SLUAAL2 Application note

SLUAAL2 june 2023 UCC256402 , UCC256403 , UCC256404

1.18.2 LLC Transformer Design

Step 1: LLC Center-Tapped Transformer Specifications

Equation 115.

Magnetizing Inductance value of the transformer L_{m} = 510 u H

Equation 116.

Turns ratio from primary to secondary n = 16.5

Equation 117.

Forward voltage drop of the secondary diode V_{f} = 0.7 V

Equation 118.

\begin{array}{l} At minimum Input voltage (365V) and at maximum output power (12V,15A): \\ Peak current of the magnetizing inductor I_{m p_\max} = 1.15 A \\ RMS current of the magnetizing inductor I_{m r m s_\max} = 0.74 A \\ Peak current of the primary winding I_{p_\max} = 1.9 A \\ RMS current of the primary winding I_{p r m s_\max} = 1.27 A \\ Peak current of the each secondary winding I_{s_\max} = 32.2 A \\ RMS current of the each secondary winding I_{s r m s_\max} = 13.65 A \\ Switching frequency f_{s w_\min} = 77 k H z \end{array}

Equation 119.

\begin{array}{l} At Rated Input voltage (390V) and at maximum output power (12V,15A): \\ Peak current of the magnetizing inductor I_{m p} = 1.1 A \\ RMS current of the magnetizing inductor I_{m r m s} = 0.68 A \\ Peak current of the primary winding I_{p} = 1.78 A \\ RMS current of the primary winding I_{p r m s} = 1.22 A \\ Peak current of the each secondary winding I_{s} = 28.9 A \\ RMS current of the each secondary winding I_{s r m s} = 13 A \\ Switching frequency f_{s w} = 88 k H z \end{array}

Step 2: Pick a Core Based on the Output Power

Equation 120.

Area Product of a core A_{p} = W_{a} A_{c} = \frac{\sum_{n = 1}^{m} \frac{I_{n} V_{n}}{J_{r m s_n}}}{4 K_{u} f_{s w} B_{m}} (m^{4})

[Equation 11.12 in Reference 16]

Equation 121.

where W_{a} is window area, A_{c} is core area

Equation 122.

I_{n} {is rms value of the current through the n}^{th} winding

Equation 123.

V_{n} {is rms value of the voltage across the n}^{th} winding

Equation 124.

J_{r m s_n} {is rms value of the current density of the n}^{th} winding: 4 t o 6 A / m m^{2}

Equation 125.

K_{u} is window utilization factor (For Litz wire it is : 0 .3 to 0 .4)

Equation 126.

B_{m} is peak flux density of the core

B_m should be chosen such that at the operating frequency, core loss power density should be less than 150 mW/cm³ for natural convection cooling.

In general, suggested magnetic materials for reducing core losses are 3C95, 3F4 from Ferroxcube (Ferroxcube Cores and Accessories) and PC47, PC90, PC95 from TDK ( TDK Cores and Accessories).

Equation 127.

For K_{u} = 0.3, J_{r m s_p} = 5 A / m m^{2}, J_{r m s_s} = 6 A / m m^{2}, B_{m} = 0.15 T, A_{p} \geq 6476.9 m m^{4}

For this design, PQ26/25 core with 3C95 material is selected. This core's effective cross section area $A_{c} = 120 m m^{2}$ (Ferroxcube PQ26/25 core data sheet) and minimum winding area, mean turn length are $W_{a} = 50.97 m m^{2}, M L T = 56.2 m m$ (Ferroxcube PQ26/25 bobbin data sheet). Cores, Bobbin, Clamp can be obtained at PQ26/25 with 3C95, PQ26/25 Bobbin, Clamp for PQ26/25 Core.

Step 3: Turns and airgap calculation

Equation 128.

N_{p} \times A_{c} \times (2 \times B_{m}) = \frac{n \times (V_{o} + V_{f})}{2 \times f_{s w}}

Equation 129.

N_{p} = \frac{16.5 \times (12 + 0.7)}{120 \times 10^{- 6} \times 2 \times 0.15 \times 2 \times 88 \times 10^{3}} = 33.0729

Equation 130.

N_{s} = \frac{N_{p}}{n} = \frac{33.07}{16.5} = 2.0044

Equation 131.

Lets consider primary number of turns (N_{p}) and secondary number of turns (N_{s}) as 33 and 2 respectively .

Equation 132.

