米6体育平台手机版

ZHCSJY0B June 2012 – June 2019 LMR12015 , LMR12020

PRODUCTION DATA.

封装选项

机械数据 (封装 | 引脚)

DSC|10
- MPDS239E

散热焊盘机械数据 (封装 | 引脚)

DSC|10
- QFND489A

订购信息

8.2.1.2.1 Inductor Calculation Example

Operating conditions for the LMR12015/20 are:

Equation 14. V_IN = 7 – 16 V

Equation 15. f_SW = 2 MHz

Equation 16. V_OUT = 3.3 V

Equation 17. V_D1 = 0.5 V

Equation 18. I_OUT = 2 A

First the maximum duty cycle is calculated.

Equation 19. D_MAX = (V_OUT + V_D1) / (V_IN + V_D1 – V_DS) = (3.3 V + 0.5 V) / (7 V + 0.5 V – 0.3 V) = 0.528

Using Figure 27 gives us a recommended ripple ratio = 0.4.

Now the minimum duty cycle is calculated.

Equation 20. D_MIN= (V_OUT + V_D1) / (V_IN + V_D1 – V_DS) = (3.3 V + 0.5 V) / (16 V + 0.5 V – 0.3 V) = 0.235

The inductance can now be calculated.

Equation 21. L= (1 – D_MIN) x (V_OUT + V_D1) / (I_OUT × r × ƒ_SW) = (1 – 0.235) × (3.3 V + 0.5 V) / (2 A × 0.4 × 2 MHz) = 1.817 µH

This is close to the standard inductance value of 1.8 µH. This leads to a 1% deviation from the recommended ripple ratio, which is now 0.4038.

Finally, we check that the peak current does not reach the minimum current limit of 2.5 A.

Equation 22. I_LPK = I_OUT × (1 + r / 2) = 2 A × (1 + 0.4038 / 2 ) = 2.404 A

The peak current is less than 2.5 A, so the DC load specification can be met with this ripple ratio. To design for the LMR12015 simply replace I_OUT = 1.5 A in the equations for I_LPK and see that I_LPK does not exceed the LMR12015 current limit of 2 A (min).