[sdiy] OT: Wisdom

[Neil Johnson]

Alan,

Needham, Alan wrote:

> | On 23 November 2010 11:27, Veronica Merryfield
> | <veronica.merryfield at shaw.ca> wrote:
> | > I wonder if the collected masses, who seem to have a decent  
> handle on
> | this sort of stuff, could take a quick look at the below ASCII art
> | schematic.
> | >
> | > Assuming all the resistors are the same, what would the output  
> be of the
> | two op-amps?
> | >
> | > Thanks
> | >
> | > 3.3v-----
> | >        |
> | >        \
> | >        /
> | >        \
> | >        /    ---/\/\/\---
> | >        |   |            |
> | >        |-----|+         |
> | >        |     |opamp  -----
> | >        \  0v-|-
> | >        /
> | >        \
> | >        /
> | >        |
> | >        |
> | >        |
> | >        \
> | >        /
> | >        \
> | >        /
> | >        |   ---/\/\/\---
> | >        |  |            |
> | >         ----|+         |
> | >             |opamp ------ ????
> | >          0v-|-
> | > _______________________________________________
>
> All the resistors are the same,
> assuming the op-amps are both reversed (as per Taku's suggestion),
> Will the bottom op-amp not drive to match the + and - inputs?
> 3.3v through 3 resistors = 1.1v each so the bottom out will be -1.1v
> The upper op-amp will be +2.2v
> This gives the same current through each of the 4 'current path'  
> resistors
>
> Or am I missing something?
> 	Alan
>

(assuming the "Taku correction")

The top op-amp will servo its output so that the -ve input is at 0V.   
This forces the bottom of the top resistor to 0V.  As the feedback  
resistor is the same value this gives an output of -1 x Vinput = -3V3.

Now the bottom op-amp has an input resistor of 2R.  But the voltage  
at the top of the middle resistor is now 0V, so there is no current  
flowing into the summing node, so the output is 0V.

Neil
--
http://www.njohnson.co.uk