I’m in the process of selecting an op-amp to act as a voltage follower for the input signal into my data acquisition card.

I'm measuring all types of signals just like an oscilloscope would.

Do I have the slew rate I need correct? I used this calculator and sampling at 1.5 MHz with a peak of 10 V I calculated that I need a slew rate of 94 V/µs.

The reason I ask is I don’t see a lot of choices of rail-to-rail op-amps, through-hole style with that high slew rate.

Slew Rate Calculator

  • \$\begingroup\$ Is your input a pure sine wave? If it is not, you will have higher frequency harmonics and will need an even higher slew rate. \$\endgroup\$ – helloworld922 Feb 25 '14 at 4:30
  • 1
    \$\begingroup\$ Im measuring all types of signals just like an oscilloscope would \$\endgroup\$ – bigshop Feb 25 '14 at 6:15
  • \$\begingroup\$ From your question I understand that the problem is the poor selection of devices with needed parameters with TH pakaging. Then why don't you consider the use of smd - th board adaptor or smd component? \$\endgroup\$ – zzz Feb 25 '14 at 8:03
  • \$\begingroup\$ My main question is - Do I have the slew rate I need correct. I used this calculator and sampling at 1.5MHz with a peak of 10v I calculated that I need a slew rate of 94 v/us \$\endgroup\$ – bigshop Feb 25 '14 at 8:44

If you are sampling at 1.5 MHz, the time taken for your analogue amplifier to slew its output has to be a bit quicker than the reciprocal of 1.5 MHz i.e. 0.667 µs. If it has to deliver a change of 10 V in this period then the slew rate, as a minimum must be:

\$\dfrac{10\ V}{0.667\ \mu s}\$ = 15 V per micro second.

You also need to look at the op-amp's settling time - this is not encapsulated within the slew rate figure, and the settle time is usually specified to within the voltage reaching its target to an accuracy of 0.1%. If you are using a 10-bit ADC, 0.1% could be 1 LSB.

Enter image description here

  • \$\begingroup\$ The image you included really helps me understand. Im curious though how did you get the 0.667us number, is there another calculation for this. Please give a not to tech an answer because im sill very much the beginner \$\endgroup\$ – bigshop Feb 26 '14 at 7:08
  • \$\begingroup\$ sampling at 1.5MHz is a sampling time of 0.6667 microseconds. \$\endgroup\$ – Andy aka Feb 26 '14 at 9:15
  • \$\begingroup\$ @bigshop This number came from the period. A 1.5MHz (frequency) signal has a period of 0.6667uS from the following math: T = 1/f (where T is period and f is frequency.) \$\endgroup\$ – jonnyd42 Aug 4 '16 at 20:37

Edit: I read bandwidth rather than "sampling at". This updated result would make more sense.

This is how you would calculate the slew rate. Assume a sinusoid with amplitude \$ A \$ and frequency \$f\$: \$A\sin(2\pi f t)\$. The Nyquist sampling theorem gives that the maximum signal frequency is half the sampling frequency, i.e. 750 MHz.

The derivative of this is: \$\frac{d A\sin(2\pi f t)}{dt} = 2\pi f A\cos(2\pi f t)\$ which has a maximum value of \$2\pi f A = 2\pi \times 750 \times 10^6 \times 10 \approx 47\$ MV/s \$ = 47 \$ V/\$\mu\$s, which is half of what the calculator says.

Assuming that you understand what the bandwidth actually means with respect to other signals than sinusoids: yes, the calculator is correct but you need to enter the largest signal bandwidth according to the Nyquist criteria, not the sample rate.

I would go for non-rail-to-rail OpAmps if I would break any of the requirements.

  • \$\begingroup\$ Be careful about confusing the signal frequency with the sample frequency here. The OP won't be able to do better than ~1/5 sample frequency for recording signal frequencies. And that includes the highest harmonics that are desired for capture. \$\endgroup\$ – gsills Feb 25 '14 at 22:24
  • \$\begingroup\$ I read bandwidth rather than "sampling at", so I updated it now. You mean 1/2 instead of 1/5, right? \$\endgroup\$ – Oscar Feb 25 '14 at 22:41
  • \$\begingroup\$ 1/2 gets you to Nyquist, but to start to get sensible information about signal shape requires more. It implies that slew rate based on Nyquist is more than is needed. For good representation, signal BW and amp slew rate req will be far below sample rate. OP may not have thought of this or aliasing yet. Can't tell from question. \$\endgroup\$ – gsills Feb 26 '14 at 1:40
  • \$\begingroup\$ That would really depend on the signal. :-) However, I do not agree with your conclusion that slew rate based on Nyquist is MORE than needed. One can possibly argue that it is less than needed, although that depends on the signal, the possible anti-aliasing before sampling and so on. Aren't you saying that the sample rate/BW factor of 2 is really 5 if you want "to get sensible information"? To me, this is clearly something new, which I more think would result from a rule of thumb rather than some theoretical reasoning, where the rule of thumb would be based on an uncommon definition of BW. \$\endgroup\$ – Oscar Feb 26 '14 at 7:06
  • \$\begingroup\$ Depend on the signal? Yes it would. That's why multiple samples/period are needed. Limit BW to Nyquist and get 2 points per period. All signals look like a triangle wave without more samples/period. No information is lost by having OpAmp SR limited at Nyquist. But that's Andy's answer, 15V/uSec. If you put that through SR=2PiFA and solve for F you'll get a frequency about 1/6 the sample rate. I said 5, but that was pushing it. After aliasing figures in it will be more (probably a lot), but that will depend on the waveform. Rules of thumb can be good, they save time.;^) \$\endgroup\$ – gsills Feb 27 '14 at 0:53

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