Onto the third page where there is a lot of explanation and expanding on the information.
So it first talks about pins 1 and 8 where when nothing is connected to them, the gain is at 20 (26dB). And if a capacitor is placed between those two pins, gain will go up to 200 (46dB) as mentioned before. Depending on what kind of resistor you place in series with the capacitor, the gain could be set between 20-200.
If you wanted to boost bass response, you’d need a series RC between pins 1 and 5. It seems that for a good and stable 6dB bass boost and assuming that Pins 1 and 8 are not connected via cap, R has to be from 10-15kohms. If Pins 1 and 8 are connected, then R can be as low as 2kohms.
And then onto Input Biasing:
So based on the schematic, both inputs (2 and 3) are biased to ground with a 50kohm resistor. As you know, the base/bias current of the input is about 250nA. If we apply Ohm’s Law to this:
Then you see that the inputs are at about 12.5mV when not connected to anything.
The datasheet then refers to “dc source resistance” and this is referring to the load resistance that is tied to the input pins - ie the circuit that is providing the signal to be amplified. And the next portion I’m covering baffled me a bit so @mgburr helped out by first going over basic resistor conformations.
Resistors in series add together. i.e. pin1 of resistor 1 to pin2 of resistor 2.
In parallel, pin1 of r1 and r2 are tied together, and p2 of r1 and r2 are tied together. In series, the resistance adds together. In parallel they will never be as much as the smallest resistor. They will become close enough, but not entirely.
Ex. Given the equation: RTotal= 1/((1/r1)+(1/r2)…)
- If R1 is 2k and R2 is 2k then total resistance would be 1k
- If R1 is 2k and R2 is 10k, then total resistance would be 1.66667k
- You can keep increasing and it will never be equal to the r1 of 2k. It might be 1.99999999999k, but never equal to.
So the input source of the signal is placed essentially in parallel when connected to pins 2 and 3. Because of this, you can think of the source’s resistance as being parallel to the chip’s resistance. Which means that even if the source’s resistance is above 250kohms, the overall resistance will never be higher than the chip’s resistance - and therefore would essentially equal that which the chip had been working with before! So there will be very little offset of the voltage in general: 2.5mV at the input, 50mV at the output.
Put another way per @mgburr:
- What they are saying in the data sheet, is that an input resistance higher than 250k makes the internal resistance close enough that it will not cause the overall biasing to be affected and will thus not cause output to be greatly affected. You can keep increasing and it will never be equal to the r1 of 2k. It might be 1.99999999999kohms (or whatever), but never equal to.
Now if you have the opposite issue where the DC source resistance is <10kohms…it acts as a short and goes to ground instead of the circuit.
So my questions were:
- Q: What “unused input” is the datasheet referring to?
So essentially, this has to do with the op-amp portion of the LM386. Essentially when you amplify a signal, whether by using an inverting or non-inverting amplifier, you only really need to have the input signal going into one pin. The other pin is often put to ground.
Assuming the input load resistance is <10kohm, and the signal is going into the + input, you will want to ground the - input. If the signal is going into the - input, you will want to ground the + input.
- Q: Why would the current go to ground if the source resistance were <10kohms?
According to @mgburr - It’s discussing the offset errors if using an input that is less than 10kohms. Each input pin is tied to ground inside the chip with a 50kohm resistor. If you use <10k on one input, it’s low enough that the other side (not connected) can’t compensate and it will start pushing offset farther and farther to where it will eventually saturate the transistors either fully open or closed. This will take the output either max high or low because the 10k is considerably less than the 50k internal.
Now between 10kohms and 250kohms…
You would want to match the input resistance as this will help prevent offset errors that will affect the output. The datasheet suggests putting a resistor from the unused input to ground equal in value to the dc source resistance. Or eliminate any offset problems by capacitively coupling the input signal to the input in.
- Wouldn’t this block the DC signal that was trying to be amplified? Sure it’d prevent offset problems, but you wouldn’t be able to have an amplified signal??
And in the last paragraph here - If you have a cap between pins 1 and 8, to prevent degradation of gain and possible instabilities, put a 0.1uf capacitor or a short to ground depending on the dc source resistance.