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wide input voltage range at the dc/dc converter(MIN. 1.8V)

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edwintsu

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Hi all,

Typically the method for ≥2.7v input voltage range is that we get fixed internal voltage for core control circuit.

But in my project the MIN. input voltage drop to 1.8v. And the sum of VTHN + |VTHP| is larger than 1.8v at some corners of 0.5u CMOS process.

Now i use charge pump voltage doubler + voltage regulator to get the fixed 2.4v internal voltage. At the limit of chip area i can't get enough load current (just 70uA, i want to at least 200uA).

My Question:
1':What is the industical method for the input voltage range?
2':Have any other little chip area solution?

Thank you for your reading.
Edwin. 09/13
 

Areky_qin

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1. If your circuit can not drive 200uA loading, then you can try your best to improve the pump power efficency, eg, apply different pump stucture to achieve your target.
2. For power supply range, for enough margin, now the most design have large margin than spec, eg, spec show the chip can work in 1.8 to 2.1v, may be when power is lower down to 1.5v, the chip can work too, Just because, the internal voltage have a constant voltage reference, bandgap outout reference, based on this reference, all internal voltage can be got, so when the input power is changed, can not influence the chip's option.

good luck.

Added after 3 minutes:

By the way, you can caculate the parameters about your pumps, the clk frequency, cap size and , pumps power supply, all of them are determined by your current loading.
 

    edwintsu

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edwintsu

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Thank you for your reply, Areky.

I implement the 2.4V regulaotr using charge pump + voltage regulator. It can drive 200uA load at 1.8V input voltage. But its ripple is too large for the internal chip.

Maybe need change other process.

B/R
Edwin.
 

ambreesh

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Re: wide input voltage range at the dc/dc converter(MIN. 1.8

can you get a resevoir cap at the output of the CP. External cap say 1uF. The ripple will be taken care off.
 

edwintsu

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Thank your, ambreesh.

But no enough pin.

Best Regards,
Edwin.
 

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