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This opamp output starts getting distorted at around 10kHz and the amplitude halves around 25kHz.

Where in the data-sheet shows the relationship between the frequency ant the attenuation? Something same with Bode plot.

EDIT:

To clarify I had to simulate the circuit what I observe via a scope:

enter image description here

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  • \$\begingroup\$ The graph on page 6 clearly shows the typical open loop gain and phase vs. frequency, \$\endgroup\$ – Peter Smith Nov 23 '18 at 13:30
  • \$\begingroup\$ Can you provide a schematic of how you have it set up? This could be a supply voltage issue, or depending on the gain you have set, it could also be a slew rate issue. It may not just be the frequency (although it may play a part) \$\endgroup\$ – MCG Nov 23 '18 at 14:15
  • \$\begingroup\$ Cannot be answered without additional information. What means "This opamp"? With or without feedback? Which amplitudes for start of distortion? Attenuation? Do you mean decreasing gain? Very vague questions! \$\endgroup\$ – LvW Nov 23 '18 at 14:53
  • \$\begingroup\$ @LvW there is a link to the op amp in the question \$\endgroup\$ – MCG Nov 23 '18 at 15:06
  • \$\begingroup\$ @LvW I simulated the circuit now, it was hard to show the actual setup on breadboard. Please see the edit. \$\endgroup\$ – HelpMee Nov 23 '18 at 15:06
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If I am understanding your question correctly, you have set up a test circuit and as you change the frequency, you are noticing your output distorting (clipping?) and getting worse as the frequency increases, leading to the amplitude decreasing?

Well, this could well be a few issues, but lets look at what could cause issues with an op amp output:

It is driving a load? If so, what is the load it is driving? As shown in the electrical characteristics, the op amp can only output a certain current depending on supply voltage:

enter image description here

Using Ohms Law, you will be able to find out the maximum load it can drive. Now, looking at one of the graphs in the datasheet, you can actually see that the output impedance changes with frequency:

enter image description here

This will again affect the output voltage/waveform depending on the load driven, the frequency, the supply voltage, and the gain.

Of course, there is the supply voltage issue which can cause clipping if the gain is too high, but as you say it seems to be OK up until 10kHz, I think it is safe to assume it isn't that.

There is also the Slew rate. If your op amp is slow, and your gain is high, then your op amp may not be able to actually reach the output voltage in time, hence a decrease in amplitude as the frequency increases. Let's use a classic 741 as an example, whioch has a slew rate of 0.55V/us. This means the output can go from 0-0.5V in 5us. So, to reach 10V on the output, it would take: 10/0.5 = 20us. At, say, 200kHz, a full cycle takes about 6us, showing the output voltage would get nowhere near the 10V output, resorting in a distorted output, and a decrease in amplitude. Of course, that was an extreme example, but you get the picture.

Looking at the datasheet for your amplifier, it does look to be rather slow. I would guess that this is the cause of the output distortion you are seeing. You either need to get a faster op-amp, or decrease the frequency

Next is the one I think is the one you are after (I added in the bits before as they are useful to know and may have something to do with what you are seeing).

Look at the Gain vs Frequency graph:

enter image description here

Using this, we can use a bit of math to see if we are going to get any distortion. I will again need to make some assumptions here as I don't know your input voltage/supply voltage/gain etc.

Let us assume you have an input voltage of 50mV pk-pk and an output voltage of 5V pk-pk, giving us a gain of 100. We can find out what that is in dB (as that is what the graph shows the gain in). To do this, we use: 20Log(100) = 40. This means your gain is 40dB. Looking at the graph, we can see that at this gain value, clipping/amplitude decrease will start to occur near the 2kHz mark:

enter image description here

This means that trying to maintain the same gain value above that frequency can't happen. The gain will actually decrease. I believe this is the graph you were interested in, however, there could be other causes, hence the rest of the information in this answer.

