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With regard to the picture below, it says with an AC signal as an input, the output is a square wave. I am familiar with the uses of op-amps like in differentiator, integrator, inverting and non-inverting, active filters etc. But those I can memorize their configuration. But this I could not understand.

Circuit diagram including op-amp

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    \$\begingroup\$ It's not a square wave: it's an infinitely high sine-wave clipped at the power-supply rails. \$\endgroup\$
    – Dan
    Dec 17, 2019 at 9:04
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    \$\begingroup\$ Well, that means it is a square wave, but you see what I mean? \$\endgroup\$
    – Dan
    Dec 17, 2019 at 9:05
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    \$\begingroup\$ No, it's not a square wave - the high/low plateaus are of different duration. \$\endgroup\$
    – Chu
    Dec 17, 2019 at 14:33
  • \$\begingroup\$ The corners are square but we reserve “square” for more or less desired 50% symmetry \$\endgroup\$ Dec 18, 2019 at 23:53

3 Answers 3

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This operates like a comparator except rather than open drain or collector is a push-pull to bipolar supplies (Vcc~Vee) minus some dropout voltage. The output is slew-rate limited to the component spec [V/us].

Remember that the (open loop) output inverts when the inputs are not equal (ignoring Vio offset) .

Other

This is not how you draw a sine wave which is more like triangular but peaks are smooth about 25% below the apex. You often see people draw sinewaves this wave with far too much "risetime". When we measure risetime of a pulse from 10 to 90% there is a formula to estimate the bandwidth BW(-3dB) BW= 0.35/Tr (10~90%)

Here is how you can verify a hand drawn wave using 3/4 of a triangle. enter image description here

Since an open loop Op Amp may saturate internally after the internal compensation , then the linear GBW does not apply and the slew rate is limited by the output stage current limit defined in the datasheet. There may be reverse recovery times specified for overdrive unlike a good comparator IC.

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That is just used as a comparator. The sine wave is either above or below the reference and that is the result of the comparison.

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  • \$\begingroup\$ Okay 😅, now I got it. Thank you 👍 \$\endgroup\$
    – f321
    Dec 16, 2019 at 21:30
  • \$\begingroup\$ Is that all?.... \$\endgroup\$ Dec 18, 2019 at 23:54
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If you consider the inverting mode amplifier with which you are familiar you should be able to spot:

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. An inverting amplifier. If Rf is omitted, then Rf = ∞ and the gain becomes the open-loop gain of the op-amp.

  1. The signal is applied to the inverting input so it's an inverting amplifier.
  2. R2 and R3 provide a reference voltage for the non-inverting input. R2 should, presumably, be connected to +VS.
  3. There is no negative feedback resistor or, to put it another way, Rf = ∞ so the gain is "infinite".

As soon as VIN- exceeds VIN+ the output will swing to the negative rail. When the situation is reversed the output will swing to the positive rail.

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  • \$\begingroup\$ no, that wouldn't make sense. \$\endgroup\$ Dec 16, 2019 at 21:45
  • \$\begingroup\$ Okay you mean of the op-amp. Ok \$\endgroup\$
    – f321
    Dec 16, 2019 at 21:49
  • \$\begingroup\$ Thank you very much for your answer \$\endgroup\$
    – f321
    Dec 16, 2019 at 21:51
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    \$\begingroup\$ @muyustan: Good! You were paying attention. \$\endgroup\$
    – Transistor
    Dec 16, 2019 at 22:11
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    \$\begingroup\$ @muyustan the LM324 would be a better choice. LM741 has minimum recommended supply voltage of 10V. Its inputs need to stay 2V away from the supply and ground for correct operation, and the output also only gets to within 1~2V of the supply rails (that's a big loss when running from 9V!). \$\endgroup\$ Dec 17, 2019 at 0:47

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