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3. Transformer for Flyback converter: Calculation example.

Required Data:

Output power: 405WVo output voltage: 27 VI output current: 15 A

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Home > SPICE Documentation > OrCAD PSpice > Modeling from Datasheet

> Magnetic Core model from Datasheet - Part 1

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Guide to design of flyback and forward transformers - YouSpice, SPICE simulation community http://www.youspice.com/ys/designswitchingtransformersSPICEmodelingcores.3sp?pageNum=2

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Grid voltage VR: 220 Vac +- 20%Grid frequency fr: 50HzSwitching frequency f: 100KHz

1. From the required data, with the aid of Table 1 is detected in ETD49 - N67the ferrite suitable for the realization of the transformer. From the catalog weobtain the magnetic properties below.

2. Through the Table 2 identifies that the maximum allowable temperaturerise for ferrites made with materials N67 is 40 C.

3. Table 3 identifies the value of the thermal resistance of the transformer forthe ETD49 in 8 C/W.

4. the maximum power dissipation for the losses in the ferrite and copper inthe windings is calculated through the (0.2)


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5. The minimum input voltage value is calculated through the (0.3)

Known Data:

Grid voltage: Vr = 220V +-20% VAC

Grid Frequency: fr = 50Hz

Max Peak Voltage: VRi max = (220 + 220*20/100) *SQRT(2) = 373V

Min Peak Voltage: VRi min = (220 - 220*20/100) *SQRT(2) = 249V

Estimated Data:

Efficiency coeffcient 0.8

Input Power : Pi = 405W/0.8 = 506W

Value of Input Capacitor: Ci= 1000uF

Determination of the minimum input voltage

We can reduce of about 10V the calculated value to take account of thevoltage drop due to the resistance of the winding and to the diodes atmaximum power.


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6. Distribution of power lost from the (0.5)

7. Calculation of power lost in specific volume for the ferrite ETD49 fromthe(0.6).

Known Data:

Power loss in square wave:

Losses single-ended vs push-pull

Considering that the volume of a pair of ferrites ETD49 is 24100mm^3, theloss of specific power per unit volume will be:


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From the diagram below, for the material chosen N67 Pv = f (f) for f =100KHz and T = 100 C we get a delta Bmax of about 180mT:

8. Determination of the maximum permitted flux delta Bmax as a function ofthe losses in the ferrite.

With the (0.7) we can determinate also by analytical way the maximum valueallowed for the induction, that for material N67 at 100 C and with afrequency of 100KHz is:


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9. Calculation of the number of turns in the primary through the (0.8) and(0.9).

Duty Cycle Tv= 0.5

Max on time:

10. Estimation of resistance in Ohm of the primary winding through (0.10)

Duty Cycle: tv = 0.4


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Maximum closing time:

Figure 3 - Parameters of support for ETD49 FERRITES and Accessories,Siemens Matsushita Components.

11. Calculatiopn of the number of secondary turns with (0.12).

To calculate the number of secondary turns is necessary to identify a goodapproximation of the real value of thevoltage applied to the primary and the secondary voltage required. For thisreason, are taken into account for the


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voltage drops to the diode of the secondary winding, estimated at Vd = 2V.

Obtaining a non-integer number of turns, it is approximated to the nearestinteger. As alternative, in an interactive way, we might recalculate thenumber of turns in the primary so that the number of secondary turns is aninteger number. Consider, however, that the number of turns in the primary isinversely proportional to the flux, to avoid that this increases, the number ofturns in the primary can be only increased.

12. Calculation of the section of the windings through (0.13).

(0.24)

(0.25)


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13. Calculation of the primary current with (0.17), (0.18), (0.19).

The primary current is assumed triangular in shape. The copper losses areconsidered to be equally divided between primary and secondary.

14. Calculation of the maximum allowed inductance of the primary from(0.20).


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15. Calculating the maximum value of the inductance factor AL through the(0.21).

If the ratio between the thickness of the winding and its width is low, thevalue of AL can be reduced; Furthermore the tolerance in the value of AL forthe type of ferrite choice, encoded as G class , is +- 2%.Considering 106nH as the maximum value we can determine the averagevalue by estimating a maximum deviation of AL of 10% (empirical data):

16. Determination of the value of the air gap from (0.22).

Known the value of AL we can find the value of air gap. From the catalog wecan find the half-core with gap close to the value of AL calculated, orestimate its value by the formula. The value can be obtained by combiningthe half-cores with different gap values. This represents the best solution,when it's possible, to contain the leakage flux and the parasitic inductance.Otherwise the value of air gap required is achieved by interposingshims between the half-cores.


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The (0.22) reminds us that the validity of the formula is restricted to gapvalues between 0.10 and 3.00 mm, then most likely the value of 5.02 mm isnot suitable for AL of 95nH. In these cases it is necessary to measure thevalue of the primary inductance and adjust the gap to obtain the calculatedvalue.

17. Maximum power transferred by the transformer (0.23).


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guide to design of flyback and forward transformers3 - youspice, spice simulation community

Documents

input power

output power

calculation of power

design of flyback

power loss

distribution of power

maximum power dissipation

minimum input voltage