IPEAK. For example, if VA 12 V, t is 0 μs to 15 μs, L 60 μH and IPEAK 3 A.The energy stored in the inductance is:L(I PEAK ) 2U , (Eq. 3)2where U is energy measured in Joules, L is in henriesand IPEAK is in amperes. In the previous example, the energystored in each pulse is 270 μJ.It is during the discharge of this stored energy that thegreatest advantage of the flyback is realized. The outputvoltage will rise to whatever level is needed to cause currentto flow, thus dissipating the stored energy. The voltage ofthe output has limits to be sure, but within the insulationstructure, transistor breakdown, and circuit design, andtaking losses into account, the voltage can rise to veryhigh levels.Though ancient technology now, the most commonexample was the flyback transformer in color TVs with acathode ray tube. These transformers could generate voltages greater than 35,000 V, voltages so high that, when theParameterDescriptionCapacitor100 μFCharge voltage2000 VdcCharge time10 secCircuitDiscontinuous-mode flybackFrequency50 kHzMaximum duty cycle45%Maximum on time9 μs ( t)Input voltage12 VdcEfficiencySee the fourth step and Fig. 3Primary inductanceTo be decidedPeak currentTo be decidedTable. Values for a typical charging circuit.to the diode polarity, current only flows in the secondaryside during discharge. During the charge cycle, energy isstored in the primary inductance by a current ramp. Thedead time shown in Fig. 2 ensures that the flyback is discontinuous. As the capacitor approaches full charge, thedead time increases.The primary current ramp (charge) follows the inductance formula:V ( t)L A, (Eq. 1) iwhere L is the inductance in henries, VA is the appliedMIAwww.powerelectronics.com27L I TA R YPPROVEDPower Electronics Technology January 2008

flyback transformerforms. In a defibrillator, pulse control will be a voltage feedback loop that fixes the number of Joules to be delivered tothe patient. During successive resuscitation attempts, thelevel will increase. For photoflashes, the charge level is fixed.The capacitor will be charged and additional pulses will onlybe applied as a refresh.In photoflash applications, the dead time may be limitedto speed up the charge time. The low dead time and variable discharge produces the characteristic of an increasinghigh-pitch sound. Variations in the pulse-control elementare almost endless. -AXIMUM EFFICIENCY %FFICIENCY %NERGY STORED *OULES Fig. 3. Energy storage efficiency in a flyback circuit is fairly linear up toabout 100 J, before it decreases asymptotically.circuits malfunctioned, the TV could generate damagingX-rays. The analysis is somewhat simplified because theTV flyback transformer performs more functions than justgenerating high voltage. The design of the flyback circuitand transformer for power transformation is well illustratedin Abraham I. Pressman’s book Switching Power SupplyDesign.[1]The block in Fig. 1 labeled “pulse control” can take manyDesign ExampleWith the background provided, we can now tackle theproblem of charging a capacitor to a given voltage in astated amount of time. Designs begin with a list of knownvalues.Let’s illustrate the process by considering the followingexample of a typical design problem, for which the valuesare listed in the table. The application is for charging a defibrillator capacitor. (Caution: The charged capacitor usedin this example can provide a lethal shock.)The first step is to determine the number of Joules required to charge the capacitor:C(VCAP ) 2 100 10 6 (2000) 2U 200 J, (Eq. 4)2228Power Electronics Technology January 2008www.powerelectronics.com

flyback transformer)0%! ! 6DC4 M&, M(figure calculated in the previous equation can be solved withan almost infinite number of solutions. The remainder of thedesign requirements and the 500-µJ calculation limits theanswer to one solution (as shown next).Eq. 1 is an inductance formula, while Eq. 3 is solved forthe inductance. The right-hand side of both equations areequal to each other. The resulting expression, with onlyone unknown, can be solved for the peak current. Havingfound the peak current, the value is entered into the original equation and solved for the inductance. For a “sanitycheck,” both formulas are solved for the inductance. Theequal results provide confidence that the calculations wereperformed correctly.2 # 6DC2 1 5 0ULSECONTROLE t i(note that i IPEAK )L Fig. 4. This completed design for a flyback transformer circuit withcomponent values is based on the calculations in this article.L L 2UIPEAK 2L where U is energy in Joules, VCAP is the capacitor voltageand C is the capacitance in Farads.The next step is to calculate the number of chargingpulses (N) in the stated time:N 10 s 50,000 pulses/s 500, 000. (Eq. 5)In the third step, calculate the energy required per charging pulse (UP):Joules200UP 400 µJ/pulse. (Eq. 6)N500, 000The fourth step is to make an estimate of the circuitefficiency and include the factor in calculating the energythat must be supplied. All calculations up to this point arebased on the assumption that there were no losses in theswitching transistor, diode and transformer (winding orcore). Switching transistor, diode, and transformer lossesare shown to a first order. Second-order losses are those forwinding capacitance and leakage inductance.Fig. 3 provides an estimation of a typical loss factor expressed as an efficiency figure. The losses must be includedin the power supplied from the dc source, as shown in thenext calculation. Note that Fig. 3 is an estimate and resultswill vary. Use the result from Eq. 4 (in this case, 200 J) tofind the efficiency value. A number of variables in circuitdesign, layout, transformer design, components and otherswill affect the result. (For very high-voltage designs, evenleakage current paths across the surface of the pc board andleakage within the capacitor must be considered.)2(500 10 6 ).IPEAK 212(9 10 6 ) 2(500 10 6 ) 2I PEAKI PEAK1.08 10 4 1 10 3 2I PEAKI PEAK2I PEAK1 10 3 I PEAK 1.08 10 4I PEAK 9.259 A.Calculate the inductance by substituting the value of IPEAKin both Eq. 1 and Eq. 3 as a check:2(500 10 6 )12(9 10 6 )L (9.259) 29.259L 11.66 µH.L 11.66 µH.Fig. 4 is the completed design with the calculated valuesincluded.A similar application to charging a defibrillator capacitor is charging a photoflash capacitor. Linear Technology’sLT3468 IC performs most functions in a small footprint.The IC is designed around the specific application with someother applications mentioned in the data sheet. One limitation of this device is the breakdown voltage of the switchingtransistor, which at 70 V at 25 C makes it better suited forlower-voltage applications.The LT3468 data sheet[2] provides design information,circuits and typical waveforms, and gives a designer goodinsight into the process, possibly sparking new avenues ofapplication. (Note: The data sheet provides a warning relatedto working with high voltage. Many of the circuits discussedcan provide lethal shocks when working properly.)For further illustration of the design process, the following low-power example was built and tested. Test data, noteson circuit operation and waveform are shown. The designL UP Joules from dc sourceEfficiency400 µJUP 500 µJ/pulse.0.8In the fifth and final step, solve for the unknowns. Thereare two unknowns and two equations previously presentedthat will provide the answer. The unknowns are the transformer primary inductance and the peak current. The 500-µJwww.powerelectronics.com12(9 10 6 ).IPEAKLIPEAK 2.2Solve for L:U 29Power Electronics Technology January 2008