DESIGN

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Y ou learnt about the major ad-vantages of a switch-modepower supply system (SMPS)

and various topologies of power con-version in my article ‘High-FrequencyPower Transformer Design’ publishedin March 2004 issue of EFY. That ar-ticle discussed about the design stepsof flyback transformer. On recommen-dation of EFY readers, I am present-ing here the transformer design usingthe forward topology.

Forward topology

The forward topology is commonlyused in the power range of 50 to 500watts. A forward converter isessentially the isolated version of abuck converter operating in the directmode and the basic single-switchversion (other versions like two-switchforward converter and push-pullconverter) can be successfullyoperated over a wide power range.Due to the transformer, the forwardtopology can be used as either an upor a down converter, although themost common application is downconversion.

A major advantage of the forwardconverter, particularly for low-output-voltage applications, is that the high-frequency output ripple is limited bythe choke connected in series with theoutput. Above 500W, the forward to-pology has some practical limitationsof operation, which we will discuss indetail.

Before jumping into the designsteps of forward transformer, let’s un-derstand operation of the forward con-verter. Consider the circuit shown in

the same time, since D2 is forward-biased, load current flows to the load(RL) through the inductor (L). D3 isreverse-biased, hence no current flowsthrough it.

When MOSFET M1 is turned off,voltages in all thewindings includ-ing the output in-ductor are re-versed and D2 isreverse-biased.Since the inductorcurrent cannotchange instanta-neously, diode D3(called flywheeldiode) conducts

Fig. 1. When power switch M1 isturned on, current builds up in the pri-mary winding (a-b, with ‘a’ positivew.r.t. to ‘b’) and energy is stored. Thisenergy is transferred to the secondary(e-f, with ‘e’ positive w.r.t. to ‘f’). At

� DINESH KUMAR

HIGH-FREQUENCYPOWER TRANSFORMERLearn how to design a high-frequency power transformer using forward topology

Fig. 2: Waveforms of primary current (Ip), switch voltage (Vsw) and inductor current (IL )

Fig. 1: Circuit diagram of forward converter

DESIGN

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the load current. Inthis case, no currentflows through thesecondary winding.

Note here thatenergy transfer fromprimary to secondarytakes place during‘on’ time only. Thisis why this topologyis called ‘forward to-pology,’ whereas inthe case of flyback to-pology energy trans-fer takes place dur-ing ‘off’ time.

To avoid satura-tion of any trans-former or inductor,volt-seconds product(V×t) during ‘on’time must be equalto reverse volt-sec-ond product during‘off’ time.

Now, considerVin volts were ap-plied across the pri-mary winding for‘on’ time Ton (whenM1 is turned on) andcurrent flows in theprimary winding

from ‘a’ to ‘b.’Volt-second product during ‘on’

time =Vin×Ton ................... Eq. (1)But when M1 is turned off, no cur-

rent flows through transformer T1, sothere is unbalancing of volt-secondproduct, which may lead to saturationof the transformer and hence completedistraction of the transformer and maybe of M1. To solve this saturation prob-lem, middle winding or tertiary wind-ing (c-d) with ultra-fast diode is used.The moment M1 is turned off, the volt-ages of tertiary winding (c-d) reverse(‘c’ point is positive w.r.t. ‘d’) and thisinduced voltage in tertiary winding ismore than Vin, hence D1 starts conduct-ing and balances volt-second productof the core. This is calleddemagnetisation or resetting of thecore.

You may be interested in knowingthe voltage across the secondary wind-ings. Assume that ‘N1’ is the number

of turns in the primary winding, ‘N2’is the number of turns in the second-ary winding and ‘N3’ is the numberof turns in the tertiary winding.

When M1 is ‘on’:Voltage induced in secondary

winding

When M1 is ‘off’:Voltage across secondary winding

Many people think that theforward converter can be used below50% duty cycle, but we will convinceourselves that it is possible to use theforward converter above 50% dutycycle, and we will see what are theissues that limit its use above 50% dutycycle.

From Eq. (1), volt-second productrequired to reset the core

where ‘Tr’ is the time required to resetthe core.

To avoid core saturation, the volt-second product must be:

If N1 = N3, we are forced to oper-ate the transformer with less than 50%duty cycle because it needs re-settingtime equal to ‘on’ time. Now what if Iam able to reduce the resetting time(Tr) to less than ‘on’ time as shown inthe waveform. In such a case, I canoperate the transformer with morethan 50% duty cycle. Let’s see how itis possible.

Assume that we have differentnumbers of turns in windings. Toreduce Tr value, I need to increasethe ratio.

