Olin Lathrop wrote:
> Marcel Birthelmer wrote:
> 
>>apologies for the n00b-ness of my question but being a first-year EE
>>student I haven't learned enough useful things yet to figure out the
>>workings of the following circuit:
>>
>>http://marcel.carrietech.com/phase.png
>>
>>This is a phase shifting circuit (Rv being variable) that I found
>>online.
> 
> 
> Not really.

Yes, sorry, I flipped the op amp accidentally. I'll assume it's right
way 'round for the rest of this email.

> 
> 
>>Could someone please explain how to calculate the output
>>voltage given the input voltage and the various components? I've
>>checked all my EE books (which are all intro level, unfortunately)
>>and have come up dry.
> 
> 
> There is a lot that can be learned from this circuit.  It will be well worth
> spending some time analysing and understanding it.  However, if I just tell
> you the answer you won't learn as much, and I don't want to do your homework
> for you anyway.
> 
> 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.

> 
> 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, so the v+ = (Rv / (Rv + Xc)) v, with Xc =
1/(jwC).

> 
> That's enough for now.  Answer the questions above to show you understand it
> to this point and we can continue.
> 

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

Thanks,
- Marcel
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