Circuit input impedance with ground in an OP Amp circuit

In summary, the op amp will have an output voltage that is the difference of the input voltages divided by the feedback resistors.
  • #1
bobg123
1
0
Homework Statement
Find input impedance of the circuit shown. Calculate impedance between the input terminal and the ”ground”. Two resistors R1 have equal resistance.
Relevant Equations
V-=V+
V=IR
V_out=-R1/R3 Vin
V_out=R1/R2+1 Vin
I've been given the following circuit and have been asked to find the input impedance and the impedance between the input terminal and ground. I've never encountered an operational amplifier configured like this.
WwSBI.png

I know that the voltages at the - and + terminals of the op amp are ideally equal. In a normal inverting/non inverting op amp, I've been told the voltages at those points are zero. Is this still the case? If so, what is the output voltage? I'm having trouble conceptually understanding what this particular circuit does.

I'm not sure how to approach this problem, but I've tried separating the circuit into an inverting and non-inverting circuit. For the inverting circuit, I know V_out=-V_in*R1/R3. This gives me V(-)=V_in+V_in/R_3. For the noninverting, I know V_out=(R1/R2+1)V_in. The voltage at V(+) should be equal to V(-), giving V_in+V_in/R_3=V(R1)-V(R2)=R1/R2V_in+V_in-V(R2). Some rearrangement gives me V_in(1/R3-R1/R2)=-V(R2). Is this the right way of approaching the problem? Where do I go from here to calculate the impedance?
 
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  • #2
bobg123 said:
I know that the voltages at the - and + terminals of the op amp are ideally equal.
That's one assumption you need to make, the other is that negligible current is drawn into those op-amp inputs. It would be wrong to think there are any virtual grounds here.

As always, input impedance = input voltage ÷ input current
and for this circuit, input current = the current in R3.
 
  • #3
Calculate ther gain of the circuit.
Then, it will not be a problem to find the current through the chain R3..R1.
 
  • #4
bobg123 said:
I've been given the following circuit and have been asked to find the input impedance and the impedance between the input terminal and ground. I've never encountered an operational amplifier configured like this.
View attachment 241071
I know that the voltages at the - and + terminals of the op amp are ideally equal. In a normal inverting/non inverting op amp, I've been told the voltages at those points are zero. Is this still the case? If so, what is the output voltage? I'm having trouble conceptually understanding what this particular circuit does.

I'm not sure how to approach this problem, but I've tried separating the circuit into an inverting and non-inverting circuit. For the inverting circuit, I know V_out=-V_in*R1/R3. This gives me V(-)=V_in+V_in/R_3. For the noninverting, I know V_out=(R1/R2+1)V_in. The voltage at V(+) should be equal to V(-), giving V_in+V_in/R_3=V(R1)-V(R2)=R1/R2V_in+V_in-V(R2). Some rearrangement gives me V_in(1/R3-R1/R2)=-V(R2). Is this the right way of approaching the problem? Where do I go from here to calculate the impedance?

Are you still working on this problem?

One problem I see is that in this:

V_in+V_in/R_3=V(R1)-V(R2)=R1/R2V_in+V_in-V(R2)

You are equating a sum of a voltage (red) and a current (blue) to a voltage (green). That's not permissible.
 
  • #5
Quote: I know V_out=-V_in*R1/R3

No - that is not correct.
This simple expression is valid in case the feedback path contains these two resistors only (negative feedback).
However - in addition, we have positive feedback via R1 and R2.
Note, that the above simple gain expression results from a general formula for feedback circuits (and idealized open-loop gain): A=-Hf/Hr.
Both expressions Hf and Hr consist of simple voltage dividers and are defined as follows
(Vd=Diff. voltage at the opamps input):
* Forward damping: Hf=Vd/Vin for Vout=0
* Return damping (feedback factor): Hr=Vd/Vout for Vin=0.

This general formula is to be used also for the given circuit.
 

Related to Circuit input impedance with ground in an OP Amp circuit

1. What is circuit input impedance with ground in an OP Amp circuit?

The circuit input impedance with ground in an OP Amp circuit refers to the resistance that the input signal encounters when it enters the circuit through the ground connection. This impedance is important because it affects the accuracy and stability of the circuit's output.

2. How does the input impedance affect the performance of an OP Amp circuit?

The input impedance of an OP Amp circuit affects its performance by influencing the amount of current that can flow through the circuit. A higher input impedance will result in less current flow, which can lead to a more accurate and stable output signal. On the other hand, a lower input impedance can cause distortion and instability in the output signal.

3. How is circuit input impedance with ground calculated?

The circuit input impedance with ground can be calculated by dividing the input voltage by the input current. This will give you the total impedance seen by the input signal as it enters the circuit through the ground connection. It is also important to take into account the impedance of any components in the circuit, such as resistors or capacitors, to get a more accurate calculation.

4. What factors can affect the input impedance of an OP Amp circuit?

There are several factors that can affect the input impedance of an OP Amp circuit. These include the type and value of components used in the circuit, the layout and design of the circuit, and the frequency of the input signal. Additionally, any external noise or interference can also impact the input impedance and should be minimized for optimal performance.

5. How can the input impedance of an OP Amp circuit be optimized?

The input impedance of an OP Amp circuit can be optimized by carefully selecting and designing the circuit components, such as resistors and capacitors, to achieve the desired impedance value. Additionally, proper circuit layout and shielding can help reduce external noise and interference, resulting in a more accurate and stable input impedance. Using high-quality components and following best practices for circuit design can also help optimize the input impedance of an OP Amp circuit.

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