Ex :


My question is since it's a single supply op amp, does this mean that we can't generate negative voltage so the transfer function would bottom at 0?

  • \$\begingroup\$ Assuming that the intersection of the axes in your graphs represents 0 V, and Vcc is not 0 V, and R1 is not infinite, and R2 is not 0 ohms, then neither graph is right. \$\endgroup\$
    – The Photon
    Oct 17, 2015 at 5:52
  • \$\begingroup\$ yea sorry realized my mistake. I recentered my question about what I am really trying to figure out. \$\endgroup\$
    – ESD
    Oct 17, 2015 at 6:06
  • 1
    \$\begingroup\$ By the way, the usual term is "single supply" not unipolar. \$\endgroup\$ Oct 17, 2015 at 8:17

1 Answer 1


Yes, the output will be approximately equal to the negative supply voltage (0 V in this case) when \$v_i < V_{cc}\frac{R2}{R1+R2}\$ and approximately equal to the positive supply voltage when \$v_i > V_{cc}\frac{R2}{R1+R2}\$.

The comparator doesn't know or care whether the negative and positive supply voltages are positive, negative, or equal to the circuit reference (ground) voltage.

  • \$\begingroup\$ To the OP: some op-amps do a better job than others of driving the output voltage close to the power rail voltages. The term you would be looking for if you want that feature is "rail-to-rail" output. \$\endgroup\$
    – user57037
    Oct 17, 2015 at 6:18

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