Continue to Site

Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

Inverting Adder

Status
Not open for further replies.

pwm_11

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

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.
 
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.

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).
 
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).

Yes, you are right. By your students you will taught.
 
Status
Not open for further replies.

Latest threads

New Articles From Microcontroller Tips

Back
Top