At 11:37 PM 3/22/2005 +0100, you wrote:
>Hi group.
>Hi Group,
>
>May I ask some questions about EE?  I'm really trying hard to learn 
>circuits and all, but it is hard.  I'm a programmer by trade and maybe I'm 
>trying to think too much like a programmer when learning this stuff?
>
>Is an AC current just a DC current that alternates?  If so, would a DC 
>current that makes a sine wave from +0 to +5v be classified as AC?

Yes.

>My ARRL Handbooks says that AC reverses polarity.  That would mean to me 
>that a -5v to +5v sine wave would be AC, but a +0 to +5 wouldn't.

A sine wave from 0 to +5 volts is just an AC signal from -2.5 volts to +2.5 
volts superimposed on a DC voltage of 2.5 volts. If you fees such a signal 
through a capacitor, only the AC gets passed through, giving you the -2.5 
volt to +2.5 volts sine wave.


>The specific reason I ask is because I got hung up trying to figure out 
>how to understand the circuit to connect a PIC to a pair of 400ST/SR sonar 
>ceramic transducers.  Then I started reading all my electronic books all 
>over again.  I just can't seem to "get" it. )-:
>
>Is basically anything that has to do with audio AC?  Can I have a 
>basically DC circuit with AC portions in it?

Many signals you deal with, audio or otherwise, are really an AC signal 
superimposed on a DC voltage. In a typical audio amplifier, the input 
signal is pure AC - no DC component - from a microphone, for example. Then 
in the amplifier stages, the various transistors that make up the amplifier 
have a DC voltage on the base and collector. The AC signal is superimposed 
on a DC voltage in these stages, and then the DC component is removed via a 
capacitor at the output to feed a speaker or headphones, for example. This 
is a very simplified version of life.

Now in the PIC world, if you want to convert an AC signal to digital format 
via the PIC's Analog to digital converter, you NEED to have the signal 
superimposed on a DC voltage, since the A/D converter in the PIC works with 
voltages between 0 and 5 volts (assuming you have a 5 volt supply). If your 
original signal is an AC signal ranging from -2 to +2 volts, you have to 
add a DC voltage of 2 volts to convert the signal range from 0 to 4 volts 
for the PIC.




>I try to think of water flowing from + through to - when trying to 
>understand a circuit.  Is this correct?  If not, how do you figure out 
>what a circuit does when you look at it?
>
>If I think this way with AC, then I get:
>
> >++++<---->++++<---->++++<----  One water Pipe
><---->++++<---->++++>---->++++  The other pipe
>
>The water goes at 60Hz first one direction then reverses and goes in the 
>other direction in on pipe, while in the other pipe it does the same but 
>in opposite directions to the first pipe.  When I imagine this the water 
>doesn't do anything.  They seem to cancel each other out.  This can't be 
>correct.
>
>With DC things seem to make sense in a way.
>
>Thanks to anyone who can help me get over this learning hump. I've been at 
>this for years and years, and I don't intend to give up until I finally 
>get it. (-:
>
>Thanks again,
>Lindy
>
>
>--
>http://www.piclist.com PIC/SX FAQ & list archive
>View/change your membership options at
>http://mailman.mit.edu/mailman/listinfo/piclist

Larry Bradley
Orleans (Ottawa), Ontario, CANADA 
-- 
http://www.piclist.com PIC/SX FAQ & list archive
View/change your membership options at
http://mailman.mit.edu/mailman/listinfo/piclist