# Understanding LM386 amplifier component selection

I am a computer engineering student and I still struggle to understand electricity. I am doing an audio amplifier project and would like to understand the math behind why these capacitor/resistor values were chosen in the LM386 operational amplifier. In concept, I understand it mostly, but I am unable to mathematically justify why these values are the way they are.

This may not be the best example of this op-amp's implementation, but its the most consistent I have seen in my research.
Also, that 10k resistor is a potentiometer.
Thank you.

• Better not call the LM386 ban OpAmp, it isn't. (If it were, this circuit would have infinite amplification, because there is no negative feedback.) – Wouter van Ooijen Apr 17 '20 at 6:37
• Yeah, so the question is wether you want to understand op-amp circuits in general or how this audio amplifier specifically works and is set up with the given circuit. – Jakob Halskov Apr 17 '20 at 6:48
• THe output 250 uF is specified to have a low impedance at the lowest frequency of interest compared to the speaker. This ensures that most of the load voltage drop across the speaker and that the frequency response is not overly affected by capacitive reactance. Zc = 1/(2.Pi.f.C) at say 100 Hz Zc = 1/(2 x 3.14 x 100 x 250E-6) ~~= 6 Ohms. That's high relative to an 8 hm speaker. At 300 Hz Zc = about 2 Ohms. Don't expect HiFi with that value. || See datasheet here/ See section 9.2.1.2 for design and the schematic on page 1 for what's inside. ... – Russell McMahon Apr 17 '20 at 7:36

## 1 Answer

The LM386 is similar to an opamp with an internal negative feedback path controlling gain.
Connecting a resistor across pins 1 & 8 (or a short circuit) changes part of the negative feedback path and alters the gain.

The output 250 uF is specified to have a low impedance at the lowest frequency of interest compared to the speaker. This ensures that most of the load voltage drop across the speaker and that the frequency response is not overly affected by capacitive reactance. Zc = 1/(2.Pi.f.C) at say 100 Hz Zc = 1/(2 x 3.14 x 100 x 250E-6) ~~= 6 Ohms. That's high relative to an 8 ohm speaker.
At 300 Hz Zc = about 2 Ohms.
Don't expect HiFi with that value.

The bypass capacitor needs to be low impedance at frequencies of interest compared to the 15K internal resistor (R1 in diagram below) that it is bypassing.

See datasheet here
See section 9.2.1.2 for design and the schematic on page 1 for what's inside.

Gain = 2 x R4 / (R3+R2)
As supplied = 2 x 15k/(150+1.35k) = 20
Short pins 1 to pin 8 = 2 x 15k/150 = 200
Connect a resistor from pin 1 to pin 8 parallels it across R3 and makes 150 < R2 + R3 effective < 1k5
making gain somewhere between 2 and 200