What Is a Flyback Transformer?

Flyback transformer is also called single-ended flyback or "Buck-Boost" converter. It gets its name because its output terminal gets energy when the primary winding is disconnected from the power supply. Flyback converters are popular among development engineers for their simple circuit structure and low cost.

Flyback transformer

Flyback transformer is also called single-ended flyback or "Buck-Boost" converter. It gets its name because its output terminal gets energy when the primary winding is disconnected from the power supply. Flyback converters are popular among development engineers for their simple circuit structure and low cost.

Flyback transformers are suitable for small power supplies and various power adapters. However, the design difficulty of the flyback converter is the design of the transformer, because the input voltage range is wide, especially at low input voltage, the transformer will work in continuous current mode under full load conditions, and under high input voltage, light load conditions Will work in discontinuous current mode again.

Chinese name

When the transistor Tr ton is switched, the primary Np of the transformer has a current Ip and stores energy in it (E = Lp * Ip ^ 2/2). Since Np and Ns have opposite polarities, at this time

Flyback transformers generally work in two modes:

1. Inductive current discontinuous mode DCM (Discontinuous Inductor Current Mode), or "full energy conversion": All energy stored in the transformer at the same time is transferred to the output during the flyback period (toff).

2. Inductor current continuous mode CCM (Continuous Inductor Current Mode) or "Incomplete Energy Conversion": A part of the energy stored in the transformer is retained at the end of toff until the beginning of the next ton cycle.

DCM and CCM are very different in terms of small-signal transfer functions, and their waveforms are shown in Figure 3. In fact, when the converter input voltage VIN changes within a larger range, or the load current IL changes within a larger range It must inevitably span two working modes. Therefore, the flyback converter is required to work stably in DCM / CCM. However, it is difficult to design. Usually we can use the DCM / CCM critical state as the design basis. Equipped with current mode control PWM. This method can effectively solve various problems in DCM, but does not eliminate the inherent instability problem of the circuit in CCM. You can adjust the control loop gain to deviate from the low frequency band and reduce the transient response speed to solve CCM Caused by the transfer function "right half plane zero".

DCM and CCM are very different in terms of small signal transfer functions.

DCM / CCM primary and secondary current waveform

In fact, when the converter input voltage VIN changes in a large range, or the load current IL changes in a large range, it must inevitably span two working modes. Therefore, the flyback converter requires DCM / CCM Both can work stably. However, it is more difficult to design. Usually we can use the DCM / CCM critical state as the design basis. It is also equipped with current mode control PWM. This method can effectively solve various problems in DCM. CCM does not eliminate the inherent instability problem of the circuit. The instability caused by the "right half-plane zero point" of the transfer function can be solved by adjusting the control loop gain to decouple from the low frequency band and reduce the transient response speed.

In a steady state, the change in the magnetic flux increment at ton must be equal to the change at "toff", otherwise the core will be saturated.

therefore,

= VIN ton / Np = Vs * toff / Ns

That is, the volts / second per turn of the transformer's primary winding must be equal to the volts / second per turn of the secondary winding.

Comparing the current waveforms of DCM and CCM in Figure 3, it can be known that during the Tr ton period in the DCM state, there is a higher primary peak current in the entire energy transfer waveform. This is because the primary inductance Lp is relatively low, which makes Ip The negative effect caused by the sharp rise is increased winding loss and ripple current of the input filter capacitor, which requires that the switching transistor must have high current carrying capacity to work safely.

In the CCM state, the primary peak current is lower, but the switching crystal has a higher collector current value in the ton state. This results in high power consumption of the switching crystal. At the same time, in order to achieve CCM, a higher transformer element is required. The side inductance Lp, the residual energy stored in the transformer core requires the transformer to be larger in size than the DCM, and other coefficients are equal.

In summary, the transformers of DCM and CCM are basically the same when designing, but there are some differences in the definition of the peak current on the primary side (Ip = Imax-Imin at CCM).

