1. G=1
2. V=9V
3. G=1
4. V=5
5. V=4.5V
6. V=4.5V
7. 0A based on ideal conditions
8. V=4.5V

is this any better?

 

Yes, that's better. Error occurs after answer 7.

I have edited the diagram and placed the correct answer on the drawing. Keep on the effort. You will soon get there.

 

1. G=1
2. V=9V
3. G=1
4. V=5
5. V=4.5V
6. V=4.5V
7. 0A based on ideal conditions
8. V=4V

so now dealing with A4... if + is 0V then doesnt the - have to be 0V as well? so if that were true, and V at point 8 is 4V, would the gain be 4?

 

Good. You have finally got point 8 right. The voltage is 4V.

so now dealing with A4... if + is 0V then doesnt the - have to be 0V as well? so if that were true, and V at point 8 is 4V, would the gain be 4?

Opamp gain are extremely high. But what determine the gain of an amplifier circuit built using opamp?

Without knowing the output voltage, can you work out the gain? How?

 

Good. You have finally got point 8 right. The voltage is 4V.



Opamp gain are extremely high. But what determine the gain of an amplifier circuit built using opamp?

Without knowing the output voltage, can you work out the gain? How?

im not exactly sure... formula for A is A=Vo/(V+-V-)
i dont know how you would find that without Vo

if V+ and V- are both 0 then the gain would be equal to whatever the output voltage would be. for example if Vo=2.5 then A would be 2.5

 

The gain of an amplifier built using an opamp are largely determined by the values of the component connected around the opamp, not the intrinsic gain of the opamp IC itself, as the gain of an opamp IC can run into millions.

You will need to make sure you fully understand why this is so and how the gain of the circuit is calculated.

 

when you connect the output of the op amp to the negative input terminal you create a voltage follower(the two on the left side of the circuit). By going through a resistor back to the negative terminal the output voltage becomes a multiple of the input voltage.

Av=(R2/R1)+1=1.33

so if Av=1.33 and the voltage at point #8 is 4V, then the input voltage is 4/1.33=3V(point 9)

calculating for the voltage drop across the 30k i would get 12V at point #11. is that correct?

 

when you connect the output of the op amp to the negative input terminal you create a voltage follower(the two on the left side of the circuit). By going through a resistor back to the negative terminal the output voltage becomes a multiple of the input voltage.

Av=(R2/R1)+1=1.33

so if Av=1.33 and the voltage at point #8 is 4V, then the input voltage is 4/1.33=3V(point 9)

calculating for the voltage drop across the 30k i would get 12V at point #11. is that correct?

You're getting closer. Your equation is for a noninverting feedback amplifier. A4 is in an inverting configuration. What is the equation for an inverting amp?
Also - point 9 is at virtual ground (0V).

 

so if point #9 is zero then G=4 at point #10.

Vout=A * Vin

 

so whatever voltage is at point #11 is equal to 0 after dropped across a 30k resistor?

 

For purposes of gain calculation, the closed loop gain of A4 is V4/V3.

 

You really need to read and understand this section of the forum's tutorials.

 

using that equation i get Av=-3 for the gain on A4

V3 would have to be -12

 

using that equation i get Av=-3 for the gain on A4

V3 would have to be -12

That's correct. Another way of looking at it is to realize that:
1. The inverting input of V4=0V, due to the negative feedback (both inputs will be at the same voltage).
2. Op amp input currents are zero. Therefore, the current through the 30k source resistor will be equal to the current through the 10k feedback resistor.
3. Since you know that V4=+4V, then V3=-12V.

I repeat: You really need to read and understand this section of the forum's tutorials, not just the equation for the gain of an inverting amplifier.

 

I dont understand how 6 is 4.5V ..... i know i/p currents to opamps are 0 .... on what basis do you say that 0.5V will drop over the 10k connected to V2?

If you say that op-amp's i/p are at same potential due to fb .... that is true only for -ve feedback ..... A3 has positive FB ...

 

I dont understand how 6 is 4.5V ..... i know i/p currents to opamps are 0 .... on what basis do you say that 0.5V will drop over the 10k connected to V2?

If you say that op-amp's i/p are at same potential due to fb .... that is true only for -ve feedback ..... A3 has positive FB ...

I know it's confusing, because the feedback is to the +pin on A3, but A3 has negative feedback, because A4 is an inverting amplifier. The signal fed back to 6 has the opposite sense of 3, i.e., if 3 goes up, 6 goes down. That's negative feedback.

 

I know it's confusing, because the feedback is to the +pin on A3, but A3 has negative feedback, because A4 is an inverting amplifier. The signal fed back to 6 has the opposite sense of 3, i.e., if 3 goes up, 6 goes down. That's negative feedback.

do you mean to say since the POSITIVE feedback to A3 is via an inverting opamp (A4) it actually becomes NEGATIVE feedback?

it is logical and makes a lot of sense .... so when analyzing i consider A3 to be -ve FB ... then it is easy 6 = 4.5V .... 0.5V across 10K and no current in i/p of A3 makes same current through the other 10k making 0.5V across it ..... making o/p of A4 = 4V, no current in i/p of A4 makes current through same making V = 12V, considering A4 is in inv mode and the o/p is +ve the V i/p must be -ve hence V3 = -12 ....

what of the currents ..... we have 0.5/10K flowing out of A2 and through the other 10K where does it disappear???


thanks for the help ...

 

Here is a list of PDFs that are useful. A couple are complete books on OpAmps.

http://mxp.physics.umn.edu/f04/Lecture/OpAmpNotes.pdf
Texas Instruments Handbook of OpAmp Applications book
Op Amp Basics. It goes a bit beyond Basics, into theory and origin, as well as the servo action of the - and + inputs.

 

do you mean to say since the POSITIVE feedback to A3 is via an inverting opamp (A4) it actually becomes NEGATIVE feedback?

it is logical and makes a lot of sense .... so when analyzing i consider A3 to be -ve FB ... then it is easy 6 = 4.5V .... 0.5V across 10K and no current in i/p of A3 makes same current through the other 10k making 0.5V across it ..... making o/p of A4 = 4V, no current in i/p of A4 makes current through same making V = 12V, considering A4 is in inv mode and the o/p is +ve the V i/p must be -ve hence V3 = -12 ....

what of the currents ..... we have 0.5/10K flowing out of A2 and through the other 10K where does it disappear???


thanks for the help ...

The current flows into the output of A4.

 

thanks ... i was wondering that can this be mathematically solved/proved WITHOUT assuming A3 has -ve fb ... I tried but w/o V4 and/or V3 ....we just end up with 1 eq having 2 variables (V4= -V3/3) ....