Basic opamp design problem ##V_{output} = 3V_1 - 2V_2##

[JC2000]

Homework Statement

I need to design an operational amplifier that performs the operation $V_{output} = 3V_1 - 2V_2$.

Relevant Equations

I realise that a difference amplifier must be used to construct this. Thus the relevant equation is :

$v_{o u t}=v_{2} *\left(\frac{R_{4}}{R_{3}+R_{4}}\right) *\left(1+\frac{R_{2}}{R_{1}}\right)-v_{1} *\left(\frac{R_{2}}{R_{1}}\right)$

Here I let $R_2 = 2$ kohms and $R_1 = 1$ kohms. Using this suggests that $R_3 = 0$ and $R_4 = 1$ kohms (?!)

The resulting circuit is

Is this the correct way to solve this?

PS : I just realized that I have interchanged $V_1$ and $V_2$ in the diagram!

 

[cnh1995]

You can directly connect V1 to the non-inverting input. No need to use the R3-R4 voltage divider.
Rest of your solution looks correct to me.

 

[Averagesupernova]

You can directly connect V1 to the non-inverting input. No need to use the R3-R4 voltage divider.
Rest of your solution looks correct to me.

Do you mean V2?
Edit: Just realized the OP made a correction.
-
Quiz question, why would a designer use a circuit such as this? Example, all four resistors being the same value.

 

[Jody]

I would think for differential pairs, but I don't see it too often outside of the classroom. I wonder why.

 

[Averagesupernova]

Anyone else? Not saying you are wrong though @Joshy .

 

[DaveE]

I've had to design similar circuits many times in the real world. The most common implementation is when you want some gain for a signal and an additional offset. In that case the "other" input is a reference voltage.

This is a great place to use the superposition of linear circuits. There is a necessary gain from V2 to the output, which can be found by setting V1=0 (so those resistors don't matter). Then you look only at the gain you need from V1 to the output, through the gain term defined by the Vo/V2 solution; so set V2=0 and solve for Vo/V1.

There is another approach that relies on knowing the solution to a multiple (arbitrary number) input balanced differential amp where each resistor is part of a pair of the same value, one on the + side, one on the - side. Kind of hard to explain in words, but the gain equations are trivial once you've solved it once. Maybe I'll post an example?

 

[Averagesupernova]

The example I'm thinking of is noise cancellation not limit to a balanced pair.

 

[JC2000]

The example I'm thinking of is noise cancellation not limit to a balanced pair.

You mean that if $R_4 = R_3$ and $R_2 = R_1$ then $v_{out} = v_2 - v_1$? Which would lead to noise cancellation if the noise was 'the same' in both inputs ?

But wouldn't using an operational amplifier without any feedback or resistors do the same ? (Except that the output for a regular amp would cancel the noise but the output would also involve gain, whereas in your design the gain is unity(?) )

 

[JC2000]

This is a great place to use the superposition of linear circuits. There is a necessary gain from V2 to the output, which can be found by setting V1=0 (so those resistors don't matter). Then you look only at the gain you need from V1 to the output, through the gain term defined by the Vo/V2 solution; so set V2=0 and solve for Vo/V1.

Sorry for being slow here but are you referring to how the equation is derived?

There is another approach that relies on knowing the solution to a multiple (arbitrary number) input balanced differential amp where each resistor is part of a pair of the same value, one on the + side, one on the - side. Kind of hard to explain in words, but the gain equations are trivial once you've solved it once. Maybe I'll post an example?

I am not sure if this is what you meant?! How would the derivation work?! I tried using the superposition theorem, but I was not sure how to proceed once I let one of the inputs be 0 V (since in the inverting difference amp with two inputs, setting one input to zero 'eliminates' one 'network' entirely whereas here if I let $V_1 =0$, I don't know how to handle $V_2 etc$...

Another question : Would the circuit I have drawn about help if a circuit involving both summing and differences had to be created ($V_{output} = (xV_A + yV_B + ... + zV_{N+N} ) - (aV_1 + bV_2 + ... + nV_N)$ ?

 

[DaveE]

First you need Rf = Rg for this to be a diff amp.

Then the way I would use superposition would be to set all but one of the inputs to zero. That doesn't remove those inputs, it just connects those resistors to ground. An open input does remove the connected resistor and needs to be done in pairs with it's mate.

I'll let you try the derivation first (since I'm a bit lazy) ask me if you run into trouble. The key to it is combining the inputs as shown in the photo. You can do it with KVL/KCL too but I think Thevenin is easier.

Finally, the answer: First define Rp as the parallel combination of all resistors connected to the inputs (R1||R2||R3...||Rn). This doesn't appear in the final equation because the resistive divider effect at the input is canceled by the gain increase from the feedback reduction.

Then the gain for any single input pair "k" (all inputs are applied as differential pairs, even if one side is 0) is Rf/Rk. Then you can add in as many inputs as you like, e.g Vo=Rf/R1(V1-Va)+Rf/R2(V2-Vb)+...

 

[JC2000]

First you need Rf = Rg for this to be a diff amp.

I'll have a go at the derivation asap but could you tell me why Rf = Rg for it to be a diff amp?

 

[DaveE]

I'll have a go at the derivation asap but could you tell me why Rf = Rg for it to be a diff amp?

It part of the requirement to have balanced (the same) gain for the positive and negative inputs. Look at your "relevant equation" in the original post and find the positive gain when R4/R3 = R2/R1 (people often satisfy this by making R3=R1 and R4=R2).

But I don't want to steer you off track. You don't need a differential amplifier. It's just a short cut that I use when circuits require different gains for different inputs, usually for more complicated problems. If you start out with my method and then simplify the circuit you will end up with your original solution, which most people wouldn't call a differential amp.

Really, they way you did it was great. Your only confusion about the value of R4 is because it could be anything except 0, including infinite. Old school designers would set R4=R1||R2 to balance the resistance "seen" by the op-amp input. But unless you are using really old op-amps, or have extreme requirements, it really doesn't matter much these days.

edit: The other good choice is to set R4 = infinite (i.e. leave it out). Then R3 can be any value. This is really what I was thinking, and should have said, when I talked about balancing inputs with R3=R1||R2.

 

[LvW]

May I mention some other applications for the shown topology?

* Replace R4 with a capacitor and make V1=V2 (only one input node): This results in a 1st-order allpass.

* Set R3=0 (remove R4) and make V1=V2. Now we have unity gain amplifier with variable loop gain (possible overshoot optimizing).

* Ground the V1 input and we have a non-inv. amplifier which offers the chance to vary the loop gain for any value for the closed-loop gain. Hence, we can use also non-compensated opamps for low closed-loop gain values.