**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. V_{i} and V_{o} are respectively the input and output of the op amp. A_{v} 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}

\]