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Op-amp slew rate - bandwidth relation?

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David_

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Hello.

While reading comments about a circuit on a site a short talk about the usable frequency range of the meter circuit, and someone said that TL074 should work to about 2MHz(slew rate = 13/µS)
and that using AD843 which has a slew rate = 250V/µS would increase the bandwidth to 20MHz.

I don't understand the relationship or how to do the numbers to figure that out, I don't "need" to know but I sure would like to know if someone could explain.

I know that slew rate is the rate of change the output can manage but how does that determine the frequency capabilities, I do see kind of how it works as higher frequency will need the voltage to change faster but how do I know that 13V/uS = 2MHz & 250V/uS = 20MHz?

Regards
 

To get teh calculation right, you have to refer to a specific output voltage. Without it, the numbers are meaningless.

The maximum slew rate of a sine voltage can be easily calculated, it's ω*Vpk.
 
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    David_

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The voltage gain of an opamp circuit also determines the bandwidth because an opamp rolls off high frequencies to avoid oscillation at a high frequency when negative feedback is added.

The datasheet for the TL07x opamp shows a bandwidth to 100Hz when the voltage gain is about 32,000 times, to 1kHz when the voltage gain is about 3,200 times and to 10kHz when the voltage gain is about 320 times.
Its slew rate allows a high level output up to 100kHz.
 

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Just for clarification:
In the data sheet there are two different bandwidth information:
* Small signal bandwidth - dependent on the small-signal gain (determined by feedback)
* Large signal bandwidth - dependent on output amplitudes and determined by the slew rate .
 
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Okey, say that we have a op-amp as a unity-gain follower that buffers a sine wave of 4Vpk-pk.
Is that plus slew rate enough data to calculate the top frequency the buffer can handle?

Thanks for ω*Vpk.

The slew rate of the sine wave of 4Vpk-pk is (2*pi*Hz)*2V?

I'm sure that if one would be good at math one would be able to figure out how to get the bandwidth with this information, but I could check the slew rate of a op-amp and then check the slew rate of the signal to make sure it wont clip. But I would do guess after guess to get the frequency that match a certain voltage peak to result in a specific slew rate.

ω*Vpk gives the slew rate for sine waves, how about squares and triangle waves?

I don't understand which is large signal bandwidth and small signal BW?

thanks for the info.
 

The slew rate limits high frequencies of high level signals.
The small signal bandwidth limits high frequencies at high gain and low levels.
 
I don't understand which is large signal bandwidth and small signal BW?

Look at the diagrams in post#3:
* The most right diagram shows the bandwidth deteremined by the slew rate (large signal bandwidth)
* The other diagram shows the frequency-dependent small-signal open-loop gain of the opamp (without feedback).
However, because of a constant product gain*bandwidth you can see that with feedback and a gain of - let`s say - 100 (40dB) the bandwidth is app. 30 kHz.
 
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Look at the poor slew rate of old low power LM358 dual and LM324 quad opamps. At high output levels they cannot even play high telephone audio frequencies:
 

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