Output voltage rule of VDD_SOC and VDD_PU

[ko-hey]

Hi all

I have question about operating range of i.MX6.

There is a following desctiption in Table 9's note of IMX6SLCEC.

VDD_SOC and VDD_PU output voltage must be set to this rule: VDD_ARM - VDD_SOC / VDD_PU < 50mV

I can't understand the meaning of description.

I understand that VDD_ARM_IN is supplied from SW1A/B of PMIC and VDD_SOC_IN and VDD_PU_IN are supplied from SW1C.

According to the Table 10 of MMPF0100's datasheet, SW1A/B and SW1C's voltage level is set to 1.375V

If I apply the above-mentioned formula, all devices can't satisfy it.

Please let me know the formula's meaning.

Ko-hey

[igorpadykov]

Hi Ko-hey

these voltages are LDO outputs: VDDARM_CAP,

VDDPU_CAP, VDDSOC_CAP, also one can look at

datasheet Figure 3. i.MX 6SoloLite SoC Power Block Diagram.

Best regards

igor

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[ko-hey]

Hi Igor

Thank you for your reply.

I understand following formula is right.

VDD_ARM_CAP - VDD_SOC_CAP / VDD_PU_CAP < 50mV      ・・・①

I found a following note in table 9 of datasheet.

Note 3: VDD_SOC_CAP and VDD_PU_CAP must be equal.

When I apply a above note to formula ①, I get a following result.

Is it correct ?

VDD_ARM_CAP < 1.05 V

Ko-hey

[igorpadykov]

Hi Ko-hey

sorry I do not understand how did you get this.

Could you provide details ?

Best regards

igor

[ko-hey]

Hi Igor

I think that the result of VDD_SOC_CAP / VDD_PU_CAP is as below When user apply a Note 3.

VDD_SOC_CAP / VDD_PU_CAP = 1  ---②

So ① change to as below.

VDD_ARM_CAP - 1 < 0.05 V

⇒ VDD_ARM_CAP < (1 + 0.05) V

⇒ VDD_ARM_CAP < 1.05 V

Is my understanding wrong ?

Ko-hey

[igorpadykov · Accepted solution]

Hi Ko-hey

meaning of "VDD_SOC_CAP / VDD_PU_CAP" is that

this should be "VDD_SOC_CAP"  or  "VDD_PU_CAP":

so formula

VDD_ARM_CAP - VDD_SOC_CAP / VDD_PU_CAP < 50mV

means:

VDD_ARM_CAP - VDD_SOC_CAP < 50mV  and

VDD_ARM_CAP - VDD_PU_CAP < 50mV

Best regards

igor

[ko-hey]

Hi Igor

Thank you for your explanation.

I understand.

Ko-hey