Stupid noob theory question about AC signals in DC circuits

Frequently asked questions regarding powering your pedal.
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EdRoperL1
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Post by EdRoperL1 »

Hope this isn't in the wrong forum, if so I apologize.

This is probably one of those 'dumb' questions that will make you question why I'm even trying to hold a soldering iron, but I'm a little unclear about how DC circuits operate with an AC guitar signal.

Since AC is alternating in direction, and DC only goes in one direction, how the heck is it that pedals even work? Wouldn't only the positive part of the signal be amplified/processed, since its alternating directions? How does this keep from chopping off half of the waveform when you turn the ac signal into a dc input?

Alternativey, when current switches the other direction, wouldnt it also flow 'backwards' through the signal path?

Kinda confused...

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deltafred
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Post by deltafred »

I can see how you are confused.

The way it works is that there is a DC bias of usually around half the battery voltage (this can very quite a lot) within the pedal.
The AC signal is superimposed on this (through a capacitor*) so it is actually varying DC.
Somewhere near the end of the circuit is a capacitor* that blocks the DC and allows the AC through to the output connector. (Very simplistic explanation.)

So it is AC in, varying DC in the circuit itself and AC out.

*Capacitors block DC but allow AC to pass.

Hope this helps.
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Post by DrNomis »

If you're powering say a simple one-transistor amplifier stage with, say +9V DC, what's basically happening is this (assuming that the transistor is biased to half the supply voltage):


When the circuit is powered-up, two resistors that connect to the base terminal of the transistor Bias it so that it operates roughly in the middle of it's transfer-curve between the two states where the transistor is either on or fully-conducting (saturated), or fully-off (cut off), you will normally see about half the supply voltage on the transistor's collector when it's Biased as such, now the input signal "adds-to" and "subtracts-from" the DC voltage on the transistor's base terminal, when the signal is adding-to the DC voltage on the transistor's base, this turns the transistor more "on" which means more current flow from the transistor's emitter to the collector (I'm using electron-flow convention here), when the input signal is subtracting-from the DC voltage on the transistor's base it turns the transistor more "off" which means less current flow from the emitter to the collector, this happens for every cycle of the input signal, if the input signal is disconnected from the input of the amplifier, the amplifier returns to what's called the "Quiescent-State" with the transistor's collector sitting at half the supply voltage..... :thumbsup


As deltafred said, a capacitor is usually connected to the transistor's collector to block DC and allow the AC-signal to pass on to another stage, if the transistor's collector was connected directly to the base of a transistor in the next stage the biasing would be upset, having said this, there are some transistor amplifier circuits where transistors are deliberately conected directly to another transistor (DC-Coupling).... :thumbsup
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deltafred
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Post by deltafred »

DrNomis wrote:
When the circuit is powered-up, two resistors that connect to the base terminal of the transistor Bias it so that it operates roughly in the middle of it's transfer-curve between the two states where the transistor is either on or fully-conducting (saturated), or fully-off (cut off), you will normally see about half the supply voltage on the transistor's collector when it's Biased as such, now the input signal "adds-to" and "subtracts-from" the DC voltage on the transistor's base terminal, when the signal is adding-to the DC voltage on the transistor's base, this turns the transistor more "on" which means more current flow from the transistor's emitter to the collector (I'm using electron-flow convention here), when the input signal is subtracting from the DC voltage on the transistor's base it turns the transistor more "off" which means less current flow from the emitter to the collector, this happens for every cycle of the input signal, if the input signal is disconnected from the input of the amplifier, the amplifier returns to what's called the "Quiescent-State" with the transistor's collector sitting at half the supply voltage..... :thumbsup
Far too complicated.

You understand it, I do, and lots of other FSBers do (if they can be bothered to wade through it all, and it takes some wading through) but to someone who is struggling with the basic idea it is too far too fast.

To understand that you first have to have a pretty good grounding in electronics.
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EdRoperL1
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Post by EdRoperL1 »

deltafred wrote:I can see how you are confused.

The way it works is that there is a DC bias of usually around half the battery voltage (this can very quite a lot) within the pedal.
The AC signal is superimposed on this (through a capacitor*) so it is actually varying DC.
Somewhere near the end of the circuit is a capacitor* that blocks the DC and allows the AC through to the output connector. (Very simplistic explanation.)

So it is AC in, varying DC in the circuit itself and AC out.

*Capacitors block DC but allow AC to pass.

Hope this helps.
Kinda sorta, but then again, if the signal coming down two wires from your guitar is alternating directions , how is it not 'losing' part of that signal?

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lolbou
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Post by lolbou »

EdRoperL1 wrote:but then again, if the signal coming down two wires from your guitar is alternating directions , how is it not 'losing' part of that signal?
Assume your guitar is outputing voltages between -5 and +5 (mV, volts, whatever). Your circuit is DC powered, so it can only work between 0 and 10. If you want your circuit to "see" the whole variation from your guitar signal, all you have to do is add a constant +5 to your signal.

This way, when your guitar plays "-5", your circuit sees "-5 +5 = 0 ".
When it plays "+5", your circuit sees "+5 +5 = 10".

Now it can work out the FULL swing of your signal; because it sees all of its variations.

Adding this +5 is called DC biasing.

The output cap removes this +5 (DC blocking) and so your pedal oputput can be seen as a guitar signal by the following unit.
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dai h.
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Post by dai h. »

maybe a building as an analogy? Instead of starting from the ground floor, start in the middle so you can go up or down (relatively speaking, so, "minus two floors, relative to the fifth floor(as starting point)", etc.). (Apologize if I'm off here since I'm still mostly a novice as far as theory.) Also check out basic tube theory.

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EdRoperL1
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Post by EdRoperL1 »

I think I now better understand the concept of biasing, and how it represents the fluctuating waveform. Still unclear on how the signal is captured since its alternating from tip to sleeve along the wire itself... I would think that since circuits just use the tip as input, when it alternate to the other wire, you dont capture/amplify that. Maybe I just need some more reading/examples or something

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