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Can't seem to find the connection between the two. I have an error amplifier being fed with an analog signal. I'm varying the GBW of the op-amp to determine how does its GBW affect the rise time of the output. From my spice simulations, what i'm only getting is that increasing the bandwidth reduces the low and high frequency voltage ripples of the output. A higher bandwidth seems to give a faster rise time but I don't understand why is that so.

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    You have given us the open-loop bandwidth of the opamp. To apply Tony's equation, we need to know the closed-loop bandwidth. Measure it, or post the circuit. – Mattman944 Apr 25 '21 at 23:03
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    Your opamp has a gain of 3, so divide your GBW by 3 to get the bandwidth of the closed loop. For the 250 Hz GWB, the BW would be 83 Hz. Looking at your bottom graph, the 25 Hz should get through mostly intact but it is attenuated quite a bit. Something isn't making sense, are you sure that the GBW is 250 for the bottom plot? If you want to understand this circuit better, analyze/simulate each input separately. – Mattman944 Apr 25 '21 at 23:49
  • Yes i am sure. Thank you, I will try your suggestion. –  Apr 26 '21 at 00:25
  • For larger signals the slew rate is often limiting, not the GBW. – Spehro Pefhany Apr 26 '21 at 01:37

3 Answers3

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There’s a clue in the name - the gain bandwidth describes the gain at any given frequency, and so higher frequency signals (or in the case of a step function, higher frequency components of the input signal) will see lower gain than low frequency components, if the GBW is the limiting factor.

Frog
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    To expand a little: with a moderate gain (say 10) and a moderate input frequency, an amplifier circuit will output 10 times the input signal, limited only by the voltage swing available at the output. At higher frequencies the output may be limited by the GBW and/or the slew rate that the amplifier can achieve. – Frog Apr 25 '21 at 22:56
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Rise time is inversely related to LPF (low pass filter) -3dB BW for a simple 1st order filter.

\$t_{10 ~to~ 90\text{%}}=\dfrac{0.35}{f_{BW-3dB}}\$

However, you cannot measure the 10% to 90% duration with a sinewave superimposed.

For a given open loop GBW product, BW = GBW / Gain closed loop.

BW can also be reduced with external CR filters with negative feedback.

For slew rate on large signals the active current limiter determines this into a specified reference cap. load

dV/dt= Ic/C

Christianidis Vasilis
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Tony Stewart EE75
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What helped me set up an intuition was thinking of signal in the frequency perspective. When you are trying to re-create a "discontinuous signal" (like a pulse) in the time domain you need higher and higher frequencies to make the slope steeper and steeper. That is why a square waves has an infinite Fourier expansion. This is from the very nature of sinusoidal oscillations, so a very quick rise in time domain means high frequency content.

Now, when you have systems with a low bandwidth, it means you are suppressing the high frequency content. And no high frequency means the edges will not be as steep and it ultimately means the rise time is slow. If your bandwidth is higher the high frequency contents will not be suppressed as much, hence faster rise-time.

So this has inherently become a rule amongst circuit designers, Low bandwidth = Slow circuits. But this however has a lot of theory and math in the background.

HyperBean
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