－Just looking at the circuit diagram, it looks like an unregulated isolated power supply; is the output really regulated?

Of course. Allow me to explain how the output of the BD7F series is regulated. The explanation will be easier using graphics, which I have prepared.

The circuit diagram on the left side of the graphic shows the simplified circuit block for the DB7F series device, an external transformer, and a secondary-side rectifying circuit. On the right side are waveforms of the voltage (Vx) and current (Ix) at each node.

How to perform regulation without feedback from the secondary side? The *flyback voltage* V_{SW} on the primary side contains V_{out} as a component. Then V_{SW} is monitored and operation is performed to indirectly detect V_{out} for regulation control.

In order to explain in a little more detail, I'll ask you to look at the waveforms and the equation within the graphic. V_{sw} is the primary-side flyback voltage, and is generated, when the internal switch (MOSFET) turns off, at the point at which one end of the transformer primary-side windings is connected to the SW pin of this IC. This V_{SW} is the result of the product of the “transformer windings ratio (primary-side Np/secondary-side Ns)” and the “sum of V_{out} and the output rectifying diode voltage V_{f}”, to which V_{in} is added. Put simply, the voltage is the result of adding V_{in} to the voltage (vout+V_{f}) multiplied by the windings ratio of the transformer. But rather than an explanation, the waveforms and equation are easier to understand. One thing must be noted though: this equation has been simplified to make it easier to grasp. The voltage appearing on the secondary side is represented at (V_{out}+V_{f}), but strictly speaking, it is necessary to add a voltage resulting from the total secondary-side impedance (the windings resistance and ESR), multiplied by I_{S}. This is an error factor, and in actual design the term must be included in calculations.

－I see that the primary-side flyback voltage is used. I understand that it is used in some kind of operation, but if it's not too complicated, could you explain just how this works?

The basic principle is not all that difficult. But, it is easier to understand using a circuit diagram and an equation, and so let's have a look at another diagram and another equation. What's important here are the two places that are circled in the circuit diagram. Details are indicated in the product data sheet, and the equation from the data sheet is not the simplified version we presented above, but includes all the terms for the secondary-side voltage, and so for consistency, from here on we will use the data sheet equation. The symbol notation (lowercase, subscripts) are somewhat different, but the meanings are the same.

The following is the basic equation. V_{SW} is the flyback voltage at the SW pin. Here “I_{S}×ESR” is added to the previous simplified equation.

V_{SW} is converted into the current I_{RFB} by the resistor R_{FB}. The voltage V_{FB} at the FB pin is roughly equal to the voltage V_{IN} due to the differential circuit with V_{IN}. This operation should be thought of and remembered as like that of the differential circuit of an op-amp, based on loop control. I_{RFB} is represented by the following equation.

V_{FB}≒V_{IN}, and therefore it is clear that the voltage obtained by subtracting V_{IN} (V_{FB}) from V_{SW}, divided by R_{FB}, gives us I_{RFB}. Here, a term with V_{OUT} appears in the equation. From here we shall determine V_{OUT}.

Because this current I_{RFB} flows through the resistor R_{REF}, the voltage V_{REF} at the REF pin is as given by the following equation.

We have already expanded I_{RFB}, and as is only to be expected, retracing gives us V_{REF}＝ R_{REF} × I_{RFB}.

Here, let's look at the circuit diagram. The REF pin voltage is input to the non-inverting pin of the comparator within the IC and is compared with the reference voltage of 0.78 V at the inverting pin. For control ICs in general, this would be an error amplifier, but in this IC the control method uses the comparator, and so the action is that of a *comparator*. Similarly, to an error amplifier, the voltage at the REF pin (the comparator inverting input) is equal to the reference voltage (0.78 V). Hence the output voltage V_{OUT} and the voltage V_{REF} at the REF pin are related as follows.

In other words, the output voltage V_{OUT} is determined by the transformer windings ratio and the ratio of the resistances R_{FB} and R_{REF}. V_{F} and I_{S}×ESR are factors resulting in output voltage errors.

－I see. I now understand the meaning of using loop control of the primary-side flyback voltage to regulate the output voltage indirectly. During design, then, the output voltage should be set based on this equation?

That is not a problem. The data sheet indicates 3.9 kΩ as the standard value for R_{REF}, and V_{REF} is 0.78 V (typ), and so the output voltage can be set using these values. Moreover, application circuit examples and recommended transformers are provided. By all means, please have a look at the data sheet.

－You have clarified the reasons why a photocoupler or auxiliary windings are unnecessary. Next, I would like to ask about the functions and features of the DB7F series of ICs.

(to be continued）