Inverting Amplifier

An inverting amplifier built around an LM741 takes a signal at its negative input terminal and delivers an amplified, phase-reversed output. Every audio mixer, signal conditioning board, and analog control system uses this configuration to set gain with just two resistors.

Core Concept

The inverting amplifier uses negative feedback. The output connects back to the inverting input through Rf. The non-inverting input is tied to ground. Because the op-amp has very high open-loop gain, the virtual ground principle applies: the inverting input is forced to 0 V by the feedback action, even though it is not physically connected to ground.

Input current flows from Vin through R1 to the virtual ground node. The same current then flows through Rf to the output, because the op-amp's input draws negligible current. The output voltage must be whatever value forces this balance. The result is Vout = -(Rf/R1) × Vin.

The negative sign means the output is 180° out of phase with the input. A 1 kHz sine at 0.5 V peak through a circuit with R1 = 10 kΩ and Rf = 47 kΩ produces a 2.35 V peak output, inverted. The input resistance of this configuration equals R1, which the driving source must be able to drive.

Key Equations

Closed-loop voltage gain: Acl = -Rf / R1. The negative sign indicates phase inversion.

Input resistance: Rin = R1. This is a key disadvantage compared to the non-inverting configuration.

Output resistance (ideal): Rout ≈ 0 Ω. In practice, for LM741, closed-loop Rout is a few ohms.

Bandwidth: BW = GBP / |Acl|. For LM741 with GBP = 1 MHz and |Acl| = 10, BW = 100 kHz.

Compensating resistor at non-inverting input: Rcomp = R1 || Rf. This minimises output error from bias currents.

Given:
  R1 = 2.2 kΩ
  Rf = 33 kΩ
  Vin = 0.3 V DC
  Op-amp: LM741, GBP = 1 MHz
  Find: Vout, closed-loop gain, bandwidth

Why this formula:
  Virtual ground gives Vout = -(Rf/R1) × Vin

Formula:
  Acl = -Rf / R1
  Vout = Acl × Vin
  BW = GBP / |Acl|

Substitution:
  Acl = -33,000 / 2,200 = -15
  Vout = -15 × 0.3 V
  BW = 1,000,000 / 15

Calculation:
  Acl = -15
  Vout = -4.5 V
  BW = 66,667 Hz ≈ 66.7 kHz

Final Answer:
  Closed-loop gain = -15 (inverting)
  Vout = -4.5 V
  -3 dB bandwidth = 66.7 kHz

Exam Tip: GATE problems often ask for input resistance of an inverting amplifier. It is R1, not infinite. Students frequently write Rin = infinity, which applies to the non-inverting configuration. Also, bandwidth = GBP / |Acl| uses the magnitude of gain. A gain of -15 gives BW = GBP/15, not GBP/(-15).

Key Properties

• Closed-loop gain magnitude is set by Rf/R1 and is independent of the op-amp's open-loop gain, as long as open-loop gain is much larger.
• Input resistance equals R1. Using R1 = 1 kΩ loads the driving source heavily; 10 kΩ is a common practical choice.
• Output phase is exactly 180° inverted relative to the input for all frequencies within the passband.
• A compensating resistor Rcomp = R1 || Rf at the non-inverting input reduces DC offset caused by op-amp bias currents flowing through Rf.
• The virtual ground node sits at 0 V but draws no current; measuring it with a voltmeter confirms the feedback is working correctly.
• For a gain of -1 (unity inverter), set R1 = Rf; the circuit then acts as a precise sign-inverter or phase-reversal stage.

Quick Revision

• Acl = -Rf / R1; negative sign means 180° phase inversion.
• Rin = R1 (not infinite, unlike the non-inverting amplifier).
• Virtual ground: inverting input held at 0 V by negative feedback.
• BW = GBP / |Acl|; higher gain means narrower bandwidth.
• Rcomp = R1 || Rf placed at the non-inverting input to cancel bias current error.
• For unity gain inversion: R1 = Rf, Acl = -1.
• LM741 output can swing to within about 1-2 V of each supply rail.
• Exam trap: students forget Rin = R1 and incorrectly state that inverting and non-inverting amplifiers both have infinite input resistance.