Inverting, non-inverting, summing and difference amplifiers

Why it matters

Four op-amp circuits do most of the everyday work in signal conditioning: the inverting amplifier scales and inverts, the non-inverting amplifier scales without loading the sensor, the summing amplifier adds offsets and mixes channels, and the difference amplifier extracts a bridge or shunt voltage from a common-mode level. Every one of them is analysed with the same two rules, so learn the method once and reuse it.

Key ideas

Method (ideal op-amp with negative feedback). (1) No current enters the inputs. (2) v₋ = v₊. Write the voltage at the non-inverting input, set the inverting input equal to it, and apply KCL at the inverting node. This handles every linear op-amp circuit in this topic.

Inverting amplifier. The input goes through R₁ to the inverting input; R_f connects output to the inverting input; the non-inverting input is grounded. The inverting input is a virtual ground, so the input current is v_i/R₁ and the same current flows through R_f. Gain = −R_f/R₁. Input resistance = R₁ (often low, which loads the source). Voltage-shunt feedback.

Non-inverting amplifier. The input drives the non-inverting input; R_f and R₁ form a divider from the output to ground whose midpoint is the inverting input. Gain = 1 + R_f/R₁ (always ≥ 1). Input resistance is that of the op-amp input itself, very high. Voltage-series feedback. With R_f = 0 and R₁ = ∞ it becomes the voltage follower (gain 1), the standard buffer.

Summing amplifier. Several inputs, each through its own resistor, feed the virtual-ground node. Because the node is at 0 V, the input currents do not interact and simply add in R_f: v_o = −R_f·Σ(v_k/R_k). Equal resistors give an inverted sum; R_f = R/n gives an inverted average. Non-inverting summers exist but their weights depend on all the resistors, so the inverting form is preferred.

Difference amplifier. One input v₁ through R₁ to the inverting node with R₂ as feedback; the other input v₂ through R₃ to the non-inverting input with R₄ to ground. By superposition v_o = −(R₂/R₁)·v₁ + (1 + R₂/R₁)·(R₄/(R₃ + R₄))·v₂. When R₄/R₃ = R₂/R₁ the circuit outputs (R₂/R₁)·(v₂ − v₁) and rejects common-mode voltage completely. Any resistor mismatch leaks common-mode voltage to the output, so CMRR depends on resistor matching (with 1 % resistors at unity gain, CMRR can fall to roughly 35–45 dB). Its input resistances are finite and unequal, which loads the source; the instrumentation amplifier of the next topic fixes this.

Practical points. Choose resistors in the 1 kΩ–100 kΩ range: too low loads the op-amp output, too high makes bias-current errors and noise grow. The closed-loop bandwidth is GBW divided by the noise gain (1 + R_f/R₁ for both the inverting and non-inverting forms; 1 + R_f/(R₁ ∥ R₂ ∥ …) for a summer).

Formulas

v_o = −(R_f/R₁)·v_i — inverting amplifier; R_in = R₁.

v_o = (1 + R_f/R₁)·v_i — non-inverting amplifier; R_in → ∞ (ideal).

v_o = v_i — voltage follower.

v_o = −R_f·(v₁/R₁ + v₂/R₂ + … + v_n/R_n) — inverting summer.

v_o = −(R₂/R₁)·v₁ + (1 + R₂/R₁)·(R₄/(R₃ + R₄))·v₂ — difference amplifier, general.

v_o = (R₂/R₁)·(v₂ − v₁) — difference amplifier when R₄/R₃ = R₂/R₁.

A_cm = (R₄·R₁ − R₂·R₃) / (R₁·(R₃ + R₄)) — common-mode gain of a mismatched difference amplifier; CMRR = |A_d/A_cm|.

Worked examples

Example 1 (standard: summer). An inverting summer has R_f = 10 kΩ, R₁ = 10 kΩ, R₂ = 20 kΩ and R₃ = 5 kΩ with inputs v₁ = 1 V, v₂ = 2 V and v₃ = 0.5 V. Find v_o.

1. v_o = −R_f·(v₁/R₁ + v₂/R₂ + v₃/R₃).
2. Weights: R_f/R₁ = 1, R_f/R₂ = 0.5, R_f/R₃ = 2.
3. v_o = −(1 × 1 + 0.5 × 2 + 2 × 0.5) = −(1 + 1 + 1) = −3 V.

Answer: v_o = −3 V.

Example 2 (GATE level: mismatched difference amplifier). R₁ = 10 kΩ and R₂ = 50 kΩ on the inverting side; R₃ = 10 kΩ and R₄ = 40 kΩ on the non-inverting side. v₁ = 1 V (inverting side) and v₂ = 2 V. Find v_o, the common-mode gain, the differential gain and the CMRR.

1. Inverting-path gain: −R₂/R₁ = −5.
2. Non-inverting-path gain: (1 + R₂/R₁)·R₄/(R₃ + R₄) = 6 × 40/50 = 4.8.
3. v_o = −5 × 1 + 4.8 × 2 = −5 + 9.6 = 4.6 V.
4. Common mode (v₁ = v₂ = v_c): A_cm = −5 + 4.8 = −0.2.
5. Differential (v₂ = +v_d/2, v₁ = −v_d/2): A_d = (4.8 + 5)/2 = 4.9.
6. CMRR = |A_d/A_cm| = 4.9/0.2 = 24.5, i.e. 27.8 dB.

Answer: v_o = 4.6 V, A_cm = −0.2, A_d = 4.9, CMRR ≈ 24.5 (≈ 28 dB). With R₄ = 50 kΩ (matched ratio) the output would be exactly 5 × (2 − 1) = 5 V.

Common mistakes

• Using −R_f/R₁ for a non-inverting stage, or forgetting the "1 +".
• Assuming the non-inverting input of a difference amplifier sees v₂ directly; it sees the divided voltage v₂·R₄/(R₃ + R₄).
• Treating the inputs of a summer as interacting; the virtual ground isolates them.
• Applying the virtual-ground idea to a non-inverting stage; there v₋ follows v_i, not 0 V.
• Ignoring resistor tolerance when asked about CMRR of a difference amplifier.
• Quoting the inverting amplifier's input resistance as very high; it is R₁.

For GATE IN

Expect direct output-voltage questions on inverting, non-inverting, summing and difference circuits, often combined in two stages, plus questions on resistor-mismatch CMRR and input resistance. Superposition with the virtual-short rule is the fastest route. Practise circuits with a reference voltage on the non-inverting input, which shifts the output by (1 + R_f/R₁)·V_ref.

Quick check

1. R_f = 47 kΩ, R₁ = 4.7 kΩ, inverting, v_i = 0.2 V. v_o?
2. Same resistors, non-inverting. v_o?
3. A unity-gain difference amplifier has v₁ = 3 V (inverting side) and v₂ = 5 V. v_o?
4. What R_f gives the average of four inputs with 40 kΩ input resistors?

Answers: 1. −2 V. 2. 2.2 V. 3. 2 V. 4. 10 kΩ (output is the negative of the average).