10-02-2014, 02:13 PM
I was asked this the other day, is a waveform such as a pulsed waveform ie: a digital stream with a base line of 0 volts an AC waveform.
Thought's...
Lawrence.
Thought's...
Lawrence.
|
AC...DC Waveform
|
|
10-02-2014, 03:33 PM
I would say no as it would be a stream of DC pulses.
the again I suppose a capacitor would treat it like AC now I am confused. Rob T
if its made by hand it can be repaired by hand
10-02-2014, 03:57 PM
Yes, can be confusing Rob, I was always thought that AC is when the current flow changes direction through a zero point.
Lawrence.
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.
10-02-2014, 04:28 PM
"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."
Surely the current is still flowing one way so maybe that's the right way to define it, the current goes more positive from zero then less positive from its peak back down to zero, the direction of flow does not change only the amplitude. Lawrence.
10-02-2014, 05:34 PM
If you conclude your reasoning at this point, then yes, I understand why you think that.
But take it a stage further: If you sent that signal through a capacitor into an oscilloscope, you'd see a square wave. Agreed? But capacitors only pass AC, right? The capacitor has passed the AC component of the signal, removing the DC component in the process. ... Another example to think about: looking at ripple on a DC power supply. I bet you've done this a fair few times... Once again, you could argue that it's a purely DC signal because there is no reversal of current*. But no, you have an AC signal - perhaps 100mV - sitting on top of a DC offset of perhaps 12V. How do you assess ripple on a power supply rail? You use a 'scope, set to AC at the input. As above, you are using a capacitor to remove the DC component so that you have the AC component in isolation. As I said initially, this takes some thinking about. But it's worth it  Mark * Actually, that's only true if the load is purely resistive.
10-02-2014, 06:06 PM
I understand what you say Mark but in the context of my original post is it an AC waveform (per as AC is defined in the text books..alternating current) if there is no zero crossing point, ie: it alternates between high and low above a zero rail.
I understand about AC with respect capacitors, inductors etc. The question arose whilst on another forum concerning data exchange in automotive electric modules. Lawrence.
10-02-2014, 07:48 PM
(This post was last modified: 10-02-2014, 07:48 PM by John M0GLN.)
Its something I've puzzled about for years, if you have a bar magnet and revolve it in front of a coil as a simple dynamo, as the N pole approaches the coil a voltage will be induced, say a +V, then as it revolves further this V would reduce to zero and as the S pole approaches the V would go to -V, this would be an AC voltage, but what if instead of revolving 360° the magnet reverses direction after 90° back to its original starting point? The induced voltage will have risen from 0V to +V and back to 0V, if this was repeated wouldn't this just be a stream of DC pulses as Lawrence asks about in his post?
John |
| Users browsing this thread: |
| 2 Guest(s) |