introduction
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 the duty cycle D and the maximum peak current. The design of the flyback transformer is to make the flyback switching power supply work. At a reasonable working point. This allows the heat to be as small as possible and the wear on the device to be as small as possible. The same chip, 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, the maximum output power will also decline, the following I say the system I design transformer Methods.
To design a transformer, you must first select a working point. At this working point, this is the most demanding point. This point is the lowest AC input voltage, which corresponds to the maximum output power. Below I will calculate an input 85V to 265V, output 5V, 2A power supply, the switching frequency is 100KHZ.
The first step is to select the primary side induced voltage VOR.
This value is set by itself, and this value determines the duty cycle of the power supply. Maybe my friends don't understand what is the primary side induced voltage. For the sake of understanding, let's start with the example shown in Figure 1 below, and slowly come.
This is a typical single-ended flyback switching power supply. Everyone is familiar with it. Let's analyze the working condition of a working cycle. When the switching transistor is turned on, the primary side is equivalent to an inductor, and the voltage is applied across the inductor. The current value does not mutate, but the linear rise, there is a rising current of the formula:
I liter = VS * Ton / L
These three items are the primary input voltage, the switch turn-on time and the primary side inductance. When the switch is turned off, the primary inductor discharges, and the inductor current decreases. Also, the above formula is obeyed. At this time, there is a dropped current:
I drop = VOR * TOFF / L
These three terms are the primary induced voltage (ie discharge voltage), switching tube turn-off time and inductance. After one cycle, the primary inductor current will return to its original value and it will not change. Therefore, there are:
VS*TON/L=VOR*TOFF/L
That is, the rise is equal to the decline, understand? So understand! In the above formula, you can use D instead of TON, and use (1-D) instead of TOFF. The shift can be obtained:
D= VOR/(VOR +VS)
This is the maximum duty cycle. For example, the transformer I designed, I chose the induced voltage to be 80V, VS is 90V, then D=80/(80+90)=0.47
The second step is to determine the parameters of the primary current waveform.
The edge current waveform has three parameters, average current, rms current, and peak current. First, the waveform of the primary current must be known. The waveform of the primary current is shown in the figure below. This is a trapezoidal wave showing the time in the horizontal direction and the current in the vertical direction. This waveform has three values, one is the average value Iav, the other is the effective value I, and the third is the peak value Ip. The average value is the area of ​​the waveform divided by Time. As shown by the horizontal line below, you must first determine this value. This value is calculated as follows.
Iav=Po/(η*VS)
Since the output power divided by the efficiency is the input power, then the input power divided by the input voltage is the input current, which is the average current. The next step is to find the current peak.
In order to find the current peak, we have to set a parameter. This parameter is KRP. The so-called KRP refers to the ratio of the maximum ripple current ΔIM to the peak current Ip (shown in Figure 2). The KRP value ranges from 0 to 1. between. This value is very important. Knowing KRP, now we need to solve the equation, we will solve the equation. This is the first application problem. Let me explain that the area of ​​this waveform is known to be S=Iav*1, and the area S of this waveform is equal to :
S=IM*KRP*D/2+IM*(1-KRP)*D
Therefore, the current average is equal to the above formula, and the peak current is solved:
IM=Iav/[(1-0.5 KRP)*D].
For example, if my output is 10W and the setting efficiency is 0.8, then the average current input is:
Iav=10/0.8*90=0.138A,
I set the value of KRP to be 0.6 and the maximum value:
IM=0.138/(1-0.5KRP).D=0.138/(1-0.5*0.6)*0.47=0.419A.
The effective value I of the ball current below, the current effective value and the average value are not the same, the definition of the effective value is still remembered, that is, the current is added to a resistor, if it is heated and another DC current is added to this If the heating effect on the resistor is the same, then the effective value of this current is equal to the current value of the DC. Therefore, the effective value of this current is not equal to the average value, generally larger than the average value, and the same average value can correspond to many The effective value, if the value of KRP is selected, the effective value will be larger, and the effective value is also related to the duty ratio D. In short, the shape of the current waveform is closely related. I will give it directly. The current formula of the value, this formula can be derived by using the integral, I will not push it, as long as you distinguish the effective value and the average value.
As shown in the circuit shown in Figure 1, the current RMS:
Therefore, corresponding to the same power, that is, when there is the same input current, the effective value is related to these parameters, and the appropriate adjustment parameters make the effective value minimum, the heat is minimized, and the loss is small. This optimizes the design.
The third step is to select the transformer core
This is based on experience. If you don't choose, you can estimate one, and the calculation will do. If it doesn't work, you can change it to a larger or smaller one. However, some data have how to select the core according to the power. Formula or zone chart, you may wish to refer to it. I usually come from experience.
The fourth step is to calculate the number of primary turns of the transformer.
The diameter of the primary side. When calculating the number of primary edges, the amplitude B of a core is selected, that is, the variation range of the magnetic induction of the core, because the magnetic induction is changed after the upper wave voltage is applied. It is because of the change that it has the effect of transformation.
