When calculating the transformer, it is necessary to choose a working point, which is the lowest AC input voltage and corresponds to the maximum output power.
The input is 85V to 265V, the output is 5V and 2A power supply, and the switching frequency is100KHz;
The first step is to select the primary induction voltage VOR, which is set by yourself and determines the duty cycle of the power supply. Maybe friends don't understand what primary induced voltage is. In fact, the primary induced voltage is a typical single-ended flyback switching power supply. Its working period can be understood as follows: when the switch tube is turned on, the primary side is equivalent to an inductor, and the voltage is applied across the inductor, so its current value does not suddenly change, but increases linearly. The formula is I=Vs*ton/L, which are the primary input voltage, the turn-on time and the primary inductance respectively. When the switch tube is turned off, the primary inductor discharges and the inductor current drops again. Similarly, the above formula method should also be respected. At this time, there is a decrease of I=VOR*toff/L, which is the primary induced voltage (discharge voltage), the off time of the switch tube and the inductance. After a period of time, the value of the primary inductor current will return to its original value, so it is impossible to change it. So there is VS*TON/L=VOR*TOFF/L, and rising equals falling. In the above formula, you can use d instead of TON and 1-D instead of TOOF to get the shift term, D=VOR/(VOR+VS). This is the maximum duty cycle. In this paper, the induced voltage is 80V and VS is 90V, so D=80/(*80+90)=0.47.
Step 2, determine the parameters of the primary current waveform.
The primary current waveform has three parameters: average current, root mean square current and peak current. First of all, we should know the waveform of primary current. The waveform of the primary current is shown in the following figure. This is a trapezoidal wave, with horizontal representing time and vertical representing current size. This waveform has three values: one is the average value; The second is effective value; The third is its peak value, and the average value is the area of this waveform divided by its time, as shown in the horizontal line below. First of all, we must determine this value, and the calculation method is as follows: average current = output power/efficiency *VS, because output power multiplied by efficiency is input power, and then input power divided by input voltage is input current, which is average current. Now the next step is to find the peak of the current. What is the peak? We must 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 following ratios are the maximum pulsating current and the peak current respectively. Is between 0 and 1. This value is very important. KRP is known, and now we need to solve the equation. It is known that the area of one period of this waveform is equal to the average current value * 1, and the area of this waveform is equal to the peak current *KRP*D+ peak current *( 1- KRP)*D, so the average current is equal to the above formula, and the peak current = average current /(65438+) is solved. For example, my output is 10W, and the setting efficiency is 0.8. Then the average input current is 10/0.8*90=0. 138A. I set the value of KRP to 0.6, and the maximum value is = 0. 138/( 1-0.5kr).
The three current parameters are the effective values of the current, which are different from the average value. The definition of effective value is to add this current to a resistor. If its thermal 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 current value of this DC, so the effective value of this current is not equal to its average value, generally greater than its average value, and the same average value can correspond to many effective values. If the value of KRP is larger, the effective value will be larger, and the effective value is also related to the duty cycle d, in short, it is closely related to the shape of its current waveform.
Effective current value = d * (krp/3-square of krp+1) Under the root sign of current peak *, for example, under the root sign of effective current value = 0.47 * (0.36/3-0.6+1) = 0.20a, effective current value = 0.4/. The third step is to start the preparatory work for designing the transformer. Given that the switching frequency is 100KHZ, the switching period is 10 microsecond and the duty cycle is 0.47, so TON is 4.7 microseconds.
The fourth step is to choose the transformer core, which is based on experience. If calculation cannot be selected,
Step 5, calculate the number of turns on the primary side of the transformer and the diameter used on the primary side.
