Electronic – Slew Rate Calculation

You only need to worry about slew rate for quickly changing signals, like audio for instance.

pH measurements change so slowly that you can ignore slew rate, and choose the amplifier on power consumption or price.

There are two things at play here:

1) Opamp bandwidth

2) Opamp slew rate

Let's say your opamp has the following transfer function (a low pass filter):

$$ H(s)=\dfrac{10}{\frac{s}{\omega_c}+1}$$

So, at dc the gain is 10 and the cutoff frequency is \$\omega_c\$.

The response of the circuit to a unit step input (just considering one half of the square wave) is:

$$ v_o=10(1-e^{-\omega_ct})$$

This is just a signal that will increase exponentially at the beginning before reaching steady state.

Let's check if the output is going to be BW-limited or SR-limited.

$$\dfrac{dv_o}{dt}= 10\omega_ce^{-\omega_ct}$$

The slope is the highest near zero, so the initial slope is:

$$\dfrac{dv_o}{dt}\bigg|_{t=0}= 10\omega_c$$

It needs to happen that \$10\omega_c\leq SR\$ so that the output is not SR-limited.

In this case, for your 1MHz cutoff, \$10(2\pi f_c)\approx63V/\mu s\$. So your output will definitely SR-limited and this is just for a unit step input (your square wave has amplitude of 5V). In fact (theoretically) the output will not be SR-limited for values of the input of about 15mV or less. But you'd still have the BW limitation, which will keep the maximum slope at:

$$\dfrac{dv_o}{dt}\bigg|_{max}=10V_{in}\omega_c \text{ for sufficiently small }V_{in}$$

And when \$V_{in}\$ is big enough so that the previous equation is greater than the SR spec—then the limitation will be the SR. For practical purposes, you'd still be SR-limited because many opamps have offset voltages in the range of the minimum input voltage found in this problem (unless you use a precision opamp) but this is homework...