Designing transformers, you should first select a work point, calculated on this work point, this is the most demanding point, this point is the lowest AC input voltage, corresponding to the maximum output power. Below I will calculate an input 85V to 265V, output 5V, 2A power supply, and the switching frequency is 100KHz.
The first step, select the original sensing voltage Vor
This value is set by ourselves, this value determines the duty cycle of the power supply. Maybe friends don't understand what is the original sensing voltage. For the convenience of understanding, we will talk slowly from the example shown below.
figure 1
This is a typical single-ended anti-excitement switch power supply. Everyone is familiar, and the work cycle work is analyzed below. When the switch tube is opened, the original side is equivalent to an inductor, and the inductance plus voltage The current value does not mutation, and the linear rising, there is a formula rose:
I liters = vs * Ton / L
These three items are the original input voltage, switch opening time and original inductance. When the switch tube is turned off, the original inductance discharge, the inductor current will decline, and also respect the above formula law, and there is a decline in current:
I drop = vor * Toff / L
These three items are the original induction voltage (ie, the discharge voltage), the switching tube shutdown time and the amount of inductance. After one cycle, the original inductance current will return to the original value, it is impossible to change, so it is:
Vs * ton / l = VOR * TOFF / L
That is, it is equal to decline, understand? I understand! In the above formula, D can be used instead of TON, and (1-d) can be used instead of TOFF. Move from:
D = VOR / (VOR + VS)
This is the maximum duty cycle. For example, I designed this transformer, I selected the sensing voltage of 80V, VS is 90V, then D = 80 / (80 + 90) = 0.47
In the second step, determine the parameters of the primary current waveform
There are three parameters, average current, effective value current, and peak current. First, we must know the waveform of the primary side current, and the waveform of the original side current is as shown below. This is a trapezoidal wave transverse indication time, longitudinal indicates current size, this waveform has three values, one is the average value IAV, and the second is the valid value I, and the three is its peak IP, the average value is to remove the area of this waveform. time. As shown in the horizontal line below, first determine this value, this value is calculated, the current average:
IAV = PO / (η * vs)
Since the output power is in addition to the input power, the input power is then input to the input voltage, 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 is that the ratio of the maximum pulsating current ΔIm and the peak current IP (shown in Figure 2), KRP ranges from 0 and 1 between. This value is very important. KRP is known, now solve the equation, will solve the equation, this is the first application question, I will know, I know this waveform, the area S = IAV * 1, this waveform is equal to :
S = IM * KRP * D / 2 + IM * (1-KRP) * D
Therefore, there is average of the current equal to the above formula, and the peak current is solved:
IM = IAV / [(1-0.5 KRP) * D].
For example, this output is 10W, the setting efficiency is 0.8, the average current input is:
IAV = 10 / 0.8 * 90 = 0.138A,
I set the value of KRP 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 valid value of the lower ball current is different, the current validity value, and the average value are different. If the definition of the effective value remembers, it is to add this current to a resistor. If it is fever and a DC current is added 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 this DC. Therefore, the effective value of this current is not equal to its average, typically larger than its average. And the same average, you can correspond to many effective values. If the value of the KRP is selected, the greater the valid value, the effective value is also related to the duty cycle D, in summary. Its current waveform is closely related. I will directly give a valid current formula. This formula can be pushed with points, I will not push it, as long as everyone is distinguished, the valid value and the average value can be
As shown in Figure 1, the current is valid:
Therefore, corresponding to the same power, it is the same input current, its valid value and these parameters are related, appropriate adjustment parameters, so that the valid value is minimized, the heating is minimized, and the loss is small. This optimizes the design.
In the third step, select the transformer magnetic core
This is experienced. If you don't choose, you will estimate it, the calculation is calculated, if you can't, you can change a big one or a small, how to choose the magic core according to the power Formula or zone line diagram, you may also wish to refer to it. I generally come with experience.