When calculating the number of turns on the primary side, it is necessary to select the amplitude b of a magnetic core, that is, the variation range of magnetic induction intensity of this magnetic core. Because the magnetic induction intensity changes after the square wave voltage is added, it is precisely because of this change that the transformer has a function, NP = vs * ton/SJ * B. These parameters are the number of turns in the primary side, the minimum input voltage, the conduction time, the cross-sectional area of the core and the amplitude of the core, which are generally taken as B. This formula comes from Faraday's law of electromagnetic induction. This law is that when the magnetic flux changes in an iron core, it will generate an induced voltage, which is equal to the change amount of magnetic flux/time t multiplied by the turns ratio. By changing the change amount of magnetic flux into the change amount of magnetic induction intensity multiplied by its area, the above formula can be derived, NP=90*4.7 microseconds /32 mm2 *0. 15. 0. 15 of 88 cycles is the selected value. Calculate the number of turns and then 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 you can choose a line of 0.25. The line area of 0.25 is 0.049 mm2 and the current is 0.2 A, so its current density is 4.08. Generally, the selected current density is 4 to 10 ampere square millimeter. If the current is large, it is better to use two or more wires to wind in parallel, because the high-frequency current has a trend effect, which can be better.
Step 6, determine the parameters, turns and wire diameter of the secondary winding.
The induced voltage on the primary side is the discharge voltage, and the primary side discharges to the secondary side through this voltage. Look at the picture, because the output power of the secondary side is too 5V, plus the Schottky tube voltage drop, it is 5.6 V, the primary side discharges 80V, and the secondary side discharges 5.6V. What is the number of turns? Of course, the transformer turns are proportional to the voltage, so the secondary voltage =NS*(UO+UF)/VOR, where UF is the Schottky tube voltage drop, and the secondary turns are equal to 88*5.6/80, so it is 6. 16. Take 6 turns as a whole, and then calculate the secondary wire diameter, and of course, calculate the secondary effective current. The figure below shows the effective current of the secondary. What doesn't stand out is d, which is just the opposite of the original side, but its KRP value is the same as the original side. The peak current is the primary peak current multiplied by its turns ratio, which is several times larger than the primary peak current.
Step 7, determine the parameters of the feedback winding.
Feedback is the flyback voltage, and its voltage is taken from the output stage, so the feedback voltage is stable. TOP's power supply voltage is 5.7 to 9V, which is wound for 7 times, so its voltage is about 6V, and its feedback voltage is flyback, and its turn ratio should correspond to the amplitude and edge. As for the wire, because the current flowing through it is very small, it is enough to use the wire wound on the original side, and there is no strict requirement. Step 8, determine the inductance.
Formula of primary current rise I = vs * ton/L Since the waveform of primary current has been drawn from above, this I is: peak current *KRP, so L=VS.TON/ peak current *KRP, which determines the value of primary inductance.
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 are respectively expressed as maximum 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 inductance coil, and flux linkage is equal to the magnetic flux multiplied by it. The magnetic flux is the magnetic induction intensity multiplied by its cross-sectional area, which is respectively substituted into it, that is, when the primary coil flows through the peak current, the magnetic induction intensity reaches the maximum at this time, and this magnetic induction intensity is calculated by the above formula. The value of BMAX is generally above 0.3T, and it can be larger if it is a good magnetic core; If it exceeds this value, it can be adjusted by increasing the number of turns on the primary side or replacing the magnetic core.
Summary:
When designing high frequency transformer, there are several parameters that need to be set by ourselves, which determine the working mode of switching power supply. The first is to set the maximum duty ratio d, which is determined by the induced voltage VOR set by yourself. Then, set the waveform of the primary current, determine the value of KRP, and set its core amplitude B when designing the transformer, which is another setting. All these settings will make the switching power supply work in the way you set, and finally adjust it constantly to make it work in the most suitable state for you. This is the design task of high frequency transformer.
Formula D=VOR/(VOR+VS) (1)
IAVE=P/ efficiency *VS (2)
IP=IAVE/( 1-0.5KRP)*D (3)
I rms = d * (square of krp/3 -krp+ 1)* root sign (4) at peak current.
NP=VS*TON/SJ*B (5)
NS=NP*(VO+VF)/VOR (6)
L = relative to TON/IP. KRP (7)
BMAX=L*IP/SJ。 NP (8)
But generally speaking, high-frequency transformer is a complicated thing, and this article is too short to explain here. Pay attention to the relationship between various parameters when calculating, because it is a whole, and everyone will be skilled if they analyze and think more. Jiangnan recommendation