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So I am using an LM741/A Op-Amp. Wondering whats the value of its Vsat? Have been looking at its datasheets but cannot find anything. Am I missing something? Is the Vsat just equal to its output voltage swing?

Wetyyjs
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2 Answers2

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This data sheet says

LM 741, +-15V supply, Output voltage swing : ±12(MIN) ±14 (TYP) for 10k ohm load

and

LM 741 (A), +-20V supply, Output voltage swing : ±16(MIN) for 10k ohm load

Ti website also says "Rail-to-rail : No"

Transistor
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AJN
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  • https://electronics.stackexchange.com/a/304522/238590 – AJN Jul 09 '20 at 06:48
  • Ok so for a 2Kohm load, the Typical output voltage swing is 13V, and the minimum is 10V with a 15V power supply.LM371's maximum supply voltage is 22V. Let's say I have another op-amp that has a typical output voltage swing for a 2Kohm as well of 12V and a minimum output swing of 10V with a 15V power supply. This other op-amp has a max supply votage of 18V. Which op-amp do I use if I want an undistorted Vout peak voltage of 12.5V? Any help is appreciated thanks. – Wetyyjs Jul 09 '20 at 06:56
  • From what you said, +-15V supply would not work with either opamp. When you say *undistorted* you might also want to specify a frequency range and choose one with larger slew rate. – AJN Jul 09 '20 at 07:11
  • Yes the Vin frequency is 50Khz and both op-amps have suitable slew rates. Just wondering which op-amp I should choose taking into account the voltage swing. Both op-amps output swing can reach over 12.5V if supply votage is 18V. So which one do I use – Wetyyjs Jul 09 '20 at 07:23
  • "*... both op-amps have suitable slew rates.*" Can you show your calculations? Mine show that 50 kHz, 12.5 V peak means you require a slew rate of at least 12.5 V in one quarter cycle at 50 kHz so \$ slew\ rate = 12.5 \times \frac 1 4 \times \frac 1 {50k} = 62.5 \ \text{V/}\mu \text s \$. The datasheet you linked to says the 741 can do 0.5 V/μs and that's only at unity gain. I think you'll struggle. – Transistor Jul 09 '20 at 19:13
  • Wouldn't the max rate expression for \$12.5 \sin(2\pi 50k\ t) \leftrightarrow 12.5 \cdot 2\pi\cdot 50k \cdot \max(cos(\cdot))\$ ? I got \$4 V/\mu s\$. – AJN Jul 10 '20 at 03:01
  • @AJN: Yes, I think you're correct. Pasting "d (12.5sin(2π50k t) )/dt" into https://www.wolframalpha.com/ gives 3926.99 k cos(100 π k t) + 0 so when cosine = 1 that equates to 4 MV/s = 4 V/μs. I assumed, for simplicity,\ a triangle wave going from the zero-cross to V-peak in one quarter cycle. I can't see where I've gone wrong though. – Transistor Jul 10 '20 at 17:34
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    This is old, but just for the record, a sine wave has a maximum dv/dt of about 1.5 times higher than the same peak voltage and frequency triangle wave. (actually π/2) – Spehro Pefhany Nov 10 '22 at 16:53
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The voltage saturation level is usually 1 or 2V less than power supply levels. If your Voltage supply is + or - 15V, then \$V_{sat}\$ is likely to be + or - 12 or 13V. The maximum voltage swing is about +13V or -13V.

Reference:

Page 133, Sedra and Smith -Microelectronic circuits 7e , Oxford university press. "An Op that is operating form power supplies of +/- 15V will saturate when the output which reaches +/- 13V".

Amit M
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