# Analyzing an op amp

Discussion in 'General Electronics Chat' started by JasonMcG, Apr 28, 2013.

1. ### JasonMcG Thread Starter New Member

Apr 27, 2013
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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

2. ### ScottWang Moderator

Aug 23, 2012
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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.

3. ### JasonMcG Thread Starter New Member

Apr 27, 2013
24
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0.9V or 1.9V ?

4. ### ScottWang Moderator

Aug 23, 2012
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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.

5. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
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You are both wrong.

6. ### ScottWang Moderator

Aug 23, 2012
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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?

7. ### JasonMcG Thread Starter New Member

Apr 27, 2013
24
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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

8. ### screen1988 Member

Mar 7, 2013
310
3
I think in your image Vin = -0.1V => output of B: 11x(-0.1)= -1.1V
Gain C = 1
1x(-1.1) = -1.1V
Vout = 0.9+11+12 = 23.9V

Apr 14, 2005
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10. ### JasonMcG Thread Starter New Member

Apr 27, 2013
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I have gone through it again and calculated Vout to be 23.9V. Is this the correct value?

11. ### Jony130 AAC Fanatic!

Feb 17, 2009
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No you are wrong. For Vcc equal to +9V LM741 won't give you more then 9V at the output.

Last edited: Apr 29, 2013
12. ### Ron H AAC Fanatic!

Apr 14, 2005
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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.

Apr 27, 2013
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14. ### Ron H AAC Fanatic!

Apr 14, 2005
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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?

15. ### JasonMcG Thread Starter New Member

Apr 27, 2013
24
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The output at the node would be 1.1V and at op amp D it would be 23.9V

16. ### Ron H AAC Fanatic!

Apr 14, 2005
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You might want to think a little harder. A sine wave is not a DC voltage.

17. ### JasonMcG Thread Starter New Member

Apr 27, 2013
24
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It would have a max peak at 55mV and a minimum at -55mV

18. ### Ron H AAC Fanatic!

Apr 14, 2005
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How did you come up with that?

19. ### JasonMcG Thread Starter New Member

Apr 27, 2013
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the output voltage being 1.1V =110mV peak to peak
meaning 50mV up and 50mv down from the origin

20. ### Ron H AAC Fanatic!

Apr 14, 2005
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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.