Maximum possible duty cycle,

So limit to operate the transformerwith more than 50% is the maximum

TABLE I

Core Area ProductCore No. A

e (mm2) A

w (mm2) A

P (mm4)

E5.3/2.7/2 2.50 4.75 11.88

E6.3/2.9/2 3.30 4.26 14.04

E8.8/4.1/2 5.00 6.90 34.51

E13/6/3 10.10 25.22 254.67

E13/6/6 20.20 25.83 521.77

E13/7/4 12.40 24.75 306.90

E16/8/5 20.10 37.62 756.16

E19/8/5 22.60 54.72 1236.67

E19/8/9 41.30 54.51 2251.27

E20/10/5 31.20 49.14 1533.17

E20/10/6 32.00 57.40 1836.80

E22/16/10 86.00 48.75 4192.50

E25/10/6 37.00 79.68 2948.16

E25/13/7 52.00 87.00 4524.00

E25/13/11 78.40 87.00 6820.80

E30/15/7 60.00 119.31 7158.60

E31/13/9 83.20 107.50 8944.00

E32/16/9 83.00 147.84 12,270.72

E34/14/9 80.70 158.76 12,811.93

E36/21/12 126.00 225.23 28,378.35

E41/17/12 149.00 167.96 25,026.04

E42/21/15 178.00 256.04 45,575.12

E42/21/20 233.00 256.04 59,657.32

E42/33/20 236.00 449.80 106,152.80

E47/20/16 234.00 203.28 47,567.52

E50/27/15 225.00 362.70 81,607.50

E55/28/21 353.00 375.55 132,569.15

E55/28/25 420.00 375.55 157,731.00

N2N1 ×Vin

= ...................... Eq. (2)

...................... Eq. (3)N2N3

–Vin×=

N1N3

Vin= ×Tr...................... Eq. (4)

N1N3

Vin×Ton=Vin×Tr

N1N3

Dmax= ×100≈60%Ton

Ton+0.66Ton

....Eq. (7)

N1N3If =1.5, Tr= =0.66 Ton

Ton

1.5......Eq. (6)

Vin×TonTr = N1

N3Vin

...................... Eq. (5)

Terms used in thetransformer design

Φ = Flux in the core (weber)B = Flux density (weber/m2)A

e= Cross-section area of the core

(m2)A

w= Total window area of the core (m2)

Vin, min

= Minimum DC input voltage (V)V

in, max= Maximum DC input voltage (V)

Np

= Number of turns in the primarywinding

Ns

= Number of turns in the secondarywinding

ap

= Cross-section of the primaryconductor (mm2)

as

= Cross-section of the secondaryconductor (mm2)

fs

= Operating frequency of thetransformer (Hz)

Iin,dc

= Average input DC current (A)Iin, rms

= rms input DC current (A)J = Current density (A/m2)V

d= Voltage drop across output diode

η = Transformer efficiencyD

max= Maximum duty cycle

Po

= Output power

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Np×ap=Aw×Kw

Primary ampere-turns,

Np×Iin,rms=J×Aw×Kw

Area product of core (AP)

Winding design

Turns ratio (N)

Vds,sat is the MOSFET turn-‘on’ volt-age drop, which is different for differ-ent MOSFETs and depends on Rds,on

and current flowing through to theMOSFET. �

The author is a principal engineer (electronics) at

Honeywell Technology Solutions Lab, Bangalore

switch voltage. If we have selected aMOSFET with a high voltage rating,we can use it with a higher duty cyclealso. Leakage induction of the wind-ings also prevents us to use the trans-former with more than 50%.

Assuming that you have under-stood the forward transformer theoryvery well, let’s start design of the mag-netics, i.e., the transformer.

Transformer core areaproduct

From Faraday’s law of induction:

Average input DC current,

Input rms current,

Current density,

Ae=V.Ton

N.∆B

Putting Ton=Dmax

ƒs

=Po Dmax√Vin,min×η

J= Iin, rms = J×ap

Iin, rmsap

⇒

Np=Aw×Kw×J

Iin,rms

=Po Dmax√

Aw×Kw×J×Vin,min×η

Ae= Aw×Kw×j×Vin,min×η×∆B׃s

Vin,min×Dmax× Po Dmax√

=Ae.Aw=Po(Dmax)

Kw×J×η×∆B׃s

3 2

Np> Vin,min×Dmax

Ae×∆B׃s

=Np

Ns

(Vin,min—Vds,sat)×Dmax

Vo+Vd

TABLE II

Wire SelectionWire SWG SWGgauge diameter areasize (mm) (mm²)

10 3.25 8.292

11 2.95 6.831

12 2.64 5.471

13 2.34 4.298

14 2.03 3.235

15 1.83 2.629

16 1.63 2.086

17 1.42 1.583

18 1.22 1.168

19 1.02 0.817

20 0.92 0.664

21 0.81 0.515

22 0.71 0.396

23 0.61 0.292

24 0.56 0.246

25 0.51 0.204

26 0.46 0.166

27 0.41 0.132

28 0.38 0.113

29 0.35 0.096

30 0.31 0.073

31 0.29 0.066

32 0.27 0.057

33 0.25 0.051

34 0.23 0.041

35 0.20 0.032

36 0.18 0.025

37 0.17 0.023

38 0.15 0.018

V.Ton=N.Ae.∆B

Iin, rms=Iin, dc Dmax√

Ns=Np

N

.Eq. (11)

Ae=Vin,min×Dmax

Np×∆B׃s

..................Eq. (8)

Iin,dc=Pin

Vin,min

=Po

Vin,min×η.....Eq. (9)

...Eq. (12)

.....Eq. (13)

=