The coolest flyback switching power supply transformer design ever

The flyback transformer is the core of the flyback switching power supply. It determines a series of important parameters of the flyback converter, such as duty cycle D and maximum peak current. The design of the flyback transformer is to make the flyback switching power supply work at a reasonable working point. This can minimize the heat generation and the wear of the device. With the same chip and the same magnetic core, if the transformer design is unreasonable, the performance of the entire switching power supply will be greatly reduced, such as the loss will increase and the maximum output power will also decrease. Let me systematically explain my method of calculating the transformer.

To calculate the transformer, you must first select a working point and calculate at this working point, which is the most demanding point. This point is the lowest AC input voltage, corresponding to the maximum output power. Below I will calculate a power supply with an input of 85V to 265V, an output of 5V, 2A, and a switching frequency of 100KHZ.

The first step is to select the primary induced voltage VOR. This value is set by yourself, and this value determines the duty cycle of the power supply. Maybe you don’t understand what the primary induced voltage is. It’s like this. You have to read it from the following, slowly.

This is a typical single-ended flyback switching power supply, which is very familiar to everyone. Let's analyze a working cycle. When the switch is turned on, the primary side is equivalent to an inductor. When the voltage is added to the two ends of the inductor, its current value will not change suddenly, but rise linearly. There is a formula of rising I=Vs*ton/L. These three items are the primary input voltage, the switch on time, and the primary inductance. When the switch is turned off, the primary inductor discharges and the inductor current will decrease again. The above formula should also be followed. At this time, there is a decrease of I=VOR*toff/L. These three items are the primary induced voltage, that is, the discharge voltage, the switch off time, and the inductance. After a cycle, the value of the primary inductor current will return to the original value and will not change. Therefore, VS*TON/L=VOR*TOFF/L. The increase is equal to the decrease. Do you understand? It is easy to understand. In the above formula, D can be used to replace TON and 1-D to replace TOOF. By moving the terms, D=VOR/(VOR+VS). This is the maximum duty cycle. For example, in the design I chose the induction voltage to be 80V and VS to be 90V, then D=80/(*80+90)=0.47

The second step is to determine the parameters of the primary current waveform.

The primary current waveform has three parameters, average current, effective current, and peak current. First, we need to know the waveform of the primary current. The waveform of the primary current is shown in the figure below. It is not well drawn, but don't laugh. This is a trapezoidal wave. The horizontal direction represents time and the vertical direction represents the current size. This waveform has three values, one is the average value, the second is the effective value, and the third is its peak value. The average value is the area of ​​this waveform divided by its time. As shown in the horizontal line below, we must first determine this value. This value is calculated as follows: current average value = output power/efficiency*VS, ​​because the output power multiplied by the efficiency is the input power, and then the input power divided by the input voltage is the input current, which is the average current. Now the next step is to find the current peak value. What is the peak value? We have to set a parameter ourselves. This parameter is KRP. The so-called KRP refers to the ratio of the maximum pulsating current to the peak current. The ratio in the figure below is the maximum pulsating current and the peak current respectively. It is between 0 and 1. This value is very important. Now that KRP is known, we need to solve the equation. Everyone knows how to solve the equation, right? This is a first-grade application question. Let me solve it. It is known that the area of ​​this waveform in one cycle is equal to the average current * 1. The area of ​​this waveform is equal to peak current * KRP * D + peak current * (1-KRP) * D, so the average current is equal to the above formula. The peak current = average current / (1-0.5KRP) * D. For example, my output is 10W, and the set efficiency is 0.8. Then the average input current is 10/0.8*90=0.138A. I set the KRP value to 0.6 and the maximum value = 0.138/(1-0.5KRP). D = 0.138/(1-0.5*0.6)*0.47=0.419A.

The three current parameters are the effective value of the current. The effective value of the current is different from the average value. Do you remember the definition of the effective value? It means that if this current is added to a resistor and the heating effect is the same as that of another DC current added to this resistor, then the effective value of this current is equal to the DC current value. Therefore, the effective value of this current is not equal to its average value, and is generally larger than its average value. Moreover, the same average value can correspond to many effective values. If the value of KRP is larger, the effective value will be larger. The effective value is also related to the duty cycle D. In short, the shape of this current waveform is closely related. I will directly give the current formula of the effective value. This formula can only be derived by integration, so I will not deduce it. As long as everyone distinguishes between the effective value and the average value, it will be fine.

Current effective value = current peak value * D under square root * (KRP square / 3-KRP + 1). For example, in my current case, current effective value = 0.419 * square root of 0.47 * (0.36/3-0.6 + 1) = 0.20A. So for the same power, that is, for the same input current, its effective value is related to these parameters. Properly adjusting the parameters to minimize the effective value, minimize the heat and reduce the loss. This optimizes the design.

The third step is to start the transformer design preparation. If the switching frequency is 100KHZ, the switching period is 10 microseconds and the duty cycle is 0.47. Then TON is 4.7 microseconds. Remember these two numbers, which will be useful in the following.

The fourth step is to select the transformer core. This is based on experience. If you don't know how to choose, just estimate it and calculate it. If it doesn't work, you can change it to a larger or smaller one. However, some materials have formulas or zone line diagrams on how to choose the core according to power. You may also refer to them. I usually rely on experience.

