Audio Amplifier Project

[jx28]

I get a circuit from internet and like to build it .

But , I need to understand some theory behind. Please see attached circuit.

C1 acts as coupling capacitor ; C11 & 12 act as bypass capacitors for power source.

Questions:-

(1) According to the author explaination , C2 is used to eliminate "interference" ,
what kind of interference ( noise ?? ) . Without C2 what will happen ?

(2) C9 & R1 are used for what purpose ? Without them what will happen ?

(3) C7 & C8 are used for what purpose ? Without them what will happen ?

(4) C3 , C4 , C5 and C6 are used for what purpose ? Without them what will happen ?

 

[OLIVE2222]

C2 filter out RF frequencies
C9 and R1 is a Boucherot cell placed to compensate the loudspeaker inductance
C7 and C8 provide bootstrap feedback
C3 and C4 provide an AC ground for the inverting s amplifiers inputs
C6 seems to be part of the IC protection
C5 ?

 

[duke37]

I have not looked up the specification of the IC you are using, so what I say is only a guess.

1. The circuit will not work! The power + and - are both connected to pin 13. Your supply will object violently.
2. C2 is to reduce interference, it is a very small capacitor to by-pass radio interference from radio stations or switching power supplies etc.
3. C9, R1 is a Zobel network (Wikipedia) to absorb high frequency power to stabilise the amplifier and reduce its power consumption. C10, R2 likewise.
4. C7, C8 are bootstrap capacitors to raise the driving voltage of the output stage to enable the output to go positive. Without these, the power output will be much less.
5. C3,4,5,6 are to by pass the signal at points in the circuit to give more gain or stability. Detailed investigation would be needed to find their effect, without them, perhaps an oscillator or perhaps low gain.
6. The lower amplifier seems to be doing nothing. Normally this configuration is used to double the output voltage so multiplying the output power by four.
7. I suggest looking up the IC makers data sheet.

 

[OLIVE2222]

Well spotted duke37, nice short circuit indeed !
A better schematic is available here:
http://www.quasarelectronics.com/kit-files/electronic-kit/3105.pdf

 

[jx28]

duke37 ,

Thanks for reply .

(1) C3,4,5,6 are bypassing what "signal" in order to make more gain ?? How it work in this way ??

(2) Don't mind to tell me little background how bootstrap capacitor work ?

By the way , thanks to olive2222 to provide correct diagram ...

 

[(*steve*)]

For those like myself who were wondering at the odd connections (or seeming lack thereof) for the lower amplifier. The answer is here.

 

[davenn]

For those like myself who were wondering at the odd connections (or seeming lack thereof) for the lower amplifier. The answer is here.

actually that didnt help. it would appear as tho the output from the top amp is permanently connected to the input of the bottom amp. which is ok if you are bridging them
so how can they ever act as separate amps in a stereo situation which it is also supposed to be able to do ?

Dave

PS .... ok so in a stereo amplifier, you actually need 2 of these IC's one for each channel ?
Why bridge 2 amps in the same chip rather than just having a single amp chip to produce the same output ? what's the advantage in the way this one is used ?

 

[duke37]

The idea of bridging amplifiers is to get twice the voltage across the speaker so four times the power. This is very useful when the supply voltage is low (12v car).

A voltage can only be defined between two points, the input is defined between the input wire and earth (ground). The amplifier however, needs the input to be above ground in a DC sense so a capacitor C3,4 is used to reference one side of the amplifier to earth.

C6 has something to do with surge protection, presumably the amplifier will shut down if it is being abused for a time, set by the charging of C6.
There is no information on C5.

There is some information on bootstrapping on Wikipedia. It would take half a page to explain it here and you would need to know how transistors operate. Suffice it to say that a npn transistor needs its base to be more positive than its emitter to turn on so, to get the output to approach the positive rail, some means must be provided to set the base above the power supply voltage. This is done by feeding the base from the power supply through a resistor but driving it also through a capacitor from the output. So, when the output voltage is high, the base is above the power supply voltage.
An alternative is to use a NPN/PNP pair in the output but NPN power transitors have been preferred.

It is always best to go to the chip makers data to get the correct information. Thanks Steve.