10-02-2014, 04:10 PM
Any waveform has an AC component, and a DC component.
Learning to think separately about these is the biggest favour you can do yourself when learning electronics. I know of many people who have been in electronics for decades, but still get confused by it, so I appreciate that it can be confusing.
Example 1: You could argue that a square wave that goes between 0 and 5 volts is entirely DC because it doesn't reverse direction to go below 0V, but that's the wrong way to think of it. Rather, it is a 5 volt peak-to-peak AC signal sitting on a 2.5V DC offset.
Example 2: When you build an audio amplifier, you establish the quiescent conditions, and then superimpose the AC signal on top. The collector of a transistor might sit at 4.5V, and be able to move anywhere between 0V and 9V - so this again looks like a DC signal because there is no zero crossing. But no, the audio signal will be AC, sitting on the DC offset of 4.5V. Of course, an output capacitor will remove this offset - it does this by charging up when you power up the circuit. It sits there with a constant 4.5V DC across it, allowing the signal to move between +4.5V and -4.5V.
When designing circuits, you always have to think of the DC conditions first. That's what "biasing" means. The AC signal simply causes the circuit to momentarily deviate from this point.
Simple, but does need to be thought about.
Learning to think separately about these is the biggest favour you can do yourself when learning electronics. I know of many people who have been in electronics for decades, but still get confused by it, so I appreciate that it can be confusing.
Example 1: You could argue that a square wave that goes between 0 and 5 volts is entirely DC because it doesn't reverse direction to go below 0V, but that's the wrong way to think of it. Rather, it is a 5 volt peak-to-peak AC signal sitting on a 2.5V DC offset.
Example 2: When you build an audio amplifier, you establish the quiescent conditions, and then superimpose the AC signal on top. The collector of a transistor might sit at 4.5V, and be able to move anywhere between 0V and 9V - so this again looks like a DC signal because there is no zero crossing. But no, the audio signal will be AC, sitting on the DC offset of 4.5V. Of course, an output capacitor will remove this offset - it does this by charging up when you power up the circuit. It sits there with a constant 4.5V DC across it, allowing the signal to move between +4.5V and -4.5V.
When designing circuits, you always have to think of the DC conditions first. That's what "biasing" means. The AC signal simply causes the circuit to momentarily deviate from this point.
Simple, but does need to be thought about.
