LM386 input stage protection

If your input signal has amplitude 5V peak-peak, then you certainly don't require gain 200 from this IC. Don't include the 10μF capacitor between pins 1 and 8. Without that capacitor, gain is 20, as shown in diagram 9-1 of the datasheet.

Scale the input signal using a resistor voltage divider. If your supply is +5V, then you are probably expecting about 3V peak-peak output (see fig. 6-3). Given that voltage gain is 20, this corresponds to \$\frac{3{\rm V}}{20}=0.15{\rm V_{PK-PK}}\$ (±75mV) at the input. Anything beyond that would cause clipping at the output.

Conveniently, this is well below the maximum 0.8V peak-peak (±0.4V) signal tolerated at the input, so if you scaled the source signal correctly, you won't ever get close to that maximum.

The proper scaling factor would be:

$$ K = \frac{0.15}{5} = 0.03 $$

This can be achieved with approximately 33:1 resistor divider, the lower of which could (if you wish) be a potentiometer for gain control:

^{simulate this circuit – Schematic created using CircuitLab}

That pair R1 and R2 attenuate according to the well-known divider formula:

$$ V_X = V_{IN} \frac{R_2}{R_1+R_2} = \frac{10\rm{k\Omega}}{330\rm{k\Omega} + 10\rm{k\Omega}} = 0.029 $$

The potential at the wiper of R2 will be some fraction between 0% and 100% of \$V_X\$, and this will be amplified (by a factor of 20) by the LM386.

If you are certain that \$V_{IN}\$ will never fall outside ±2.5V, then you could safely use a larger factor \$K\$, and still never exceed \$V_X=\pm 0.4V\$:

$$ K = \frac{0.8}{5} = \frac{1}{6.25} $$

This would require a resistor ratio of \$(6.25 - 1):1=5.25:1\$, which when using easily available E12 resistor values might look like this:

^{simulate this circuit}

This permits adjustable overall gain between zero and \$20\times \frac{10}{10+56}=3\$, in case you ever needed that much, without exceeding the ±0.4V input constraint for the LM386.

Just a small caveat - there's a 50kΩ resistance to ground from each IC input to ground, inside the IC, which is effectively in parallel with the the 10kΩ potentiometer. This will make my calculations above a little inaccurate, since it will necessarily alter the equivalent resistance of R2 from 10kΩ down to 8.3kΩ. If that's a problem, just reduce R1 until you obtain the correct R1:(R2∥50kΩ) ratio for the overall gain you desire.

This little gem of an IC was a bread-and-butter learning tool for me back when I was young. It employs a mix of concepts from DC/AC coupling, through biasing, to negative feedback. By studying the internal transistor schematic (from the datasheet), this little marvel taught me more than I could ever thank anyone for. I hope you can learn from it as much as I did.