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This suggested edit was approved and applied to the post almost 2 years ago by Olin Lathrop‭.

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Is it possible to calculate the rise/fall time of an operational amplifier ?
  • Hi guys so i am planning to use LM318 operational amplifier to generate a pwm signal and i need to know how to calculate the rise/fall time thank you.
  • [LM318/LM218/LM118 datasheet](https://www.google.com/url?sa=t&source=web&rct=j&url=https://www.ti.com/lit/gpn/LM318-N&ved=2ahUKEwjN7LqX6L77AhUYVaQEHYrCCPQQFnoECA0QAQ&usg=AOvVaw2teDo15HVgRulQWeLnMu54)
  • **Edit:**
  • -The slew rate is 70v/ųs max.
  • -Slew rate can reach 150v/ųs
  • But with a condition of"feedforward compensation will boost the slew rate to over
  • 150V/μs and almost double the bandwidth."
  • **Edit2:**
  • As suggested from Mr. Olin here my calculation:
  • the LM318 offer 70v/ųs and the output voltage that i need is 18v so:
  • (18 V) / (70V/µs) = 0.257143 ųs = 257.143 ns.
  • And then i did this:
  • (257.143/5)×4 = 205.714 ns.
  • Because rise/fall time start at 10% and end at 90%.
  • And in case of 150v/ųs this gonna be:
  • (18 V) / (150 V/µs) = 0.12 µs = 120 ns
  • (120/5)×4 = 96ns.
  • Is my calculation correct? Thank you.
  • I am planning to use the LM318 operational amplifier to generate a PWM signal and I need to know how to calculate the rise/fall time.
  • [LM318/LM218/LM118 datasheet](https://www.google.com/url?sa=t&source=web&rct=j&url=https://www.ti.com/lit/gpn/LM318-N&ved=2ahUKEwjN7LqX6L77AhUYVaQEHYrCCPQQFnoECA0QAQ&usg=AOvVaw2teDo15HVgRulQWeLnMu54)
  • -The slew rate is 70 V/µs maximum
  • -Slew rate can reach 150 V/µs
  • But with a condition of "feedforward compensation will boost the slew rate to over
  • 150 V/μs and almost double the bandwidth."
  • As suggested from Mr. Olin, here is my calculation:
  • the LM318 offers 70 V/µs and the output voltage that I need is 18 V so:
  • (18 V) / (70 V/µs) = 0.257143 µs = 257.143 ns.
  • And then I did this:
  • (257.143/5)×4 = 205.714 ns.
  • Because the rise/fall time start at 10% and end at 90%.
  • And in case of 150 V/µs this is going to be:
  • (18 V) / (150 V/µs) = 0.12 µs = 120 ns
  • (120/5)×4 = 96 ns.
  • Is my calculation correct?

Suggested almost 2 years ago by Peter Mortensen‭