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I used two stage differential opamp as comparator but I'm getting the output -ve voltage range.

I have to get high (1.8 V) when it is greater than reference voltage and low (0 V) when less than reference voltage. But as of now I'm getting high -6 V and low -1.75 V.

(net4 is reference voltage and net6 is the input voltage which is a sine wave.)

Schematic schematic

vss=-1.8V , vdd-1.8V vss=-1.8V , vdd-1.8V

Result result

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  • \$\begingroup\$ Can you ask a specific question? \$\endgroup\$
    – Voltage Spike ♦
    Commented Apr 18 at 13:01
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    \$\begingroup\$ You cannot get a -6 V outpput with only +/- 1.8 V power supplies. \$\endgroup\$
    – Peter Bennett
    Commented Apr 19 at 6:25
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    \$\begingroup\$ I'm sorry but the sinus wave gets right up my nose... :-D \$\endgroup\$
    – TonyM
    Commented Apr 19 at 9:13
  • \$\begingroup\$ @TonyM Sorry, my bad \$\endgroup\$
    – Velvet
    Commented Apr 19 at 11:45
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    \$\begingroup\$ @TonyM On the contrary; it made me smile :) \$\endgroup\$
    – Velvet
    Commented Apr 19 at 13:12

1 Answer 1

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From your schematic we cannot see what is "net6" that you use in the waveform plot, but I assume that it is V2, where the sine wave source is present. We can then further assume that "net4" is the reference voltage.

In the waveform plot the output is high when "net6" is higher than the reference voltage, and low when it is less than the reference voltage, which is exactly what you describe as the desired behavior in your question. So it is unclear why you are not satisfied!

Nevertheless, you can of course change the behavior to the opposite, by using the opposite input terminals for reference voltage and signal input. That is the essence of having a differential amplifier.

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