# Question regarding slew-rate

I have a question regarding SR effect.

Let's say that we have some simple amplifier with op-amp with the following parameters:

$G=+10 \ V/V, \ f_c=1MHz, \ SR =1V/\mu s, \ V_{supply}=\pm 15V$

The source signal is a square wave:

$V_{in}=5V, \ \overline{V_{in}}=0V, \ f=10kHz$

And I want to ask about checking if SR effect will occur - it is easy to notice that the output voltage will be limited by $|V_{outMAX}|\approx 13.5V$.

So should I check if

(1) $2\pi\cdot\frac{f_c}{G}\cdot V_{outMAX}> SR$

or

(2) $2\pi\cdot\frac{f_c}{G}\cdot V_{out}> SR$ where $V_{out}=V_{in}\cdot G$

Edit: I assume that amplifier will produce something like this (red color) and this is what I am asking about

• An ideal square wave has infinite risetime, so the slew rate and bandwidth will always affect the signal. What's the risetime of your non-ideal squarewave? Commented Jul 16, 2018 at 1:08
• @JohnD It is just a homework problem, so we can assume, that the square wave is ideal. Commented Jul 16, 2018 at 1:12
• One word answers are not a good fit for this site. We prefer to give answers that will be helpful for future readers. Commented Jul 16, 2018 at 1:26
• The risetime in your graph is limited by the external capacitor C and resistor R, not by the internal behavior of the op-amp. Commented Jul 16, 2018 at 1:43

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...

For an ideal square wave, there's no need to check any equations. Regardless of the frequency or amplitude, slew rate limiting will distort the output.

An op-amp with limited slew rate (which means any real op-amp) can not produce an instantaneous transition from one level to another, which is required to produce an ideal square wave.

• But the point is that op-amp will not produce an ideal square (even with infinite SR), but an exponential function, due to the existence of capacitance inside the op-amp. And this can be - or not - limited by SR. Commented Jul 16, 2018 at 1:35
• For that particular case (a square wave for academic purpose), you can also simply go for a trapezoid model instead of the slightly more complicated exponential function. It carries the same concept that voltage can't change to any value instantaneously while being more simplistic to understand (especially with a slewrate expressed in V/us). Finally, I would also add that most of the time, it doesn't really matter if the square wave frequency is a lot smaller than the actual slew rate frequency. Commented Jul 16, 2018 at 1:52