Slew rate is defined as the maximum rate of change of the op amp output voltage per unit of time in the closed-loop configuration under large-signal condition. Slew rate indicates how rapidly the op amp output can change in response to change in the input frequency. It is given as
\[
\quicklatex{color=”#000000″ size=20}
\boxed{Slew\; rate=\frac{\mathrm{d} V_o}{\mathrm{d} t}\mid_{max}}
\]
The Slew rate of an op-amp is fixed. If the slew rate is $10\frac{V}{\mu sec}$ then it means that the op-amp output voltage can change maximum by 10 V in 1 $\mu sec$.
The ideal value of the slew rate is infinite. Practically it is a finite value and it is due to the following reasons
- saturation of internal stages and
- the finite time constant of internal capacitors
Consider figure 1. Vi and Vo are respectively the input and output of the op amp. Av is the closed-loop voltage gain.
$V_{i}=V_{m}\sin \omega _ot$
Therefore, the output voltage $V_o=A_vV_i$
\[\Rightarrow V_{o}=(A_vV_{m})\sin \omega _ot\]
Now,
\[\frac{\mathrm{d} V_o}{\mathrm{d} t}=A_v\frac{\mathrm{d} V_i}{\mathrm{d} t}\]
$\Rightarrow \frac{\mathrm{d} V_o}{\mathrm{d} t}=A_vV_{m}\omega _o\cos \omega _ot$
Therefore,
$\Rightarrow \frac{\mathrm{d} V_o}{\mathrm{d} t}\mid_{max}=A_vV_{m}\omega _o\times 1$
\[\frac{\mathrm{d} V_o}{\mathrm{d} t}\mid_{max}=A_vV_{m}\omega _o\]
For producing undistorted op amp output, $\frac{\mathrm{d} V_o}{\mathrm{d} t}\mid_{max}$ must be less than the slew rate i.e.
\[
\quicklatex{color=”#000000″ size=20}
\boxed{\frac{\mathrm{d} V_o}{\mathrm{d} t}\mid_{max}\;<Slew\; rate}
\]
