See image:
http://i1362.photobucket.com/albums/r687/JasonMcG1/Circuit_zps98e971ed.jpg

I have tried to decipher what i think is correct. Using the gain on each amplifier and multiplying it by the voltages i acquired Vout to be +1.9 volts. Need to know if i am correct. Also need to know what to do for part ii. Also where do you connect the battery in this circuit? Any and all help is much appreciated. Thanks.

Jason M

 

The 0.9V output voltage is right.

How to connecting the batteries to the 741, where is the +9V of battery that it will be connecting to which is the +9V on the circuit, and where is the -9V of battery that it will be connecting to which is the -9V on the circuit.

 

0.9V or 1.9V ?

 

0.9V or 1.9V ?

The gain of inverting amplifier = -(9V*(Rf/Ri)) = -(9V * 0.1) = -0.9V
The gain of noninverting amplifier = 9V*((Ri+Rf)/Ri) = 9V*(11/10) = 9.9V.

The +V only 9V, so the output could not reach to 9.9V.

 

See image:
http://i1362.photobucket.com/albums/r687/JasonMcG1/Circuit_zps98e971ed.jpg

I have tried to decipher what i think is correct. Using the gain on each amplifier and multiplying it by the voltages i acquired Vout to be +1.9 volts. Need to know if i am correct. Also need to know what to do for part ii. Also where do you connect the battery in this circuit? Any and all help is much appreciated. Thanks.

Jason M

The gain of inverting amplifier = -(9V*(Rf/Ri)) = -(9V * 0.1) = -0.9V
The gain of noninverting amplifier = 9V*((Ri+Rf)/Ri) = 9V*(11/10) = 9.9V.

The +V only 9V, so the output could not reach to 9.9V.

You are both wrong.

Jason, show your work.

 

You are both wrong.

Jason, show your work.

When I got the answer, I just felt the values is little strange, but that is according to the formular, so what's wrong with the formular or those values?

 

I have renamed the components for easier referencing.
See image:
http://i1362.photobucket.com/albums/r687/JasonMcG1/Ciruit_3_zpsac535a8d.jpg

Amp A: inverting
Gain A = R2/R1 = -0.1
-0.1x9V= -0.9V

Amp B: non-inverting
Gain B = 1+(R4/R3) = 11V
11x0.1 = 1.1V

Amp C: non-inverting 'buffer'
Gain C = 1
1x1.1 = 1.1V

Amp D: 'Summing amp'
G1=-(R7/R5) = +0.9V
G2=-(R7/R6) = -11V
R5//R6
G3=1+(100/9.091) =1+11=12
12x1 = 12V

Therefore Vout = 0.9-11+12 = 1.9V

 

Amp B: non-inverting
Gain B = 1+(R4/R3) = 11V
11x0.1 = 1.1V

I think in your image Vin = -0.1V => output of B: 11x(-0.1)= -1.1V

Amp C: non-inverting 'buffer'
Gain C = 1
1x1.1 = 1.1V

Gain C = 1
1x(-1.1) = -1.1V

Therefore Vout = 0.9-11+12 = 1.9V

Vout = 0.9+11+12 = 23.9V

 

I have renamed the components for easier referencing.
See image:
http://i1362.photobucket.com/albums/r687/JasonMcG1/Ciruit_3_zpsac535a8d.jpg

Amp A: inverting
Gain A = R2/R1 = -0.1
-0.1x9V= -0.9V

Amp B: non-inverting
Gain B = 1+(R4/R3) = 11V
11x0.1 = 1.1V

But Vin =-0.1V!!!
In other words, it is negative 0.1V!

 

I have gone through it again and calculated Vout to be 23.9V. Is this the correct value?

 

I have gone through it again and calculated Vout to be 23.9V. Is this the correct value?

No you are wrong. For Vcc equal to +9V LM741 won't give you more then 9V at the output.

 

I have gone through it again and calculated Vout to be 23.9V. Is this the correct value?

If you had an ideal op amp with no limits on output voltage, that would be correct. As jony130 says, the output can never exceed the rail voltages. With a 741, and ±9V supplies, your output will never exceed ≈±8V.

That is either a trick problem, or a very poorly designed one.

 

Okay thanks.
I need help for part two also please. Not sure what to do or how to go about it.

http://i1362.photobucket.com/albums/r687/JasonMcG1/Ciruit_3_zpsac535a8d.jpg

 

Okay thanks.
I need help for part two also please. Not sure what to do or how to go about it.

http://i1362.photobucket.com/albums/r687/JasonMcG1/Ciruit_3_zpsac535a8d.jpg

For now, just imagine that the RC network is not there, and the sine wave appears unattenuated at the output of op amp C. What would the voltage be at that node? What would the output of op amp D be?

 

The output at the node would be 1.1V and at op amp D it would be 23.9V

 

The output at the node would be 1.1V and at op amp D it would be 23.9V

You might want to think a little harder. A sine wave is not a DC voltage.

 

It would have a max peak at 55mV and a minimum at -55mV

 

It would have a max peak at 55mV and a minimum at -55mV

How did you come up with that?

 

the output voltage being 1.1V =110mV peak to peak
meaning 50mV up and 50mv down from the origin

 

the output voltage being 1.1V =110mV peak to peak
meaning 50mV up and 50mv down from the origin

I regret to inform you that 1.1V=1100mV, NOT 110 mV.
Do you really think carefully before replying? I'm starting to feel like a babysitter.