Flyback transformer design considerations

1. Energy storage capacity. When the transformer works in CCM mode, because of the DC component, AIR GAP needs to be added to tilt the magnetization curve toward the H axis, so that the transformer can withstand a larger current and transfer more energy.

Ve: effective volume of the core and air gap.

or P = 1 / 2Lp (Imax2-Imin2)

In the formula, Imax, Imin ---- are the corresponding current values at the beginning and end of the conduction period.

Since the flyback transformer core only works in the first quadrant hysteresis loop, the BH effect of the core under AC and DC is closely related to the size of the AIR GAP, as shown in Figure 4. The air gap has no effect on Bac under AC current. The effect is changed, but the Hac will be greatly increased, which is a positive side, which can effectively reduce the effective permeability of CORE and reduce the inductance of the primary winding.

The addition of an air gap under DC current can make CORE withstand a larger DC current to generate HDC, but BDC remains unchanged, so it can effectively prevent the magnetic core from being saturated under a large DC bias, which is an energy storage and transfer Both are beneficial. When the flyback transformer works in CCM, there is a considerable DC component, and then there must be an air gap.

The additional volt-second value, the number of turns, and the core area determine the Bac value on the B axis; the average DC current value, the number of turns, and the magnetic path length determine the position of the HDC value on the H axis. Bac corresponds to the range of Hac values. It can be seen that the larger the air gap Hac is, the larger the core air gap must be to prevent saturation and smooth the DC component.

Flyback transformer design steps

For example: input voltage: AC90-264V output voltage: 19V output current: 3.16A output power: 60W frequency: 70K

Step1. Select CORE material and confirm B

This example is ADAPTER DESIGN. Due to the poor heat dissipation effect of this type of machine, the choice of CORE material should consider high Bs, low loss and high i materials. In combination with cost considerations, Ferrite Core is selected here. PC40 or PC44 from TDK is preferred. TDK DATA BOOK, we can know the unit density of PC44 material

The relevant parameters are as follows: i = 2400 ± 25% Pvc = 300KW / m2 @ 100KHZ, 100

Bs = 390mT Br = 60mT @ 100 Tc = 215

In order to prevent the transient saturation effect of X'FMR, this example is designed with low B.

Choose B = 60% Bm, that is, B = 0.6 * (390-60) = 198mT 0.2 T

Step2 Determine the Core Size and Type.

1> Find the core AP to determine the size

AP = AW * Ae = (Pt * 104) / (2B * fs * J * Ku)

= [(60 / 0.83 + 60) * 104] / (2 * 0.2 * 70 * 103 * 400 * 0.2) = 0.59cm4

Where Pt = Po / + Po transferred power;

J: current density A / cm2 (300 ~ 500); Ku: winding coefficient 0.2 ~ 0.5.

2> Shape and specifications are determined.

The shape is determined by the external size, which can be determined in accordance with BOBBIN, EMI requirements, etc.The specifications can be determined by referring to the AP value and shape requirements. Based on the above principles and referring to the DATA BOOK of TDK, we know that RM10, LP32 / 13, and EPC30 can meet the above requirements, but The available winding volume of RM10 and EPC30 is less than LP32 / 13. Here we choose LP32 / 13 PC44, its parameters are as follows:

Ae = 70.3 mm2 Aw = 125.3mm2 AL = 2630 ± 25% le = 64.0mm

AP = 0.88 cm4 Ve = 4498mm3 Pt = 164W (forward)

Step3 Estimate the critical current IOB (DCM / CCM BOUNDARY)

In this example, the transformer is designed with a critical point when the IL reaches 80% Iomax.

That is: IOB = 80% * Io (max) = 0.8 * 3.16 = 2.528 A

Step4 Find the turns ratio n

n = [VIN (min) / (Vo + Vf)] * [Dmax / (1-Dmax)] VIN (min) = 90 * 2-20 = 107V

= [107 / (19 + 0.6)] * [0.5 / (1- 0.5)]

= 5.5 6

The turns ratio n can be 5 or 6, here 6 is taken to reduce the iron loss, but the copper loss will increase.

