Op Amp Circuit design...help !

Hi,
I am trying to come up with a design of an operational Amplifier circuit design which will provide an output voltage C = 2A + 3B, where
A and B are time varying input voltages....
I would greatly appreciate any help on this !!
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
snipped-for-privacy@gmail.com snipped-for-privacy@gmail.com wrote previously in alt.engineering.electrical:

It's called a summing amplifier
Read here: http://www.ecircuitcenter.com/Circuits/opsum/opsum.htm http://en.wikipedia.org/wiki/Operational_amplifier_applications
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
Yes,.. It is a summing amplifier. But I need the C to be positive.

Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
snipped-for-privacy@gmail.com wrote:

Solution without inspiration: add an inverting amplifier after the summer.
Solution with inspiration: combine the concepts of the summing amplifier and non-inverting amplifier. Textbooks usually show summing only into the inverting input, but you can sum into the non-inverting input instead (or even sum into both inputs at once). You can achieve C = 2A + 3B with one opamp.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
This is the the design I came up w/ according to the textbooks -- non- inverting summer.
http://www.prism.gatech.edu/~gth881f/OpAmp.JPG
Where Vi1 = A , Vi2 = B and Vo = C
However, I do not know what possible values should be the resistances...

Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
snipped-for-privacy@gmail.com wrote:

Good job! I'm glad you chose inspiration.
Time to use what we know about an ideal opamp:
1. The differential input voltage is zero. 2. Neither input draws or supplies any current.
Write an equation for V0 when V2 = 0 (shorted). Write another equation for V0 when V1 = 0. Superposition is valid for this linear circuit, so the total output V0 is the sum of the two output voltages you just found. The rest is algebra.
For an ideal amplifier, the input offset current is zero so resistor can be eliminated. In that case, the resistances could be 1, 2, 3, and 4 to achieve C = 2A + B.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
Bryce wrote:

AAAargh! That SHOULD say "resistor R3 can be eliminated".
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
I am another student trying to obtain this Op Amp Circuit and saw your post as I was posting on google. I am not getting the correct answer when doing the algebra from your steps. Is there any way you can post this or also further explain how those resistances achieve C* + B? I'd greatly appreciate it. Thanks.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
Jillian Lewis wrote:

Original poster needed C = 2A + 3B. That's what I designed. Alas, my usenet message proofreader is on vacation. Thanks for straightening me out.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
Bryce wrote:

PMJI. The difference between summing at the + and - inputs is that feedback keeps the - input reasonably still, which prevents the inputs from talking to each other. (If you move the + input around, the + signals will also talk to the - signals). Some sources don't mind that, but others do--you can get all sorts of nasty intermodulation and so on.
If the circuit needs to be used with different sources, I'd spend another dime and use a dual op amp. Sum the + signals in the first amp, and send its output into the (-) input of the second amp, along with the - signals.
Cheers,
Phil Hobbs
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
snipped-for-privacy@electrooptical.net says...

3. The differential gain is infinite.

Also need a reference or bipolar supplies (with ground being the reference).
--
Keith

Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

Hmm. As I understand it, one advantage of the inverting summer configuration is that the summing node (the inverting input) is actively held at ground voltage, so each of the individual inputs sees a resistive load to ground (just the input resistor) which does not change with input voltage on that input, and it also unaffected by the voltage on any of the other inputs.
It seems that with the circuit layout you're suggesting, the summing node voltage will vary in proportion to the output voltage (i.e. the sum being calculated). Thus the current in each input resistor will depend on the state of the other inputs, and the apparent input impedance will vary between infinity (with all inputs at equal voltage) and a value somewhat lower than the individual input resistors.
Now, if the inputs are actually coming from voltage sources (e.g. other opamp outputs, the varying input impedance wouldn't matter, and you would get the summed value that you wanted. But if the sources have their own finite impedance, how does the unpredictable input impedance of the summer affect the result? You could always avoid this by adding a buffer between each input and the summer input resisters, but an extra opamp for each input is now more complex than an inverting summer plus one extra opamp to invert its output.
    Dave     (not a student or an EE, but an electronics hobbyist)
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

Polytechforum.com is a website by engineers for engineers. It is not affiliated with any of manufacturers or vendors discussed here. All logos and trade names are the property of their respective owners.