Calculating the unknown voltages in ideal op-amp circuits

In summary, the conversation discusses the correct calculations for two op-amp circuits. The first calculation involves using the voltage divider formula to relate the non-inverting "feeding" voltage (V1) to the actual voltage at the non-inverting node (V+). It is explained that the fraction of V1 dropped across 10R is 10/11, and the final answer for V1 is 1V. The second op-amp calculation is confirmed to be correct with an output voltage of 6V. The concept of virtual Earth's and the input impedance of the amp are also discussed.
  • #1
peasngravy
72
6
Homework Statement
FIGURE 1 shows two amplifier circuits, each of which has one unknown
voltage (shown highlighted). Select, from TABLE A, the most
appropriate value for the unknown voltages for each circuit. Assume that
the op-amps are ideal and that the magnitude of their output voltages is
less than their maximum peak output voltage swing (VOM).
Relevant Equations
V+ = V1/(R+10R)

(V2 - V-)/R = -(Vo - V-)/10R
1603881915089.png1603884922052.pngHi - I have made an attempt at both of these circuits, but I am not 100% sure I am correct. Would it be possible to check if I have this right, and if not, please give me some pointers as to where I have gone wrong? Thanks

First op-amp
Given data:
Vo = 5V
I have called the voltage feeding into the non-inverting input V1 = 0.5V
voltage feeding into the inverting input = V2

For non inverting input V+ = V1/(1+10)

For inverting input V- :
(V2 - V-)/R= -(Vo - V-)/10R
V2 - V- =(-Vo + V-)/10
10(V2 - V-)-V- = -Vo
10V2 - 11V- = -Vo
Plug in values
(10*0.5) - 11V- = -5
-11V- = -10
V- =10/11 = 0.909V

As it is an ideal op-amp:

V+ = V- = 0.909V

Therefore:

V+ =V1/(10+1)

V1 = V+ * 11
V1 = 0.909*11 = 10V

Second op-amp

At non-inverting input:

V+ =Vo/(R+R)
V+=12/2=6V

As it is an ideal op-amp:

V+ = V- =6V
Due to short from inverting input to output, output voltage (V) is 6V
 
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  • #2
peasngravy said:
For non inverting input V+ = V1/(1+10)
You are relating the non-inverting "feeding" voltage (V1) to the actual voltage at the non-inverting node (V+) right?
In that case, you have incorrectly applied the voltage divider formula here.
 
  • #3
cnh1995 said:
You are relating the non-inverting "feeding" voltage (V1) to the actual voltage at the non-inverting node (V+) right?
In that case, you have incorrectly applied the voltage divider formula here.

Yep that's what I was trying to do. What have I done wrong with the formula?
 
  • #4
peasngravy said:
Yep that's what I was trying to do. What have I done wrong with the formula?
Your V+ is the voltage across 10R.
The total series resistance is 11R.
How are V1 and V+ related then?
 
  • #5
cnh1995 said:
Your V+ is the voltage across 10R.
The total series resistance is 11R.
How are V1 and V+ related then?

So should it be
(0.5-V2)/R = (V2-5)/10R
?
 
  • #6
No.
You have input voltage V1 and two series resistances R and 10R forming a voltage divider.
What fraction of V1 is dropped across 10R?

You have calculated that fraction. You need to find V1 from that fraction.
 
  • #7
cnh1995 said:
No.
You have input voltage V1 and two series resistances R and 10R forming a voltage divider.
What fraction of V1 is dropped across 10R?

You have calculated that fraction. You need to find V1 from that fraction.

I see - so it is 10/11?
 
  • #8
peasngravy said:
I see - so it is 10/11?
Yes.
 
  • #9
Thank you - so

V1= 10/(11*0.91) = 0.99V?
 
  • #10
peasngravy said:
V1= 10/(11*0.91) = 0.99V?
Sure, but if you keep all the terms in fraction form till the end, you will get V1= 1V, which is the accurate answer.
 
  • #11
Thank you so much for helping me to understand.

Does the second op-amp look Ok to you?
 
  • #12
peasngravy said:
Does the second op-amp look Ok to you?
Yes 6V looks correct to me.
 
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  • #13
Thanks again for your help
 
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Likes berkeman
  • #14
If the circuit is working (ie the amp is in its linear working range and the output is not driven to either rail voltage) then the difference between the -ve and the +ve inputs will be Vout / Amp gain. As the Amp gain is very, very high, the difference will be effectively zero. That is why they are called "virtual Earth's".

The input resistance (actually impedance) of the amp will be very, very high so you can ignore it.

Everything else then follows.

Then do a final check - do the voltages you calculate give the output voltage?
 

Related to Calculating the unknown voltages in ideal op-amp circuits

1. How do you calculate the unknown voltages in an ideal op-amp circuit?

The unknown voltages in an ideal op-amp circuit can be calculated using the basic equation Vout = A*(V+ - V-), where A is the open-loop gain of the op-amp and V+ and V- are the input voltages at the non-inverting and inverting terminals, respectively.

2. What is the open-loop gain of an op-amp and how is it determined?

The open-loop gain of an op-amp is the ratio of the output voltage to the input voltage when there is no feedback in the circuit. It is typically a very large value and is determined by the internal components and design of the op-amp.

3. Can the open-loop gain of an op-amp change in different circuits?

No, the open-loop gain of an op-amp is a characteristic of the op-amp itself and does not change in different circuits.

4. What is the difference between the non-inverting and inverting terminals of an op-amp?

The non-inverting terminal is the input terminal that is connected to the positive side of the power supply, while the inverting terminal is the input terminal that is connected to the negative side of the power supply. The output voltage of an op-amp is determined by the difference between these two input voltages.

5. How does feedback affect the calculation of unknown voltages in an op-amp circuit?

Feedback in an op-amp circuit can affect the open-loop gain and thus change the calculation of unknown voltages. In a negative feedback circuit, the output voltage will adjust in order to reduce the difference between the input voltages, resulting in a more accurate calculation of the unknown voltages. In a positive feedback circuit, the output voltage will increase and can potentially cause instability in the circuit.

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