Design method of power supply high frequency transformer

2019-05-15 14:30:41 JUKE CHINA ODM OEM Transformer factory Read

Design method of power supply high frequency transformer

The function of the power transformer is power transmission, voltage conversion and insulation isolation. As a main soft magnetic electromagnetic component, it is widely used in power supply technology and power electronics technology. According to the size of the transmission power, the power transformer can be divided into several files: 10kVA or more for high power, 10kVA to 0.5kVA for medium power, 0.5kVA to 25VA for low power, and 25VA for micropower.
Power high frequency transformer.jpg

Designing a High frequency transformer is a difficult point in the power supply design process. The following is a feedback current-type discontinuous power supply high-frequency transformer as an example to introduce a design method of a power supply high-frequency transformer.
Design goals:
The power input AC voltage is between 180V and 260V, the frequency is 50Hz, the output voltage is DC 5V, 14A, the power is 70W, and the power supply operating frequency is 30KHz.
Design steps:
Calculate the primary peak current Ipp of the high frequency transformer
Because it is a current discontinuous power supply, when the power tube is turned on, the current will reach a peak value, which is equal to the peak current of the power tube. The current and voltage relationship of the inductor V=L*di/dt shows that:
Input voltage: Vin(min)=Lp*Ipp/Tc
Take 1/Tc=f/Dmax, then the above formula is:
Vin(min)=Lp*Ipp*f/Dmax
Where: Vin: DC input voltage, V
Lp: high frequency transformer primary inductance value, mH
Ipp: transformer primary peak current, A
Dmax: maximum duty cycle factor
f: power supply operating frequency, kHz
In a current discontinuous power supply, the output power is equal to the energy stored in each cycle at the operating frequency, which is:
Pout=1/2*Lp*Ipp2*f
Dividing it from the inductor voltage gives:
Pout/Vin(min)=Lp*Ipp2*f*Dmax/(2*Lp*Ipp*f)
Therefore:
Ipp=Ic=2*Pout/(Vin(min)*Dmax)
Among them: Vin (min) = 1.4 * Vacin (min) -20V (DC chopping and diode pressure drop) = 232V, taking the maximum duty cycle coefficient Dmax = 0.45. then:
Ipp=Ic=2*Pout/(Vin(min)*Dmax)=2*70/(232*0.45)=1.34A
When the power tube is turned on, the collector should be able to withstand this current.
Find the minimum duty cycle factor Dmin
In a feedback current discontinuous power supply, the duty cycle factor is determined by the input voltage.
Dmin=Dmax/[(1-Dmax)*k+Dmax]
Where: k=Vin(max)/Vin(min)
Vin(max)=260V*1.4-0V (DC chopping)=364V, if 10% error is allowed, Vin(max)=400V.
Vin(min)=232V, if 7% error is allowed, Vin(min)=216V.
Therefore:
k=Vin(max)/Vin(min)=400/216=1.85
Dmin=Dmax/[(1-Dmax)*k+Dmax]=0.45/[(1-0.45)*1.85+0.45]=0.31
Therefore, when the input DC voltage of the power supply is between 216V and 400V, the duty cycle coefficient D is between 0.31 and 0.45.
Calculate the primary inductance value of the high frequency transformer Lp
Lp=Vin(min)*Dmax/(Ipp*f)=216*0.45/(1.34*30*103)=2.4mH
The product Aw*Ae of the winding area Aw and the core effective area Ae is calculated, and the core size is selected.
If we only go around the primary line, then:
Aw*Ae=(6.33*Lp*Ipp*d2)*108/Bmax
Where: d is the diameter of the insulated wire and Bmax is chosen to be Bsat/2.
In a feedback transformer, if the primary part accounts for 30% of the winding area, the remaining 70% is the secondary and insulating space. Therefore, Aw*Ae must be multiplied by 3 times. For safety reasons, we increase it to 4 times. . therefore,
Aw*Ae=4*(6.33*Lp*Ipp*d2)*108/Bmax
This formula is only an estimate, and finally the choice of core and wire frame can be changed.
Assuming that the current density of the winding winding is 400c.m/A, then 400c.m/A*1.34A=536(cm), we refer to the American Wire Gauge (AWG) and take the NO.22AWG wire with a value of 0.028IN. . We choose the E-E core of TDKH7C1 material, which is Bsat=3900G and Bmax=Bsat/2=1950G at 100°C.
Aw*Ae=4*(6.33*Lp*Ipp*d2)*108/Bmax=4*(6.33*2.4*103*1.34*0.0282)*108/1950=3.27cm4
The core of the EE42*42*15 and the wire frame Aw*Ae=1.83cm2*1.83cm2=3.33cm4 can be found in the TDK catalogue. Therefore, we choose this core.
Calculate the air gap length Lg
In a feedback power transformer, the flux changes only in the first quadrant, and the current flux does not appear negative. In order to prevent the transformer from saturating, a large-volume core can be used or an air gap can be used on the magnetic flux path. To reduce the power supply volume, we use the gap method to make the hysteresis loop flat, and reduce the working magnetic force under the same DC bias. Through density.
In the magnetic flux path, the air gap produces a large magnetic reluctance. Therefore, most of the energy stored in the transformer is in the air gap. The air gap volume is Vg and the length is Lg. by,
1/2*Lp*Ipp2=1/2*Bmax*H*Vg*108
Vg=Ae*Lg
u0H=Bmax/(0.4*π)
U0: air permeability = 1
Available:
Lg=0.4*π*Lp*Ipp2*108/(Bmax2*Ae)=0.4*π*2.4*10-3*1.342*108/(19502*1.82)=0.078cm
Therefore, we split the gap of 0.078 cm in the E-E core center column.
Calculate the transformer primary coil Np
Np=Lp*Ipp*108/(Ae*Bmax)=2.4*10-3*1.34*108/(1.82*1950)=90(T)
Calculate the transformer secondary winding Ns
The transformer secondary winding Ns is calculated when the input voltage is minimum and the duty cycle is maximum.
Vout+Vd=Vin(min)*Dmax/(1-Dmax)*Ns/Np (Vd is the rectifier pressure drop)
Ns=Np*(Vp+Vd)*(1-Dmax)/(Vin(min)*Dmax)=90*(5+1)*(1-0.45)/(232*0.45)=2.84(T)
Take an integer of 3 turns.
The input current of the power supply is 14A, 400c.m/A*14A=5600(c.m). Considering the skin effect and the convenient production operation, we use 1400c.m, 4 wires are connected in parallel, and AWGNO:18.