Marcel Birthelmer wrote:
> Yes, sorry, I flipped the op amp accidentally.

That changes everything of course.

> I'll assume it's right
> way 'round for the rest of this email.

>> Look at the circuit as two parts, the part controlling the opamp
>> positive input, and the part controlling the negative input.  Analyse
>> what happens if the input is driven by a symmetric sine wave around
>> ground and the negative input were held at ground.  For example, let's
>> say it's a perfect opamp with +5V and -5V supply and the sine peaks
>> also go from -5V to +5V.  Try to figure out what the opamp will do.
>> If you get stuck, look up "positive feedback" or "hystersis".
>
> Well with V+ tied to ground, it's a negating amplifier, with closed-loop
> gain = -1.

Yes.  Of course my comments about hysterisis don't apply after flipping the
opamp inputs.

>> Now forget about what the opamp is doing and look only at the signal
>> presented to the negative input.  It's a perfect opamp, so the negative
>> input has infinite impedence.  In other words, it's not there for
>> purposes of analysing what the voltage will be.  That leaves a
>> capacitor in series with the input and a resistor to ground.  What
>> will the result look like for a very low frequency sine wave input?  A
>> high frequency sine wave input?
>
> That's a high-pass filter,

Yes.

> So looking at this I can just combine those two to subtract v- from v+?

It's not that easy.  An ideal opamp has infinite gain.  I was trying to lead
you somewhere else that would have been valid for a hysterisis amp.  That
would not have operated linearly.  This one however does.

In this case pretend the two inputs weren't tied together.  You can analyse
what happens as each input is wiggled separately while the other is held
fixed, like at ground.  You should then be able to figure out how signals on
both inputs combine to make the output signal.  To make this a little
easier, think of what happens with very low frequencies and very high
frequencies.  At low frequencies, the + input will be at ground.  At high
frequencies, the + input will be the AC component of the input signal.


*****************************************************************
Embed Inc, embedded system specialists in Littleton Massachusetts
(978) 742-9014, http://www.embedinc.com
-- 
http://www.piclist.com PIC/SX FAQ & list archive
View/change your membership options at
http://mailman.mit.edu/mailman/listinfo/piclist