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I am currently struggeling with the following problem. I need to find an amplifier circuit which implements the following equation:

$$ U_a=U_1+2U_2-2U_3 $$

I decided to go with the circuit shown below. I calculated the following equation for the output:

$$ U_a=-\frac{R_F}{R_3}U_3+\frac{R_1 || R_4}{R_2+R_1||R_4}\frac{R_F+R_3}{R_3}U_2+\frac{R_2 || R_4}{R_1+R_2||R_4}\frac{R_F+R_3}{R_3}U_1 $$

Now I can formulate all the conditions necessary in order for the equations to fit.

$$ I: \frac{R_F}{R_3}=2 \\ II: \frac{R_1 || R_4}{R_2+R_1||R_4}\frac{R_F+R_3}{R_3}=3\frac{R_1 || R_4}{R_2+R_1||R_4}=2\\ III: 3\frac{R_2 || R_4}{R_1+R_2||R_4}=1 $$ From the conditions I tried to calculate specific values for the resistors, starting with the 2nd condition.

$$ II: R_2=\frac{R_1||R_4}{2}\\ R_2R_1+R_2R_4=\frac{R_1R_4}{2}\\ R_1=\frac{R_2R_4}{\frac{R_4}{2}-R_2} $$

This result I plugged into a condition following from III.

$$ III: R_1=2(R_2||R_4)\\ \frac{R_2R_4}{\frac{R_4}{2}-R_2}=\frac{2R_2R_4}{R_2+R_4} \\ \frac{1}{\frac{R_4}{2}-R_2}=\frac{2}{R_2+R_4}\\ R_2+R_4=R_4-2R_2 \\ 3R_2=0 $$

Well ... crap, R2 is supposed to be 0, so would be R1. And the value for R4 does not matter? I guess I did something wrong but I cannot figure out what. What is wrong with my calculation? Is it possible to implement the given equation with the circuit I drew?

EDIT: I forgot to mention that it is only allowed to use one op-amp.

enter image description here

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The way I would approach this is to observe that Rf/R3 = 2 (by inspection)

The gain from the non-inverting op-amp input is thus 2 + 1 = 3.

So you need a divider with gain 1/3 from V1 and 2/3 from V2.

Set R1||R2||R4 = R

Then R/R2 = 2/3 and R/R1 = 1/3

Since 1/R4 = 1/R - 1/R2 - 1/R1 = 1/R - 2/3R - 1/3R = 0 So R4 = \$\infty\$

You can pick two more values arbitrarily- say R = 10K which leads to R2 = 15K, R1 = 30K

It would make sense to constrain RF||R3 = 10K to balance the offset from bias currents, so RF = 30K R3 = 15K.


Edit:-

General principle for a voltage divider with n resistors R1...Rn

Vout/Vin = R * (V1/R1+V2/R2+...+VN/RN)

Where R = R1||R2||..||RN

(It's easy to prove this by superposition)

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  • \$\begingroup\$ Can you explain me more why you set R=R1||R3||R4 and then say R/R3=1/3 and R/R1 = 2/3? Ok, as you mentioned, condition II and III in my question show that for U2 a divider 2/3 is needed and for U1 a divider 1/3, because (RF+R3)/R3 equals 3. But, R/R3 is not the same as (R2||R4)/(1+R2||R4), which would need to be 1/3 (condition III). \$\endgroup\$ – Daiz May 7 '16 at 17:04
  • \$\begingroup\$ I meant (R2||R4)/(R1+R2||R4) \$\endgroup\$ – Daiz May 7 '16 at 17:26
  • \$\begingroup\$ Sorry- I couldn't read your numbers that well, so wrote it down and transcribed it a bit strangely should be correct now. I'll add another edit to show the principle. \$\endgroup\$ – Spehro Pefhany May 7 '16 at 18:32
  • \$\begingroup\$ This problem is a bit contrived - you might want to think about what you have to change if R4 worked out to be negative. \$\endgroup\$ – Spehro Pefhany May 8 '16 at 4:41
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Your question doesn't state that you are only allowed use one op-amp. If you are allowed use two then use two inverting summing amplifiers. Use the first to sum all the positive voltages and the second to sum the inversion of those and the negative voltages.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Two stage inverting summer.

  • OA1 sums U\$_1\$ and 2U\$_2\$ voltages but inverts the output giving -U\$_1\$ - 2U\$_2\$.
  • OA2 sums (-U\$_1\$ - 2U\$_2\$) with +U\$_3\$ to give U\$_1\$ + 2U\$_2\$ - 2U\$_3\$.

It looks like homework and you seem well able to do the calculations. If you want to post your results as an answer we can check it.

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  • \$\begingroup\$ I am sorry, I forgot to mention that .. actually the problem states to only use one op-amp. This is not homework, I am preparing for an exam in electrical measurement and this one is a sample problem. Still, thank you for the offer of checking the results :) Do you have an idea how to solve the problem with one op-amp? \$\endgroup\$ – Daiz May 7 '16 at 16:18
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    \$\begingroup\$ Spehro's answers are trustworthy. ;^) \$\endgroup\$ – Transistor May 7 '16 at 16:26
  • \$\begingroup\$ ok, working it through right now ^^ \$\endgroup\$ – Daiz May 7 '16 at 16:27
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To achieve an arbitrary linear combination of inputs given by N ground-referenced voltage sources V1 ... VN you can think as follows:

  1. A linear combination of positive terms is easy to achieve: just use N resistors from each source to a common node Vp. A suitable choice gives you the desired ratio which can be scaled down with an additional resistor Rs or scaled up with a non-inverting op-amp.

  2. A linear combination of negative terms is the same, but requires an inverting op-amp.

  3. In a mixed linear combination you treat the positive terms as in 1, the negative ones together as in 2 and use a differential amplifier structure

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