Voltage regulator and battery

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saga_1136906
Level 3
Level 3
10 replies posted 10 questions asked 10 sign-ins

Hi all,

I have a circuit that uses a battery type NCR18650PF with the relative charger mounted on breakout 03962A which uses the 4056E regulator.

I have to supply a CYBLE-012011-00 type module. I was thinking of two solutions.

The first: put the 03962A output, between out + and out-, a low consumption linear regulator, let's say 1.8V output. In this case I would have a current consumption as well as additional components to add.

Second solution: supply the CYBLE module directly between out + and out-, since the charging voltages are within the tolerance limits of the module. So I would have the advantage of not having added current consumption and components.

Can anyone give me some opinion?

Thank you!

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Len_CONSULTRON
Level 9
Level 9
Beta tester 500 solutions authored 1000 replies posted

saga,

I don't understand why you don't drive the CYBLE-012011-00 directly from the battery.  The part is capable all the way to VDD=5.5V.

If you want to supply 1.8V it sounds like you will need a LDO analog VREG as an output.  Sadly as you pointed out, there is current consumption usually in the range of 1 to 20 uA even if the CYBLE-012011-00 is in low power mode.

I think you would be more power efficient running the CYBLE-012011-00 at about 5V nominal (during charging the NCR18650PF might see 4.4V).  At VDD=5V, the CYBLE-012011-00 in very low power mode should be > 10uA.  Besides starting with a higher VDD allows you to set a Brownout voltage detect to a reasonable value to allow you to shutdown better should you need to.

Len

Len
"Engineering is an Art. The Art of Compromise."

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2 Replies
Vasanth
Moderator
Moderator
Moderator
250 sign-ins 500 solutions authored First question asked

Hi Savio,

I was not exactly clear about what is the difference between the two options ? Are you planning to avoid the regulator in the second case ? If that is the case, having the regulator could be helpful for applications where a constant supply is critical( may be an ADC using VDD as reference). Also isolation and protection will also come in picture. If the specs are within the device tolerance, you can use it directly too.

Best Regards,
Vasanth

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Len_CONSULTRON
Level 9
Level 9
Beta tester 500 solutions authored 1000 replies posted

saga,

I don't understand why you don't drive the CYBLE-012011-00 directly from the battery.  The part is capable all the way to VDD=5.5V.

If you want to supply 1.8V it sounds like you will need a LDO analog VREG as an output.  Sadly as you pointed out, there is current consumption usually in the range of 1 to 20 uA even if the CYBLE-012011-00 is in low power mode.

I think you would be more power efficient running the CYBLE-012011-00 at about 5V nominal (during charging the NCR18650PF might see 4.4V).  At VDD=5V, the CYBLE-012011-00 in very low power mode should be > 10uA.  Besides starting with a higher VDD allows you to set a Brownout voltage detect to a reasonable value to allow you to shutdown better should you need to.

Len

Len
"Engineering is an Art. The Art of Compromise."
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