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Here is the three stage stepping up architecture,and in the paper,the author said that when \$ V_{out} \$ increase up to 1V,shunt connecting the capacitors \$ C_{DDi} \$and \$ C_{out} ...Finally,the ZCS-controlled boost converter is gated and regulated by the on-chip reference circuitry, providing a 1.2-V regulated voltage at the output .

My question is when \$ V_{out} \$ up to 1V,\$ C_{DDi} \$and \$ C_{out} \$ will be shunt connected,then,why will the \$ V_{out} \$ = 1.2V in the end,not 1V?

Also,i want to make sure something,for the 1. mark in the first picture,it said the internal voltage \$ V_{DDi} \$ is boosted with an enhanced charging current into \$ C_{DDi} \$.Dose it mean that the \$ C_{DDi} \$ charge to boost the \$ V_{DDi} \$?

enter image description here enter image description here

This paper is from the :ieeexplore.ieee.org/document/6484920/

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  • \$\begingroup\$ "why will the Vout = 1.2V in the end" - why wouldn't it? \$\endgroup\$
    – Bruce Abbott
    Feb 11, 2018 at 7:37
  • \$\begingroup\$ Why will the Vout = 1.2V,but not 1V ? The author said when Vout =1V,it will shunt connect that two capacitor,so does that two capacitor discharge and let the Vout become from 1V to 1.2V? \$\endgroup\$
    – Shine Sun
    Feb 11, 2018 at 7:41
  • \$\begingroup\$ The paper says that when the ZCS booster is activated it produces 1.2V at Vout. Since CDDi is 'shorted' to Cout, that means Vddi should also be 1.2V. \$\endgroup\$
    – Bruce Abbott
    Feb 11, 2018 at 7:59
  • \$\begingroup\$ Oh!I misunderstand the meaning,now i get it \$\endgroup\$
    – Shine Sun
    Feb 11, 2018 at 8:03
  • \$\begingroup\$ By the way.The paper said the internal voltage \$ V_{DDi} \$ is boosted with an enhanced charging current into \$ C_{DDi} \$.Dose it mean that the \$ C_{DDi} \$ charge to boost the \$ V_{DDi} \$?or discharge? \$\endgroup\$
    – Shine Sun
    Feb 11, 2018 at 8:05

1 Answer 1

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when Vout up to 1V,CDDi and Cout will be shunt connected,then,why will the Vout = 1.2V in the end,not 1V?

Vout is 1.2V because the ZCS controlled converter boosts the TEG voltage and regulates it at 1.2V, and when the shunt is turned on VDDI also becomes 1.2V.

it said the internal voltage VDDi is boosted with an enhanced charging current into CDDi. Does it mean that the CDDi charge to boost the VDDi?

Yes. The voltage on CDDI is first boosted to 300mV by the Low-Voltage Starter. This is sufficient to operate the logic circuit, which then activates the Auxiliary Step-up Converter to boost the voltage to 800mV. At this point the ZCS-Controlled Boost Converter is activated, which charges COUT while the logic circuits are still being powered by the Auxiliary Step-up Converter. During this time COUT is isolated from CDDI so it can charge up without dragging down VDDI.

When the voltage on COUT gets high enough to power the entire circuit the 'shunt' is turned on, connecting VOUT to VDDI so the logic circuit gets 1.2V. The Auxiliary Step-up Converter is now no longer required so it can be turned off. The Low-Voltage Starter also turns off because the extra load of the ZCS-Controlled Boost Converter causes the TEG's output voltage to drop below the Low-Voltage Starter's operating voltage.

This 3 stage stepping up procedure is required because the TEG cannot produce enough voltage to run the ZCS-Controlled Boost Converter (or even the Auxiliary Step-up Converter) directly. However when lightly loaded it can produce sufficient voltage to operate the Low-Voltage Starter, which can only produce enough voltage to operate the Auxiliary Step-up Converter. Once the Step-up or Boost Converter gets going it can use its own output voltage to power its control circuit, making the previous stage redundant.

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