Designing Isolated Flyback Converter Circuits: Selecting Critical Components ? Output Rectifier and Cout

In this section, the rectifying diode D6 and the output capacitors (Cout) C7 and C8, provided on the secondary side of the transformer T1 as the output, are explained.

As the overall flow, as explained in the section [Isolated Flyback Converter Basics: Switching AC-DC Conversion] , the input voltage from the AC power supply is first rectified by a diode bridge and converted to a DC voltage. This DC voltage is cut (chopped) into the required power amounts through switching of a MOSFET controlled by a power supply IC to again be converted to AC, and a rectification and smoothing circuit in the output stage again converts this to the desired DC voltage, which in this design is 12 V DC. For an explanation of the entire circuit, please see the section [Designing Isolated Flyback Converter Circuits].

Output rectifying diode D6

With a margin included, the diode selected is 87 V÷0.7＝124 V ⇒ 200 V

The diode losses (estimated) are: Pd＝VF×Iout＝1 V×3 A＝3 W

In general, it is recommended that a diode be used at 70% or lower than the rated voltage and at 50% or lower than the rated current. In the circuit shown, a RF1001T2D ROHM fast-recovery diode (200 V, 10 A, TO-220F package) is used.

Finally, increases in temperature are checked with the diode incorporated into the circuit, and if necessary the component is reviewed, and the possibility of heat dissipation by a heat sink is studied.

Output capacitors (Cout) C7, C8

The impedance of electrolytic capacitors (low-impedance products) for general switching power supplies is stipulated to be 100 kHz, and so:

In other words, when a ripple voltage of 200 mVpp is taken to be the allowed value, a capacitor with an impedance of 0.01 Ω or lower must be selected.

Next the ripple current is determined, and on the basis of the ripple current, the ripple current rating of the capacitor is studied.

The rated voltage is as a rule set to about twice the output voltage. It should be at least Vout×2＝12V×2＝24V ⇒ 25V

In the circuit shown, two low-impedance capacitors (35 V, 1000 μF) for a switching power supply are used in parallel.

In general, electrolytic capacitors are used as the output capacitors. Electrolytic capacitors have a service lifetime, and if large ripple currents are passed, the lifetime is shortened. The capacitor manufacturer provides methods for calculating lifetime and related stipulations, and so the information provided by the manufacturer of the capacitor to be used should be consulted.

The output ripple voltage and ripple current of an actual circuit should be checked.