Question

For the given operational amplifier circuit R₁ = 120 Ω, R₂ = 1.5 k and V_s = 0.6 V, then the output current I₀ is ____mA.

Answer 1

Correct Answer: 5

1. Identify the Circuit Configuration

The given circuit is a non-inverting amplifier because the input voltage source (Vs​) is connected to the non-inverting terminal (+) of the operational amplifier.

Given Values:

R1​=120Ω

R2​=1.5kΩ=1500Ω

Input Voltage (Vs​) = 0.6V

2. Determine the Voltage at the Inverting Terminal

For an ideal operational amplifier operating in the linear region, the virtual short concept applies. This means the voltage at the inverting terminal (V−​) is equal to the voltage at the non-inverting terminal (V+​).

V+​=Vs​=0.6V

Therefore, V−​=Vs​=0.6V

3. Calculate the Current through R1​

Let I1​ be the current flowing through resistor R1​. Since one end of R1​ is grounded (0V) and the other is at V−​, the current flows from the node V−​ to ground.

I1​=R1​V−​−0​=1200.6​

I1​=0.005A=5mA

4. Calculate the Output Current (I0​)

In an ideal op-amp, the input impedance is infinite, meaning no current flows into the inverting terminal input of the op-amp itself. Therefore, the current flowing through the feedback resistor R2​ is the same as the current flowing through R1​.

The problem asks for the "output current I0​". Looking at standard op-amp analysis, this usually refers to the load current or the feedback current if no load is specified. However, in this specific diagram configuration without an external load resistor RL​, the current I0​ typically refers to the current flowing out of the op-amp output node, which goes through the feedback path (R2​) to ground via R1​.

Since the current I1​ flows from the node to ground, and that same current must have come from the output node V0​ through R2​:

Ifeedback​=I1​=5mA

Therefore, the current supplied by the output (I0​) to the feedback network is:

I0​=5mA

(Note: Just to be thorough, let's check the output voltage V0​. For a non-inverting amp, V0​=Vs​(1+R1​R2​​)=0.6(1+1201500​)=0.6(1+12.5)=0.6(13.5)=8.1V. The current leaving V0​ through the feedback branch is (V0​−V−​)/R2​=(8.1−0.6)/1500=7.5/1500=0.005A, which confirms the result.)

Final Answer

The output current I0​ is 5 mA.