CHECK Dmax:

Dmax = n (Vo + Vf) / [VINmin + n (Vo + Vf)] = 6 * (19 + 0.6) / [107 + 6 * (19 + 0.6)] = 0.52

Step5 Find CCM / DCM Pro

ISB = 2IOB / (1-Dmax) = 2 * 2.528 / (1-0.52) = 10.533

Step6 Calculate the secondary inductance Ls and the primary inductance Lp

Ls = (Vo + Vf) (1-Dmax) * Ts / ISB = (19 + 0.6) * (1-0.52) * (1/70000) / 10 = 12.76uH

Lp = n * n * Ls = 6 * 6 * 12.76 = 459.4 uH 460

This inductance value is critical inductance.If the circuit needs to work in CCM, you can increase this value.If you need to work in DCM, you can adjust this value appropriately.

Step7 Find the secondary side peak current isp in CCM

Io (max) = (2Is + ISB) * (1- Dmax) / 2 Is = Io (max) / (1-Dmax)-(ISB / 2)

Isp = ISB + Is = Io (max) / (1-Dmax) + (ISB / 2) = 3.16 / (1-0.52) + 10.533 / 2 = 11.85A

Step8 Find the primary peak current Ipp at CCM

Ipp = Isp / n = 11.85 / 6 = 1.975 A

Step9 Determine Np, Ns

1> Np

Np = Lp * Ipp / (B * Ae) = 460 * 1.975 / (0.2 * 70.3) = 64.6 Ts

Because the calculation result is fractional turns, it is considered to take into account the number of turns of the primary and secondary windings, so that the primary and secondary windings of the transformer have the same ampere-turn value, so adjust Np = 60Ts OR Np = 66Ts

Considering that when the turns ratio n is set, the existing copper loss increases. In order to balance Pfe and Pcu as much as possible, first select Np = 60 Ts.

2> Ns

Ns = Np / n = 60/6 = 10 Ts

3> Nvcc

Find the volts per turn Va Va = (Vo + Vf) / Ns = (19 + 0.6) / 10 = 1.96 V / Ts

Nvcc = (Vcc + Vf) / Va = (12 + 1) /1.96=6.6

Step10Calculate AIR GAP

lg = Np2 * o * Ae / Lp = 602 * 4 * 3.14 * 10-7 * 70.3 / 0.46 = 0.69 mm

Step11 Calculate the wire diameter dw

1> dwp

Awp = Iprms / J Iprms = Po / / VIN (min) = 60 / 0.83 / 107 = 0.676A

Awp = 0.676 / 4 J take 4A / mm2 or 5A / mm2

= 0.1 (take 0.35mm * 2)

2> dws

Aws = Io / J = 3.16 / 4 (1.0 mm)

Volume winding and skin effect, using multiple wires and winding, single wire should not be larger than 0.4, Aw of 0.4 = 0.126mm2, then 0.79 (that is, Ns uses 0.4 * 6)

3> dwvcc Awvcc = Iv / J = 0.1 / 4

The above winding wire diameter is calculated by 4A / mm2 to reduce copper loss.If the wire package is too fat when the structure is designed, the value of J can be adjusted appropriately.

4> Estimate copper occupancy.

0.4Aw Np * rp * (1 / 2dwp) 2 + Ns * rs * (1 / 2dws) 2 + Nvcc * rv * (1 / 2dwv) 2

0.4Aw 60 * 2 * 3.14 * (0.35 / 2) 2 + 10 * 6 * 3.14 + (0.4 / 2) 2 + 7 * 3.14 * (0.18 / 2) 2

11.54 + 7.54 + 0.178 = 19.26

0.4 * 125.3 = 50.12

50.12> 19.26 OK

Step12 Estimate loss and temperature rise

Calculate the wire length of each winding. Calculate the RDC and Rac of each winding @ 100 Calculate the loss power of each winding and sum the power loss of each winding (total value). For example: Np = 60Ts, LP32 / 13BOBBIN winding average Turn length 4.33cm

