How to Approach an Op-Amp Homework Problem with Earthed Terminals?

[thereddevils]

Homework Statement

See attachment.

Homework Equations

The Attempt at a Solution

I am not familiar with questions like this. The usual op-amp i see have either of the terminals earthed so i need some hints to get started. Thanks.

 

[gneill]

Hint: The important things to remember about the (ideal) op-amp is that
1. No current flows into or out of the input terminals.
2. In a negative feedback situation, both input terminals will always be at the same voltage (no voltage difference).

That being the case, what can you say about the voltage at the (-) input?
How about the current in the resistors?

 

[thereddevils]

Hint: The important things to remember about the (ideal) op-amp is that
1. No current flows into or out of the input terminals.
2. In a negative feedback situation, both input terminals will always be at the same voltage (no voltage difference).

That being the case, what can you say about the voltage at the (-) input?
How about the current in the resistors?

Thanks Gneill for your reply.

Ok, since the input impedance of the op-amp is very high, the current flowing through the 20 k ohm resistor is the same as the one through the 100 k ohm resistor.

I am not sure about point 2. Why are the voltage of the input terminals the same? How do i set up the equations? V- = 8V and V+ = 4V ??

 

[gneill]

Thanks Gneill for your reply.

Ok, since the input impedance of the op-amp is very high, the current flowing through the 20 k ohm resistor is the same as the one through the 100 k ohm resistor.

Good.

I am not sure about point 2. Why are the voltage of the input terminals the same? How do i set up the equations? V- = 8V and V+ = 4V ??

8V is not the same as 4V.

Because the op-amp has very high gain (infinite for the ideal op-amp), any difference in voltage between the V- and V+ input terminals would be magnified infinitely at the output terminal. In practical terms, the output would swing towards one of the power supply rails.

This doesn't happen in a properly configured amplifier circuit with feedback, because the feedback path acts to keep the differential voltage at zero. Thus V- = V+. In your circuit, one of the inputs is tied to a fixed voltage supply. What should you conclude?

 

[thereddevils]

8V is not the same as 4V.

Because the op-amp has very high gain (infinite for the ideal op-amp), any difference in voltage between the V- and V+ input terminals would be magnified infinitely at the output terminal. In practical terms, the output would swing towards one of the power supply rails.

I understand this.

This doesn't happen in a properly configured amplifier circuit with feedback, because the feedback path acts to keep the differential voltage at zero. Thus V- = V+. In your circuit, one of the inputs is tied to a fixed voltage supply. What should you conclude?

The difference between V+ and V- is zero if connected with a negative feedback resistor? Is V- = 8V and V+ = 4V? I am still confused with the potential difference across the positive and negative input terminals.

 

[gneill]

The difference between V+ and V- is zero if connected with a negative feedback resistor? Is V- = 8V and V+ = 4V? I am still confused with the potential difference across the positive and negative input terminals.

What's to be confused about? For an ideal op-amp in a circuit with negative feedback, the difference is zero. 8V and 4V is not a difference of zero volts, it's a difference of 4 volts.

Set the difference between the inputs to zero. One of the inputs is tied to +4v by a fixed supply voltage. What must the voltage at the other input terminal be?

 

[thereddevils]

What's to be confused about? For an ideal op-amp in a circuit with negative feedback, the difference is zero. 8V and 4V is not a difference of zero volts, it's a difference of 4 volts.

Set the difference between the inputs to zero. One of the inputs is tied to +4v by a fixed supply voltage. What must the voltage at the other input terminal be?

What a miss!

Obviously, the voltage of the upper input terminal is 4V and having the same current flow,

(8-4)/20k = (4-Vo)/100k

Vo=-16V

Am i right?

 

[gneill]

Bravo!

 

[thereddevils]

Bravo!

Thanks for all the help! Cheers.