The clearest forward magnetic component design ever

In addition to the transformer, the magnetic component of the forward converter also has an inductor, namely the choke. The general information starts from the transformer, but I think it is better to start from the inductor, which is clearer and can make the thinking clearer. Because the forward converter originated from the BUCK converter, and the power heart of the BUCK converter is the energy storage inductor, therefore, the power heart of the forward converter is the choke, not the transformer. The transformer is only responsible for changing the voltage and has no other functions. The power transmission relies on the inductor. Of course, it is not wrong to start from the transformer in general books, but this way of thinking is not very clear and it is not easy for readers to understand. Below I will demonstrate my algorithm, hoping to help readers.

Inductor Design

First, we will study the filter inductor as the research object. In one cycle, when the switch tube is turned on, a voltage is added to both ends of the filter inductor, and its current does not change suddenly, but rises linearly. There is a formula I=V*TON/L, which respectively represents the increment of the inductor current, input voltage, turn-on time, and inductance. This voltage is released by the secondary side of the transformer. When the switch tube is turned off, the inductor discharges at a constant voltage, and its current will decrease linearly, also following this formula, that is, I=Vo*TOFF/L. In one cycle, the discharge current is equal to the charging current, so the above two formulas are equal. Then use 1-D instead of TOFF, and D instead of TON, so Vo=V*D is obtained from the above two formulas. Draw the voltage and current waveforms at both ends of the inductor as shown in the figure below.

Voltage and current waveforms across the inductor

The upper part shows the current waveform, and the lower part shows the voltage waveform. Therefore, the first step in my design is to determine the waveform of the primary current.

The first step is to determine the inductor charging voltage value. First, determine the voltage V applied to the two ends of the inductor when the switch tube is turned on. This voltage is set by the designer. After selecting this voltage, the maximum duty cycle D is determined.

The second step is to set the pulsating value IR of the inductor current . You can draw a graph of the inductor current yourself, which is similar to the above. Then select a pulsating current value, that is, the value of the rising current or the falling current. Because the output power and output voltage are known, the average current value IO is also known.

The third step is to determine the waveform of the current based on the above conditions. To determine this waveform, you need to know its peak value IP. The above conditions are sufficient to calculate this peak value. There is an equation IR/2+(IP-IR)=IO, and the solution is IP=IO+IR/2

Step 4: Set the inductance. According to the waveform of the primary current, calculate the inductance, L=V*TON/IR. Do you understand this formula? It is the same as the one above. Don't say you don't understand it.

The fifth step is to determine the effective value IRMS of this current. This step is used to determine the wire diameter. Note that the effective value, not the average value, is used to determine the wire diameter. The effective value formula of this current waveform is: IRMS=IP*root square〈(KRP square/3-KRP+1)*D〉+IP*root square〈(KRP square/3-KRP+1)(1-D)〉. The derivation of this formula requires integration, which is quite complicated, so I will not explain it. Just remember to use it. After calculating the current value, you can determine the wire diameter. It is very important to make the effective value current density between four amperes per square millimeter and ten amperes per square millimeter. Everyone must remember this.

The above steps complete the design of the inductor, and the above steps determine some important parameters, which will be the basis for the next step of transformer design.

Design of high frequency transformer

In general, the biggest difference between a forward transformer and a flyback transformer is that a forward transformer does not need an air gap and requires its inductance to be as large as possible. There is also current on the primary side of the forward transformer, but this current is not stored by itself through the input voltage, but is induced from the secondary inductance. Knowing this, the forward transformer is easy to design.

The first step is to determine the number of primary turns. Of course, you have to choose a magnetic core first. Assume that the lowest input voltage of the primary side is VS, the conduction time is represented by TON, and you also need to set a magnetic core amplitude. Generally, I take 0.2 to 0.25T, because the forward transformer does not need a DC component, so this value can be larger than the flyback. The number of primary turns NP = VS*TON/AE*B, where AE is the cross-sectional area of ​​the magnetic core.

The second step is to determine the number of turns on the secondary side, because when the switch is turned on, the secondary side needs to discharge at a voltage of V, and this V value has been determined when setting the duty cycle of the switch, so the number of turns on the secondary side NS=NP*V/VS

The third step is to draw the primary current waveform and calculate the effective value of the primary current waveform to determine the wire diameter. As shown in the figure below, because the current waveform is induced from the secondary side, its waveform is the part of the inductor current waveform where the switch tube is turned on.

The peak value of this current waveform is the inductor current peak value divided by the turns ratio. You can calculate this, right? So the effective value of this current waveform = (IP*V/VS)* square root of 〈(KRP squared/3-KRP+1)*D>. Then select the line based on this current value. The current density is the same as above.

The fourth step is to determine the waveform of the secondary current and find the effective value of the secondary current waveform. The waveform of the secondary current is the part of the inductor current when the switch is turned on. This waveform is similar to the waveform of the primary current, because the waveform of the primary current is induced by this. I will not draw it. Its effective value = IP* square root of 〈(KRP square/3-KRP+1)*D〉. Select the line accordingly.

The fifth step is to determine the self-feeding winding. Generally, it is opposite to the same-name terminal of the primary side. Use magnetic reset to release the voltage and induce the voltage. This is how I do it. There are some other solutions. Experts can study it by themselves.

The above is the whole process of my design of the magnetic element of the forward converter. It is closely linked to each other. I think the idea is relatively clear, avoiding complicated formulas and making things easier. I hope my colleagues can give me some suggestions.