The air-gap length is l_{g} = \frac{μ_{o} A_{c} N_{p}^{2}}{L_{m}} = \frac{4 π \times 10^{- 7} \times 120 \times 10^{- 3} \times 33^{2}}{510 \times 10^{- 6}} = 0.32 m m

Step 4: Wire selection for both primary and secondary

Equation 133.

\begin{array}{l} The effective cross-sectional area of the bare winding of the primary should be \\ A_{w p} = \frac{I_{p r m s}}{J_{r m s}} = \frac{1.22}{5} = 0.244 m m^{2} \end{array}

Equation 134.

AWG23 has a copper area about 0 .2558 m m^{2} which is closest to required copper area .

Equation 135.

In general, for high frequency (around
                            100kHz) designs, AWG38-42 should be selected for each strand .

Equation 136.

The skin depth of copper at 88 k H z is δ_{w} = \frac{66.2}{\sqrt{f}} (m m) = \frac{66.2}{\sqrt{88, 000}} = 0.2232 m m

[Equation 10.148 in Reference 16]

Equation 137.

Here AWG 38 is chosen with 30 strands for following reasons:

Equation 138.

1 . Its over all copper area (0.2432 m m^{2}) is closest to required copper area

Equation 139.

2 . Each strand diameter is much less than skin depth so that current through the each strand will be uniform

Equation 140.

3 . And its readily available

(Litz Wire Data from MWS Wire Industries)

Equation 141.

So the actual current density would be J_{p_a c t} = \frac{1.22}{0.2432} = 5.01 \frac{A}{m m^{2}}

Equation 142.

With insulation, the overall diameter (d_{o p}) of the S E R V E D L I T Z W I R E of 38 A W G ​ with 30 strands is 0 .7874mm

Equation 143.

Area required by the primary winding = N_{p} \times \frac{π \times d_{o p}^{2}}{4} = 33 \times \frac{π \times {0.7874}^{2}}{4} = 16.0692 m m^{2}

Equation 144.

\begin{array}{l} For the each secondary winding, the effective cross-sectional area of the bare winding \\ of the each secondary should be A_{w s} = \frac{I_{s r m s}}{J_{r m s}} = \frac{13}{6} = 2.167 m m^{2} \end{array}

Equation 145.

AWG14 has a copper area about 2 .082 m m^{2} which is closest to required copper area .

Equation 146.

Here AWG38 with 260 strands are considered for each secondary winding which has a bare copper about 2 .1078 m m^{2}

Equation 147.

So the actual current density of the each secondary winding would be J_{s_a c t} = \frac{13}{2 .1078} = 6.16 \frac{A}{m m^{2}}

Equation 148.

With insulation, the overall diameter (d_{o s}) of the U N S E R V E D L I T Z W I R E of 38 A W G ​ with 260 strands is 2 .286mm

(Litz Wire Data from Remington Industries, Digikey Link)

Equation 149.

Area required by the each secondary winding = N_{s} \times \frac{π \times d_{o s}^{2}}{4} = 2 \times \frac{π \times {2.286}^{2}}{4} = 8.2087 m m^{2}

Equation 150.

Window utilization factor K_{u} = \frac{A r e a o c c u p i e d b y e a c h w i n d i n g}{O v e r a l l W i n d o w A r e a} = \frac{16.0692 + 8.2087 + 8.2087}{50.97} = 0.637

Step 5: Wire Loss Calculation with Interleaved Winding

Equation 151.

The winding width of the bobbin is 13 .56mm

(Ferroxcube PQ26/25 bobbin datasheet)

Equation 152.

\begin{array}{l} In primary, since there are 33 bundle turns with 0 .7874mm over all diameter, total number of \\ bundled winding layers (N_{p l}) would be 2 . \end{array}

Equation 153.

Each bundle has a total number of strands k_{p} = 30

Equation 154.

The strands in each bundle are modelled as a square with \sqrt{k_{p}} = \sqrt{30} ≅ 6 strands on each side of the bundle

Equation 155.

\begin{array}{l} The total number of strands in each layer is given as N_{p s l} = \\ the number of bundle turns in a layer \times number of strands of each side of the square bundle \\ = 17 \times 6 = 102 \end{array}

Equation 156.

The mean length of each turn is l_{T} = 56.2 m m

(Ferroxcube PQ26/25 bobbin datasheet)

Equation 157.

\begin{array}{l} The DC resistance of the single AWG38 strand is R_{w D C s} = l_{T} \times (A W G 38 DC resistance per m) \\ = 56.2 m m \times 2.1266 Ω / m = 119.5 m Ω \end{array}

Equation 158.

