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I'm trying to implement the following schematic of a simple analog computer

Analog Computer

The inputs are x and y, according to the text where I found this diagram the circles represent voltage dividers or potentiometers.

There are some things I dont get, for instance the divider in the feedback loop is \$ \frac{1}{k}\$ and it also states that k is between 0 and 1, so that means that \$ \frac{1}{k}>1\$, that implies that the divider should actually be an amplifier instead?

This is how I implemented it:

schematic

simulate this circuit – Schematic created using CircuitLab

With the resistor values I selected for the circuit, a=b=0.5, and \$\frac{1}{k}=5\$, which means that the final equation is supposed to be \$-0.2(0.5x+5y) \$. However, if one analyzes the circuit using superposition, we have:

$$v_o=-(ax+10by+\frac{1}{k}v_o)$$

or

$$v_o=\frac{-k(ax+10by)}{k+1}$$

What am I doing wrong? how is the circuit supposed to be to in order to get the correct output \$v_o=-k(ax+10by)\$?

winny
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S.s.
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3 Answers3

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Your circuit is way too complicated. You need to study how an inverting gain op amp circuit works. Once you understand, the gain for voltage on the left side of R6 is -1 (-R3/R6) and the gain for voltage on left side of R2 is -10 (-R3/R2).

You can eliminate everything other than OA1, R3, R2 and R6. R2 and R6 will connect directly to Y and X power supplies.

The gain of the resulting circuit will be -R3/R6 * x + -R3/R2 * y.

Now rearrange original equation to -kax + -10bky.

R6 = R3/ka and R2 = R3/10bk

Michael
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Heres how I'll implement it using simple adder circuit

schematic

simulate this circuit – Schematic created using CircuitLab

The overall transfer function is $$z = -R_3\left(\frac{x}{R_1} + \frac{y}{R_2}\right)$$ For \$ a=b=0.5;\ k=0.2\$ you can use resistor values mentioned in the figure. Other two OpAmps are to avoid input overloading. You can replace them with some kind of filtering mechanism which is typically required if you are taking data from analog sensors.

Yogesh
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You can apply superposition only on sources, not for parts of the circuit as you do for the OA2 path.

Also 1/k is not 5 but 6 because you have to take R3 into account, 5 for OA2 path and 1 for R3 path.

With that said everything fits just fine.

ax = -5Voa - Voa -> Voa = -ax/6 = -axk

10bx = -5Vob - Vob -> Vob = -10by/6 = -10byk

Vo = Voa + Vob = -(ax + 10by)k

I can also notice that you could simply use R3 = k * 100kohm and leave OA2 path with the same result.

Thanks to @Arsenal for pointing the sign error.

Dorian
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  • there is a difference in your solution and the expected solution in the question (a minus is missing in your solution). – Arsenal Aug 28 '18 at 09:30
  • Can you elaborate on why 1/k is 6 and not 5? I dont get what you are saying that you have to consider R3 path. Also your nomenclature is not clear, what is Vox? – S.s. Aug 28 '18 at 13:57
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    Actually I don't see why R3 is there, in the first schematic is not. Only 1/k feedback which is through OA2. R3 adds a feedback that is not in your first schematic and adds a unit to 1/k factor. Vox and Voy are the partial results of Vo using superposition, Voa = Vo when y = 0 , Vob = Vo when x = 0. – Dorian Aug 28 '18 at 13:59
  • Ok so if Vox is the partial result when y=0 then what is Voa?, I thought Voa was the partial result. – S.s. Aug 28 '18 at 14:00
  • Sorry , edited, Voa and Vob – Dorian Aug 28 '18 at 14:02
  • Got it, by the way you can do the superposition the way I did (check https://electronics.stackexchange.com/questions/107435/superposition-principle-dependent-sources-treated-as-independent-sources), in fact if you consider 1/k=5 like I did, my equation equals yours. – S.s. Aug 28 '18 at 14:09
  • Uh.. no, let's take the good schematic, without R3. What Vo do you get if you make OA2 output zero? A negative infinite voltage. The apparently good result is just coincidental. – Dorian Aug 28 '18 at 14:25
  • You are correct about the circuit not having R3, the text has weird symbols, the rounded amplifier actually means opamp while the squared amplifier means adder. Thanks Dorian – S.s. Aug 28 '18 at 14:35
  • @A.J. Is questionable that what Alfred does in the chosen answer is really a superposition. It's something wrong with my answer, is something missing? – Dorian Aug 28 '18 at 15:02
  • Seems that what you consider as k is in my equation k/(k+1) if you substitute 1/k=6 in your equation or 1/k=5 in mine, it gives you the same result – S.s. Aug 28 '18 at 16:00