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Original question:

TPS54122: Wrong formula?

TPS54122: Inductor selection

Carlos Silva85
Expert 4915 points
Part Number: TPS54122

Hi Team,

I am working on a design utilizing the TPS54122 DC-DC/LDO device. The desired operating conditions are listed below.

Parameter

Value

Vin (typical) (v)

5v ±10%

DC-DC Out (v)

3.6v

LDO Out (v)

3.3v

Iout (Max) (A)

0.5A

Switching frequency

500KHz


These have been selected to give good LDO efficiency and to closely match the simplified design table in the data sheet(I used 1A in some of the calculations when trying to replicate results from the datasheet). My issue is the inductor selection, the suggested value is 3.3uH which is also the suggested value for the 3A configuration. When using equation 10 I don’t get the 3.3uH that the table suggests, even when using K=0.3. Ideally I would like a lower output ripple as the device been powered is very noise sensitive. So using a K value of 0.1 is also preferred but the inductor value starts to increase when this is selected. Is there an advantage to selecting a inductor value based on a max current that is closer to the current actually been taken or should I use the max the device is capable of instead of the 0.5A been used.

Any advice in the component selection for this device would be useful, the design procedure document for the device doesn’t seem to have any new information over the data sheet. 

  • 0
    TI__Guru**** 191445 points
    Hi Carlos,

    You have touched on an interesting topic. As you know, the standard inductor calculation is:

    lout = (Vout / Vin) * (Vin - Vout) / (Fsw * K * Iout)

    where k is the percent of ripple current relative to the output current.

    So calculating for current at the IC specified maximum will give a lower inductor value than at a lower output current. The switching converter inside TPS54122 uses peak current mode control. For peak current mode control, the switch or inductor current acts as the PWM ramp for the control circuit. If the inductor p-p ripple current (ramp) is small (large inductor value), you may encounter higher than desired jitter in your switching waveform. So I generally use something close to the IC rated current for inductor calculation. typically, I will not go lower than 25% of the rated IC current for a design but often design for full rated current.

    There is an added side benefit for using a smaller value inductor. For the same inductor package size, the DCR of the smaller value will be lower than for a larger inductor. Even though the switching losses will be higher for higher RMS current of the small inductor value, the conduction loss savings will generally dominate, so the lower inductor value design can have better efficiency. If you are looking to optimize efficiency, you need to look at that trade off closely.

    Anyway, there is no absolute answer. There is always an engineering trade off to be made to balance jitter efficiency and size. Hope this helps. let me know if you have further questions.