The AC power loss of a single layer is given by P_{a c} = DC power loss of a single layer \times φ \times Q^{'} (φ, m)

[Equation 10.80 in Reference 9]

Equation 159.

\begin{array}{l} Here φ = \sqrt{η} \sqrt{\frac{π}{4}} \frac{d_{s}}{δ} \\ where η is porosity factor, d_{s} is strand bare wire diameter, δ is skin depth for a given frequency \end{array}

[Equation 10.74 in Reference 9]

Equation 160.

Q^{'} (φ, m) = (2 m^{2} - 2 m + 1) G_{1} (φ) - 4 m (m - 1) G_{2} (φ)

[Equation 10.81 in Reference 9]

Equation 161.

\begin{array}{l} G_{1} (φ) = \frac{\sinh (2 φ) + \sin (2 φ)}{\cosh (2 φ) - \cos (2 φ)} \\ G_{2} (φ) = \frac{\sinh (φ) \cos (φ) + \cosh (φ) \sin (φ)}{\cosh (2 φ) - \cos (2 φ)} \end{array}

[Equation 10.76 in Reference 9]

Equation 162.

\begin{array}{l} m for each layer can be found by m = \frac{m m f (h)}{m m f (h) - m m f (0)} \\ where h is thickness of each layer \end{array}

[Equation 10.81 in Reference 9]

Equation 163.

\begin{array}{l} In this design example, the DC power loss of a single layer of the primary is given as = \\ square of the RMS current through each strand \times DC resistance of a single strand \times total number of strands in a single layer \\ = {(\frac{I_{r m s}}{k})}^{2} \times R_{w D C s} \times N_{p s l} = {(\frac{1.22}{30})}^{2} \times 119.5 m Ω \times 102 = 20.6 m W \end{array}

Equation 164.

{Porosity factor η}_{p} = \frac{Number of strands per layer (N_{p s l}) \times S trand diameter with insulation}{width of the bobbin} = \frac{102 \times 0.124 m m}{13 .56mm} = 0.933

Equation 165.

\begin{array}{l} For each of the secondary winding, since there are 2 bundle turns each with 2 .286mm over all diameter, \\ total number of bundled winding layers (N_{s 1 l}, N_{s 2 l}) would be 1 . \end{array}

Equation 166.

Each secondary winding bundle has a total number of strands k_{s} = 260

Equation 167.

The strands in each bundle are modelled as a square with \sqrt{k_{s}} = \sqrt{260} ≅ 16 strands on each side of the bundle

Equation 168.

\begin{array}{l} The total number of strands in each layer of the secondary is given as N_{s s l} = \\ The number of bundle turns in a layer \times number of strands of each side of the square bundle \\ = 2 \times 16 = 32 \end{array}

Equation 169.

\begin{array}{l} The DC resistance of the single AWG38 strand of the secondary is R_{w D C s} = l_{T} \times (A W G 38 DC resistance per m) \\ = 56.2 m m \times 2.1266 Ω / m = 119.5 m Ω \end{array}

Equation 170.

\begin{array}{l} The DC power loss of a single layer of the secondary is given as = \\ square of the RMS current through each strand \times DC resistance of a single strand \times total number of strands in a single layer \\ = {(\frac{I_{r m s}}{k})}^{2} \times R_{w D C s} \times N_{s s l} = {(\frac{13}{260})}^{2} \times 119.5 m Ω \times 32 = 9.56 m W \end{array}

Equation 171.

{Porosity factor η}_{s} = \frac{Number of strands per layer (N_{s s l}) \times S trand diameter with insulation}{width of the bobbin} = \frac{32 \times 0.124 m m}{13 .56mm} = 0.3

Equation 172.

\begin{array}{l} So, layer sequence would be \\ 1 . 1st bundle layer of the primary winding \\ 2 . Secondry-1 bundle layer \\ 3 . Secondry-2 bundle layer \\ 4 . 2nd bundle layer of the primary winding \end{array}

Equation 173.

For Primary, φ_{p} = \sqrt{η} \sqrt{\frac{π}{4}} \frac{d_{s}}{δ} = \sqrt{0.933} \sqrt{\frac{π}{4}} \frac{0.1007}{0.2232} = 0.386

Equation 174.

