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.
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.
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.
