Construct circuit ??? which uses power supplies +12V and -12V and gives output voltage:



Sketch output voltage Viz(t) diagram for given input voltages:



Here is circuit:



Any idea?
I know that ??? contains OPAMP (or OPAMPS). We see from first term in expression of Viz that it's response of basic inverting OPAMP circuit and from second term that it is response of integrator circuit.

 

I am not sure I am reading the integral part right, that -1/10ms seems odd to me, probably me not understanding what you are trying to say there.

The first part seems simple. You are sending input to the inverting input of the op amp, like you said. You also providing a gain of 2. So this particular part is just V3 into inverting input of the op amp, none inverting input connected to ground, 1 kOhm input resistor and 2 kOhm feedback resistor to provided gain of 2.

 

I am not sure I am reading the integral part right, that -1/10ms seems odd to me, probably me not understanding what you are trying to say there.

That's because you aren't used to seeing a problem treat units properly (since it is such a rare thing).

\(
V_{iz} \: = \: -2 V_3 \: - \: \frac{1}{10 \: ms} \: \int V_4 dt
\)

The integral produces a result that has units of (volt)(time), so to get the second term to have units of (volt), you need to divide by something that has units of (time).

The one thing that the problem missed and requires the reader to guess are the limits of integration.

\(
V_{iz} \: = \: -2 V_3 \: - \: \frac{1}{10 \: ms} \: \int_{0}^{t} V_4 dt
\)

 

Construct circuit ??? which uses power supplies +12V and -12V and gives output voltage:

View attachment 92454

Sketch output voltage Viz(t) diagram for given input voltages:

View attachment 92458

Here is circuit:

View attachment 92456

Any idea?
I know that ??? contains OPAMP (or OPAMPS). We see from first term in expression of Viz that it's response of basic inverting OPAMP circuit and from second term that it is response of integrator circuit.

You basically have a circuit that needs to do three things -- all of which are directly indicated by the expression you are trying to implement.

1) You need a circuit that scales V3.
2) You need a circuit that integrates V4.
3) You need a circuit that sums the results of the prior two circuits.

You could tackle the problem exactly like that with three separate opamp circuits connected together (and possibly a fourth to invert the overall output if necessary).

Or you could see if you can come up with a circuit using a single opamp that does that. For this approach, start with the complicated part, which is the integrator. Get a circuit that integrates V4. Now ask what happens if you introduce V3 into the other input of the opamp. Does it make it so that it only affects the offset of the output, or does it interact with the operation of the integral? If the former then you have a good chance of making it work, if the latter then you might be better off using separate circuits.

 

Thanks WBahn for reply.
I didn't find way to solve this problem using only one OPAMP.
Related to your suggestion to use 3 (or 4 opamp):



i'm not sure I know how to sum it. I analysed OPAMP summer but how to apply that circuit here?

 

Call the output of the top circuit Va and call the output of the bottom circuit Vb. How do you use an opamp to sum two signals, Va and Vb?

 

Well here is opamp summer scheme:





So to sum Va and Vb, we should connect opamps on this way:



Right?

 

Close, but now work out the final relationship for this circuit, being particularly careful to keep track of the signs.

 

I got this:
\(v_i_z(t)=\frac{R_FR_2}{R_1_1R_1}v_3(t)+\frac{R_F}{R_2_2R_3C}\int V_4dt\).
I need now to choose value of capacitance and values of resistances to match relations:
\(\frac{R_FR_2}{R_1_1R_1}=2\), \(\frac{R_F}{R_2_2R_3C}=\frac{1}{10*10^{}-3}\). And finally to add one more OPAMP to change signs, right?

 

And I would like to see solution using only one opamp

 

Here is solution using single opamp:

 

Here is output voltage of single opamp circuit:


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:
\(R1 = \frac{10*10^{}(-3)[seconds]}{470*10^{}(-6)[F]} = 21.2766 [\frac{seconds}{Farads}] ?\)

 

From Q = CV you know that 1F = 1C/1V

So seconds/farad = s/(C/V) = s·V/C = V/(C/s) = V/A = Ω.

Your equation should simply be

\(
R_1 C_2 \; = \; 10 \times 10^{-3} \: s
\)

The R_1 and C_2 carry their own units, so putting additional units next to the symbolic symbols is incorrect.

But it's sure refreshing to see someone paying attention to units. What a change! Thank you.