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ZHCSJK0A April 2019 – September 2019 UC1843B-SP

PRODUCTION DATA.

封装选项

机械数据 (封装 | 引脚)

KGD|0
HKU|10
- MCDF015C

散热焊盘机械数据 (封装 | 引脚)

订购信息

8.2.2.2 Transformer

The transformer of the design consists of two major values, turns ratio and primary side inductance. There is no minimum limit to the turns ratio of the transformer, just a maximum limit. The equation below will give the turns ratio as a function of duty cycle, which if you put in the maximum duty cycle of the converter will give you a maximum turns ratio. The UC1843B-SP design targeted a duty cycle of 50%, which is somewhat low for this controller. The suggested value would be around 70% duty cycle to take advantage of the fact the UC1843B-SP has full duty cycle range. The equation of the turns ratio of the transformer is Equation 3.

Equation 3.

N_{p s M A X} = \frac{V_{i n M I N} \times D_{l i m}}{(V_{o u t} + V_{D i o d e}) \times (1 - D_{l i m})}

Equation 4.

N_{p s M A X} = \frac{20 V \times 0.5}{(5 V + 0.7 V) \times (1 - 0.5)} = 3.5

Often the turns ratio will slightly change in design due to how the transformer is manufactured. For the UC1843B-SP design a turns ratio of 3.33 was used. Another turns ratio that is important is the turns ratio of the auxiliary winding. The auxiliary winding is found by figuring out what positive voltage is needed from the auxiliary winding. Picking what voltage the auxiliary winding should have lets one pick the turns ratio from the secondary to the auxiliary winding, which in turn allows for the turns ratio from primary to auxiliary to be found. The equation for the turns ratio for the auxiliary winding is Equation 5.

Equation 5.

N_{p a} = \frac{N_{p s} \times (V_{o u t} + V_{D i o d e})}{V_{a u x}}

Equation 6.

N_{p a} = \frac{3.33 \times (5 V + 0.7 V)}{13 V} = 1.46

An auxiliary winding of 1.43 was used for the UC1843B-SP design due to manufacturing constraints. The primary inductance of the transformer is found from picking an appropriate ripple current. A higher inductance will often mean reduced current ripple, thus lower EMI and noise, but a higher inductance will also increase physical size and limit the bandwidth of the design. A lower inductance will do the opposite, increasing current ripple, lowering EMI, lowering noise, decreasing physical size, and increasing the limited bandwidth of the design. The percent ripple current can be anywhere from 20% to 80% depending on the design. The equation for finding the primary inductance from the percentage ripple current is Equation 7.

Equation 7.

L_{P R I} = \frac{{V_{i n M A X}}^{2} \times {D_{M I N}}^{2}}{V_{o u t} \times I_{o u t} \times f_{o s c} \times %_{R i p p l e}}

Equation 8.

\frac{(40 V)^{2} \times 0 . 25^{2}}{5 V \times 10 A \times 200 k H z \times 0.4} = 25 μ H

There are quite a few physical limitations when making transformers, so often this inductance will change slightly. For the UC1843B-SP design a primary inductance of 21 µH. This corresponds to a percent ripple of around 0.475. The peak and primary currents of the transformer are also generally useful for figuring out the physical structure of the transformer. See the following equations for proper calculations.

Equation 9.

I_{R i p p l e} = \frac{V_{o u t} \times I_{o u t} \times %_{R i p p l e}}{V_{i n M A X} \times D_{M I N}}

Equation 10.

I_{R i p p l e} = \frac{5 V \times 10 A \times 0.475}{40 V \times 0.25} = 2.375 A

Equation 11.

I_{P r i P e a k} = \frac{V_{o u t} \times I_{o u t}}{V_{i n M I N} \times D_{M A X} \times η} + \frac{I_{R i p p l e}}{2}

Equation 12.

I_{P r i P e a k} = \frac{5 V \times 10 A}{20 V \times 0.5 \times 0.8} + \frac{2.375}{2} = 7.44 A

Equation 13.

I_{P r i R M S} = \sqrt{D \times (\frac{V_{o u t} \times I_{o u t}}{V_{i n} \times D})^{2} + \frac{{I_{R i p p l e}}^{2}}{3}}

Equation 14.

I_{P r i R M S} = \sqrt{0.5 \times (\frac{5 V \times 10 A}{20 V \times 0.5})^{2} + \frac{(2.375 A)^{2}}{3}} = 3.79 A

Equation 15.

I_{S e c R M S} = \sqrt{(1 - D) \times {I_{o u t}}^{2} + \frac{(I_{R i p p l e} \times N_{p s})^{2}}{3}}

Equation 16.

I_{S e c R M S} = \sqrt{0.5 \times (10 A)^{2} + \frac{(2.375 A \times 3.33)^{2}}{3}} = 8.42 A