Differential amplifier and current mirrors

Why it matters

Industrial sensors such as strain-gauge bridges and thermocouples produce millivolt signals riding on volts of common-mode voltage and mains pickup. The differential amplifier amplifies the wanted difference and rejects what is common to both wires; it is the input stage of every op-amp and instrumentation amplifier. Current mirrors are its partners: they bias the pair with a stable tail current and act as high-resistance active loads that give one stage thousands of gain.

Key ideas

Differential and common-mode signals. For inputs v₁ and v₂ define the differential input v_d = v₁ − v₂ and the common-mode input v_cm = (v₁ + v₂)/2. A real amplifier responds to both: v_o = A_d·v_d + A_cm·v_cm. The ideal differential amplifier has A_cm = 0.

The emitter-coupled (long-tailed) pair. Two matched transistors share an emitter node fed by a tail current I_EE (from a resistor R_EE or, better, a current source). With no differential input each transistor carries I_EE/2.

• A differential input steers current from one side to the other while the total stays I_EE. Each side behaves like a CE stage driven by v_d/2, so the single-ended gain is g_m·R_C/2 and the gain between the two collectors is g_m·R_C.
• A common-mode input moves both bases together; the tail resistance R_EE gives strong negative feedback (each half sees 2R_EE in its emitter), so the common-mode gain is small, about R_C/(2R_EE).
• The pair stays linear only for |v_d| up to a few V_T (about ±50 mV for a BJT pair); beyond that all the tail current goes to one side. Emitter-degeneration resistors widen the linear range at the cost of gain.

CMRR. The common-mode rejection ratio CMRR = |A_d/A_cm|, usually quoted in dB. Raising the tail resistance raises CMRR, which is why the tail resistor is replaced by a current source with output resistance of hundreds of kΩ to MΩ. Mismatch between the two halves converts common-mode input into a differential output and ultimately limits CMRR.

Current mirrors. A diode-connected transistor Q1 carrying a reference current sets a V_BE (or V_GS); an identical Q2 with the same V_BE then carries the same current. With finite β the base currents steal a little: I_out = I_ref/(1 + 2/β). For MOSFETs there is no gate current and I_out/I_ref equals the ratio of W/L, which makes scaled mirrors easy.

Mirror output resistance. The output of a simple mirror is r_o of Q2, so I_out rises slightly with output voltage (Early effect). The Widlar mirror adds an emitter resistor to make small currents with modest resistors; the Wilson and cascode mirrors use feedback or stacking to raise output resistance by about β/2 or g_m·r_o and to reduce the β error.

Active load. Replacing the collector resistors of a differential pair with a current mirror converts the differential output to single-ended without losing half the signal, and the load resistance becomes r_o of the transistors, so the gain becomes about g_m·(r_o2 ∥ r_o4), often several hundred to thousands.

Formulas

v_d = v₁ − v₂, v_cm = (v₁ + v₂)/2, v_o = A_d·v_d + A_cm·v_cm.

g_m = (I_EE/2) / V_T — each transistor of a BJT pair (S).

A_d = g_m·R_C (output between collectors), A_d = g_m·R_C / 2 (single-ended).

A_cm ≈ −R_C / (2·R_EE) — single-ended common-mode gain with tail resistance R_EE (Ω).

CMRR = |A_d / A_cm|, CMRR(dB) = 20·log₁₀|A_d / A_cm|. For the single-ended BJT pair CMRR ≈ g_m·R_EE.

R_id = 2·r_π — differential input resistance of a BJT pair.

I_out = I_ref / (1 + 2/β) — simple BJT mirror; I_ref = (V_CC − V_BE)/R for a resistor-set reference.

I_out = I_ref·(W/L)₂ / (W/L)₁ — MOSFET mirror.

R_out = r_o = V_A / I_out — simple mirror output resistance.

Worked examples

Example 1 (standard: BJT pair). A BJT differential pair has I_EE = 1 mA, R_C = 10 kΩ, tail resistance R_EE = 50 kΩ, β = 100 and V_T = 25 mV. Find g_m, the single-ended differential gain, the common-mode gain, CMRR in dB and R_id.

1. Each transistor carries 0.5 mA: g_m = 0.5 mA/25 mV = 20 mS.
2. A_d(single-ended) = g_m·R_C/2 = 0.020 × 10 000/2 = 100 (200 between collectors).
3. A_cm ≈ R_C/(2R_EE) = 10 kΩ/100 kΩ = 0.1 in magnitude.
4. CMRR = 100/0.1 = 1000, i.e. 20·log₁₀(1000) = 60 dB.
5. r_π = β/g_m = 5 kΩ, so R_id = 2r_π = 10 kΩ.

Answer: g_m = 20 mS, A_d = 100, |A_cm| = 0.1, CMRR = 60 dB, R_id = 10 kΩ.

Example 2 (GATE level: current mirror). A simple npn mirror has V_CC = 10 V and reference resistor R = 9.3 kΩ; V_BE = 0.7 V, β = 100 and V_A = 100 V. Find I_ref, the ideal-β output current, the output resistance, and the output current when the output transistor sits at V_CE = 10 V.

1. I_ref = (V_CC − V_BE)/R = 9.3/9300 = 1.000 mA.
2. I_out = I_ref/(1 + 2/β) = 1/1.02 = 0.980 mA (at V_CE2 = V_BE).
3. r_o = V_A/I_out = 100/0.980 mA = 102 kΩ.
4. Early effect: V_CE2 exceeds V_CE1 by 9.3 V, so I_out ≈ 0.980 × (1 + 9.3/100) = 1.072 mA.

Answer: I_ref = 1 mA, I_out ≈ 0.98 mA, r_o ≈ 102 kΩ, I_out ≈ 1.07 mA at V_CE = 10 V. The 9 % rise from the Early effect is far bigger than the 2 % β error, which is why Wilson and cascode mirrors matter.

Common mistakes

• Using the full tail current instead of I_EE/2 when computing g_m of each transistor.
• Quoting the collector-to-collector gain when the circuit takes a single-ended output (it is half).
• Writing v_cm = v₁ + v₂ instead of the average.
• Dividing A_cm by A_d for CMRR, or forgetting to convert to dB when asked.
• Assuming the pair is linear for large v_d; it saturates beyond a few V_T.
• Ignoring the base-current error and the Early effect in BJT mirrors.

For GATE IN

Expect numericals on A_d, A_cm and CMRR (in dB), the output of a differential amplifier given v₁, v₂, A_d and CMRR, tail-current and mirror-current calculations including finite β, and conceptual questions on active loads and why current sources improve CMRR. Practise splitting arbitrary v₁, v₂ into v_d and v_cm.

Quick check

1. v₁ = 5.01 V and v₂ = 4.99 V. Find v_d and v_cm.
2. A_d = 2000 and CMRR = 80 dB. What is |A_cm|?
3. An amplifier has A_d = 100 and CMRR = 60 dB. With v_d = 10 mV and v_cm = 1 V, what output does each component produce?
4. A MOSFET mirror has (W/L)₂ = 4·(W/L)₁ and I_ref = 50 µA. What is I_out?

Answers: 1. v_d = 20 mV, v_cm = 5 V. 2. 0.2. 3. 1 V from v_d and 0.1 V from v_cm (A_cm = 0.1). 4. 200 µA.