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  • \$\begingroup\$ ...."anything above that frequency with that gain value..." ??? Gain values above that frequency cannot assume "that gain value" - the gain decreases continuously ! These characteristics have nothing to do with distortion!! Theay are small.signal characteristics! It is the large-signal characteristic which matters only (slew rate induced distortions). \$\endgroup\$ – LvW Nov 23 '18 at 14:59
  • \$\begingroup\$ @LvW Yeah..... I suppose the wording isn't very good there! I'll have a think about how to word it lol. I also used distortion in the sense that the output won't be as expected.... I'll clean up my wording. Thanks for pointing it out \$\endgroup\$ – MCG Nov 23 '18 at 15:00
  • \$\begingroup\$ @LvW tidied up the wording. Hopefully that is better. When something makes sense in my head (because I know what I'm thinking) I tend to just type away without proof reading! \$\endgroup\$ – MCG Nov 23 '18 at 15:04
  • \$\begingroup\$ @ MCG Delightful answer. You might explore distortion, because the input differential voltage must rise at higher frequencies, to sustain a 1volt output; this decade per decade increase in Vin lets the Taylor-series model of the input diffpair predict how 2nd and 3rd order products will increase. With bipolars that are not resistively linearized, this is a fine tool to explain distortion. MOSFETS have other degrees of freedom in the diffpair, but the same general relationship (albeit different Intercept Points) still apply. Again ..... a fine answer. \$\endgroup\$ – analogsystemsrf Nov 23 '18 at 15:06
  • \$\begingroup\$ Please see the edit, I tried to simulate the actual circuit. In this case 60kHz to see the distortion and attenuation clearly. It starts far before indeed. \$\endgroup\$ – HelpMee Nov 23 '18 at 15:07
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Slew rate is the issue: -

enter image description here

The data sheet specifies a figure of typically 60 mV per micro second and, your signal changes about 500 mV in about 8 us and this of course is 62.5 mV per micro second i.e. right on the limit of what the table in the data sheet tells you.

enter image description here

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@user1234...as I have expected: The triangle form of the output signal (shown in your edited answer) is a typical indication for slew-induced distortions.

That means: The limited slew rate of the opamp - if it has negative feedback ! - allows sinusoidal output forms up to a certain level only. And this level is frequency-dependent.

This maximum output signal (sinus !) as a function of frequency is given as "large signal characteristics" in the data sheet. For small frequencies, it is a horizontal line and then, suddenly it goes down.

Explanation: Slew rate is defined and is measured WITH feedback only - and the input signal must drive the first diff. stage of the opamp into saturation. In this case, the first stage acts as a current source which can charge the following MILLER capacitance with a limited speed only. This determines the so-called "slew rate".

The open-loop characteristic (gain and phase vs. frequency) has NOTHING to do with the distortions you have observed.

EDIT/UPDATE: The LT1490 datasheet gives the large-signal characteristics in the first graph on page 10.

You can find the slew-rate determined large-signal bandwidth Bsr (maximum frequency for maximum sinusoidal signal at the output) by yourself:

Bsr=SR/(2*Pi*Vomax)

For a typical value (datasheet) SR=0.06 V/µs and (estimated) Vomax=4V (supply 5V) we get Bsr=2.4 kHz (see the graph on page 10)

(Slewrate SR, Vomax: max, possible output voltage - determined by the supply)

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  • \$\begingroup\$ Where in the data sheet should I look at for slew rate and how to relate to my desired max freq.? Lets say I want no distorion up to 1MegHz, where in data sheet we can refer 1Meg will not be alright because of slew rate? Thanks \$\endgroup\$ – HelpMee Nov 23 '18 at 15:41
  • \$\begingroup\$ If you do not want to have such distortions, you must take care that the output amplitude is not larger than given in the large -signal charcteristic. Hence, you must decrease the gain or the input amplitude correspondingly. \$\endgroup\$ – LvW Nov 23 '18 at 15:43
  • \$\begingroup\$ @user1234 the slew rate is at the bottom of page 3, in the electrical characteristics section. And if you want no distortion up to 1M, you will have to either decrease your gain, or lower your pk-pk input voltage \$\endgroup\$ – MCG Nov 23 '18 at 15:44
  • \$\begingroup\$ For the LT1490 the slew rate is pretty bad: Just 0.03V(s app. And the corresponding large-signal characteristic is on page 10 (first graph). \$\endgroup\$ – LvW Nov 23 '18 at 15:49
  • \$\begingroup\$ The mentioned value of 0.03V/us is a worst case value. Typically: SR=0.06V/us. \$\endgroup\$ – LvW Nov 23 '18 at 16:15

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