For Secondary, φ_{s} = \sqrt{η} \sqrt{\frac{π}{4}} \frac{d_{s}}{δ} = \sqrt{0.3} \sqrt{\frac{π}{4}} \frac{0.1007}{0.2232} = 0.219

Equation 175.

MMF due to the each layer of the primary = current through each strand \times number of strands in a layer = \frac{I_{p r m s}}{k_{p}} \times N_{p s l}

Equation 176.

S o, layer 1's m value can be determined by m = \frac{(\frac{I_{p r m s}}{k_{p}} \times N_{p s l})}{(\frac{I_{p r m s}}{k_{p}} \times N_{p s l}) - 0} = 1

Equation 177.

Layer 2's m value can be determined by m = \frac{2 (\frac{I_{p r m s}}{k_{p}} \times N_{p s l})}{2 (\frac{I_{p r m s}}{k_{p}} \times N_{p s l}) - (\frac{I_{p r m s}}{k_{p}} \times N_{p s l})} = 2

Equation 178.

\begin{array}{l} Similarly, m value for other adjacent layers increases by 1 . Since there are 6 layers for the first \\ bundled primary, m value increases upto 6 . \end{array}

Equation 179.

Same method has been followed to find out the m value for other layers

Equation 180.

\begin{array}{l} The copper loss of all the layers in the primary is given by P_{p w} \\ = DC power loss of a single layer of the primary \times [\sum_{m = 1}^{6} φ Q^{'} (φ, m) + \sum_{m = - 8.043}^{- 3.043} φ Q^{'} (φ, m)] = 295 m W \end{array}

Equation 181.

\begin{array}{l} The copper loss of all the layers in the secondary1 is given by P_{s w 1} \\ = DC power loss of a single layer of the secondary1 \times [\sum_{m = - 11.763}^{3.237} φ Q^{'} (φ, m)] = 158 m W \end{array}

Equation 182.

\begin{array}{l} The copper loss of all the layers in the secondary2 is given by P_{s w 2} \\ = DC power loss of a single layer of the secondary2 \times [\sum_{m = 4.237}^{19.237} φ Q^{'} (φ, m)] = 170 m W \end{array}

Equation 183.

{Total copper losses P}_{w} = P_{p w} + P_{s w 1} + P_{s w 2} = 295 m W + 158 m W + 170 m W = 0.623 m W

Step 6: Flux Density and Core-Loss Calculation

Equation 184.

The amplitude of the core magnetic flux density at rated input
                            voltage is B_{m} = \frac{L_{m} I_{m p}}{N_{p} A_{c}} = \frac{510 \times 10^{- 6} \times 1.1}{33 \times 120 \times 10^{- 6}} = 0.142 T

Equation 185.

The core loss per unit volume (P_{v}) at 0.142 T, 88 k H z is 130 m W / c m^{3}

Equation 186.

The total core loss is P_{C} = V_{C} P_{v} = 130 m W / c m^{3} \times 6530 m m^{3} = 848 m W

Equation 187.

The amplitude of the core magnetic flux density at minimum input
                            voltage is B_{m_\max} = \frac{L_{m} I_{m p_\max}}{N_{p} A_{c}} = \frac{510 \times 10^{- 6} \times 1.15}{33 \times 120 \times 10^{- 6}} = 0.148 T

Equation 188.

At worst case current, B_{m_\max} is less than saturation flux density of the ferrite material .

Step 7: Temperature rise and Bobbin Fit Calculations

Equation 189.

Total power loss of the transformer is P_{w c} = P_{w} + P_{C} = 1.472 W

Equation 190.

The surface area of  PQ26/25 core is A_{t} = 32.6 c m^{2}

Table 3-39

Equation 191.

The surface power loss density is ψ = \frac{P_{w c}}{A_{t}} = \frac{1.472 W}{32.6 c m^{2}} = 0.045 W / c m^{2}

Equation 192.

The temperature rise of the transformer is Δ T = 450 ψ^{0.826} = 450 \times {(0.045 W / c m^{2})}^{0.826} = 34.7^{o} C

[Equation 10.193 in Reference 16]

Equation 193.

\begin{array}{l} The actual core window utilization factor is K_{u} = \frac{A r e a o c c u p i e d b y e a c h w i n d i n g}{O v e r a l l W i n d o w A r e a} \\ = \frac{16.0692 + 8.2087 + 8.2087}{50.97} = 0.637 \end{array}