0
\$\begingroup\$

I have a design problem to solve. The problem statement goes as follows: "Design a clocked comparator resolving input voltages close to vdd dissipating less than 100 uA".

What does "resolving input voltages close to vdd" mean? Does it mean that if the input voltage is close to vdd then the output should go to vss, depending on the reference voltage? It would be nice if you guys could explain it to me with an example.

\$\endgroup\$
  • 1
    \$\begingroup\$ Is this something your professor ask you? If so, have you asked them? \$\endgroup\$ – KingDuken Mar 5 at 18:52
  • \$\begingroup\$ Take a look at a comparator datasheet. Any comparator datasheet. \$\endgroup\$ – Toor Mar 5 at 19:01
  • \$\begingroup\$ @KingDuken Yes it is a homework problem, however i dont have access to the proff now, it is too late, and the deadline is tomorrow. \$\endgroup\$ – RAN Mar 5 at 19:19
  • \$\begingroup\$ @Toor Went through TI comparator datasheet, they all talk about resolution which i understand and the error, nothing about resolving voltages \$\endgroup\$ – RAN Mar 5 at 19:20
  • \$\begingroup\$ Pay more attention to the common-mode input range. You're probably unknowingly making sweeping false assumptions about what it is for every op-amp. If my eyes can distinguish billions of colours but get blinded by a night light, I ain't distinguishing nothin'. \$\endgroup\$ – Toor Mar 5 at 19:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.