[sdiy] Using Opamps as INV or NON-INV amplifiers

[Ian Fritz]

At 03:25 PM 6/21/2006, Aaron Lanterman wrote:

>On Wed, 21 Jun 2006, Ian Fritz wrote:
>
>>At one point I worked out the equations for the general  case of arbitrary
>
>Dave - check out my op-amp lectures:
>
>http://users.ece.gatech.edu/~lanterma/ece4803/
>
>on 1/23 and 1/30 - I thinkat some point I work out the general case Ian 
>describes.


Hmmm ... well, close.  That's a very nice presentation and shows the 
general methods to use.  But the more general circuit I was refering to has 
an arbitrary number of sources feeding each input.

At first, this sounds like a fairly complicated calculation.  But there is 
a trick to simplify it.  The trick is to use the method of 
superposition.  In this method, you start by grounding all the 
inputs.  Then you lift each one individually, connect it to its input 
voltage and calculate the output voltage.  Finally, you add together all 
the individual results.  (This works because the system is linear.)

Of course, there are only two distinct cases to consider, a (-) input and a 
(+) input, so there are really just two calculations to do before adding 
everything up.  The result turns out to be amazingly simple.  In addition 
to the input resistors, only three additional quantities appear in the 
result.  These are the effective parallel resistances seen looking out from 
each of the opamp terminals, and the value of the feedback resistor.

If the circuit is kept balanced so that the effective input resistances are 
equal, then the output voltage is just the feedback resistance times the 
(signed) sum off all the input voltages divided by their corresponding 
input resistors.  What's odd to me, is that these "input voltages divided 
by their corresponding input resistors" are not actual physical currents, 
since the opamp terminals are not at zero potential!  But there you have it.

If the circuit is unbalanced, then the result is modified multiplying the 
sum of the (+) terms by the ratio of the two equivalent input resistances.

It's well worth working through this calculation, both because of the 
amazingly simple result and to understand the utility and power of the 
superposition method.

   Ian