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any idea how to obtain a 1.8V for Vref where in fact the supply runs from 1.2V to 3V?

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allennlowaton

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I have a journal paper now. The input runs from 1.2V to 3V.
But it has a 1.8V for Vref in its comparator for the whole range.
I don't know if it's internally or externally produced. It's not stated in the paper anyway.

Assuming that it's produced internally, is it possible to have a Vref that is higher than the Vdd? (in this case, 1.2V for Vdd against 1.8V for Vref)...

Any ideas are appreciated here...
Thanks.
 

1. Use a charge pump to achieve a 2xVdd rail, and obtain your 1.8V reference.
2. Create a 0.9V reference, and attenuate the comparator input by half.
 
checkmate, I'm interested with your suggestion number 2.
correct me, if I'm wrong please. Since, I reduced the Vref by half that means I will also
reduced my Vin by half and I can do that with a simple voltage divider using resistors.

---------- Post added at 18:46 ---------- Previous post was at 18:45 ----------

checkmate, I'm interested with your suggestion number 2.
correct me, if I'm wrong please. Since, I reduced the Vref by half that means I will also
reduced my Vin by half and I can do that with a simple voltage divider using resistors.
 

allennlowaton said:
checkmate, I'm interested with your suggestion number 2.
correct me, if I'm wrong please. Since, I reduced the Vref by half that means I will also
reduced my Vin by half and I can do that with a simple voltage divider using resistors.
Click to expand...

That will depend on the source impedance of the "vin" generator. The divider will also contributre additional offset, and the attenuation will further degrade your SNR. So it all depends on your requirements.
 
checkmate said:
That will depend on the source impedance of the "vin" generator. The divider will also contributre additional offset, and the attenuation will further degrade your SNR. So it all depends on your requirements.
Click to expand...

I will use that technique for my first stage (the input sensing comparator) for my charge pump. The pump frequency is at 500kHz.
With that technique, I believe that the problem will be the power loss and additional chip area for the resistors.

---------- Post added at 14:53 ---------- Previous post was at 14:46 ----------

charge pump for power management not for the PLL.
 

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