Spectrum analyser, understand this project

Hello,

I'm trying to understand this project: https://www.waitingforfriday.com/index.php/Real-Time_Audio_Spectrum_Analyser
It's about a spectrum analyser.

My question is about the operational amplifier of the schematic: **broken link removed**

What is the function of this amplifier ? It amplify the signal ok, but the goal is to work with saturation or not ? What are the roles of the capacitors C1 and C2 ?


Thanks!

LM386 is not operational amplifier it is low voltage audio power amplifier.

Oh sorry, effectively it's not.

I have an other question, the Resistor R4 and capacitor C2 wich are in serie after the lm386 are for adapte impedance, isn't it ? How did he choose these valors for adapting the PIC ?

Thank

The values of the so-called boucherot cell C2/R4 are specified in the LM386 and haven't to do with connected PIC. https://en.wikipedia.org/wiki/Boucherot_cell

Seriously speaking, the ADC input is effectively open circuit for the amplifier. There's no understanddable reason why they used an audio power amplifier instead of a simple OP.

FvM said:

The values of the so-called boucherot cell C2/R4 are specified in the LM386 and haven't to do with connected PIC. https://en.wikipedia.org/wiki/Boucherot_cell

Seriously speaking, the ADC input is effectively open circuit for the amplifier. There's no understanddable reason why they used an audio power amplifier instead of a simple OP.

Click to expand...

Stated at the website:

'In order to correctly sample the signal we have to do two things. Firstly we need to amplify the signal to ensure we can use as much of the 0-5V range as possible. Secondly we have to move the signal's ground (of 0 volts) to a 'virtual ground' of 2.5Vs. This will allow the PIC to sample both the positive and the negative sides of the input signal. To do this the demonstration board uses a simple amplifier IC (the LM386-1). Since the IC is powered from a 0V and 5V power supply it has the handy side-effect of also moving the signal into the middle of our required power range. The LM386-1 was used because it is cheap and simple, however you could use a rail-to-rail opamp to achieve the same thing with a few more external components.'

An opamp that is biased at +2.5V with 2 resistors can be used with an input coupling capacitor to replace the LM386 power amplifier. The LM386 has internal biasing and has internal feedback resistors for a voltage gain of 20. Two feedback resistors added to an opamp can make its voltage gain 20.

The output of an LM386 with a 5V supply and feeding a high resistance has an output swing of from about +0.6V to about +4.4V which is not rail-to-rail. Many opamps have an output swing that is the same.

Interesting, I didn't known these cell. Another question, when in schematics they write for a resistor value 10R, What is the R ? Means 10 Ohm ?

Thanks

10R is 10 ohms. I don't know why some people say 10R BESIDE AN OBVIOUS SYMBOL OF A RESISTOR.