Then INP = 60 * 4.33 = 259.8 cm Ns = 10Ts

Then INS = 10 * 4.33 = 43.3 cm

Nvcc = 7Ts

Then INvc = 7 * 4.33 = 30.31cm

Check the wire resistance meter: 0.35mm WIRE RDC = 0.00268 / cm @ 100

0.40mm WIRE RDC = 0.00203 / cm @ 100

0.18mm WIRE RDC = 0.0106 / cm @ 100

R @ 100 = 1.4*R@20

Find the current value of the secondary side. Known Io = 3.16A.

Secondary peak average current: Ispa = Io / (1-Dmax) = 3.16 / (1- 0.52) = 6.583A

Effective DC secondary current: Isrms = ((1-Dmax) * I2spa) = (1- 0.52) * 6.5832 = 4.56A

Effective secondary AC current: Isac = (I2srms-Io2) = (4.562-3.162) = 3.29A

Find the current values of the primary side:

Np * Ip = Ns * Is

Average primary peak current: Ippa = Ispa / n = 6.58 / 6 = 1.097A

Primary DC effective current: Iprms = Dmax * Ippa = 1.097 * 0.52 = 0.57A

Primary effective AC current: Ipac = D * I2ppa = 1.097 * 0.52 = 0.79A

Find the AC and DC resistance of each winding.

Primary side: RPDC = (lNp * 0.00268) / 2 = 0.348

Rpac = 1.6RPDC = 0.557

Secondary side: RSDC = (lNS * 0.00203) / 6 = 0.0146

Rsac = 1.6RSDC = 0.0243

Vcc winding: RDC = 30.31 * 0.0106 = 0.321

Calculate the AC and DC losses of each winding:

Secondary side DC loss: PSDC = Io2RSDC = 3.162 * 0.0146 = 0.146W

AC loss: Psac = I2sac * Rsac = 3.292 * 0.0234 = 0.253W

Total: Ps = 0.146 + 0.253 = 0.399 W

Primary DC loss: PPDC = Irms2RPDC = 0.572 * 0.348 = 0.113W

AC loss: Ppac = I2pac * Rpac = 0.792 * 0.557 = 0.348W

Ignore Vcc winding losses (because of its low current) Total Pp = 0.461W

Total coil loss: Pcu = Pc + Pp = 0.399 + 0.461 = 0.86 W

2> Calculate iron loss PFe

Checking TDK DATA BOOK shows that when B = 0.2T of PC44 material, Pv = 0.025W / cm2

Ve for LP32 / 13 = 4.498cm3

PFe = Pv * Ve = 0.025 * 4.498 = 0.112W

Ptotal = Pcu + PFe = 0.6 + 0.112 = 0.972 W Estimated temperature rise t According to the empirical formula t = 23.5P / Ap = 23.5 * 0.972 / 0.88 = 24.3

The estimated temperature rise t is less than SPEC, and the design is OK.

Step13 structure design

Check the winding width of LP32 / 13 BOBBIN is 21.8mm.

Considering the safety distance, the creepage distance is not less than 6.4mm.

In order to reduce LK and improve efficiency, a sandwich structure is adopted, and its structure is as follows:

X'FMR structure:

Np	#1	3.2 / 3.2	2-A	0.35 * 2	30	1L
SHI	#2	3.2 / 3.2	SHI- 4	2mils * 12	1	3L
Ns	# 3	3.2 / 3.2	8.9-6.7	0.4 * 6	10	3L
SHI	# 4	3.2 / 3.2	SHI- 4	2mils * 12	1	1L
Np	# 5	3.2 / 3.2	A-1	0.35 * 2	30	1L
Nvcc	# 6	3.2 / 3.2	3-4	0.18	7	2L
	# 7			Connect two A points	2L