mm air gap 90 primary turns will result in about 970

µH primary inductance. 14 secondary

turns give a 6.43 turns ratio, very close to the target. Estimating the thermal resistance of the

ETD29 equal to 26°C/W, the maximum power dissipation (supposed to be on copper only) for

30°C hot-spot temperature rise will be 1.15W (half will be allocated to the primary and half to

the secondary). This requires the resistance of the primary to be no more than 1.26

Ω and the

secondary’s no more than 40 m

Ω. An AWG27 (∅ ≈ 0.4 mm) wire for the primary and a strand

of 5xAWG27 for the secondary will meet the requirement.The primary winding will be split in

two halves of 45 turns each, series connected, and the secondary will be sandwiched in be-

tween to reduce leakage inductance.

6. MOSFET selection

∆V = 373 + 100 + 70 = 543V

There is margin to select a 600 V device. This will minimise gate drive and capacitive losses.

Assuming that the MOSFET will dissipate 5% of the input power, that losses are due to con-

about 2

7. Catch diode selection

=

n

373

6.41

+ 15 = 73.2V

A 100V Schottky diode will minimise conduction losses. As to its current rating, a tentative

could be the STPS8H100D. From the relevant datasheet, the power dissipation is estimated

2

8. Output Capacitor Selection

The minimum capacitance value that meets the specification on the 100/120 Hz ripple is:

=

1

⋅

H2

F2

⋅

=

0.108

⋅ 2

3.14

⋅ 50 ⋅ 0.254 ⋅ 1

= 5417µF

Three 2200

µF electrolytic capacitors will have an ESR low enough to consider the high fre-

quency ripple negligible as well as sufficient AC current capability.

9. Clamp network

With a proper construction technique, the leakage inductance can be reduced to as much as

2% of the primary inductance, that is 20

µH in the present case. A transil clamp is selected.

∆V = 100 + 70 = 170V. The steady-state power dissipa-

tion is estimated to be about 2W. A P6KE170A transil is selected. The blocking diode is an

STTA106.

10. Multiplier bias and sense resistor selection

⋅ 88/264 = 0.8V which, multiplied by the

maximum slope of the multiplier, 1.65, gives 1.32V peak voltage on current sense (CS, pin 4).

Since the linearity limit (1.6V) is not exceeded, this is acceptable. The divider ratio will then be

2.4/(

be 20k

Ω, and the upper one 3MΩ.The sense resistor will not exceed 1.32/2.32 = 0.57Ω (0.5Ω

is selected), while its power rating will be 0.5

2

11. Feedback and Control Loop

The selected optocoupler is a 4N35 from Toshiba. 1 mA quiescent collector current is se-

lected. From opto’s datasheet, with 1mA collector current, the diode current can be between 1

and 2 mA approximately (0.5 < CTR < 1).

AN1059 APPLICATION NOTE

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