Hello!
http://images41.fotosik.pl/134/2106477a4e79a5d9.jpg
Can anybody help me to solve this exercise about operational amplifiers. I tried to redraw it in a little bit simpler way, then calculate Zz, which is equivalent impedance for three resistors and one capacitor above second amplifier. I am not sure which resistors I can erase (treat them as short) and why. If I would know which resistors I can erase, I would be able to see what op-amps these are (inverting, non-inv., summing, differential, voltage comparator with or without hysteresis). Then it should be relatively easy to calculate K(j omega). But what about (c) determining the allowed range for Uin if the output voltage of both op-amps cannot exceed +-Usat=15V?
Greetings!

 

Erasing resistors is not always the same as treating them as a short.

Assuming your opamps are ideal, and that Uin is an ideal voltage source (zero output impedance), you can erase R1 (treat as an open circuit) because it's just an extra load on Uin and has no effect on the operation of the rest of the circuit.

You can erase (open circuit) R4 because the first opamp has zero output impedance, so R4 is just an extra load on the first opamp.

You can erase R5 (treat as a short circuit) because the input impedance of the second opamp is infinite.

You cannot erase R8, because it's a part of the second opamp's feedback network.

The first opamp is just a non-inverting amplifier with a voltage gain of 2.

But now you have to derive the transfer function of the second opamp. I think I would use the nodal method of circuit analysis to do that.

For part c of the problem you need the DC gain of the second opamp; you can just leave C out of the transfer function expression for the second opamp and you will have the DC gain. Then just determine what input voltage will drive either opamp output to 15 volts.

 

Thank you very much for the answer, The Electrician!
I try to solve it with the use of nodal analysis (http://images45.fotosik.pl/139/b2af29e3b002bb73.jpg) but it looks like I've got some difficulties :/.
Greetings

 

You can also do it without using the nodal method. Just treat the network on the output of opamp2 as a pair of voltage dividers.

If the output of opamp2 is designated Ua2, the voltage at the junction of R7 and R8 as Ud1, and the voltage at the junction of R6 and R7 as Ud2, then we have:

               (R7||C + R6)||(R9)
Ud1 = Ua2 * ------------------------
             R8 + (R7||C + R6)||(R9)

and:

                 R6
Ud2 = Ud1 * ------------
             R6 + R7||C

Then the transfer function, TF2, from the + input of opamp2 to the output of opamp2 will be Ua2/Ud2.

Just for a hint, Ud1 is Ua2 * (15000 jωC + 30)/(28000 jωC + 53)

Then the transfer function from the + input of opamp2 to Uout is just:

         1       (R7||C + R6)||(R9)
TFout = --- * ------------------------
        TF2    R8 + (R7||C + R6)||(R9)

Don't forget the gain of 2 in the first opamp when calculating your overall transfer function.

 

I appreciate Your help very much.
I try to understand this solution and this is my question:
part1: http://images48.fotosik.pl/139/19864abe23453cd9.jpg
part2: http://images41.fotosik.pl/135/ea87ef2d0c116941.jpg
Greetings!

 

You ask: Why is it (1) and not (2)?

In (1), you show Ud2 at the junction of R7 and R8, but that's not where it is.

Think of the network as two voltage dividers. The output of the first divider is at Ud1. The second divider has the voltage at Ud1 as its input, and its output is at Ud2 (which is the junction of R6 and R7). R8 is taken into account by the first divider, but R8 isn't part of the second divider.

In (1), what you have labeled Uout2 is what I defined as Ua2. Keep in mind that the output for which the problem asks for a transfer function is not Uout2, but is the other end of R8, which is the same as Ud1.

Does this help?

 

Zdzislavv co, ten układ jest za trudny dla "naszego" forum ?
This problem is to had for "our" forum
http://elportal.pl/forum/viewtopic.php?p=54991#54991

 

http://images40.fotosik.pl/135/9100154fce79467c.jpg
Please, explain me this part:

Then the transfer function from the + input of opamp2 to Uout is just:

                 (R7||C + R6)||(R9)
TFout = TF2 * ------------------------
               R8 + (R7||C + R6)||(R9)

Is TFout=Ua2/Ud1 ? What is TF2?
Greetings!

 

Since Ud2 (which is the - input of opamp2) is equal to Ua2 multiplied by some transfer function, we can let that transfer function be TF2 and say that Ud2 = Ua2 * TF2, or TF2 = Ud2/Ua2. Then the transfer function from the + input of opamp2 to its output is 1/TF2.

Then TFout = (Ud1/Ua2) * (1/TF2) = (Ud1/Ua2) * (Ua2/Ud2) = Ud1/Ud2 = (R6 + R7||C)/R6

I made an error in the expression for TFout; it should be:

         1       (R7||C + R6)||(R9)
TFout = --- * ------------------------
        TF2    R8 + (R7||C + R6)||(R9)

Sorry. :-(

 

Hello!

Why cannot I erase R9 in this example: http://images46.fotosik.pl/151/22844edb4a2981c5.jpg

I tried to use analogy to this exercise http://images40.fotosik.pl/146/99ab60bf7b16ed1c.jpg and its solution http://images39.fotosik.pl/147/30682b6d46b44b2a.jpg (and by the way, where does 0,16 in numerator come from?).

Greetings!

 

Hello!

Why cannot I erase R9 in this example: http://images46.fotosik.pl/151/22844edb4a2981c5.jpg

Because in this circuit, R9 is not connected directly to the opamp output; it's part of a voltage divider.

I tried to use analogy to this exercise http://images40.fotosik.pl/146/99ab60bf7b16ed1c.jpg

In this circuit, R8 is connected directly to the opamp output, and since the (ideal) opamp output impedance is zero ohms, R8 has no effect.

and its solution http://images39.fotosik.pl/147/30682b6d46b44b2a.jpg (and by the way, where does 0,16 in numerator come from?).

It comes from 1/(2*pi). Remember that ω = 2*pi*f