NP=VS*TON/SJ*B
These parameters are the primary side turns, the minimum input voltage, the on-time, the cross-sectional area of ​​the core and the core amplitude. Generally, the value of B is between 0.1 and 0.2. The smaller the transformer, the smaller the iron. The smaller the loss, the larger the volume of the corresponding transformer. This formula is derived from Faraday's law of electromagnetic induction. This law says that in a core, when the flux changes, it produces an induced voltage. = The amount of change in flux/time T is multiplied by the turns ratio. The amount of change in flux change to the amount of change in magnetic induction multiplied by its area can be used to derive the above formula. Simple. My NP=90*4.7 μS/32mm2*0.15==88.15, taking an integer of 88åŒ.
Calculate the number of turns and then determine the wire diameter. Generally speaking, the larger the current, the more easily the wire is heated, and the thicker the wire is needed. The required wire diameter is determined by the current effective value, not the average value. The effective value has been calculated above, so the line is selected. I can use the 0.25 line, with a line of 0.25, the area is 0.049 square millimeters, the current is 0.2 amps, so the current density is 4.08. Generally, the current density is 4~10A/mm2. It is important to remember this. In addition, because the high-frequency current has a tendency, if the current is large, it is best to use two or more wires and wrap around. Better results.
The fifth step is to determine the number of parameters and the wire diameter of the secondary winding.
Remember the primary side induced voltage, this is a discharge voltage, the primary side is to discharge this voltage to the secondary side, see the picture above, because the secondary side output voltage is 5V, plus the pressure drop of the Schottky tube, there is 5.6V, the primary side is discharged with a voltage of 80V, and the secondary side is discharged with a voltage of 5.6V. What is the number of turns? Of course, it follows the law that the number of transformer turns is proportional to the voltage. Therefore, the number of turns of the secondary side is:
NS=NP*(UO+UF)/VOR,
Among them, UF is the Schottky tube pressure drop. If my secondary side turns is equal to 88*5.6/80, I get 6.16 and take 6åŒ.
To calculate the line diameter of the secondary side, first calculate the rms current of the secondary side. Will the waveform of the secondary current be drawn? Let me show it to you. The time for the protrusion is 1-D, and the time without the protrusion is D. Just the opposite of the original side, but the value of KRP is the same as the original side. Now, how do you know the effective value of this waveform? Oh, let me remind you that this peak current is the primary peak current multiplied by its turns ratio, which is several times larger than the primary peak current.
Sixth, step to determine the parameters of the feedback winding
The feedback is the voltage of the flyback. The voltage is taken from the output stage, so the feedback voltage is stable. The power supply voltage of TOP is 5.7 to 9V. When it is wound around 7åŒ, the voltage is about 6V. This is fine. Remember, the feedback voltage is flyback, and the turns ratio corresponds to the width. Do you know what it means? As for the line, because the current flowing through it is very small, it can be wound around the original side. Strict requirements.
The seventh step is to determine the inductance
Remember the current rise formula of the primary side? I liter = VS * TON / L. Because you have drawn the waveform of the primary current from above, this I liter = IM * KRP, so:
L=VS.TON/(IM*KRP)
Did you know, from then on, the value of the primary inductance is determined.
Step 8, verify the design
That is to verify whether the maximum magnetic induction exceeds the allowable value of the core, there is BMAX = L * Ip / SJ * NP.
These five parameters represent the maximum flux, primary inductance, peak current, core cross-sectional area, and primary turns. This formula is derived from the conceptual formula of inductance L because The current flowing through the inductor, the flux linkage is equal to the magnetic flux multiplied by its number of turns, and the magnetic flux is the magnetic induction multiplied by its cross-sectional area, which is substituted respectively, that is, when the primary coil flows through the peak current, the magnetic core To achieve the maximum magnetic induction, the magnetic induction is calculated by the above formula. The value of BMAX generally exceeds 0.3T. If it is a good core, it can be larger. If it exceeds this value, it can increase the number of primary turns, or Change the magnetic core to adjust.
to sum up
Design high-frequency transformers, there are several parameters to set, these parameters determine the working mode of the switching power supply, the first is to set the maximum duty cycle D, this duty cycle is set by yourself The induced voltage VOR is determined, and then the waveform of the primary current is set to determine the value of KRP. When designing the transformer, the core amplitude B is also set, which is a setting, and all these settings are made. This switching power supply works under the mode you set. To constantly adjust and work under the best condition for you, this is the design task of high frequency transformer.
Laptop Holder is upgraded high-strength aluminum alloy, all-aluminum alloy is durable and has a delicate texture. Laptop Holder Vertical is more stable and more durable. The size is enlarged and widened. Laptop Holder Portable is suitable for all laptops. Full non-slip silicone surface care, portable and foldable, six-level height adjustment. Laptop Holder Rose Gold is lightweight, portable and easy to store, stable without shaking, hollowed out heat dissipation design at the bottom. Laptop Holder Mount is new posture, just for more comfortable, ergonomically improved sitting posture.
Skyzonal Aluminum Laptop Stand, Portable Adjustable Aluminum Laptop Stand, Amazon Aluminum Laptop Stand
Shenzhen ChengRong Technology Co.,Ltd. , https://www.laptopstandsuppliers.com