The fifth step is to calculate the number of primary turns of the transformer and the diameter used by the primary side. When calculating the number of primary turns, it is necessary to select an amplitude B of the magnetic core, that is, the range of change of the magnetic induction intensity of the magnetic core, because after the square wave voltage is added, the magnetic induction intensity changes. It is because of the change that it has the function of voltage transformation, NP=VS*TON/SJ*B. These parameters are the number of primary turns, the minimum input voltage, the conduction time, the cross-section area of ​​the magnetic core and the core amplitude. Generally, the value of B is between 0.1 and 0.2. The smaller it is, the smaller the iron loss of the transformer, but the corresponding transformer volume will be larger. This formula comes from Faraday's law of electromagnetic induction. This law states that in an iron core, when the magnetic flux changes, it will generate an induced voltage. This induced voltage = the change in magnetic flux/time T multiplied by the turns ratio, and the change in magnetic flux The above formula can be derived by converting the quantity into the change of magnetic induction intensity and multiplying it by its area. It's simple, right? My NP=90*4.7 microseconds/32 square millimeters*0.15, and I get 88 turns. 0.15 is the value I chose. After calculating the number of turns, determine the wire diameter. Generally speaking, the larger the current, the hotter the wire, so the thicker the wire is needed. The required wire diameter is determined by the effective value, not the average value. The effective value has been calculated above, so let's choose the wire. I can use 0.25 wire. The area of ​​0.25 wire is 0.049 square millimeters, and the current is 0.2 amps, so its current density is 4.08. It's OK. Generally, the current density is 4 to 10 amps per square millimeter. Remember this, it's very important. If the current is very large, it's best to use two or more strands of wire and wind them together, because high-frequency current has a trending effect, which is better.

Step 6: Determine the parameters of the secondary winding, the number of turns and the wire diameter. Remember the primary induced voltage, which is a discharge voltage. The primary discharges to the secondary with this voltage. Look at the figure above. Because the secondary output voltage is 5V, plus the voltage drop of the Schottky tube, there is 5.6V. The primary discharges with a voltage of 80V and the secondary discharges with a voltage of 5.6V. So what is the number of turns? Of course, it follows the law that the number of turns of the transformer is proportional to the voltage. So the secondary voltage = NS *(UO+UF)/VOR, where UF is the voltage drop of Schottky tube. For example, the number of turns of my secondary side is equal to 88*5.6/80, which is 6.16, and the whole number is 6 turns. Then calculate the wire diameter of the secondary side, and of course, calculate the effective value current of the secondary side. Can you draw the waveform of the secondary current? I will draw it for you to see. It is not symmetrical, but it doesn’t matter. As long as you know the meaning, it will be fine. The time with protrusion is 1-D, and the time without protrusion is D, which is just opposite to the primary side, but its KRP value is the same as the primary side. Now you know how to calculate the effective value of this waveform. Oh, let me remind you that this peak current is the peak current of the primary side multiplied by its turns ratio, which is several times larger than the peak current of the primary side.

The seventh step is to determine the parameters of the feedback winding. Feedback is the voltage of flyback, and its voltage is taken from the output stage, so the feedback voltage is stable. The power supply voltage of TOP is 5.7 to 9V. After winding 7 turns, its voltage is about 6V. This is OK. Remember, the feedback voltage is flyback, and its turns ratio should correspond to the amplitude side. Do you understand what it means? As for the wire, because the current flowing through it is very small, it can be wound with the wire wound around the original side. There is no strict requirement.

Step 8, determine the inductance. Remember the formula for the current rise on the primary side? I=VS*TON/L. Because you have drawn the waveform of the primary current from above, this I is: peak current*KRP, so L=VS.TON/peak current*KRP. Do you know? From now on, the value of the primary inductance is determined.

The ninth step is to verify the design, that is, to verify whether the maximum magnetic induction intensity exceeds the allowable value of the magnetic core. BMAX=L*IP/SJ*NP. These five parameters represent the maximum value of magnetic flux, primary inductance, peak current, and primary turns. This formula is derived from the conceptual formula of inductance L, because L=flux linkage/current flowing through the inductor coil, flux linkage equals magnetic flux multiplied by its turns, and magnetic flux is magnetic induction intensity multiplied by its cross-sectional area. Substitute them into the above, that is, when the peak current flows through the primary coil, the magnetic core reaches the maximum magnetic induction intensity at this time, and this magnetic induction intensity is calculated using the above formula. The value of BMAX is generally more than 0.3T. If it is a good magnetic core, it can be larger. If it exceeds this value, you can increase the number of primary turns or change to a larger magnetic core to adjust.

To summarize:

When designing a high-frequency transformer, there are several parameters that need to be set by yourself. These parameters determine the working mode of the switching power supply. The first is to set the maximum duty cycle D, which is determined by the induction voltage VOR set by yourself. Then set the waveform of the primary current and determine the value of KRP. When designing a transformer, you also need to set its core amplitude B, which is another setting. All these settings make the switching power supply work in the way you set. You need to make continuous adjustments and work in a state that is best for you. This is the design task of the high-frequency transformer. Summary

Formula D = VOR / (VOR + VS ) (1)

IAVE=P/efficiency*VS (2)

IP=IAVE/(1-0.5KRP)*D (3)

I effective value = current peak value * D under square root * (KRP square / 3-KRP + 1) (4)

NP=VS*TON/SJ*B (5)

NS=NP*(VO+VF)/VOR (6)

L=VS.TON/IP.KRP (7)

BMAX=L*IP/SJ.NP (8)

But in general, high-frequency transformers are a relatively complicated thing, and my short article is not enough to explain it here. I worked hard for two months to learn high-frequency transformers. I finally felt that I had a clue. Maybe I didn't explain some things clearly, so you can figure it out by yourself. If you don't understand, just ask me in the group, and I will try my best to answer. When learning, pay attention to the relationship between the various parameters, because this thing is originally a whole. If you analyze and think more, I think you will be proficient. Tell me what you have discovered.