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Here is a schematic I'm planning to make (right click on the image and 'open image in new tab' to see it better'): Schematic for amplifier

I have a transformer and AC to DC converter that makes +/-12V.

(1) Does 'power supply 1' correspond to the +12V wire and 'power supply 2' to the -12V wire?

(2) When I see the symbol 'ground' on the schematic does that mean that I need to connect those to the -12V line? I have no 0 volt line in the circuit so what is the ground line?

(3) Why are C3, C5 non-polarized capacitors and why are the other ones polarized? The current flows from negative to positive. Does that mean that the polarized capacitors need to be placed with their positive legs pointing to the left?

I already asked the grounding question but I got only answers that were theory. I would like specifically to know how do I deal with grounding in this circuit and why there is no grounding symbol in an arbitrary "powering an LED with a 9V battery" circuit.

Thank you very much. This will clarify a lot of things for me.

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(1) Does 'input 2' correspond to the +12V wire and 'input 1' to the -12V wire?

NO. This is connected to the power supply connector. (Pin 1 is the positive, pin 2 the ground or 0V). The INPUT connector is where the audio signal goes (2) is the live and (1) is the ground.

(2) When I see the symbol 'ground' on the schematic does that mean that I need to connect those to the -12V line? I have no 0 volt line in the circuit so what is the ground line?

All voltages are relative to each other. If you only have TWO wires from the AC/DC converter then the most positive is taken as the positive and the other one is the OV or ground. A simple check with a voltmeter will determine which is which.

(3) Why are C3, C5 non-polarized capacitors and why are the other ones polarized? The current flows from negative to positive. Does that mean that the polarized capacitors need to be placed with their positive legs pointing to the left?

C3 and C5 are small value capacitors (0.1uF). These can be easily made as physically small, non-electrolytic types. They have the advantage of being able to decouple (short out) the higher frequencies. The larger values (uFs) are made as electrolytics as these can be made with high values in small physical packages. They are generally much poorer at handling the high frequencies. By combining an electrolytic with a non-electrolytic capacitor in parallel (eg C3, C6) you get a much better response over a wider range of frequencies. In this case they are used for 'smoothing' the supply voltage, preventing hum and hiss. The positive plate of an electrolytic is shown as an open rectangle but left and right (or up, down) have no meaning in terms of connection as this will be determined by board layout. Conventionally current is taken to flow positive to negative.

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  • \$\begingroup\$ I mixed up 'input' with 'power supply' but I meant power supply 1 to +12V and power supply 2 to -12V. I actually have three wires coming out of the transformer, and I noticed that using the the first and the third gives more power so I hooked up accordingly with the rectifier and capacitor. I omitted the middle one from the circuit. What was the middle one for? Also, how do I decide how to place the polarized capacitors then without relying on the schematic? Could you give me an explanation on that? Thank you. \$\endgroup\$ – Tamás Jun 22 '14 at 18:16
  • \$\begingroup\$ I think I'm getting now the pattern. One more thing though: Why does the volume control needs to be connected to the ground? \$\endgroup\$ – Tamás Jun 22 '14 at 18:36
  • \$\begingroup\$ @Tamás This completes the input signal circuit (loop). The input voltage (a few mV) is dropped across the variable resistor. The wiper can then 'tap off' a fraction of this voltage (between 0 (no signal) and 1 (full signal) to the amplifier input through a decoupling capacitor (C1) which allows only the AC signal to get through. \$\endgroup\$ – JIm Dearden Jun 22 '14 at 19:45
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The TDA2003 datasheet indicates that this part can operate from a single supply of 8 - 18 volts. It does not use a negative supply.

From your desctiption, your power supply produces +/-12 volts, or 24 volts between your "first" and "third" wires, which is more than the part is designed for, so you should use just the first and second wire - you will need a voltmeter, or the power supply documentation, to determine which wire is positive.

All the ground symbols on the schematic must be connected together, and to the middle terminal of the power supply. All the ground symbols are your "zero volt" line.

All electrolytic capacitors must be installed with their positive terminal connected to the more positive part of the circuit. C1 and C4 should be installed with their positive terminal towards the TDA2003 (I think C1 is drawn backwards).

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(1) Does input 2 correspond to the +12V wire and input 1 to the -12V wire?

No. On the POWER SUPPLY connector, pin 1 is the +12V input from the supply and pin 2 is the -12V input from the supply.

(2) When I see the symbol 'ground' on the schematic does that mean that I need to connect those to the -12V line aka. input 1?

I don't know what you mean by "aka. input 1"; the power supply -12V is supposed to be connected to pin 2 of the POWER SUPPLY connector.

I have no 0 volt line in the circuit so what is the ground line?

Everything connected to pin 2 of the POWER SUPPLY connector.

(3) Why are C3, C5 non-polarized capacitors and why is C5 not?

C5 is ground-referred, and the signal exiting C4 is real plus-and-minus-about-zero-volts AC, so the signal at the top of C5 will swing both positive and negative with respect to ground.

The current flows from negative to positive.

It does in the world of physics, but in this one it's called "Conventional current flow" and it's considered to flow from positive to negative, thanks to Benjamin Franklin.

Does that mean that the polarized capacitors need to be placed with there positive legs >pointing to the left?

In the drawing, the lighter-shaded electrolytic capacitor plates should be marked with a "+" sign.

I already asked the grounding question but I got only answers that were theory. I would like specifically to know how do I deal with grounding in this circuit and why there is not grounding symbol in an arbitrary "powering an LED with a 9V battery" circuit.

In this circuit, you deal with grounding by connecting everything with a ground symbol to the negative side of the supply, and that's usually considered to be the "zero-volt" reference for whatever in the outside world is plugged into the circuit.

In a stand-alone LED circuit powered with a 9V battery, the ground reference/symbol isn't important because nothing else will be connected to the circuit.

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