# Construct Opamp circuit for given inputs and output

Construct circuit ??? which uses power supplies +12V and -12V and gives output voltage: where Here is circuit: I solved it using 4 opamps:

First opamp scales V3, second integrates V4, third sums the results of the prior two circuits and fourth inverts signs in sum (output of third opamp). Can this problem be solved using single opamp?

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Baluncore
2019 Award
How do you interpret the factor “ –1/100ms ” in the equation for Viz ? “ –(1/100) ms ” ?

The LTspice schematic "filename.asc" is a text file.
You can append a .txt extension to make "filename.asc.txt" and attach it to your post.

• etf
To be honest, I don't know exactly what they mean with ms.

Baluncore
2019 Award
Because dt is time, the integral of a constant voltage is a ramp with units of volts * time.
Before summing as a voltage it must be converted to the same dimensions, by multiplication by 1 / time.

Maybe the constant of integration is –1k / (100* sec) = – 10 / sec = – 10 * sec-1 ?

The op-amp sums a ramp with a sinewave, then inverts the result. Can V4= +1V be a floating voltage source ?

• etf
I don't think they made mistake in original formulation...
I solved it using single opamp. Here is schematic: Here is relation for V1 (Viz in original formulation): We can choose now C1 and C2 so C1 / C2 = 2 (for example C1= 2*470uF, C2=470uF), but I'm confused about second term, more precisely, about units.

If we solve equation $$R1[Ohm]*C2[Farad]=10*10^{}(-3)[seconds]$$ for R1, we get:
$$R1 = \frac{10*10^{}(-3)[seconds]}{470*10^{}(-6)[F]} = 21.2766 [\frac{seconds}{Farads}] ?$$

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Baluncore
2019 Award
Well done. Seems like a sensible solution.
There needs to be initial conditions of V3 = Vc1 = Vc2 = zero.

The mention of +/– 12V supplies suggests the real world. That may require a low value resistor be switched across C2 to bring the output back up to zero following an integration.

Most people do not consider potential dividers made out of capacitors, probably because stray capacitance becomes a problem when the signal currents are minimised. Oscilloscopes often use a resistance of 1M with about 12pF of parallel capacitance. For a broad band flat response, all R||C elements of the divider need to have the same time constant.

• etf
Baluncore
2019 Award
We can choose now C1 and C2 so C1 / C2 = 2 (for example C1= 2*470uF, C2=470uF), but I'm confused about second term, more precisely, about units.
-1/100ms. The units are reciprocal seconds which will cancel with the ( integrated V .dt ) = volts * seconds

The coefficient wanted is -1 / (R1 * C1)
Therefore R1*C1 = 100 m = 0.1
If C2 = 470 uF, and C1 = (470 + 470) uF
Since R1 * C1 = 0.1
R1 = 0.1 / C1 = 106.38 ohms

But that low resistance and huge capacitance will require very high currents.

Use 3 identical capacitors from the same batch to make the two capacitors needed.
Make the 2*C by paralleling two, as series connection reduces capacitance and so wastes space and dollars.

Use the smallest cap that will work with R1 = 100k. Above 100k, surface leakage starts to become important.
So now scale the capacitance by a factor of 1000 to get smaller stable capacitors of say; 940 nF and 470 nF.
R1 is now 0.1 / 0.940u
R1 = (0.1/0.940) Meg ohm
R1 = 106.38 k ohm.
The integration current has fallen from 10 mA to 10 uA. That's better.

But what is the order of magnitude of the current through the capacitive potential divider ?
C = Q / V; Q = I * t; C = I * t / V; therefore C = I * t / V; I = C * V / t
For C = 1uF, t = 1/100 sec and V = 2 we get I = 200uA which is now OK.
Add half more again for the output side; 300 uA. That is certainly better than 300mA.

• etf
Baluncore
2019 Award
Attached is the LTspice simulation. Not sure why the output starts at -3V.

#### Attachments

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-1/100ms
You mean -1/10ms ?

Baluncore
2019 Award
You mean -1/10ms ?
You are correct. I should have checked the OP.
Everything I did needs to be recalculated, but I know you know what you are doing, because you picked up my error.

• etf
Baluncore