reset switch in preamplifier

Status
Not open for further replies.
O

Osawa_Odessa

Banned
Joined
Dec 31, 2012
Messages
168
Helped
1
Reputation
2
Reaction score
1
Trophy points
1,298
Visit site
Activity points
0
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.
 

Attachments

  • A 42 mW 2 GSs 4-bit flash ADC in 0.18-μm CMOS.pdf
    326.4 KB · Views: 46
  • Preamplifier reset switch.JPG
    89 KB · Views: 72

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.
 

Status
Not open for further replies.

Similar threads

Menu
Log in
Register
Cookies are required to use this site. You must accept them to continue using the site. Learn more…