In the fourth step, calculate the number of turns of the transformer
The line diameter used by the original side. When calculating the number of turns, the amplitude B of the core is selected, that is, the magnetic induction of the magnetic core of this magnetic core, because when the square wave voltage is added, this magnetic induction intensity is changed, because of the change, so It has the effect of transformer,
NP = vs * ton / sj * b
These parameters are the number of turns, minimum input voltage, on time, magnetic core cross-sectional area and magnetic core amplitude, typically the value of B is between 0.1 and 0.2, the smaller, the micro-transformer The smaller the damage, but the volume of the corresponding transformer will be large. This formula is derived from the Law of Fara, which is said that in one core, when the magnetic flux changes, it produces an induced voltage, this induction voltage = the amount of magnetic flux / time T multiplied by turns Comparison, the amount of change in magnetic flux changes to magnetic induction strength can be launched by its area, simple. My NP = 90 * 4.7μs / 32mm2 * 0.15 == 88.15, the total number of pieces is 88.
Forget the number of turns, then determine the wire diameter. In general, the larger the current, the easier the wire is used, the thicker the wire required, and the needed wire diameter is determined by the current valid value, not the average. The valid value has been calculated above, so come to the line. I can use 0.25 lines, with 0.25 line, the area is 0.049 square millimeters, and the current is 0.2, so its current density is 4.08. Generally, the current density is 4 ~ 10A / mm2. Remember this is important. In addition, because the high frequency current has the effect, if the current is very large, it is best to use two or more or more or more lines, which is better.
The fifth step is determined to determine the number of parameters and line dialators of the secondary winding.
Remember the original induction voltage, this is a discharge voltage, the original side is the second side of this voltage discharge, watching the picture, because the secondary output voltage is 5V, plus the Schott branch pressure drop, there is 5.6V, the original side is discharged from 80V, the subside is discharged by 5.6V voltage, then the number of turns is, of course its law complies with the number of transformers and the voltage proportional to the voltage. So the number of hemps:
NS = np * (uo + uf) / VOR,
Where uf is a Schottky pressure drop. As I am equipped with the number of turns equal to 88 * 5.6 / 80, get 6.16, pay 6.
It is necessary to calculate the secondary line diameter, first calculate the valid value of the second side, the waveform of the secondary current will draw, I painted it to everyone, there is a protrusion time is 1-D, no protrusions are D. , Just opposite the primary side, but its KRP is the same as the original side, what is the effective value of this waveform? Oh, then reminded that this peak current is the primary side of the peak current by its number of turns, and it is much longer than the original peak current.
image 3
Sixth, determine the parameters of the feedback windings
Feedback is a flying voltage, its voltage is taken from output stage, so the feedback voltage is stable, TOP's power supply voltage is from 5.7 to 9V, to the 7th, then the voltage is about 6V, this is OK. Remember, the feedback voltage is flying, its number of turns corresponds to the right side, what does it mean? As for the line, because the current flowing is small, it is useful to use the line around the original side, no Strict requirements.
Step 7, determine the amount of inductor
Remember the current of the original to rise formula I liters = vs * Ton / L. Because you have drawn from above, the waveform of the primary current, this I liter = IM * KRP, so:
L = vs.ton / (iM * KRP)
I know, I have determined the values of the original inductor.
Eighth step, verification design
That is, if the maximum magnetic induction strength is not exceeding the allowable value of the core,
BMAX = L * IP / SJ * NP.
This five parameters represent the maximum magnetic flux, the original edge inductor, the peak current, the cross-sectional area of the core, the original side, the formula is pushed from the concept formula of the inductance amount L, because L = magnetic chain / The current flowing through the inductor coil, the magnetic chain is equal to the magnetic flux multiplying the number of turns, and the magnetic flux is that the magnetic induction intensity is multiplied by its cross-sectional area, which is substituted to the upper, that is, when the original side coil flows through the peak current, at this time the magnetic core To achieve the maximum magnetic induction strength, this magnetic induction strength is calculated using the above formula. The value of Bmax is usually more than 0.3t. If it is a good magnetic core, it can be large. If it exceeds this value, you can increase the number of turns, or Change the magnetic core to adjust. Read more
Our other product:
