What time the Op amp voltage follower takes to settle?

Hi, all

I have designed a 2 stage differential input and differential output op-amp with gain close to 45dB and phase margin of 60 degrees. I am trying to connect it as voltage follower based on Op Amp Voltage Follower Uses (source https://www.apogeeweb.net/electron/what-is-the-use-of-op-amp-voltage-follower.html) to check its response for a step input. I am connecting a large resistor between the (+)input and (-)output and vice versa. With VDD = 0.9V and VSS = -0.9V, and differential input 200mV, I am expecting the same differential output voltage.

However, I get an amplified output voltage (Almost -0.8V to +0.8V). I checked that the op-amp is not getting saturated. What am I missing? I want to check what time the output takes to settle.

by Robertlee
June 28, 2021

You have to check the output voltage swing value of the op amp. As example, for a LM741, with a + and - 15V supplied, the maximum output could be as bad as no more than + and - 12 V, but typically + and - 14V. That means that, typically, you can expect to "lose" 1 volt from "the rail" , from what is the supply. Note that the maximum that your output can be also depends on the "load" seen by the output. There are OpAmps which can go (almost) up to the supplied voltage, they are said "rail to rail", but they MAY bring other problems of their own.

For the LM741, you can check at its datasheet from Texas Instrument, among others, here, section 6.5 LMt41_MIL . Note that we lose an extra volt, in this case, when the output resistance drop from 10k to 2k. Other manufacturers may call that caracteristic under an another name than "output voltage swing", but this value is generally close, but less (in magnitude) than, the supplied voltage to the chip.

by vanderghast
June 28, 2021

I typed something wrong. Try the following link instead of the one that I supplied before (and if it does not work, try a search on "LM741 datasheet": LM741

by vanderghast
June 28, 2021

Oh, and a possible solution? Try to increase the supplied voltage to the chip, leaving the "input" signal voltage at a one volt, in magnitude, at least, from the "rails". And if your chip supplied "ground" is a your circuit ground, then, too bad, the output cannot reach ... zero volt difference to your ground, indeed (frequent problem first times we use OpAmps).

by vanderghast
June 28, 2021
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