Inverting Adder

Hello! I am trying to answer the following question:

Design an circuit with operational amplifiers that has the following output function:

Vo=2*V1+10*V2+11*V3-V4

Shoud i use the inverting adder?

Using three different voltage sorces V1,V2,V3 and V4 ,an OpAmp and a 4 resistors(R1 to R4) and a feedbak resistor.

So the expression for the output of and adding inverter is given by:

V0=-(Rf/R1)-(Rf/R2)-(Rf/R3)-(Rf/R4)

So i could use Rf=10k ,R1=5k,R2=1k,R3=1,1k and R4=10k

The problem is then Vo=-2V1-10V2-11V3-V4....Shoud i use and non-inverting adder,conecing the resistors to the non-inverting input?

Thanks

I think you could use one opamp to create 11*(V3-V4) and another to do the summing.

MikeMl said:

I think you could use one opamp to create 11*(V3-V4) and another to do the summing.

Click to expand...

The problem is that in the function that i am looking for all terms are positive except for V4...So using an inverting adder i will get -2V1-10V3-11V3....How can i make the terms positive?Is there another configuration of a non inverting adder?

Thanks

It is just simple algebra. The final output you want is 2V1+10V2 + 11V3 - 11V4.

That is equivalent to 2V1 + 10V2 +11V3 + 11*(-V4), so use an opamp with a gain of -1 to create -V4. Now the main opamp summer happens to invert the entire sum [i.e. -(2V1 + 10V2 +11V3 + 11*(-V4)) ], so use another opamp with a gain of -1 to undo the summer's inversion to get the final answer. Requires a total of three opamps.

Hi pwm_11,

I think you can do it with only two opamps.

*Use an INVERTING adder (classical configuration) for V1, V2 and V3 and feed the output into a second inverter (gain -1) - thus, you have two inversions for a positive sign.
*Feed V4 through another suitable resistor (equal to the feedback R) into the inverting input of the second opamp.

Winterstone said:

Hi pwm_11,

I think you can do it with only two opamps.

*Use an INVERTING adder (classical configuration) for V1, V2 and V3 and feed the output into a second inverter (gain -1) - thus, you have two inversions for a positive sign.
*Feed V4 through another suitable resistor (equal to the feedback R) into the inverting input of the second opamp.

Click to expand...

Wouldn't the second adder have to have one input with a gain of -1 and the other with a gain of -11?

Hi Mike,

I suppose there is a misunderstanding on your side , see you post#2: 11*(V3-V4)

However, in the original question (post#1) there are no brackets; that means: only V4 is to be subtracted (gain of -1).

Winterstone said:

Hi Mike,

I suppose there is a misunderstanding on your side , see you post#2: 11*(V3-V4)

However, in the original question (post#1) there are no brackets; that means: only V4 is to be subtracted (gain of -1).

Click to expand...

Yes, you are right. By your students you will taught.