reset switch in preamplifier

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Osawa_Odessa

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Hi, could anyone help me? I want to understand how the reset switch works in the preamplifier.

The paper says that the reset switch is inserted between the two output nodes for fast overdrive recovery.
Could you explain it more detail?
Also when the Vclk is high, the reset switch is turned on to erase the residual voltage from the previous sample.
How the output be erased? When the two nodes are shorted out then the voltage between them is zero. But I think there is a mechanism that is more complex in which output voltage is erased not simple as I said above.
This method improves the bandwidth.
I also can't explain why. If possible please help me with this one.
Here I will attach the paper just in case you need to have a closed look on this.
 

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  • A 42 mW 2 GSs 4-bit flash ADC in 0.18-μm CMOS.pdf
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  • Preamplifier reset switch.JPG
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Hi,
1.Let's assume in the previous operation has gone to rails Voutp-->vdd & Voutn-->0(Vinp>Vinn) then during the next transition if (Vinp< Vinn) then the circuit takes bit long time for the output to reverse, if not reset-ed using the switch. That is if the circuit is not given enough time for it to settle, then the output will have a component of previous transition too., ie, present vout=vout(1-exp(t/rc))-vout(previous)(exp(-t/rc)).
2.the output being reset-ed, ie, the differential voltage being '0'.
3. Intuitively it makes sense, since now the transition happens much faster and no need for large settling time-->looks as though the bandwidth as increased.
 

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