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Summing Amplifier Problem

Mark123

May 14, 2013
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Hi there.

I have been instructed to build an amplifier to preform the following mathematical function:

Vout= (Vin1 + Vin2 + Vin3 + Vin4)/4

It is virtually the same function as a summing amplifier except for the divide by 4 part. I would like to ask those out there who are learned in this field, how can I achieve this? I have constructed a working summing op amp, but how do i get the divide by 4?

Thanks in advance
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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you have a summing amplifier which does v1 + v2 v3 + v4.

What is the difference between that and (v1 + v2 v3 + v4)/4.

How can you implement that difference?

Can you do it using the summing amplifier?
 

Harald Kapp

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Also note that the classical OpAmp based summing amplifier delivers -sum(xi) (note the sign). According to the task you've described that needs to be taken care of.
 

Mark123

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Yeah, I forgot the part where it changes to negative. So that raises another problem which I'm not sure how to tackle.

As for the dividing part, could i use a basic voltage divider?
 

(*steve*)

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Yeah, I forgot the part where it changes to negative. So that raises another problem which I'm not sure how to tackle.

Inverting is a gain of -1 isn't it?

What could possibly do that?

Is your summing amplifier already doing it -- is that a hint?

As for the dividing part, could i use a basic voltage divider?

Hmmm... Can you change the gain on the summing amplifier?
 

Mark123

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I don't know. I have never worked on op amps before and electronics is not my strongest or favorite field, but this has become quite frustrating. I have done as much reading as I can but nothing has clicked. I know you are not supposed to give me the answer so to say, but could you at least point me to somewhere I can figure the answer out?

I looked at other amplifiers and I saw that on the non inverting amp, Vin goes in on the + side of the op amp. Would it work if I fed all the voltages in on the + side of the summing amp? Would this prevent my output from inverting?

I also found this formula:

Vout = - (V1x Rf/R1 + V2x Rf/R2 + V3x Rf/R3 + V4x Rf/R4)

Would it work if i just divided the voltage by 4 using Rf?
For example if R1=R2=R3=R4, and i make Rf a quarter of the resistance, would that quarter the output voltage?
 
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Harald Kapp

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I looked at other amplifiers and I saw that on the non inverting amp, Vin goes in on the + side of the op amp. Would it work if I fed all the voltages in on the + side of the summing amp? Would this prevent my output from inverting?
There's a reason for having different circuits for inverting and non-inverting amplifiers.
The basic idea of most any amplifier circuit is to have negative feeedback to stabilize the amplifier and/or set the gain. Will you still have negative feedback if you exchange "+" and "-" inputs of the OpAmp while leaving the rest of the circuit as it is?

How could you invert Vout of the summing amplifier? Vout2=-Vout=-(-sum(Vin)/4)= +sum(Vin) ("-" * "-" cancels).


Would it work if i just divided the voltage by 4 using Rf?
For example if R1=R2=R3=R4, and i make Rf a quarter of the resistance, would that quarter the output voltage?

Insert that Rf into the equation and evaluate. What happens to Vout?
 
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duke37

Jan 9, 2011
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The basic inverting amplifier is also known as a virtual earth amplifier.
The input terminals take negligible current so the current flowing through the input resistance is the same as that flowing through the output resistance. The voltage produced can be calculated by Ohms law.
The advantage of the virtual earth amplifier is that each input is not affected by other inputs.
 

Laplace

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Check out the attached application note from http://www.edn.com/design/analog/43...ch-for-designing-positive-summing-amplifiers-

What I was able to make of it is in the attached pdf. Note the node equations for the voltage at the '+' input of the op amp, Vi, and at the '-' input of the op amp, also Vi because of the negative feedback. In this circuit the op amp input terminals have a wide common mode range, so select the op amp appropriately.
 

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