How can we effectively design and manufacture high-frequency transformers for power supplies

Designing a high-frequency transformer is a difficult point in the power supply design process. Taking the feedback current discontinuous power supply high-frequency transformer as an example, we will introduce a design method for a power supply high-frequency transformer.

Design goal: 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 operating frequency is 30KHz.

Design steps:

Calculate the primary peak current Ipp of the high-frequency transformer

Since it is a discontinuous current 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. From the current and voltage relationship of the inductor V=L*di/dt, we can know 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: primary inductance of high-frequency transformer, mH

Ipp: transformer primary peak current, A

Dmax: Maximum duty cycle factor

f: power supply operating frequency, kHz

In a discontinuous current 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 this by the inductor voltage gives:

Pout/Vin(min)=Lp*Ipp2*f*Dmax/(2*Lp*Ipp*f)

From this we can get:

Ipp=Ic=2*Pout/(Vin(min)*Dmax)

Where: Vin(min)=1.4*Vacin(min)-20V(DC ripple and diode voltage drop)=232V, take 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 must be able to withstand this current.

Find the minimum duty cycle coefficient Dmin

In a feedback current discontinuous power supply, the size of 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 ripple)=364V. If 10% error is allowed, Vin(max)=400V.

Vin(min)=232V. If 7% error is allowed, Vin(min)=216V.

From this we can get:

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 Lp of the high-frequency transformer

Lp=Vin(min)*Dmax/(Ipp*f)=216*0.45/(1.34*30*103)=2.4mH

Calculate the product Aw*Ae of the winding area Aw and the effective core area Ae, and select the core size.

If we only wind the primary on the bobbin, then:

Aw*Ae=(6.33*Lp*Ipp*d2)*108/Bmax

Where: d is the diameter of the insulating wire, and Bmax is selected as Bsat/2.

In a feedback transformer, if the primary occupies 30% of the winding area, the remaining 70% is the secondary and insulation space, so Aw*Ae must be multiplied by 3. For safety reasons, we increase it to 4 times. Therefore,

Aw*Ae=4*(6.33*Lp*Ipp*d2)*108/Bmax

This formula is just an estimate, and the final choice of core and bobbin can be changed.

Assuming that the current density of the winding wire 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, whose value is 0.028IN. We choose the EE type core of TDKH7C1 material, which has Bsat=3900G and Bmax=Bsat/2=1950G at 100℃, and we can get:

Aw*Ae=4*(6.33*Lp*Ipp*d2)*108/Bmax=4*(6.33*2.4*103*1.34*0.0282)*108/1950=3.27cm4

From the TDK catalog, we can find that the EE42*42*15 core and wire rack Aw*Ae=1.83cm2*1.83cm2=3.33cm4, so we choose this core.

Calculate the air gap length Lg

In the feedback power transformer, the magnetic flux changes only in the first quadrant, and the current magnetic flux will not be negative. To prevent the transformer from saturating, a large core can be used or an air gap can be used on the magnetic flux path. To reduce the volume of the power supply, we use the gap method to make the hysteresis loop flat, which can reduce the working magnetic flux density under the same DC bias.

In the magnetic flux path, the air gap will produce a large magnetic resistance, so most of the energy stored in the transformer is in the air gap, the volume of the air gap is Vg, and the length is Lg. From,

1/2*Lp*Ipp2=1/2*Bmax*H*Vg*108

Vg=Ae*Lg

u0H=Bmax/(0.4*π)

u0: air magnetic permeability = 1

We can get:

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 divide a gap of 0.078cm in the center column of the EE core.

Calculate the transformer primary winding 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

Calculate the transformer secondary winding Ns 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 voltage drop)

Ns=Np*(Vp+Vd)*(1-Dmax)/(Vin(min)*Dmax)=90*(5+1)*(1-0.45)/(232*0.45)=2.84(T)

Take the integer of 3 circles.

The power input current is 14A, 400c.m/A*14A=5600(cm). Considering the skin effect and the convenience of production operation, we use 1400c.m, 4 wires in parallel, AWGNO: 18.

There are many ways to design high-frequency transformers . I hope this example will be helpful to everyone.